ch09

9-1

Fully developed, laminar flow of a viscous fluid (µ = 2.17 N·s/m2) flows between horizontal parallel plates 1-m long that are spaced 3.0-mm apart. The pressure drop is 1.25 kPa. Determine the volumetric flow rate (per unit width) through the channel (in m3/s m).

Approach: Pressure drop can be calculated for this non-circular duct with the combination of Eq. 9-31 and Eq. 9-32 and using the hydraulic diameter. The velocity is unknown; once it is determined the volume flow rate can be calculated. We need to evaluate the friction factor. We combine all the expressions and solve for velocity.

Assumptions: 1. The system is steady.

Solution: Volume flow rate is hydraulic diameter: ∆P = hL = f ρg

V = VAx = VHW or per unit width V W = VH . Pressure drop can be calculated using the L V2 Dh 2 g

For fully develop laminar flow between infinite parallel plates, f = 96 Re = 96 µ ρ VDh . Substituting this into the pressure drop equation, simplifying, solving for the average velocity V: 2 Dh2 ∆P V = 96µ L The hydraulic diameter with W → ∞ , Dh = 4 Ax pwetted = 4 HW ⎡⎣ 2 ( H + W ) ⎤⎦ = 2 H = 2 ( 0.003m ) =0.006m 2 ( 0.006m ) (1.25 kN m 2 ) (1000 N kN ) 2

V =

96 ( 2.17 Ns m

2

) (1m )

=0.000432

m s

V = VA = VHW V m⎞ m2 m3 ⎛ = VH = ⎜ 0.000432 ⎟ ( 0.006m ) =2.59×10-6 =2.59×10-6 W s ⎠ s ms ⎝

9- 1

Answer

9-2

Journal bearings are constructed with concentric cylinders with a very small gap between the two cylinders; the gap is filled with oil. Because of the very small gap, the flow in the gap is laminar. Consider a sealed journal bearing with inner and outer diameters of 50- and 51-mm, respectively, and a length of 75 mm. The shaft (inner cylinder) rotates at 3000 RPM. At start-up the torque needed to turn the shaft is 0.25 N-m. Determine the viscosity of the oil (in N·s/m2). After an hour of operation will the torque have increased or decreased? Explain.

Approach: Because the gap is small compared to the diameter, we can analyze the flow as if it were between infinite parallel plates. Viscosity is defined with Newton’s law of viscosity, Eq. 9-2.

Assumptions: 1. The system is steady. 2. The flow is fully developed between infinite parallel plates. 3. Properties are constant.

Solution: Shear stress at a solid wall, using Newton’s law of viscosity, is: dV τ =µ dy Because we can analyze this flow as between infinite parallel plates, we know that with the inner shaft rotating and the outer shaft stationary, the velocity profile is linear, so the velocity gradient is; τ ( r2 − r1 ) V τ =µ µ= → r2 − r1 V Velocity is V = r1ω . Torque is ℑ = Fr1 = τ Ar1 . Substituting these expressions into the shear stress equation and solving for visocity: ℑ ( r2 − r1 ) ℑ ( r2 − r1 ) ℑ ( r2 − r1 ) µ= = = 2 2 Ar1 ω ( 2π r1 L ) r1 ω 2π Lr13ω =

( 0.25 N m )( 0.0255m-0.025m )( 60s 1min ) Ns =0.054 2 3 m 2π ( 0.025m ) ( 0.075m )( 3000 rev min )( 2π rad rev )

Answer

Comments: After an hour of operation, torque will decrease. Assuming that the bearing has little heat loss, the viscous friction will raise the oil temperature. Because viscosity for a liquid decreases with increasing temperature, torque will decrease with time.

9- 2

9-3

Consider laminar water flow at 20 °C between two very large horizontal plates. The lower plate is stationary and the upper plate moves to the right at a velocity of 0.25 m/s. For a plate spacing of 2 mm, determine the pressure gradient and its direction required to produce zero net flow at a cross section.

Approach: Because the plates are very large, we assume the flow is fully developed between infinite parallel plates. For this flow, we can perform an analysis similar to what was done in Section 9.3 for flow in a circular tube to find the relationship between flow and pressure gradient.

Assumptions: 1. 2. 3.

The system is steady. The flow is fully developed between infinite parallel plates. Properties are constant.

Solution: For steady, fully developed flow, conservation of momentum in the x-direction reduces to: ∑ Fx = 0 Evaluating the forces on the differential element shown above: P1 A1 − P2 A2 + τ T AT − τ B AB = 0

With length, dx, along the channel and a width W: P1 (Wdx ) − P2 (Wdy ) + τ T (Wdx ) − τ B (Wdx ) = 0 The width cancels.

τ =µ

Shear stress at a solid wall, using Newton’s law of viscosity, is: Therefore,

τT = µ

d VT dy

and

τB = µ

d VB dy

( P1 − P2 ) dy − µ

Substituting into the force balance and simplifying:

( d VT dy − d VB dy ) P1 − P2 =µ dx dy

dV dy

→

d VT d VB dx + µ dx = 0 dy dy

dP d2 V =µ dx dy 2

Recognizing that for fully developed flow, dP dx = constant , we separate variables and integrate twice: 1 dP 2 y + C1 y + C2 2µ dx The boundary conditions are: 1) at y = 0, V = 0; 2) at y = b, V = Vw Applying the boundary conditions: 1 dP 2 2) VW = b + C1b 1) 0 = 0 + 0 + C2 so C2 = 0 2µ x V =

Substituting into the general velocity equation and simplifying:

V =

so C1 =

VW b dP − b 2µ x

1 dP 2 y y − by ) + VW ( b 2µ dx

Flow rate is obtained by integration: b ⎛ 1 dP y⎞ −b3W dP VW bW V = ∫ VdA = ∫ VWdy = ∫ ⎜ y 2 − by ) + VW ⎟Wdy = + ( 0 2 µ dx b⎠ 12 µ dx 2 ⎝  Solving this equation for zero flow ( V = 0 ) and from Appendix A-6 for water at 20 ºC, µ = 9.85 × 10−4 Ns m 2 -4 2 dP 6 µ VW 6 ( 9.85×10 Ns m ) ( 0.25 m s ) N = = =369 2 2 dx m m b2 0.002m ( )

Pressure must increase in the x-direction to obtain zero flow. 9- 3

Answer

9-4

In the ¾-in. pipe shown below, oil flows downward at 6 gal/min. The oil has a specific gravity of 0.87 and a dynamic viscosity of 0.4 lbm/ft·s. The specific gravity of the manometer fluid is 2.9. Determine the manometer defection, h (in ft).

Approach: The pressure drop between points 1 and 2 can be calculated with the steady, incompressible flow energy equation. The manometer equation is used to calculate the manometer deflection once the pressure drop is determined.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

The steady, incompressible flow energy equation is:

The areas at 1 and 2 are equal, so V1 = V2 . There is no pump or turbine, so hP = hT = 0 , and minor losses are

The velocity is

L V2 ρ D 2 4 ( 6 gal min ) ( 0.1337 ft 3 gal ) (1min 60s )

P1 − P2 = ρ g ( z2 − z1 ) + f

neglected. Therefore, V =

V 4V = = A π D2

The Reynolds number is:

π ( 0.0625ft )

Re =

2

=4.36

ft s

3 ρ VD ( 0.87 ) ( 62.4 ft lbm ) ( 4.36ft s )( 0.0625ft ) = = 37 0.4 lbm fts µ

For fully developed laminar flow in a straight circular tube: Substituting this into the pressure drop equation:

f = 64 Re = 64 37 = 1.73

lbm ⎞ ⎛ ft ⎞ lbm ⎞ ( 4.26 ft s ) ⎛ ⎛ 15 ⎞ ⎛ P1 − P2 = ( 0.87 ) ⎜ 62.4 3 ⎟ ⎜ 32.2 2 ⎟ ( -15ft ) + (1.73) ⎜ ⎟ ( 0.87 ) ⎜ 62.4 3 ⎟ ft ⎠ ⎝ s ⎠ ft ⎠ 2 ⎝ ⎝ 0.0625 ⎠ ⎝

2

2

lbm ⎞ ⎛ lbf s 2 ⎞ ⎛ 1 ⎞ lbf ⎛ = ⎜178,300 2 ⎟ ⎜ ⎟⎜ ⎟ =38.5 2 fts ⎠ ⎝ 32.2ft lbm ⎠ ⎝ 12in. ⎠ in. ⎝ For the manometer, stepping through the various legs: P1 − PA = 0 PA − PB = − ρ o ghAB PB − PC = 0 PC − PD = + ρ m ghCD PD − PE = 0 PE − PF = − ρ o ghEF PF − P2 = 0 P1 − P2 = − ρ o ghAB + ρ m ghCD − ρ o ghEF = − ρ o g ( hAB + hEF ) + ρ m ghCD

From the geometry of the manometer and pipe: L = hAB − hCD + hEF → hAB + hEF = L + hCD

Substituting into the manometer equation: P1 − P2 = − ρ o g ( L + hCD ) + ρ m ghCD hCD =

2 3 2 P1 − P2 + ρ o gL 178,300 lbm fts + ( 0.87 ) ( 62.4 lbm ft )( 32.2 ft s ) (15ft ) = =50.1ft g ( ρm − ρo ) ( 32.2 ft s2 )( 62.4 lbm ft 3 ) ( 2.9-0.87 )

9- 4

Answer

9-5

In Problem P 9-4 if the flow is upward instead of downward, determine the manometer deflection, h (in ft).

Approach: The pressure drop between points 1 and 2 can be calculated with the steady, incompressible flow energy equation. The manometer equation is used to calculate the manometer deflection once the pressure drop is determined.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

The steady, incompressible flow energy equation is:

The areas at 1 and 2 are equal, so V1 = V2 . There is no pump or turbine, so hP = hT = 0 , and minor losses are

The velocity is

L V2 ρ D 2 4 ( 6 gal min ) ( 0.1337 ft 3 gal ) (1min 60s )

P1 − P2 = ρ g ( z2 − z1 ) + f

neglected. Therefore, V =

V 4V = = A π D2

The Reynolds number is:

π ( 0.0625ft )

Re =

2

=4.36

ft s

3 ρ VD ( 0.87 ) ( 62.4 ft lbm ) ( 4.36ft s )( 0.0625ft ) = = 37 µ 0.4 lbm fts

For fully developed laminar flow in a straight circular tube: Substituting this into the pressure drop equation:

f = 64 Re = 64 37 = 1.73

lbm ⎞ ⎛ ft ⎞ lbm ⎞ ( 4.26 ft s ) ⎛ ⎛ 15 ⎞ ⎛ P1 − P2 = ( 0.87 ) ⎜ 62.4 3 ⎟ ⎜ 32.2 2 ⎟ (15ft ) + (1.73) ⎜ ⎟ ( 0.87 ) ⎜ 62.4 3 ⎟ ft ⎠ ⎝ s ⎠ ft ⎠ 2 ⎝ ⎝ 0.0625 ⎠ ⎝

2

2

lbm ⎞ ⎛ lbf s 2 ⎞ ⎛ 1 ⎞ lbf ⎛ = ⎜ 230,700 2 ⎟ ⎜ ⎟⎜ ⎟ =49.8 2 fts ⎠ ⎝ 32.2ft lbm ⎠ ⎝ 12in. ⎠ in. ⎝ For the manometer, stepping through the various legs: P1 − PA = 0 PA − PB = + ρ o ghAB PB − PC = 0 PC − PD = − ρ m ghCD PD − PE = 0 PE − PF = + ρ o ghEF PF − P2 = 0 P1 − P2 = ρ o ghAB − ρ m ghCD + ρ o ghEF = ρ o g ( hAB + hEF ) − ρ m ghCD

From the geometry of the manometer and pipe: L = hAB − hCD + hEF → hAB + hEF = L + hCD

Substituting into the manometer equation: P1 − P2 = ρ o g ( L + hCD ) − ρ m ghCD hCD =

2 3 2 P1 − P2 + ρ o gL 230,700 lbm fts - ( 0.87 ) ( 62.4 lbm ft )( 32.2 ft s ) (15ft ) =-50.1ft = g ( ρm − ρo ) ( 32.2 ft s2 )( 62.4 lbm ft 3 ) ( 0.87-2.9 )

Answer

Comments: The minus sign indicates the manometer deflection is in the opposite direction than what is shown in the figure. 9- 5

9-6

Data are read from and written to spinning computer disks (3600 rpm) by small read-write heads that float above the disk on a thin (0.5µm) film of air. Consider a 10-mm by 10-mm head located 55 mm from the disk centerline. For air at 25 °C, and assuming the flow is similar to that between infinite parallel plates, determine: a. the Reynolds number based on the gap dimension b. the power required to overcome the viscous shear (in W).

Approach: Reynolds number can be obtained from its definition. Power is torque times rotational speed, and torque is forces times distance. The force is caused by the shear stress, which can be determined with Newton’s law of viscosity, Eq. 9-2.

Assumptions: 1. 2. 3.

The system is steady. The flow is fully developed between infinite parallel plates. Properties are constant.

Solution: a) The Reynolds number is defined as: ρ Vb Vb = Re =

µ

υ

V = rω , and air viscosity from Appendix A-7 at 25 ºC is υ = 15.4 × 10−6 m 2 s ,

Velocity is

µ = 1.83 × 10−5 Ns m 2 Re =

( 0.055m )( 3600 rev min )(1min

60s )( 2π rad rev ) ( 5×10-7 m )

15.4×10-6 m 2 s

= 0.67

b) Power is defined as: W = ℑω Torque is ℑ = Fr = τ Ar = τ LDr . Assuming Newtonian flow between parallel walls with one wall moving and using Newton’s law of viscosity: dV V rω τ =µ =µ =µ dy b b Substituting all the expressions into the power expression and simplifying: 2 LD 2 2 ⎛ Ns ⎞ ( 0.01m )( 0.01m ) rad ⎤ ⎛ 1Ws ⎞ 2 ⎡⎛ 3600 ⎞ W = µ r ω = ⎜1.83×10-5 2 ⎟ 0.055m 2π ( ) ⎜ ⎟ ⎢⎜ 60 ⎟ b m ⎠ 5×10-7 m s ⎥⎦ ⎝ Nm ⎠ ⎝ ⎠ ⎣⎝ =1.57W

Answer

Comments: This is only part of the power required to spin the disk. Shear forces on other parts of the rotating disk would increase the power required to spin the disk.

9- 6

9-7

Skimmers are used to remove viscous fluids, such as oil, from the surface of water. As shown on the diagram below, a continuous belt moves upward at velocity Vo through the fluid and the more viscous liquid (with density ρ and viscosity µ) adheres to the belt. A film with thickness h forms on the belt. Gravity tends to drain the liquid, but the upward belt velocity is such that net liquid is transported upward. Assume the flow is fully developed, laminar, with zero pressure gradient, and zero shear stress at the outer film surface where air contacts it. Determine an expression for the velocity profile and flow rate. Use a differential analysis similar to that used for fully developed laminar flow through an inclined pipe. Clearly state the velocity boundary conditions at the belt surface and at the free surface.

Approach: The amount of oil that is picked up by the moving belt is a result of a force balance between gravity, which causes the oil to flow downward, and shear stress, which causes the oil to flow upward. We apply a force balance in the x-direction to obtain the velocity profile, and then we integrate it to find volume flow rate.

Assumptions: 1. The system is steady. 2. The flow is fully developed between infinite parallel plates. 3. Properties are constant.

Solution: A force balance in the x-direction on the differential element shown above, assuming fully developed laminar flow with constant properties, is: ∑ Fx = 0 = τ T AT − τ B AB − dW where the weight is W = ρVg . Note there is no net pressure force. Atmospheric pressure acts all along the surface of the oil, and the infinitesimal difference in hydrostatic pressure is negligible. Letting D be the depth of the plane into the page: τ T Ddx − τ B Ddx − ρ Ddxdyg = 0 Canceling Ddx and rearranging:

τT − τ B dy

= ρg dτ = ρg dy

Recognizing the left hand side is a derivative:

τ =µ

Using Newton’s law of viscosity Separating variables and integrating twice:

Boundary conditions are: 1) at y = 0, V = VB Applying the boundary conditions: VB = 0 + 0 + C2 → C2 = VB 0=

Therefore:

ρg h + C1 µ V =

→

C1 = −

dτ d ⎛ dV ⎞ d2 V = ⎜µ = ρg ⎟=µ dy dy ⎝ dy ⎠ dy 2 d V ρg ρ g y2 V = y + C1 → = + C1 y + C2 dy µ µ 2

dV dy

2)

→

at y = h there is no shear, so d V dy = 0

ρg h µ

⎞ ρ g y2 ρ g ρ g ⎛ y2 − hy + VB = ⎜ − hy ⎟ + VB µ 2 µ µ ⎝ 2 ⎠

Volume flow rat is 2 ⎤ h h ⎡ ρg ⎛ y ⎞ − hy ⎟ + VB ⎥ Ddy V = ∫ VdA = ∫ VDdy = ∫ ⎢ ⎜ 0 0 ⎠ A ⎣ µ ⎝ 2 ⎦ 3  V ρ gh =− + VB h Answer 3µ D

9- 7

9-8

Consider a fully developed laminar flow of 20 °C water down an inclined plane that is 20° to the horizontal. The water thickness is 1-mm. The water is exposed to atmosphere everywhere, and the air exerts zero shear on the water. Using a differential analysis similar to that used for fully developed laminar flow through an inclined pipe, determine the volume flow rate per unit width (in m3/s m).

Approach: The water flows down the inclined plane due to the influence of gravity. A force balance between gravity and shear forces is solved to find the velocity profile, and then integrated to find the volume flow rate.

Assumptions: 1. 2. 3.

The system is steady. The flow is fully developed. Properties are constant.

Solution: A force balance in the x-direction on the differential element shown above, assuming fully developed laminar flow with constant properties, is: ∑ Fx = 0 = τ T AT − τ B AB + dW sin θ where the weight is W = ρVg . Note there is no net pressure force. Atmospheric pressure acts all along the surface of the oil, and the infinitesimal difference in hydrostatic pressure is negligible. Letting D be the depth of the plane into the page: τ T Ddx − τ B Ddx + ρ Ddxdyg sin θ = 0 Canceling Ddx and rearranging:

τT − τ B dy

= − ρ g sin θ dτ = − ρ g sin θ dy

Recognizing the left hand side is a derivative:

τ = µ d V dy

Using Newton’s law of viscosity d ⎛ dV ⎞ d V = − ρ g sin θ ⎜µ ⎟=µ dy ⎝ dy ⎠ dy 2 2

ρg dV =− sin θ + C1 µ dy

Separating variables and integrating twice: Boundary conditions are: 1) at y = 0, V = 0 Applying the boundary conditions: 0 = 0 + 0 + C2 → C2 = 0 0=−

Therefore:

ρg sin θ h + C1 µ V =−

→

→

V =−

ρg y2 sin θ + C1 y + C2 2 µ

at y = h there is no shear, so d V dy = 0

2)

ρg sin θ h µ

C1 =

⎛ ρg y2 ρ g ρg y2 ⎞ + sin θ sin θ hy = sin θ ⎜ hy − ⎟ 2 2 ⎠ µ µ µ ⎝

Volume flow rate is h h ⎡ ρg ⎛ y 2 ⎞⎤ V = ∫ VdA = ∫ VDdy = ∫ ⎢ ty − ⎟ ⎥Ddy ⎜ 0 0 2 ⎠⎦ A ⎣ µ ⎝ V ρ g sin θ h3 = D 3µ

For water at 20 ºC from Appendix A-6, µ = 9.85 × 10−4 Ns m 2 , ρ = 998.2 kg m3 3 2 o 2 V ( 998.2 kg m )( 9.81m s ) sin ( 20 ) ( 0.001m ) ( Ns kgm ) m3 = =0.00113 D sm 3 ( 9.85×10-4 Ns m 2 ) 3

9- 8

Answer

9-9

A biomedical device start-up company is developing a liquid drug injection device. The device uses compressed air to drive the plunger in a piston-cylinder assembly that will push the drug (viscosity and density similar to water at 10 °C) through the hypodermic needle (inside diameter 0.25 mm and length 50 mm). If the flow must remain laminar in the hypodermic needle, determine: a. the maximum flow possible (in cm3/s) b. the required air pressure for the maximum flow if the pressure at the end of the needle must be 105 kPa (in kPa). (Assume fully developed flow.)

Approach: Using a transition Reynolds number of 2100 and the Reynolds number definition, we can calculate the maximum allowable velocity and, hence, flow rate. We can use Eq. 9-13, which relates pressure drop and velocity, to calculate the required air pressure.

Assumptions: 1. The system is steady. 2. The flow is fully developed laminar flow. 3. Properties are constant.

Solution: a) Volume flow rate is defined as: V = VAx = V π D 2 4 Reynolds number is defined as: ρ VD Re =

µ

Using Re = 2100 for the transition between laminar and turbulent flow, and water properties from Appendix A-6 at 10 ºC, µ = 12.9 × 10−4 Ns m 2 , ρ = 999.6 kg m3 : V =

−4 2 2 µ Re (12.9 × 10 Ns m ) ( 2100 ) ( kgm Ns ) m = =10.8 3 ρD s 999.6 kg m 0.00025m ) ( )(

m ⎞ π ( 0.00025m ) m3 ⎛ V = ⎜ 10.8 ⎟ =5.30×10-7 s ⎠ 4 s ⎝ 2

Answer

b) From Eq. 9-13 for a horizontal flow: 8µ L V 32µ L V ∆P = = R2 D2 32 (12.9 × 10−4 Ns m 2 ) ( 0.05m )(10.8 m s )(1kN 1000N ) 32 µ L V P2 = P1 + = 105kPa+ 2 D2 ( 0.00025m ) = 105kPa+357kPa=462kPa

Answer

9- 9

9-10

The viscosity of liquids is measured with a capillary viscometer, in which a laminar flow is maintained in a small diameter tube, and the pressure drop and flow rate are measured. If the flow is fully developed, then Eq. 9-13 can be used to calculate the liquid viscosity. However, entrance effects often are present. Consider the flow of a liquid (SG = 0.92) through a tube 450-mm long and 0.75-mm in diameter. A flow of 1 cm3/s is obtained when the pressure drop is 65 kPa. Determine: a. the viscosity if the flow is fully developed (in Ns/m2) b. the viscosity if the pressure drop in the entrance length is twice that for the same length of fully developed flow (in N·s/m2).

Approach: With the given information, Eq. 9-13 is used to calculate the viscosity. If entrance effects are present, a correction to the effective length of the tube must be made, and that corrected length used in Eq. 9-13. Laminar entrance length can be estimated with Eq. 9-39.

Assumptions: 1. The system is steady. 2. The flow is laminar. 3. Properties are constant.

Solution:

V = VAx = V π D 2 4 a) Volume flow rate is defined as: Pressure drop for laminar flow of a Newtonian fluid in a circular tube is obtained from Eq. 9-13: ∆P = 8µ L V R 2 = 32µ L V D 2

Noting that A = π D 2 4 , combining these three equations, and solving for viscosity:

µ=

π D 4 ∆P 128LV

π ( 0.00075m ) ( 65 kN m 2 ) (100 cm m ) (1000 N 1kN ) 4

=

3

128 ( 0.45m ) (1cm s ) 3

=1.122×10-3

Ns m2

Answer

b) If entrance effects are taken into account (assuming ∆Pent = 2∆Pwithout ) and from Eq. 9-39 Lent = 0.065 ReD ent

∆Ptot = ∆Pent + ∆Pfully

developed

π D 4 ∆Ptot 32 µV ⎛ 32µ Lent V ⎞ 32 µ ( L − Lent ) V 32µV 2 Lent + L − Lent ) = L + L) µ = ∆Ptot = 2 ⎜ + = ⎟ 2 2 2 ( 2 ( ent D D D D 128 ( Lent + L ) V ⎝ ⎠ Because the entrance length is a function of Reynolds number, which depends on viscosity, an iterative solution is required. The procedure is: guess the viscosity, calculate the Reynolds number, calculate the entrance length, and then calculate viscosity and compare to the guessed value. Guess µ = 1.122 × 10−3 Ns m 2 ρ VD 4 ρV 4 ρV Reynolds number is defined as: Re = and V = Re = → 2 µ πD πµ D 4 ( 0.92 ) (1000 kg m 3 )(1cm3 s ) (1m 100cm ) ( Ns 2 kgm ) 3

Re =

π (1.122×10-3 Ns m 2 ) ( 0.00075m )

=1392

Lent = 0.065 (1392 )( 0.00075m ) =0.0679m π ( 0.00075m ) ( 65 kN m 2 ) (100 cm m ) (1000 N 1kN ) 4

µ=

3

128 ( 0.45m+0.0679m ) (1cm s ) 3

=9.75×10-4

Ns m2

This does not match the guessed value, so continuing the iteration until it converges: µ = 0.953 × 10−4 Ns m 2 Re = 1640 Lent = 0.080m Answer Therefore, the error caused if entrance effects are not taken into account is: 1.122 × 10 −3 − 9.53 × 10 −4 error= × 100 = 17.7% Answer 9.53 × 10−4 9- 10

9-11

A machine tool manufacturer is considering using gravity flow to supply cutting oil (SG = 0.87, µ = 0.003 N·s/m2) to the tool and workpiece. The vertical 5-mm diameter tube connecting the oil reservoir to the workpiece is very long so the flow can be assumed fully developed; in addition, the depth of oil in the reservoir is negligible compared to the tube length. The pressure is atmospheric at the exit of the tube and at the surface of the reservoir. Determine the volumetric flow rate of the oil (in cm3/s).

Approach: Flow is caused by a balance between gravity, pressure, and shear stress/viscosity forces. Equation 9-13 was developed for this situation, so we will apply it directly. Because that equation uses the average velocity, volume flow rate is obtained easily from its definition.

Assumptions: 1. The system is steady. 2. The flow is fully developed and laminar. 3. Properties are constant.

Solution:

Volume flow rate is defined as: V = VAx = V π D 2 4 Pressure drop for fully developed laminar flow of a Newtonian fluid in a circular tube is obtained from Eq. 9-13: 8µ L V 32µ L V ∆P = + ρ gL sin θ = + ρ gL sin θ 2 R D2 For the present problem θ = 90o . With atmospheric pressure at the surface of the reservoir and at the exit of the tube, ∆P = 0 . Using this information and solving for velocity: − ρ g sin θ D 2 V = 32 µ Note that in the development of Eq. 9-13, the x-direction was in opposition to the gravity force, so the negative sign indicates a downward flow. Using the given information: 3 o − ρ g sin θ D 2 - ( 0.87 ) (1000 kg m ) sin ( 90 ) ( 0.005m ) m = =0.227 V = 2 2 32µ s 32 ( 0.003 Ns m )( kgm Ns ) 2

Checking the Reynolds number: ρ VD 4 ρV and V = Re = µ π D2 Re =

( 0.87 ) (1000 kg

→

Re =

Answer

4 ρV πµ D

m 3 ) ( 0.227 m s )( 0.005m ) ( Ns 2 kgm )

0.003 Ns m 2 This is laminar flow, so our assumption checks out. Therefore, the volume flow rate is: 2 m ⎞ π ( 0.005m ) m3 ⎛ =4.46×10-6 V = ⎜ 0.227 ⎟ s ⎠ 4 s ⎝

=329

Answer

9- 11

9-12

A manometer, with pressure taps 25 ft apart, is used to measure the pressure drop of oil (SG = 0.82) flowing in a 1.5-in. pipe with a volumetric flow rate of 4 ft2/min. The manometer fluid is mercury (SG = 13.6). From the lower pressure tap to the surface of the mercury highest in the manometer is 2 ft, and the distance from the upper pressure tap to the same height in the mercury is 4 ft. For a manometer deflection of 4 in., determine: a. the flow direction b. the friction factor c. whether the flow is laminar or turbulent d. the oil viscosity (in lbm/ft s).

Approach: Flow direction can be determined by inspection, and the friction factor can be calculated using the steady, incompressible flow energy equation. Whether the flow is laminar or turbulent must be determined by assuming one or the other, using a correlation for the friction factor, calculating the viscosity, and then checking the Reynolds number. The manometer equation is used to calculate the pressure drop.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: a) Hydrostatic pressure increases farther down in a fluid. The manometer indicates that P2 < P1 . The only way this could occur would be if flow was from point 1 to point 2 with friction losses. b, c, d) The steady, incompressible flow energy equation is: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

Answer

The areas at 1 and 2 are equal, so V1 = V2 . There is no pump or turbine, so hP = hT = 0 , and minor losses are neglected. Therefore,

P1 − P2 L V2 + ( z1 − z2 ) = f ρo g D 2g

From the definition of volume flow rate, the velocity is

V =

→

f =

⎤ 2 Dg ⎡ P1 − P2 + ( z1 − z2 ) ⎥ 2 ⎢ L V ⎣ ρo g ⎦

4 ( 4 ft 3 min ) (1min 60s ) V 4V ft = = =5.43 2 s A π D2 π ( 0.125ft )

For pressure drop, using the manometer equation and stepping through the various legs: P1 − PA = − ρ o gh1 A PA − PB = 0 PB − PC = ρ m ghBC PC − P2 = ρ o ghC 2 P1 − P2 = − ρ o gh1 A + ρ m ghBC + ρ o ghC 2 Substituting this into the friction factor expression: 2 ⎤ 2 ( 0.125ft ) ( 32.2 ft s ) ⎡ ρm 2 Dg ⎡ 13.6 ⎤ f = h h h z z − + + + − = ( 1 2 )⎥ ⎢ 1A 2 ⎢ -4.33ft+ 0.82 ( 0.33ft ) +2ft+2ft ⎥ =0.0562 ρ o BC C 2 LV 2 ⎣ ⎦ ( 25ft )( 5.43ft s ) ⎣ ⎦ Assuming the flow is fully developed laminar flow: 3 ρ VDf ( 0.82 ) ( 62.4 ft lbm ) ( 5.43ft s )( 0.125ft )( 0.0562 ) 64 f = → µ= = = 0.0305lbm fts Answer 64 64 Re

Check the Reynolds number:

Re =

3 ρ VD ( 0.82 ) ( 62.4 ft lbm ) ( 5.43ft s )( 0.125ft ) = = 1140 0.0305lbm fts µ

which is laminar, so our assumption is valid. 9- 12

9-13

Develop an expression for the velocity profile for fully developed laminar flow between stationary infinite parallel plates. Use an approach similar to that applied in Section 9.3 for a circular tube.

Approach: Flow through this infinite parallel plate channel results from a balance among viscous, pressure, and gravity forces. A force balance on a differential element is used to determine the velocity profile, similar to what was done for the circular tube.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: A force balance in the x-direction on the differential element shown above, assuming fully developed laminar flow with constant properties, is: ∑ Fx = 0 = P1 A1 − P2 A2 − 2τ As − mg sin θ Let D be the depth of the plane into the page: Using Newton’s law of viscosity

τ = µ d V dn where n is the direction from the wall.

To put this in terms of y (from the channel centerline), y = H − n

→

dy = d ( H − n ) = − dn .

Substituting these expressions into the force balance, and letting A1 = A2 = 2 Dy, As = DL, and m = ρ D 2 yL : dV DL − ρ L 2 yDg sin θ = 0 dy Canceling 2D, and simplifying: P1 − P2 µ dV − ρ g sin θ = − L y dy Separating variables and integrating: 2 1 ⎡P − P 1 ⎡P − P ⎤ ⎤y V = − ⎢ 1 2 − ρ g sin θ ⎥ + C1 − ∫ ⎢ 1 2 − ρ g sin θ ⎥ ydy = ∫ d V → µ ⎣ L µ⎣ L ⎦ ⎦ 2 The boundary condition is: at y = H, V = 0 . Applying the boundary condition: 2 1 ⎡P − P ⎤H C1 = ⎢ 1 2 − ρ g sin θ ⎥ µ⎣ L ⎦ 2 Therefore: 2 1 ⎡ P1 − P2 H 2 ⎡ P1 − P2 ⎤ 2 ⎤⎡ ⎛ y ⎞ ⎤ 2 − − = − − V =− ρ θ ρ θ sin sin 1 g H y g ) 2µ ⎢ L ⎥( ⎥⎢ ⎜ H ⎟ ⎥ 2 µ ⎢⎣ L ⎦ ⎣ ⎦ ⎣⎢ ⎝ ⎠ ⎦⎥

( P1 − P2 ) 2 Dy + 2µ

Comments: Note the parabolic velocity profile, which is similar to that in a circular tube.

9- 13

Answer

9-14

In an inclined 50-mm diameter pipe, a fluid (SG = 0.88) flows with a volumetric flow rate of 0.003 m3/s. The gage pressure at the pipe inlet is 720 kPa. The pipe outlet is at atmospheric pressure and is 15 m above the inlet. Determine the head loss between the inlet and outlet (in m).

Approach: We can use Eq. 9-41 directly to find head loss.

Assumptions: 1. 2. 3. 4.

The system is steady. The flow is fully developed. Properties are constant. Neglect minor losses.

Solution: The steady, incompressible flow energy equation is: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g The areas at 1 and 2 are equal, so V1 = V2 . There is no pump or turbine, so hP = hT = 0 , and minor losses are neglected. Therefore,

∑h

L

=

P1 − P2 + ( z1 − z2 ) ρg

kN ⎛ 1000kg m ⎞ ⎜ ⎟ m 2 ⎝ kN s 2 ⎠ + ( -15m ) =83.4-15=68.4m h = ∑ L kg ⎞ ⎛ m⎞ ⎛ ( 0.88) ⎜1000 3 ⎟ ⎜ 9.81 2 ⎟ m ⎠⎝ s ⎠ ⎝

( 720-0 )

9- 14

Answer

9-15

The pipe exit in Problem P 9-14 is lowered to the same elevation as the inlet. Determine the inlet pressure for this new condition (in kPa).

Approach: We can use the steady flow energy equation to find the head loss in the inclined pipe. The frictional head loss is the same in a horizontal pipe, so we can again use the steady flow energy equation, but this time we will determine the new inlet pressure.

Assumptions: 1. 2. 3. 4.

The system is steady. The flow is fully developed. Properties are constant. Neglect minor losses.

Solution: The steady, incompressible flow energy equation is: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g The areas at 1 and 2 are equal, so V1 = V2 . There is no pump or turbine, so hP = hT = 0 , and minor losses are neglected. Therefore, for when the outlet is 15 m above the inlet

∑h

L

=

P1 − P2 + ( z1 − z2 ) ρg

kN ⎛ 1000kg m ⎞ ⎜ ⎟ m 2 ⎝ kN s 2 ⎠ + ( -15m ) =83.4-15=68.4m h Answer = ∑ L kg ⎞ ⎛ m⎞ ⎛ ( 0.88) ⎜1000 3 ⎟ ⎜ 9.81 2 ⎟ m ⎠⎝ s ⎠ ⎝ If the pipe is in a horizontal orientation, z1 = z2 , with the same flow rate, and assuming the same total head loss, solving the energy equation for P1: P1 = P2 + ∑ hL ρ g = 0 + ( 68.4m )( 0.88 ) (1000 kg m3 )( 9.81m s 2 )( kN s 2 1000kg m ) =591kPa Answer

( 720-0 )

9- 15

9-16

An air conditioning duct is 25-cm square and must convey 25 m3/min of air at 100 kPa, 25 °C. The duct is made of sheet metal that has a roughness of approximately 0.05 mm. Determine the pressure drop for 25-m of horizontal duct run (in kPa and mm of water).

Approach: Pressure drop is calculated directly with the steady, incompressible flow energy equation

Assumptions: 1. 2. 3. 4.

The system is steady. The flow is fully developed. Properties are constant. The duct is smooth.

Solution: The steady, incompressible flow energy equation is: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g The duct is constant area and horizontal, so V1 = V2 and z1 = z2 There is no pump or turbine, so hP = hT = 0 , and there are no minor losses: P1 − P2 L V2 L V2 = f → P1 − P2 = f ρ D 2g D 2g ρg The friction factor is a function of Reynolds number and roughness. For air from Appendix A-7 at 100 kPa, 25 ºC, µ = 1.83 × 10−5 Ns m 2 , ρ = 1.169 kg m3 . Velocity is :

3 V ( 25 m min ) (1min 60s ) m V = = =6.67 A s ( 0.25m )( 0.25m )

Because this is a non-circular duct, the hydraulic diameter must be used: 4 ( 0.25m )( 0.25m ) 4 Ax =0.25m Dh = = 4 ( 0.25m ) Pwetted

ρ VDh (1.169 kg m ) ( 6.67 m s )( 0.25m ) =106,500 = µ 1.83×10-5 Ns m 2 3

Re =

This is turbulent flow so with a smooth duct: ⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0 ⎞1.11 1 6.9 ⎤ 1.8log = −1.8log ⎢⎜ + = − ⎥ ⎢⎜ ⎥ → f = 0.0176 ⎟ ⎟ + Re ⎦⎥ 106,500 ⎦⎥ f ⎣⎢⎝ 3.7 ⎠ ⎣⎢⎝ 3.7 ⎠ Substituting known values into the pressure drop equation: kg ⎞ ( 6.67 m s ) ⎛ 1kN s 2 ⎞ ⎛ 25 ⎞ ⎛ P1 − P2 = ( 0.0176 ) ⎜ 1.169 ⎜ ⎟ =0.0457kPa ⎟⎜ ⎟ 2 m3 ⎠ ⎝ 0.25 ⎠ ⎝ ⎝ 1000kg m ⎠ Using the manometer equation ∆P = ρ w gh 2

Answer

2 2 ∆P ( 0.0457 kN m )(1000kgm 1kNs ) hP = = =0.00466m of water = 4.66mm of water ρw g (1000 kg m3 )( 9.81m s2 )

9- 16

Answer

9-17

A manufacturer develops a new type of flow control valve. Before it can be advertised and sold, its loss coefficient must be determined. The valve is installed in a 6-in. pipe and 2 ft3/s of water flows through it. The pressure drop is measured with a manometer whose fluid has a specific gravity of 1.3. The manometer deflection is 7.5 in. Determine the loss coefficient for the valve.

Approach: We can use the definition of head loss for minor losses to calculate the loss coefficient. The pressure drop (head loss) can be determined from the manometer equation.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: The head loss due to minor losses is given by: V2 P −P hL = 1 2 hL = K L and 2g ρg Combining the expressions and solving for the loss coefficient: 2 ( P1 − P2 ) KL = ρw V 2

4 ( 2 ft 3 s ) 4V ft V V = = The velocity in the upstream pipe is =10.2 = 2 2 s A πD π ( 0.5ft ) For the manometer, stepping through the various legs: P1 − PA = − ρ w gh1 A PA − PB = 0 PB − PC = ρ m ghBC PC − P2 = + ρ w ghC 2 P1 − P2 = − ρ w ghAB + ρ m ghBC + ρ w ghC 2 = − ρ w g ( hBC + hC 2 ) + ρ m ghBC + ρ w ghC 2 = ( ρ m − ρ w ) ghBC

Substituting this expression into the loss coefficient equation: 2 ( ρ m − ρ w ) ghBC 2 ( SGm − 1) ghBC KL = = ρw V 2 V2 KL =

2 (1.3-1) ( 32.2 ft s 2 ) ( 7.5 12 ft )

(10.2 ft s )

2

= 0.116

Answer

9- 17

9-18

When pumping a fluid, the pressure at the entrance to the pump must never drop below the saturation pressure of the fluid. If the pressure does drop below the saturation pressure, cavitation (the forming of vapor bubbles) occurs which can damage the pump impeller. Consider the system shown below constructed of commercial steel pipe and threaded connections. For water at 10 °C, determine the maximum possible flow rate without cavitation occurring (in m3/s).

Approach: Cavitation will occur if the pressure at point 2 falls below the saturation pressure of 10 ºC water. We can solve for the flow rate using the steady, incompressible flow energy equation. The pressure at 2 is the saturation pressure.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

The steady, incompressible flow energy equation is:

There is no pump or turbine, so hT = hp = 0 . We assume the area at 1 large, so V1 ≈ 0 , and the pressure at 1 is

atmospheric, so P1 = 101.3 kPa. At point 2, from the saturated water table at 10 ºC, Psat (10 o C ) = 1.23kPa . The losses include one entrance, two bends, and line loss:

∑h

L

2 ⎛ L ⎞V = ⎜ f + K ent + 2 K bend ⎟ . Therefore, ⎝ D ⎠ 2g

2 P1 − P2 ⎛ L ⎞V + ( z1 − z2 ) = ⎜ f + K ent + 2 K bend + 1⎟ ρg ⎝ D ⎠ 2g The friction factor depends on flow, so an iterative solution is required. From Figure 9-15, for an zero area ratio at the entrance Kent = 0.5. From Table 9-3, for a regular 90º threaded bend, Kbend = 1.5. Velocity is: 4 (V m3 s ) 4V m V =226.4V = (1) V = = 2 2 s A πD π ( 0.075m )

For commercial steel pipe ε =0.045mm , and for water from Appendix A-6 at 10 ºC, µ = 12.9 × 10−4 Ns m 2 , ρ = 999.6 kg m3 . The friction factor is a function of Reynolds number and roughness. 3 ρ VD ( 999.6 kg m )( 226.2V m s ) ( 0.075m ) Re = = = 1.316 × 106 V (2) µ 12.9 × 10−4 Ns m 2 The volume flow will be large enough so that the flow will be turbulent, so ⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0.045 75 ⎞1.11 6.9 ⎤ 1 1.8log (3) = −1.8log ⎢⎜ + = − ⎥ ⎢ ⎥ ⎟ ⎜ ⎟ + Re ⎥⎦ Re ⎥⎦ f ⎢⎣⎝ 3.7 ⎠ ⎢⎣⎝ 3.7 ⎠ Substituting known values into the pressure drop equation: (101.3-1.23) kN m 2 (1000 kg m kN s 2 ) V2 13 ⎡ ⎤ + -5m = f +0.5+2 1.5 +1 ( ) ( ) ⎢ 0.075 ⎥ 2 9.81m s 2 ⎣ ⎦ ( ( 999.6 kg m3 )( 9.81m s 2 ) ) 102 = ( 4.5 + 173.3 f ) V 2

(4)

Iterating on the four equations given above, we obtain: V = 0.0147 m3 s V = 3.33m s f = 0.0271

Re = 19, 200

Answer

Comments: When the water enters the pump, it accelerates more. Hence, the actual minimum pressure at 2 would need to be greater than 1.23 kPa. The pump manufacturer would specify the minimum required pressure. 9- 18

9-19

Fire codes mandate that the pressure drop in horizontal runs of commercial steel pipe must not exceed 1.0 lbf/in.2 per 150-ft of pipe for flows up to 500 gal/min. For a water temperature of 50 °F, determine the minimum pipe diameter required (in in.). Is the number you calculated feasible?

Approach: Pressure loss depends on velocity, which depends on pipe diameter. The friction factor is a function of velocity. Hence, an iterative solution is required. The steady, incompressible flow energy equation is used to determine the minimum pipe diameter.

Assumptions: 1. 2. 3. 4.

The system is steady. The flow is fully developed. Properties are constant. Neglect minor losses.

Solution: The steady, incompressible flow energy equation is: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g The areas at 1 and 2 are equal, so V1 = V2 . The pipe is horizontal, so z1 = z2 . There is no pump or turbine, so hP = hT = 0 , and minor losses are neglected. Therefore,

P1 − P2 L V2 = f ρa g D 2g

4 ( 500 gal min ) ( 0.1337ft 3 gal ) (1min 60s ) 1.419 ft 4V V = = = (1) 2 2 A π D2 π ( Dft ) ( Dft ) s The friction factor is a function of Reynolds number and roughness. From Table 9-2 for commercial steel pipe, ε =0.00015ft , and for water from Appendix B-6 at 50 ºF, µ = 88 × 10−5 lbm fts , ρ = 62.4 lbm ft 3 :

Velocity is : V =

Re =

3 2 ρ VD ( 62.4 lbm ft )(1.419 D ft s ) ( Dft ) 100, 600 = = µ 88 × 10−5 lbm fts D

(2)

For any reasonable size diameter, the flow will be turbulent, so ⎡⎛ 0.00015 D ⎞1.11 6.9 ⎤ 1 = −1.8log ⎢⎜ (3) ⎥ ⎟ + 3.7 Re ⎦⎥ f ⎠ ⎣⎢⎝ Substituting known values into the pressure drop equation: (1lbf in.2 )( 32.2 ft lbm lbf s2 )(144in.2 ft 2 ) = f ⎛ 150 ⎞ ( V ft s )2 → 0.991 = f V 2 (4) ⎜ ⎟ 62.4 lbm ft 3 2 D ⎝ D ⎠ An iterative solution can be found using the above four equations. The procedure to use is: assume a diameter D, calculate the velocity V, Reynolds number Re, and friction factor f, and then use equation 4 to calculate the diameter. Compare the calculated diameter with the guessed one; iterate until converged. Performing the iteration: D = 0.513ft = 6.2 in. V = 5.39 ft s Re=196,000 f =0.0175 Answer

Comments: The next large standard pipe size would be used.

9- 19

9-20

The owners of a luxurious mountain resort want to install a fancy water fountain. The artist’s initial design uses 75-m of 7.5-cm diameter commercial steel pipe ending in a nozzle with a diameter of 3.75-cm with a 40-kW pump to pull water from a lake above the resort at a flow rate of 0.05 m3/s. To save operating costs, the owners want to remove the pump and rely only on gravity head to power the fountain. Assuming the friction factor is 0.016 for both cases and neglecting minor losses, determine: a. the flow rate if the pump is removed from the system (in m3/s) b. the height of the water jet with and without the pump if the nozzle is pointed vertically up (in m).

Approach: The elevation difference between the lake surface and the nozzle exit is not given and must be determined. The steady, incompressible flow energy equation can be used to calculate the elevation difference and the height of the fountain.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant and evaluated at 10 ºC.

Solution: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

a) The steady, incompressible flow energy equation is:

There is no turbine, so hT = 0 . At points 1 and 2 the pressure is atmospheric and the area at 1 is very large, so V1 = 0 and P1 = P2 . We will assume no minor losses. Therefore, 2

2 L V pipe Vnozzle + D 2g 2g    P = ρVgh The pump head can be evaluated from WP = mgh P . For water from Appendix A-10 at 10 ºC,

z1 − z2 = −hP + f

ρ = 999.6 kg m3 hP =

( 40,000W ) ( N s 2 kg m ) W P =81.6m =  ρVg ( 999.6 kg m3 )( 0.05 m3 s )( 9.81m s2 ) The velocity at the nozzle exit is:

Likewise, for the velocity in the pipe,

Vnozzle = V pipe =

4 ( 0.05 m3 s ) V 4V m =45.3 = = A π Dnozzle 2 π ( 0.0375m )2 s

4 ( 0.05 m3 s ) V 4V m = = =11.3 A π D 2 pipe π ( 0.075m )2 s

Note that Vnozzle = 4 V pipe . Therefore, z1 − z2 = −81.6m +

( 45.3m s )

⎛ 75 ⎞ (11.3m s ) + ( 0.016 ) ⎜ ⎟ 2 2 ( 9.81m s ) ⎝ 0.075 ⎠ 2 ( 9.81m s 2 ) 2

2

= −81.6 + 104 + 104 = 127.4m Now using the energy equation without the pump: 2 2 2 L V pipe Vnozzle L ⎞ V pipe ⎛ z1 − z2 = f + = ⎜ 16 + f ⎟ D 2g D ⎠ 2g 2g ⎝ Solving for the velocity in the pipe: ⎡ 2 ( 9.81m s 2 ) (127.4m ) ⎤ ⎥ V pipe =⎢ ⎢⎣16+ ( 0.016 )( 75 0.075 ) ⎥⎦ Therefore, the flow rate without the pump is: 2 V = VA = ( 8.83m s )( π 4 )( 0.075m ) =0.039 m3 s ⎡ 2 g ( z1 − z2 ) ⎤ =⎢ ⎥ ⎣ 16 + f L D ⎦

0.5

9- 20

0.5

=8.83

m s

Answer

b) To determine the height of the fountain, apply the energy equation between points 2 and 3. The pressure is the same and the velocity at 3 is zero, so V3 = 0 and P2 = P3 : V2 V22 + z 2 = 3 + z3 2g 2g

→

z3 − z 2 = z3 − z 2 =

With the pump:

V22 2g

( 45.3m s )

2

2 ( 9.81m s 2 )

=104.6m

Without the pump, the velocity at the nozzle exit is 4 ( 0.039 m3 s ) m =35.3 Vnozzle = 2 s π ( 0.0375m ) z3 − z 2 =

( 35.3m s )

2

2 ( 9.81m s 2 )

=63.6m

Answer

Comments: Even without the pump, the fountain still reaches an impressive height.

9- 21

Answer

9-21

At an oil tank farm, a vandal opens a valve at the end of a 5-cm diameter, 50-m long horizontal pipe from the bottom of a large diameter oil tank. The oil tank is open to the atmosphere, and the oil depth is 6.5 m. The oil has a SG = 0.85 and a kinematic viscosity of 6.8×10-4 m2/s. Neglecting minor losses, determine the initial flow rate from the tank (in m3/s).

Approach: The flow rate depends on a balance between the head supplied by the oil depth and the head loss in the drain pipe. The steady, incompressible flow energy equation can be used to determine the flow. An iterative solution may be required because the friction factor is a function of the flow (velocity).

Assumptions: 1. 2. 3. 4.

The system is steady. The flow is fully developed. Properties are constant. Neglect minor losses.

Solution: The steady, incompressible flow energy equation is: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g The areas at 1 and 2 are large, so V1 = V2 ≈ 0 . The pressures at 1 and 2 are both atmospheric, so P1 = P2 . There is no pump or turbine, so hP = hT = 0 , and minor losses are neglected. Therefore, L V2 D 2g The friction factor is a function of Reynolds number and roughness. Assuming a smooth pipe: ρ VD VD ( V m s )( 0.05m ) = = = 73.5 V Re = µ ν 6.8×10-4 m 2 s For any reasonable velocity, the flow will be laminar (check this). Therefore, for fully developed laminar flow in a straight circular tube: f = 64 Re = 64ν VD Substituting this into the pressure drop equation and solving for velocity: 2 g ( z1 − z2 ) D 2 64ν L V 2 z1 − z2 = V = → 64ν L VD D 2 g z1 − z2 = f

V =

2 ( 9.81m s 2 ) ( 6.5m )( 0.05m ) 64 ( 6.8×10-4 m 2 s ) ( 50m )

2

=0.147

m s

Answer

Checking the Reynolds number: ( 0.147m s )( 0.05m ) Re = = 10.8 6.8×10-4 m 2 s Laminar flow, so our assumption is correct.

Comments: Note that the velocity calculated above is the initial velocity, assuming that the transient start-up of the flow is brief and the flow is quasi-steady. As the oil level drops, then the velocity will decrease.

9- 22

9-22

In a large convention center, heated air at 85 °F must be conveyed from the furnace room to the display rooms through a 500-ft smooth duct. The required flow rate is 7500 ft3/min. If the pressure loss must not exceed 2.5 in. of water, determine: a. the minimum diameter required (in in.) b. pumping power required (in hp).

Approach: Pressure loss depends on velocity, which depends on pipe diameter. The friction factor is a function of velocity. Hence, an iterative solution is required. The steady, incompressible flow energy equation is used to determine the minimum pipe diameter.

Assumptions: 1. 2. 3. 4.

The system is steady. The flow is fully developed with constant properties. Neglect minor losses. Air is an ideal gas.

Solution: The steady, incompressible flow energy equation is: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g The areas at 1 and 2 are equal, so V1 = V2 . The pipe is horizontal, so z1 = z2 . There is no pump or turbine, so P1 − P2 L V2 = f ρa g D 2g The pressure drop can be obtained from the manometer equation: ρ ∆P ∆P = ρ a g ∆ha = ρ w g ∆hw → = ∆ha = w ∆hw ρa g ρa Assuming air is an ideal gas at one atmosphere: 2 2 2 lbm PM (14.7 lbf in. ) ( 28.97 lbm lbmol ) (144in. ft ) =0.0728 3 ρa = = 1545ft lbf lbmR 85+460 R RT ft ( )( )

hP = hT , and minor losses are neglected. Therefore,

∆P 62.4 lbm ft 3 1ft ⎞ = ( 2.5in.) ⎛⎜ ⎟ =178.6ft 12in. ρ a g 0.0728lbm ft 3 ⎝ ⎠

4 ( 7500 ft 3 min ) (1min 60s ) 159.2 ft 4V V = = = (1) 2 2 A π D2 π ( Dft ) ( Dft ) s The friction factor is a function of Reynolds number and roughness. Assuming a smooth pipe, and for air from Appendix B-7 at 85 ºF, µ = 1.259 × 10−5 lbm fts :

Velocity is :

V =

3 2 ρ VD ( 0.0728ft lbm )(159.2 D ft s ) ( Dft ) 920, 600 Re = = = 1.259 × 10−5 lbm fts D µ For any reasonable size diameter, the flow will be turbulent, so with ε = 0

⎡⎛ ε D ⎞1.11 6.9 ⎤ = −1.8log ⎢⎜ ⎥ ⎟ + Re ⎦⎥ f ⎣⎢⎝ 3.7 ⎠ Substituting known values into the pressure drop equation: 2 f V2 ⎛ 500 ⎞ ( V ft s ) D → = 178.6 ft = f ⎜ ⎟ 23.0 ⎝ D ⎠ 2 ( 32.2 ft s 2 ) 1

(2)

(3)

(4)

An iterative solution can be found using the above four equations. The procedure to use is: assume a diameter D, calculate the velocity V, Reynolds number Re, and friction factor f, and then use equation 4 to calculate the

9- 23

diameter. Compare the calculated diameter with the guessed one; iterate until converged. Performing the iteration: D = 1.70ft V = 55.1ft s Re=541,500 f =0.0129 Answer The next large standard duct size would be used.   b) The pumping power is W = mgh = ρ Vh a

a

a

⎛ ⎛ lbfs 2 ⎞ ⎛ 1hp ⎞ ft ⎞ ⎛ ft ⎞ ⎛ 1min ⎞ ⎛ ft ⎞ W = ⎜ 0.0728 ⎟ ⎜ 7500 ⎟⎜ ⎟⎜ ⎟ ⎟ ⎜ 32.2 2 ⎟ (178.6ft ) ⎜ lbm min 60s s 32.2ft lbm 550 ftlbf s ⎠⎝ ⎠ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎝ ⎠⎝ =2.96hp Answer 3

3

9- 24

9-23

Consider a heat exchanger that has 1000 2.5-cm diameter smooth tubes in parallel, each 6-m long. The total water flow of 1 m3/s at 10 °C flows through the tubes. Neglecting entrance and exit losses, determine: a. the pressure drop (in kPa) b. the pumping power required (in kW) c. the pumping power for the same flow rate if solid deposits from the water build up on the inner surface of the pipe with a thickness of 1-mm and an equivalent roughness of 0.4 mm.

Approach: Because the tubes operate in parallel, all tubes have the same pressure drop, so we only need to calculate the pressure drop in one tube. The steady, incompressible flow energy equation is used.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

a) The steady, incompressible flow energy equation is:

There is no pump or turbine, so hT = hP = 0 . At points 1 and 2 the pipe areas are the same, so V1 = V2 . L V2 ρ D 2 The friction factor is a function of Reynolds number and roughness. For smooth pipe ε =0 , and for water from Appendix A-6 at 10 ºC, µ = 12.9 × 10−4 Ns m 2 , ρ = 999.6 kg m3 . P1 − P2 = f

Therefore,

The velocity is in one tube is:

V =

4 (1.0 m3 s ) 4V m V =2.04 = = 2 2 A Nπ D s (1000 ) π ( 0.025m )

ρ VD ( 999.6 kg m ) ( 2.04 m s )( 0.025m ) = = 39,500 12.9 × 10−4 Ns m 2 µ 3

Re =

⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0 ⎞1.11 6.9 ⎤ = −1.8log ⎢⎜ + = − 1.8log ⎥ ⎢ ⎥ ⎟ ⎜ ⎟ + 39,500 ⎦⎥ Re ⎦⎥ f ⎣⎢⎝ 3.7 ⎠ ⎣⎢⎝ 3.7 ⎠

1

This is turbulent, so

→

f = 0.0219

kg ⎞ ( 2.04 m s ) ⎛ N s 2 ⎞ ⎛ 1kN ⎞ ⎛ 6 ⎞⎛ P1 − P2 = ( 0.0219 ) ⎜ 999.6 ⎜ ⎟⎜ ⎟⎜ ⎟ ⎟ =10.9kPa m3 ⎠ 2 ⎝ 0.025 ⎠ ⎝ ⎝ kg m ⎠ ⎝ 1000N ⎠ 2

Therefore,

Answer

⎛ m3 ⎞ ⎛ kN ⎞ ⎛ kW s ⎞ b) Pumping power for the total flow rate is: W P = V ∆P = ⎜ 1 Answer ⎟ ⎜ 10.9 2 ⎟ ⎜ ⎟ =10.9kW s m ⎠ ⎝ kN m ⎠ ⎝ ⎝ ⎠ c) The new pipe diameter is D = 0.025m-2 ( 0.001m ) =0.023m . Therefore, ε D = 0.4 23 = 0.0174

V = 2.04 ( 25 23) = 2.41m s 2

⎡⎛ ε D ⎞ = −1.8log ⎢⎜ ⎟ f ⎢⎣⎝ 3.7 ⎠

1.11

1

+

→

Re = ( 999.6 )( 2.41)( 0.023) (12.9 × 10 −4 ) = 43, 000

⎡⎛ 0.0174 ⎞1.11 6.9 ⎤ 6.9 ⎤ ⎥ = −1.8log ⎢⎜ ⎥ ⎟ + 43, 000 ⎥⎦ Re ⎥⎦ ⎢⎣⎝ 3.7 ⎠

→

f = 0.0472

kg ⎞ ( 2.41m s ) ⎛ N s 2 ⎞ ⎛ 1kN ⎞ ⎛ 6 ⎞⎛ P1 − P2 = ( 0.0472 ) ⎜ ⎜ ⎟⎜ ⎟⎜ 999.6 3 ⎟ ⎟ =35.7kPa m ⎠ 2 ⎝ 0.023 ⎠⎝ ⎝ kg m ⎠ ⎝ 1000N ⎠ 2

⎛ m3 ⎞ ⎛ kN ⎞ ⎛ kW s ⎞ W P = V ∆P = ⎜ 1 ⎟ ⎜ 35.7 2 ⎟ ⎜ ⎟ =35.7kW m ⎠ ⎝ kN m ⎠ ⎝ s ⎠⎝

Answer

Comments: Fouling can have a significant effect on flow rate. Note, however, that with centrifugal pumps, the motor power is fixed, so the flow rate would decrease. 9- 25

9-24

The piping system that connects one reservoir to a second reservoir consists of 150-ft of 3-in. cast iron pipe that has four flanged elbows, a well-rounded entrance, and sharp-edged exit, and a fully open gate valve. For 75 gal/min of water at 50 °F, determine the elevation difference between the two reservoirs (in ft).

Approach: The flow rate is set by the balance between the head caused by the difference in reservoir elevations and the frictional losses in the pipe. The steady, incompressible flow energy equation applied between points 1 and 2 is used to determine the elevation difference.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: The steady, incompressible flow energy equation is:

P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

There is no pump or turbine, so hT = hp = 0 . We assume the surface of the reservoirs are large, so V1 ≈ V2 ≈ 0 , and the pressure at 1 and 2 are atmospheric, so P1 = P2. Therefore, taking into account all the minor losses: 2 ⎛ L ⎞V z1 − z2 = ⎜ f + K ent + K valve + K exit + 4 K bend ⎟ ⎝ D ⎠ 2g From Table 9-3, Figure 9-13 and Figure 9-14, Kent = 0.04, Kbend = 0.3, and Kvalve = 0.15. For cast iron pipe ε =0.00085ft , and for water from Appendix B-6 at 50 ºF, µ = 88 × 10−5 lbm ft s , ρ = 62.4 lbm ft 3 . The velocity is determined from conservation of mass for incompressible flow: 4 ( 75gal ft 3 )( 0.1337ft 3 gal ) (1min 60s ) 4V ft V =3.40 = V = = 2 A π D2 s π ( 0.25ft ) The friction factor is a function of Reynolds number and roughness. 3 ρ VD ( 62.4 lbm ft ) ( 3.40 ft s )( 0.25ft ) Re = = = 60, 400 88 × 10−5 lbm ft s µ The flow is turbulent, so ⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0.00085 0.25 ⎞1.11 1 6.9 ⎤ = −1.8log ⎢⎜ + = − 1.8log ⎥ ⎢⎜ ⎥ → f = 0.0289 ⎟ ⎟ + 3.7 60, 400 ⎦⎥ Re ⎦⎥ f ⎠ ⎣⎢⎝ 3.7 ⎠ ⎣⎢⎝ 150 ⎡ ⎤ ( 3.40 ft s ) z1 − z2 = ⎢( 0.0289 ) + 0.04 + 0.15 + 1.0 + 4 ( 0.3 ) ⎥ = 3.54ft 0.25 ⎣ ⎦ 2 ( 32.2 ft s 2 ) 2

9- 26

Answer

9-25

Vandals open the drain valve on a water tower that is 10-m in diameter with a water depth of 8 m. The water flows out a sharp-edged opening into a horizontal 30-m long pipe, both of which are 10-cm in diameter; the gate valve in the pipe is half opened. Assume the friction factor is 0.016. Determine: a. the time required for the tank to drain (in min) b. the time required for the tank to drain if only the sharp-edged opening and the valve are present (in min) c. the appropriateness of the friction factor value used.

Approach: Conservation of mass is needed to determine the time to drain the pool. The driving pressure head decreases with time, so the head as a function of time is needed. The open system conservation of mass equation and the steady, incompressible flow energy equation are used to calculate the time.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: a) The open system conservation of mass equation is for the control volume defined as the water in the pool: dm dt = m in − m out . Water only leaves, and expressing the mass in terms of volume and flow rate in terms of velocity: d ( ρV ) d ( ρ A1 z ) = = − ρ A2 V2 dt dt Density cancels so rearranging the equation: 2

⎛D ⎞ A dz = − 2 V2 = − ⎜ 2 ⎟ V2 dt A1 ⎝ D1 ⎠ We can obtain an expression for velocity by assuming (at any instant in time) that the steady, incompressible flow energy equation is applicable with a constant friction factor: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

At points 1 and 2 the pressure is atmospheric, so P1 = P2 . There is no pump or turbine, so hP = hT = 0 . Let z2 = 0 (the datum) and drop the subscript on z1, the depth of water in the tank. The losses include an entrance, a V1 2 V2 ⎛ L⎞V2 + z = 2 + ⎜ K ent + K valve + f ⎟ 2 2g 2g ⎝ D ⎠ 2g The tank velocity can be described in terms of the pipe velocity by conservation of mass. Any mass flow rate from the tank must equal the mass flow rate through the pipe: 2 m 1 = m 2 → ρ V1 A1 = ρ V2 A2 → V1 = ( D2 D1 ) V2

valve, and line loss, so

Substituting this into the energy equation and solving for velocity: 0.5

⎡ ⎤ 2 gz V2 = ⎢ ⎥ 4 ⎢⎣1 − ( D2 D1 ) + K ent + K valve + f L D ⎥⎦ Substituting this in the conservation of mass equation and separating variables: 2 ⎤ ⎛ D2 ⎞ ⎡ dz 2g = − ⎥ ⎜ ⎟ ⎢ 0.5 4 z ⎝ D1 ⎠ ⎢⎣1 − ( D2 D1 ) + K ent + K valve + f L D ⎥⎦ We integrate this from initial height zo to 0:

9- 27

0.5

dt

0

∫

zo

2 t ⎤ ⎛ D2 ⎞ ⎡ dz 2g = − ⎢ ⎥ ⎜ ⎟ z 0.5 ∫0 ⎝ D1 ⎠ ⎢⎣1 − ( D2 D1 )4 + K ent + K valve + f L D ⎥⎦

0.5

dt

0.5

−2 z

0.5 o

2 ⎤ ⎛D ⎞ ⎡ 2g = −⎜ 2 ⎟ ⎢ ⎥ t 4 ⎝ D1 ⎠ ⎢⎣1 − ( D2 D1 ) + K ent + K valve + f L D ⎥⎦ 0.5

2 4 ⎛ D ⎞ ⎡1 − ( D2 D1 ) + K ent + K valve + f L D ⎤ t = 2z ⎜ 1 ⎟ ⎢ ⎥ 2g ⎝ D2 ⎠ ⎣⎢ ⎦⎥ From Figure 9-14, Knit = 1.0, and from Table 9-3, Kvalve = 2.1. 0.5 o

⎡1- ( 0.1 10 )4 +1.0+2.1+ ( 0.016 )( 30 0.10 ) ⎤ ⎢ ⎥ t = 2 ( 8m ) 2 ( 9.81m s 2 ) ⎢⎣ ⎥⎦ b) If the 30 m long pipe is not taken into account: 0.5

⎛ 10 ⎞ ⎜ ⎟ ⎝ 0.1 ⎠

2

0.5

=38,100 s=635 min=10.6hr

Answer

0.5

⎡1- ( 0.1 10 )4 +1.0+2.1 ⎤ ⎢ ⎥ =25,900 s=432 min=7.2 hr t = 2 ( 8m ) 2 ( 9.81m s 2 ) ⎢⎣ ⎥⎦ c) We can evaluate whether the chosen friction factor is appropriate by calculating an “average” velocity, Reynolds number, and friction factor. The velocity is: 0.5

⎛ 10 ⎞ ⎜ ⎟ ⎝ 0.1 ⎠

2

V z π D12 4 ( 8m ) π (10m ) 4 m3 = = 0.0165 V ≈ tot = o 38,100s s t t 2

4 ( 0.0165 m3 s ) 4V m V V2 ≈ = = = 2.1 2 2 s A π D2 π ( 0.10m )

The friction factor is a function of Reynolds number and roughness. For a commercial steel pipe ε =0.045mm , and for water from Appendix A-6 at 10 ºC, µ = 12.9 × 10−4 Ns m 2 , ρ = 999.6 kg m3 . The Reynolds numbers are: Re =

3 ρ VD ( 999.6 kg m ) ( 2.1m s )( 0.10m ) = = 162, 700 12.9 × 10−4 Ns m 2 µ

The flow is turbulent, so ⎡⎛ 0.045 100 ⎞1.11 1 6.9 ⎤ = −1.8log ⎢⎜ ⎥ → f = 0.0187 ⎟ + 3.7 162, 700 ⎥⎦ fA ⎠ ⎢⎣⎝ Using this friction factor, we obtain t = 39,800 s = 11.1 hr.

Answer

Comments: Because an “average” flow rate was used, there is some uncertainty in this answer.

9- 28

9-26

An oil transporter truck is filled from the top with 15 m3 of fuel oil (SG = 0.86, µ = 5.3 × 10-2 N·s/m2) from a reservoir that is 4 m below the truck top. A 10-m long flexible hose 6-cm in diameter whose surface roughness is equivalent to that of galvanized iron connects the truck to the reservoir. A one-third closed ball valve and two bends that are equivalent to 90° threaded elbows are in the hose. For a filling time of 15 min, and a pump mechanical efficiency of 75%, determine the required pump power (in kW).

Approach: The steady, incompressible flow energy equation is used to determine the required pump head. Once that is known, the power can be calculated.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: The steady, incompressible flow energy equation is:

P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

There is no turbine, so hT = 0 . At points 1 and 2 the areas are large and the pressure is atmospheric, so V1 = V2 = 0 and P1 = P2 . Therefore, 2 ⎛ L ⎞V hP = z2 − z1 + ⎜ f + K exit + K ent + K valve + 2 K bend ⎟ ⎝ D ⎠ 2g where, from Table 9-3 and Figure 9-13: K exit = 1.0 , Kent = 1.0, K bend = 1.5, K valve = 5.5 . The friction factor is a function of Reynolds number and roughness. The volume flow rate is: V 15m3 m3 V = = =0.0167 t (15min )( 60s min ) s

The velocity is:

3 m V 4 ( 0.0167 m s ) =5.89 V = = 2 s A π ( 0.06m )

ρ VD ( 0.86 ) (1000 kg m ) ( 5.89 m s )( 0.06m ) Re = = = 5740 5.3 × 10−2 Ns m 2 µ The flow is turbulent, so using ε =0.15 mm for galvanized iron. 3

⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0.15 60 ⎞1.11 6.9 ⎤ = −1.8log ⎢⎜ + = − 1.8log ⎥ ⎢ ⎥ ⎟ ⎜ ⎟ + Re ⎦⎥ 5740 ⎦⎥ f ⎣⎢⎝ 3.7 ⎠ ⎣⎢⎝ 3.7 ⎠

1

→

f = 0.0387

⎡ ⎤ ( 5.89 m s ) ⎛ 10 ⎞ = 34.0m hP = 4m+ ⎢( 0.0387 ) ⎜ ⎟ +1.0+1.0+5.5+2 (1.5 ) ⎥ 2 ⎝ 0.06 ⎠ ⎣ ⎦ 2 ( 9.81m s ) 2

Therefore, Power is

  P ρVgh mgh P = = W P =

ηP

=6.37kW

ηP

( 0.86 ) ⎛⎜1000 ⎝

⎛ kN s 2 ⎞ kg ⎞ ⎛ m3 ⎞ ⎛ m⎞ 0.0167 ⎟ ⎜ 9.81 2 ⎟ ( 34.0m ) ⎜ ⎟ 3 ⎟⎜ s ⎠⎝ m ⎠⎝ s ⎠ ⎝ 1000kg m ⎠ 0.75

Answer

9- 29

9-27

Large office buildings use circulating hot water systems to ensure that hot water is available instantly in all restrooms. Consider a system that consists of 200-m of 2.5-cm commercial steel pipe. It has 15 90° regular threaded elbows, two fully open gate valves, three half open gate valves and one three-quarter closed gate valve. For water at 50 °C and a pump with a mechanical efficiency of 75%, determine: a. the power required if the water velocity is 2 m/s (in kW) b. the power required if the water velocity is 1 m/s (in kW).

Approach: The required pump head can be determined with the steady, incompressible flow energy equation must be used. Once the head is known, the power can be calculated.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: a) Because this is a closed loop, we can use the steady, incompressible flow energy equation from point 1 to point P1 V2 P V2 + 1 + z1 + hP = 1 + 1 + z1 + + hT + ∑ hL 1: ρ g 2g ρ g 2g There is no turbine, so hT = 0 , and the types of losses are given in the problem statement, so that: ⎛ L ⎞V2 hP = ⎜ f + 2 K valve + 15K bend + 3K valve half + 1K valve quarter ⎟ open open open ⎝ D ⎠ 2g From Table 9-2 for the pipe ε =0.045m , and for water from Appendix B-6 at 50 ºC, µ = 5.29 × 10−4 Nm s 2 ,

ρ = 988 kg m3 . From Table 9-3, Kvalve open = 0.15, Kvalve half open = 2.1, Kvalve quarter open = 17, Kbend = 1.5. ρ VD ( 988 kg m ) ( 2 m s )( 0.025m ) = = 93, 400 5.29 × 10−4 Nm s 2 µ 3

Re =

The flow is turbulent, so ⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0.045 25 ⎞1.11 1 6.9 ⎤ 1.8log = −1.8log ⎢⎜ + = − ⎥ ⎢ ⎥ → ⎟ ⎜ ⎟ + Re ⎦⎥ 93, 400 ⎦⎥ f ⎣⎢⎝ 3.7 ⎠ ⎣⎢⎝ 3.7 ⎠

f = 0.0245

200 ⎡ ⎤ (2 m s) hP = ⎢( 0.0245 ) +2 ( 0.15 ) +15 (1.5 ) +3 ( 2.1) +1(17 ) ⎥ =49.4m 0.025 ⎣ ⎦ 2 ( 9.81m s 2 ) 2

3 2  P ( 988 kg m ) ( 2 m s )(π 4 )( 0.025m ) ( 9.81m s ) ( 49.4m )( Ns kgm ) mgh =626W W = = 0.75 ηP b) If the velocity is 1 m/s, the same calculations are made as in part (a): Re = 46, 700 f = 0.0260

Answer

2

hP = 12.9m W = 82.1W

Answer Answer

9- 30

Answer

9-28

To ensure adequate water supplies to a town, a municipal water department developed a second reservoir and wants to connect the new reservoir to the old one using a concrete pipe. The reservoirs are 1.5 miles apart with a difference in surface elevations of 25 ft. Determine the minimum pipe diameter needed to carry 10 ft3/s of water at 50 °F?

Approach: The head caused by the difference in reservoir elevations is balance by the frictional head loss, which depends on the velocity and diameter. Applying the steady, incompressible flow energy equation between points 1 and 2, we can determine the pipe diameter with an iterative solution.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: The steady, incompressible flow energy equation is:

P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

There is no pump or turbine, so hP = hT = 0 . The areas of the reservoirs are large, and the pressure are atmospheric, so V1 = V2 = 0 and P1 = P2 . Taking into account line losses only: z1 − z2 = f

L V2 D 2g

The friction factor is a function of Reynolds number and roughness. For concrete pipe ε =0.005 ft (estimated), and for water from Appendix B-6 at 50 ºF, µ = 88 × 10−5 lbm ft s , ρ = 62.4 lbm ft 3 . The velocity is: Re =

V =

4 (10 ft 3 s ) 12.73 ft V 4V = 2 = = 2 A πD 2 D s π ( Dft )

3 2 ρ VD ( 62.4 lbm ft )(12.73 D ft s ) ( Dft ) = = 903, 000 D µ 88 × 10−5 lbm ft s

For any reasonable size pipe, the flow is turbulent, so ⎡⎛ ε D ⎞1.11 6.9 ⎤ 1 = −1.8log ⎢⎜ ⎥ ⎟ + Re ⎦⎥ f ⎣⎢⎝ 3.7 ⎠ Substituting known quantities into the pressure drop equation: 2 ⎡ (1.5mi )( 5280 ft mi ) ⎤ ( V ft s ) 25ft = f ⎢ ⎥ 2 Dft ⎣ ⎦ 2 ( 32.2 ft s ) f V2 D The above four equations are solved iteratively. Doing so: D = 2.64ft Answer V = 4.81ft s f = 0.0232 The next large standard size of concrete pipe would be used. 0.2033 =

9- 31

(1) (2)

(3)

(4)

9-29

The reservoir behind a dam is connected to a hydroelectric power plant with a penstock (a large pipe to convey the water). At a particular plant, the elevation difference between the reservoir surface and the hydroturbine is 50 m, and the penstock is constructed of 150-m of 1-m diameter cast iron pipe. The turbine has a mechanical efficiency of 78% and the electric generator has an efficiency of 94%. For a 1 m3/s flow of 10 °C water, determine: a. the power output from the plant (in kW) b. the power output if a fully open gate valve and two long radius 45° flanged elbows also are in the pipe (in kW).

Approach: The steady, incompressible flow energy equation is used to determine the turbine head. Once that is known, the power can be calculated.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: a) The steady, incompressible flow energy equation is:

P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

There is no pump, so hP = 0 . At points 1 and 2 the areas are large and the pressure is atmospheric, so V1 = V2 = 0 and P1 = P2 . Including the exit loss where, K exit = 1.0 .: 2 ⎛ L ⎞V hT = z1 − z2 − ⎜ f + K exit ⎟ ⎝ D ⎠ 2g The friction factor is a function of Reynolds number and roughness, and from Table 9-2 ε =0.26 mm . For water from Appendix A-6 at 10 ºC, µ = 12.9 × 10−4 Ns m 2 , ρ = 999.6 kg m3 .

The velocity is:

V =

4 (1m3 s ) V 4V m =1.27 = = 2 A π D2 s π (1m )

ρ VD ( 999.6 kg m ) (1.27 m s )(1m ) = = 987, 000 12.9 × 10−4 Ns m 2 µ 3

Re =

The flow is turbulent, so ⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0.26 1000 ⎞1.11 1 6.9 ⎤ = −1.8log ⎢⎜ ⎥ = −1.8log ⎢⎜ ⎥ ⎟ + ⎟ + Re ⎥⎦ 3.7 987, 000 ⎥⎦ f ⎠ ⎢⎣⎝ 3.7 ⎠ ⎢⎣⎝

→

f = 0.0152

⎡ ⎤ (1.27 m s ) ⎛ 150 ⎞ hT = 50m- ⎢( 0.0152 ) ⎜ = 49.7m ⎟ +1.0 ⎥ 2 1 ⎝ ⎠ ⎣ ⎦ 2 ( 9.81m s ) 2

Therefore,

Power extracted from the water is ⎛ kN s 2 ⎞ kg ⎞ ⎛ m3 ⎞ ⎛ m⎞ ⎛   T = ρVgh 999.6 1 9.81 49.7m WT = mgh = ( ) ⎟⎜ ⎜ ⎟ =488kW T ⎜ ⎟⎜ ⎟ m3 ⎠ ⎝ s ⎠ ⎝ s2 ⎠ ⎝ ⎝ 1000kg m ⎠ Taking into account the turbine and electric generator efficiencies: Welec = ηT η genWT = ( 0.78 )( 0.94 )( 488kW ) =358kW Answer

b) Adding a gate valve, K gate = 0.15 , and 2 long radius 45º bends, K bend = 0.2 ⎡ ⎤ (1.27 m s ) ⎛ 150 ⎞ hT = 50m- ⎢( 0.0152 ) ⎜ = 49.7m ⎟ +1.0+0.15+2 ( 0.2 ) ⎥ 2 ⎝ 1 ⎠ ⎣ ⎦ 2 ( 9.81m s ) 2

Welec = 358kW

Answer

Comments: Other losses dominate the process so adding a gate valve has essentially no effect. 9- 32

9-30

The drain at the bottom of a swimming pool (10-m in diameter and 2-m deep) is well rounded and is connected to a 5-cm diameter, 20-m long plastic pipe. The water is at 20 °C. For a friction factor of 0.021, determine: a. the time required to drain the pool (in min) b. if this value of friction factor is appropriate.

Approach: Conservation of mass is needed to determine the time to drain the pool. The driving pressure head decreases with time, so the head as a function of time is needed. The open system conservation of mass equation and the steady, incompressible flow energy equation are used to calculate the time.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: a) The open system conservation of mass equation is for the control volume defined as the water in the pool: dm dt = m in − m out . Water only leaves, and expressing the mass in terms of volume and flow rate in terms of velocity: d ( ρV ) d ( ρ A1 z ) = = − ρ A2 V2 dt dt Density cancels so rearranging the equation: 2

⎛D ⎞ A dz = − 2 V2 = − ⎜ 2 ⎟ V2 dt A1 ⎝ D1 ⎠ We can obtain an expression for velocity by assuming (at any instant in time) that the steady, incompressible flow energy equation is applicable with a constant friction factor: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

At points 1 and 2 the pressure is atmospheric, so P1 = P2 . There is no pump or turbine, so hP = hT = 0 . Let z2 = 0 (the datum) and drop the subscript on z1, the depth of water in the tank. The losses include an exit and line V1 2 V2 ⎛ L⎞V2 + z = 2 + ⎜ K ent + f ⎟ 2 2g 2g ⎝ D ⎠ 2g The pool velocity can be described in terms of the pipe velocity by conservation of mass. Any mass flow rate from the pool must equal the mass flow rate through the pipe: 2 m 1 = m 2 → ρ V1 A1 = ρ V2 A2 → V1 = ( D2 D1 ) V2

loss, so

Substituting this into the energy equation and solving for velocity: ⎡ ⎤ 2 gz V2 = ⎢ ⎥ 4 ⎣⎢1 − ( D2 D1 ) + K ent + f L D ⎦⎥

0.5

Substituting this in the conservation of mass equation and separating variables: 2 ⎤ ⎛ D2 ⎞ ⎡ 2g dz = −⎜ ⎥ ⎟ ⎢ 0.5 4 z ⎝ D1 ⎠ ⎢⎣1 − ( D2 D1 ) + K ent + f L D ⎥⎦ We integrate this from initial height zo to 0:

0.5

dt

9- 33

0

∫

zo

2 t ⎤ ⎛ D2 ⎞ ⎡ 2g dz = − ⎢ ⎥ ⎜ ⎟ z 0.5 ∫0 ⎝ D1 ⎠ ⎢⎣1 − ( D2 D1 )4 + K ent + f L D ⎥⎦

0.5

dt

0.5

−2 z

0.5 o

2 ⎤ ⎛D ⎞ ⎡ 2g = −⎜ 2 ⎟ ⎢ ⎥ t 4 ⎝ D1 ⎠ ⎢⎣1 − ( D2 D1 ) + K ent + f L D ⎥⎦

2 4 ⎛ D ⎞ ⎡1 − ( D2 D1 ) + K ent + f L D ⎤ t = 2z ⎜ 1 ⎟ ⎢ ⎥ 2g ⎝ D2 ⎠ ⎣⎢ ⎦⎥ From Figure 9-13, Kent = 0.04

0.5

0.5 o

t = 2 ( 2m )

0.5

⎛ 20 ⎞ ⎜ ⎟ ⎝ 0.05 ⎠

2

⎡1- ( 0.05 10 )4 +0.04+ ( 0.021)( 20 0.05 ) ⎤ ⎢ ⎥ 2 ( 9.81m s 2 ) ⎢⎣ ⎥⎦

0.5

=314,000s=87.2hr

Answer

b) We can evaluate whether the chosen friction factor is appropriate by calculating an “average” velocity, Reynolds number, and friction factor. The velocity is: V z π D12 4 ( 2m ) π (10m ) 4 m3 = = 0.00050 V ≈ tot = o t t 314, 000s s 2

V2 ≈

4 ( 0.00050 m3 s ) V 4V m = = = 0.255 2 A π D22 s π ( 0.05m )

The friction factor is a function of Reynolds number and roughness. For smooth pipe ε =0 , and for water from Appendix A-6 at 20 ºC, µ = 9.85 × 10−4 Ns m 2 , ρ = 998.2 kg m3 . The Reynolds numbers are:

ρ VD ( 998.2 kg m ) ( 0.255 m s )( 0.05m ) Re = = = 12,900 9.85×10-4 Ns m 2 µ 3

The flow is turbulent, so ⎡⎛ 0 ⎞1.11 1 6.9 ⎤ = −1.8log ⎢⎜ ⎥ → f = 0.0288 ⎟ + 12,900 ⎥⎦ fA ⎢⎣⎝ 3.7 ⎠ Using this friction factor, we obtain t = 362,000 s and 100.6 hr.

Answer

Comments: The answer for part (b) is significantly different than that in (a), so the new estimate is probably better.

9- 34

9-31

If the pool in Problem P 9-30 has a sharp-edged entrance and two 90° regular threaded elbows, determine the time required to drain the pool (in min).

Approach: We use the same equations as derived in Problem P 9-30. Additional minor losses are included.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: The equation for the draining time, including the additional minor losses is: 0.5

2 4 ⎛ D ⎞ ⎡1 − ( D2 D1 ) + K ent + 2 K bend + f L D ⎤ t = 2z ⎜ 1 ⎟ ⎢ ⎥ 2g ⎝ D2 ⎠ ⎣⎢ ⎦⎥ From Table 9-3, Kbend = 1.5 and from Figure 9-13, Kent = 1.0. Using the friction factor from part (b) in Problem 931: 0.5 o

⎡1- ( 0.05 20 )4 +1+2 (1.5 ) + ( 0.0288 )( 20 0.05 ) ⎤ ⎢ ⎥ t = 2 ( 2m ) 2 ( 9.81m s 2 ) ⎢⎣ ⎥⎦ =415,300 s=6922 min=115.4hr Answer 0.5

⎛ 20 ⎞ ⎜ ⎟ ⎝ 0.05 ⎠

2

0.5

Comments: As can be seen the addition of a small amount of minor loss increased the draining time from about 100 hr to 115 hr. Note, however, we used “average” values of velocity and friction factor, so there is uncertainty in this answer.

9- 35

9-32

A pipe connects two reservoirs at different elevations. The pipe is constructed of 12-in. diameter commercial steel with flanged fittings. The gate valve is one-fourth closed. The water temperature is 50 °F. Determine the required elevation difference between the two reservoirs to produce a water flow rate of 10 ft3/s (in ft).

Approach: The head caused by the difference in reservoir elevations is balance by the frictional head loss. Applying the steady, incompressible flow energy equation between points 1 and 2, we can determine the required elevation difference.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: The steady, incompressible flow energy equation is:

P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

There is no pump or turbine, so hP = hT = 0 . The areas of the reservoirs are large, and the pressure are atmospheric, so V1 = V2 = 0 and P1 = P2 . Taking into account all losses: 2 ⎛ L ⎞V z1 − z2 = ⎜ f + K ent + K valve + K exit + 2 K bend ⎟ ⎝ D ⎠ 2g The friction factor is a function of Reynolds number and roughness. For commercial steel pipe ε =0.00015 ft , and for water from Appendix B-6 at 50 ºF, µ = 88 × 10−5 lbm ft s , ρ = 62.4 lbm ft 3 .

The velocity is:

V =

4 (10 ft 3 s ) V 4V ft =12.7 = = 2 A πD 2 s π (1ft )

ρ VD ( 62.4 lbm ft ) (12.7 ft s )(1ft ) = = 903, 000 µ 88 × 10−5 lbm ft s 3

Re =

The flow is turbulent, so ⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0.00015 1 ⎞1.11 1 6.9 ⎤ 1.8log = −1.8log ⎢⎜ + = − ⎥ ⎢⎜ ⎥ ⎟ ⎟ + Re ⎦⎥ 3.7 903, 000 ⎦⎥ f ⎠ ⎣⎢⎝ 3.7 ⎠ ⎣⎢⎝

→

f = 0.0141

From Table 9-3 and Figures 9-13 and 9-14, K ent = 0.6, K valve = 0.26, K exit = 1.0, K bend = 0.3 Therefore, 2 ⎡ ⎛ 310 ⎞ ⎤ (12.7 ft s ) z1 − z2 = ⎢0.5 + 0.26 + 1.0 + 2 ( 0.3) + ( 0.0141) ⎜ = 16.9ft ⎟⎥ ⎝ 1 ⎠ ⎦ 2 ( 32.2 ft s 2 ) ⎣

9- 36

Answer

9-33

A liquid (SG = 0.93, µ = 0.00068 N·s/m2) is contained in a vertical 2-cm diameter pipe. At one elevation the fluid pressure is 230 kPa; at an elevation 10-m higher, the pressure is 110 kPa. Determine: a. if the flow is moving and in what direction b. the flow velocity if it is flowing (in m/s).

Approach: If the fluid is moving, there will be frictional losses. The sign on the frictional head loss will tell us the flow direction. The frictional head loss can be determined with the steady, incompressible flow energy equation.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: a) Assuming the flow is from point 1 to point 2, the steady, incompressible flow energy equation is: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + + hT + ∑ hL ρ g 2g ρ g 2g There is no pump or turbine, so hP = hT = 0 , the pipe diameter is constant, V1 = V2 , and there are no minor losses: (110-230 ) kN m 2 (1000kg m kN s 2 ) P1 − P2 + (10m ) =-3.15m + ( z1 − z2 ) = hL = ρg ( 0.93) (1000 kg m3 )( 9.81m s 2 ) The frictional head loss is negative, the assumed flow direction is incorrect. Answer The flow is upward. L V2 . We do not know the velocity or the friction factor, so we will need to iterate D 2g to find a solution. The friction factor depend on Reynolds number, so 3 ρ VD ( 0.93) (1000 kg m ) ( V m s )( 0.020m ) Re = = = 27, 400 V (1) µ 0.00068 Nm s 2 For any reasonable velocity the flow will be turbulent. We assume the flow is turbulent and that the pipe is smooth, ε =0 , so ⎡⎛ ε D ⎞1.11 6.9 ⎤ 1 (2) = −1.8log ⎢⎜ ⎥ ⎟ + Re ⎥⎦ f ⎢⎣⎝ 3.7 ⎠ Substituting known values into the head loss equation: 0.5 2 ⎛ 0.0928 ⎞ m ⎛ 10m ⎞ ( V m s ) V (3) 3.15m = f ⎜ → = ⎜ ⎟ ⎟ ⎝ 0.015m ⎠ 2 ( 9.81m s 2 ) ⎝ f ⎠ s

b) The head loss is hL = f

The iterative procedure is to: guess a velocity, calculate the Reynolds number and the friction factor, solve equation (3) for velocity, and then compare to the guessed value. Continue until converged. V (m/s) 0.5 1.81 2.11 2.15

Re 13,700 49,600 57,800 58,900

f 0.0284 0.0208 0.0201 0.0200

9- 37

V (m/s) 1.81 2.11 2.15 2.16

Answer

9-34

The designers of a large shopping mall install 18-in. diameter smooth concrete storm sewers to channel away runoff after heavy rainstorms. Each storm sewer will need to carry a flow of 10 ft3/s. The pressures at the entrance and exit of the sewer are atmospheric. If the sewers are 200 ft long before they join with larger pipes, determine the required elevation change per 100 ft of pipe (in ft).

Approach: Because we can solve for elevation change directly, the steady, incompressible flow energy equation can be used to determine the required slope.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: The steady, incompressible flow energy equation is: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g Let locations 1 and 2 be just outside the pipe ends, and both are at atmospheric pressure, so P1 = P2 . The duct is constant area, so V1 = V2 . There is no turbine or pump, so hT = hP = 0 . For the minor losses at the entrance and exit, assume the area ratios are zero, so from Figure 9-15, Kent = 0.5 and Kexit = 1.0. L⎞V2 ⎛ z1 − z2 = ⎜ K ent + K exit + f ⎟ D ⎠ 2g ⎝ The friction factor is a function of Reynolds number and roughness. For water from Appendix B-6 at 50 ºF, µ = 88 × 10−5 lbm fts , ρ = 62.4 lbm ft 3 . Velocity is :

V =

4 (10 ft 3 s ) V 4V ft =5.66 = = A π D 2 π (1.5ft )2 s

ρ VD ( 62.4 lbm ft ) ( 5.66 ft s )(1.5ft ) = = 602, 000 µ 88 × 10−5 lbm fts This is turbulent flow. From Table 9-2 for smooth concrete ε = 0.001ft, ε D = 0.00067 : 3

Re =

⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0.00067 ⎞1.11 6.9 ⎤ 1.8log = −1.8log ⎢⎜ + = − ⎥ ⎢ ⎥ ⎟ ⎜ ⎟ + Re ⎦⎥ 602, 000 ⎦⎥ f ⎣⎢⎝ 3.7 ⎠ ⎣⎢⎝ 3.7 ⎠

1

→

f = 0.0184

200 ⎤ ( 5.66 ft s ) ⎡ z1 − z2 = ⎢0.5+1.0+ ( 0.0184 ) = 1.97ft 1.5 ⎥⎦ 2 ( 32.2 ft s 2 ) ⎣ 2

Therefore, the required negative slope is −1.97 200 = -0.985 100ft of pipe

Comments:

Note that the angle is θ = sin −1 ( −0.00985 ) = −0.56o .

9- 38

Answer

9-35

In mountainous regions, tunnels often are used for cars, trucks, and trains. If the tunnel is too long, ventilation air must be supplied to dilute and purge vehicle exhaust gases from the tunnel. Consider a 3-ft diameter, 2500-ft long duct constructed of commercial steel pipe that carries air at 45 °F, 14.1 psia with a flow rate of 10,000 ft3/min. Determine: a. the pressure drop (in in. of water) b. the power required (in hp).

Approach: The pressure drop is determined with the steady, incompressible flow energy equation. The pump power can be calculated with the same equation but applied just across the pump.

Assumptions: 1. 2. 3. 4.

The system is steady. The flow is fully developed. Properties are constant. Air is an ideal gas.

Solution: a) The steady, incompressible flow energy equation is:

P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

The pipe diameter and elevation are constant, so V1 = V2 and z1 = z2 . There is no pump or turbine, so P1 − P2 L V2 = ha = f ρg D 2g

hP = hT = 0 . There are no minor losses. Therefore,

ρ=

Assuming air is an ideal gas:

2 2 2 PM (14.1lbf in. ) ( 28.97 lbm lbmol ) (144in. ft ) lbm =0.0754 3 = RT ft (1545ft lbf lbmR )( 45+460 ) R

The friction factor is a function of Reynolds number and roughness. For air from Appendix B-7 at 45 ºF,

µ = 1.188 × 10−5 lbm fts . For commercial steel pipe, ε = 0.00015ft, ε D = 0.00015 3 = 0.00005 . Velocity is :

V =

4 (10000 ft 3 min ) (1min 60s ) V 4V ft =23.6 = = 2 2 A πD s π ( 3ft )

ρ VD ( 0.0754 lbm ft ) ( 23.6 ft s )( 3ft ) = = 4.49 × 105 µ 1.188×10-5 lbm fts 3

Re =

This is turbulent flow so ⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0.00005 ⎞1.11 1 6.9 ⎤ 1.8log = −1.8log ⎢⎜ + = − ⎥ ⎢ ⎥ → f = 0.0139 ⎟ ⎜ ⎟ + Re ⎦⎥ 4.49 × 105 ⎦⎥ f ⎣⎢⎝ 3.7 ⎠ ⎣⎢⎝ 3.7 ⎠ Substituting known values into the pressure drop equation: ⎛ 2500 ⎞ ( 23.6 ft s ) ha = ( 0.0139 ) ⎜ = 100 ft of air ⎟ ⎝ 3 ⎠ 2 ( 32.2 ft s 2 ) 2

For pressure in inches of water: ∆P = ρ w ghw = ρ a gha

→

hw =

Answer

ρa ⎛ 0.0754 ⎞ ⎛ 12in. ⎞ ha = ⎜ ⎟ (100ft ) ⎜ ⎟ =1.45in. of water ρw ⎝ 62.4 ⎠ ⎝ 1ft ⎠

b) Applying the same energy equation between points 1 and 2: The pumping power is

hP =

P2, g − P1, g

ρg

=

Answer P2, g

ρg

 ( ∆P ρ g ) = V ∆P = V ρ gh W = ρVg a a

2 ⎛ ⎞⎛ ⎞ ft 3 ⎞ ⎛ lbm ⎞ ⎛ ft ⎞ 1hp ⎛ 1min ⎞ ⎛ lbfs 0.0754 32.2 100ft = ⎜ 10000 ( ) ⎟⎜ ⎟⎜ ⎟ =2.28hp ⎜ ⎟⎜ 3 ⎟⎜ 2 ⎟ min ⎠ ⎝ ft ⎠ ⎝ s ⎠ ⎝ 60s ⎠ ⎝ 32.2ftlbm ⎠ ⎝ 550 ft lbf s ⎠ ⎝

9- 39

Answer

9-36

Water at 10 °C flows from a lake at a flow rate of 0.1 m3/s. A 15-cm diameter, 100-m long galvanized iron pipe connects the lake to a building in which either a pump or a turbine is located. The elevation difference between the lake surface and the building is 10 m. Determine: a. if the device in the building is a pump or a turbine b. the power of the device (in W).

Approach: The steady, incompressible flow energy equation has terms for pump and turbine power. We will assume the device inside the building is a pump. Applying the energy equation, if the power calculated is positive, then our assumption is correct. If it is negative, then our assumption was wrong, and the device actually is a turbine.

Assumptions: 1. 2. 3. 4.

The system is steady. The flow is fully developed. Properties are constant. The device is a pump.

Solution: a) The steady, incompressible flow energy equation is:

P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

We assume the device is a pump, so hT = 0 . At points 1 and 2 the areas are large and the pressure is atmospheric, so V1 = V2 = 0 and P1 = P2 . The losses are line, exit, and entrance losses. Therefore, 2 ⎛ L ⎞V hP = z2 − z1 + ⎜ f + K exit + K ent ⎟ ⎝ D ⎠ 2g where K exit = 1 and Kent = 0.5. The friction factor is a function of Reynolds number and roughness. For galvanize iron pipe ε =0.15 mm , and for water from Appendix A-6 at 10 ºC, µ = 12.9 × 10−4 Ns m 2 , ρ = 999.6 kg m3 .

The velocity is:

4 ( 0.1m3 s ) V 4V m =5.66 V = = = 2 2 A πD s π ( 0.15m )

ρ VD ( 999.6 kg m ) ( 5.66 m s )( 0.15m ) Re = = = 657, 700 12.9 × 10−4 Ns m 2 µ 3

The flow is turbulent, so ⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0.15 150 ⎞1.11 1 6.9 ⎤ = −1.8log ⎢⎜ ⎥ = −1.8log ⎢⎜ ⎥ ⎟ + ⎟ + Re ⎥⎦ 657, 700 ⎥⎦ f ⎢⎣⎝ 3.7 ⎠ ⎢⎣⎝ 3.7 ⎠

→

f = 0.0201

⎡ ⎤ ( 5.66 m s ) ⎛ 100 ⎞ hP = −10m+ ⎢( 0.0201) ⎜ = 14.3m ⎟ +0.5+1⎥ 2 0.15 ⎝ ⎠ ⎣ ⎦ 2 ( 9.81m s ) 2

Therefore,

Answer Because this is positive, the device is a pump. b) Power is ⎛ N s2 ⎞ kg ⎞ ⎛ m3 ⎞ ⎛ m⎞ ⎛   P = ρVgh 0.1 W P = mgh ⎟ ⎜ 9.81 2 ⎟ (14.3m ) ⎜ ⎟ =14,020W=14.0kW P = ⎜ 999.6 3 ⎟⎜ m ⎠⎝ s ⎠⎝ s ⎠ ⎝ ⎝ kg m ⎠

9- 40

Answer

9-37

Ski resorts pump water to make snow when the weather does not cooperate. Consider a resort that uses 100 gal/min of 35 °F water. It is pumped from the water holding pond through a 4-in. diameter, 3000-ft steel pipe to the top of the mountain. The elevation difference is 950 ft. The gage pressure required at the nozzle at the end of the pipe is 150 lbf/in.2. Determine the required pumping power (in hp).

Approach: By applying the steady, incompressible flow energy equation between points 1 and 2, we can determine the required pump power.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

The steady, incompressible flow energy equation is:

There is no turbine, so hT = 0 . At point 1 the area of the pond is large, and the gage pressure is atmospheric, so V1 = 0 and P1 = 0 . Ignoring entrance loss, the losses are line and exit. Therefore, hP =

2 P2 ⎛ L ⎞V + ( z2 − z1 ) + ⎜ f + K exit ⎟ ρg ⎝ D ⎠ 2g

where K exit = 1 . The friction factor is a function of Reynolds number and roughness. For commercial steel pipe ε =0.00015 ft , and for water from Appendix B-6 at 35 ºF, µ = 114 × 10−5 lbm ft s , ρ = 62.4 lbm ft 3 . The velocity is:

V =

4 (100 gal min ) ( 0.1337 gal ft 3 ) (1min 60s ) V 4V ft =2.61 = = 2 A πD 2 s π ( 0.33ft )

ρ VD ( 62.4 lbm ft ) ( 2.61ft s )( 0.33ft ) = = 47, 060 µ 114 × 10−5 lbm ft s 3

Re =

The flow is turbulent, so ⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0.00015 0.33 ⎞1.11 1 6.9 ⎤ = −1.8log ⎢⎜ ⎥ = −1.8log ⎢⎜ ⎥ ⎟ + ⎟ + Re ⎥⎦ 3.7 47, 060 ⎥⎦ f ⎠ ⎢⎣⎝ 3.7 ⎠ ⎢⎣⎝ Therefore,

hP

(150 lbf =

in.2 ) (12in. ft ) ( 32.2 ft lbm lbf s 2 ) 2

( 62.4 lbm ft )( 32.2 ft s ) 3

2

→

f = 0.0223

⎡ ⎛ 3000 ⎞ ⎤ ( 2.61ft s ) +950ft+ ⎢( 0.0223) ⎜ ⎟ +1⎥ ⎝ 0.33 ⎠ ⎦ 2 ( 32.2 ft s 2 ) ⎣ 2

=346+950+21.5=1318ft

Power is ⎛ lbf s 2 ⎞ ⎛ 1hp s ⎞ lbm ⎞⎛ gal ⎞ ⎛ gal ⎞ ⎛ 1min ⎞ ⎛ ft ⎞ ⎛   P = ρVgh W P = mgh 100 ⎟⎜ P = ⎜ 62.4 ⎟ ⎜ 0.1337 3 ⎟ ⎜ ⎟ ⎜ 32.2 2 ⎟ (1318ft ) ⎜ ⎟ 3 ⎟⎜ ft ⎠⎝ min ⎠ ⎝ ft ⎠ ⎝ 60s ⎠ ⎝ s ⎠ ⎝ ⎝ 32.2ft lbm ⎠ ⎝ 550 ft lbf ⎠ =33.3 hp

Answer

9- 41

9-38

In the western United States, many crops are irrigated, and water must be pumped long distances. Consider a system that consists of a 1-m diameter, 2-km long steel pipe, which connects a river to an irrigation canal. The canal’s elevation is 50 m higher than that of the river. For water at 15 °C, a pump with a mechanical efficiency of 80%, and neglecting minor losses, determine the power required to pump 2.5 m3/s of water (in kW).

Approach: The steady, incompressible flow energy equation is used to calculate the required pump head. Once that is known, the required pumping power can be calculated.

Assumptions: 1. 2. 3. 4.

The system is steady. The flow is fully developed. Properties are constant. Neglect minor losses.

Solution: The steady, incompressible flow energy equation is: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g The areas at points 1 and 2 the areas are large and the pressure is atmospheric, so V1 = V2 = 0 and P1 = P2 . There is no turbine, so hT = 0 , and minor losses are neglected. Therefore,

hP = z2 − z1 + f

L V2 D 2g

4 ( 2.5 m3 s ) V 4V m =3.18 = = 2 s A πD 2 π (1m ) The friction factor is a function of Reynolds number and roughness. For the pipe ε =0.045mm , and for water from Appendix A-6 at 15 ºC, µ = 11.2 × 10 −4 Ns m 2 , ρ = 999 kg m3 . The Reynolds numbers are:

The velocity is:

V =

ρ VD ( 999 kg m ) ( 3.18 m s )(1m ) = = 2.81× 106 11.2 × 10−4 Ns m 2 µ 3

Re =

The flow is turbulent, so ⎡⎛ 0.045 1000 ⎞1.11 1 6.9 ⎤ = −1.8log ⎢⎜ ⎥ ⎟ + 3.7 2.81× 106 ⎦⎥ f ⎠ ⎣⎢⎝

→

f = 0.0113

Therefore, ⎛ 2000 ⎞ ( 3.18 m s ) hP = 50m + ( 0.0113) ⎜ = 61.7m ⎟ ⎝ 1 ⎠ 2 ( 9.81m s 2 ) 2

Answer

Pumping power, taking into account pump efficiency, is: 3 3 2 2  P ( 999 kg m )( 2.5 m s ) ( 61.7m ) ( 9.81m s )( Ns kgm ) mgh =1,890kW = W = ηP ( 0.80 )(1000W kW )

9- 42

9-39

Air at 105 kPa and 25 °C flows from a 7.5-cm circular duct into a 22.5-cm circular duct. The downstream pressure is 6.5-mm of water higher than the upstream pressure. Determine: a. the average air velocity approaching the expansion (in m/s) b. the volumetric flow rate (in m3/s) c. the mass flow rate (in kg/s).

Approach: Applying the steady, incompressible flow energy equation between points 1 and 2, we can determine the flow rate through the expansion. The loss coefficient across the expansion must be evaluated.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: a) The steady, incompressible flow energy equation is:

P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

There is no pump or turbine, so hT = hp = 0 . The pipe is horizontal, so z1 = z2 . The only loss is due to the sudden expansion. Solving for the pressure rise: V2 ρ P2 − P1 = ( V1 2 − V22 ) − K exp ρ 1 2 2 From conservation of mass, m 1 = m 2 →

V2 = V1 ( A1 A2 ) = V1 ( D1 D2 )

2

Substituting this into the pressure drop equation and solving for velocity: 0.5

⎧ ⎫ 2 ( P2 − P1 ) ⎪ ⎪ V1 = ⎨ ⎬ 4 ⎪ ρ ⎡⎣1 − ( D1 D2 ) − K exp ⎤⎦ ⎪ ⎩ ⎭ Assuming an ideal gas, the air density is: 2 PM (105 kN m ) ( 28.97 kg kmol ) kg ρ= = =1.228 3 RT m (8.314 kJ kgK )( 25+273) K

Using Figure 9-15b for a sudden expansion, with A1 A2 = ( D1 D2 ) = ( 7.5 22.5 ) = 0.111 → 2

2

K exp ≈ 0.76

The pressure rise is obtained from the given information and water density: P2 − P1 = ρ gh = (1000 kg m3 )( 9.81m s 2 ) ( 0.0065m ) ( kN s 2 1000 kg m ) =0.0638 kN m 2 0.5

⎧ 2 0.0638 kN m 2 1000kg m kN s ⎫ )( ) ⎪ =21.4 m ⎪ ( V1 = ⎨ ⎬ 4 3 s ⎪ (1.228 kg m ) ⎡⎣1- ( 7.5 22.5 ) -0.76 ⎤⎦ ⎪ ⎩ ⎭ b) The volume flow rate is: 2 V = VA = ( 21.4 m s )( π 4 )( 0.075m ) =0.0944 m3 s

c) Mass flow rate is:  (1.228 kg m3 )( 0.0944 m3 s ) = 0.116 kg s  ρ V= m=

9- 43

Answer

Answer Answer

9-40

In some high rise buildings, water is stored in an elevated tank on the roof to minimize pressure fluctuations in the system. Consider water that is pumped through a 10-cm steel pipe to the roof of a 200-m tall building; the pump is on the ground floor. For a water temperature of 10 °C and a flow of 0.02 m3/s, what is the pressure at the pump discharge (in kPa)?

Approach: The steady, incompressible flow energy equation is used directly to calculate the pressure at point 1.

Assumptions: 1. 2. 3. 4.

The system is steady. The flow is fully developed. Properties are constant. Neglect minor losses.

Solution: The steady, incompressible flow energy equation is: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g The area at point 2 is large and the pressure is atmospheric, so V2 = 0 and P2 = 101.3kPa . There is no pump or turbine, so hT = hP = 0 , and all minor losses except the exit are neglected. Therefore, V1 2 L⎞ V2 ⎛ + ρ g ( z2 − z1 ) + ⎜ K exit + f ⎟ ρ 1 D⎠ 2 2 ⎝ = 1 , so

P1 = P2 − ρ

Note that K exit

L V1 2 ρ D 2   4 ( 0.02 m3 s ) 4V m V The velocity is: V = = = =2.55 2 s A πD 2 π ( 0.10m ) The friction factor is a function of Reynolds number and roughness. For the pipe ε =0.045mm , and for water from Appendix A-6 at 10 ºC, µ = 12.9 × 10−4 Ns m 2 , ρ = 999.6 kg m3 . The Reynolds number is: P1 = P2 + ρ g ( z2 − z1 ) + f

ρ VD ( 999.6 kg m ) ( 2.55 m s )(1m ) = = 198, 000 12.9 × 10−4 Ns m 2 µ 3

Re =

The flow is turbulent, so ⎡⎛ 0.045 100 ⎞1.11 1 6.9 ⎤ = −1.8log ⎢⎜ ⎥ ⎟ + 3.7 198, 000 ⎦⎥ f ⎠ ⎣⎢⎝

→

f = 0.0184

Therefore, 2 ⎛ kN s 2 ⎞ kg ⎞ ⎛ ⎛ 200 ⎞ ( 2.55 m s ) ⎛ kN s ⎞ P1 = 101.3kPa + ⎜ 999.6 3 ⎟ ( 9.81m s 2 ) ( 200m ) ⎜ + 0.0184 ( ) ⎟ ⎜ ⎟ ⎜ ⎟ m ⎠ 2 ⎝ ⎝ 0.10 ⎠ ⎝ 1000 kg m ⎠ ⎝ 1000 kg m ⎠ = 101.3 + 1960 + 120 = 2181kPa ( absolute ) =2080kPa (gage) Answer 2

9- 44

9-41

A fluid flows by gravity down an 8-cm galvanized iron pipe. The pressures at the higher and lower locations are 120 kPa and 140 kPa, respectively. The horizontal distance between the two locations is 30 m, and the pipe has a slope of 1m rise per 10 m of run (horizontal distance). For a fluid with a kinematic viscosity of 10-6 m2/s and a density of 900 kg/m3, determine the flow rate (in m3/s).

Approach: Flow is set by a balance among elevation difference, pressure difference, and line losses. Line losses depend on velocity, so this maybe an iterative solution depending on whether this is a laminar or a turbulent flow. The steady, incompressible flow energy equation can be used to determine the flow.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + + hT + ∑ hL ρ g 2g ρ g 2g

The steady, incompressible flow energy equation is:

There is no pump or turbine, so hP = hT = 0 , the pipe diameter is constant, V1 = V2 , there are no minor losses, and with hL = ( f L D ) V 2 2 g :

P1 − P2 L V2 = ( z2 − z1 ) + f ρg D 2g

Note that ( z2 − z1 ) ≠ −3m . From the problem geometry:

θ = tan −1 (1 10 ) = 5.71o

→

z2 − z1 = −30sin ( 5.71o ) = −2.985m

The friction factor depend on Reynolds number, so ρ VD VD ( V m s )( 0.08m ) Re = = = = 80, 000 V 10−6 m 2 s µ ν

(1)

For any reasonable velocity the flow will be turbulent, so with ε =0.15 mm, ε D = 0.15 80 = 0.00188 ⎡⎛ ε D ⎞1.11 6.9 ⎤ = −1.8log ⎢⎜ ⎥ ⎟ + Re ⎥⎦ f ⎢⎣⎝ 3.7 ⎠ Substituting known values into the pressure drop equation: kN ⎛ 1000kgm ⎞ (120-140 ) 2 ⎜ 2 ⎟ m ⎝ kN s 2 ⎠ ⎛ 30m ⎞ ( V m s ) =-2.985m+f ⎜ ⎟ ⎝ 0.08m ⎠ 2 ( 9.81m s 2 ) ( 900 kg m3 )( 9.81m s2 ) 1

-2.265m=-2.985m+19.11f ( V m s )

2

(2)

(3)

The above three equations can be solved iteratively to determine velocity and, hence, flow rate. Performing the iteration: V (m/s) = 1.24 m/s Re=98,900 f = 0.0247 For volume flow rate m ⎞π m3 2 ⎛ V = VA = ⎜1.24 ⎟ ( 0.08m ) =0.00623 Answer s ⎠4 s ⎝

9- 45

9-42

A new factory is to be built that requires 0.03 m3/s of water. The water main from which the water will be obtained is 150-m from the factory. The water main pressure is 400 kPa (gage), and the factor needs 100 kPa (gage) water at a location 10-m above the water main. Assuming that galvanized steel pipe will be used, determine the minimum pipe diameter needed (in m).

Approach: Friction pressure loss depends on the friction factor and velocity, both of which depend on the pipe diameter. An iterative solution is required. The steady, incompressible flow energy equation is used to find the required diameter.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: From point 1 to point 2, the steady, incompressible flow energy equation is: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + + hT + ∑ hL ρ g 2g ρ g 2g There is no pump or turbine, so hP = hT = 0 . The pipe diameter is constant, so V1 = V2 . Only line losses are present Therefore, P1 − P2 L V2 = ( z2 − z1 ) + f ρg D 2g 4 ( 0.3m3 s ) 0.0382 m 4V V = = = (1) 2 A πD 2 D2 s π ( Dm ) The friction factor is a function of Reynolds number and roughness. For water from Appendix A-6 at 10 ºC (assumed), µ = 12.9 × 10−4 Ns m 2 , ρ = 999.6 kg m3 . The Reynolds numbers are:

The velocity is:

Re =

V =

3 2 ρ VD ( 999.6 kg m )( 0.0382 D m s ) ( Dm ) 29, 600 = = 12.9 × 10−4 Ns m 2 D µ

(2)

For any reasonable size pipe, the flow will be turbulent, and from Table 9-2, with ε =0.15mm , so ⎡⎛ ε D ⎞1.11 6.9 ⎤ 1 (3) = −1.8log ⎢⎜ ⎥ ⎟ + Re ⎦⎥ f ⎣⎢⎝ 3.7 ⎠ Substituting known values into the pressure drop equation 2 ( 400-100 ) kN m 2 (1000kgm kNs 2 ) ⎛ 150 ⎞ ( V m s ) =10m+ f ⎜ ⎟ ⎝ D ⎠ 2 ( 9.81m s ) ( 999.6 kg m3 )( 9.81m s2 ) f V2 f V2 → D= D 2.694 The above four equations are solved iteratively: D = 0.104 m V = 3.54 m/s Re = 285,000 f = 0.0223 The next standard pipe size greater than 0.104 would be used.

2.694 =

9- 46

(4)

Answer

9-43

For the system shown below, a water flow rate of 3 m3/s is to be pumped from the lower to the upper reservoir through a 1-m diameter commercial steel pipe. The pump has a mechanical efficiency of 80%. Neglecting minor losses, determine the power required (in kW).

Approach: The required pump head can be obtained directly from the steady, incompressible flow energy equation. Once the head is determined, the pumping power can be calculated.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: From point 1 to point 2, the steady, incompressible flow energy equation is: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + + hT + ∑ hL ρ g 2g ρ g 2g There is no turbine, so hT = 0 . The areas at 1 and 2 are large, V1 = V2 = 0 , and pressure is atmospheric, so P1 = P2 . The losses include inlet, exit, and line losses. Therefore, L⎞V2 ⎛ hP = ( z2 − z1 ) + ⎜ K ent + K exit + f ⎟ D ⎠ 2g ⎝

4 ( 3m3 s ) 4V m V =3.82 The velocity is: V = = = 2 2 s A πD π (1m ) The friction factor is a function of Reynolds number and roughness. For water from Appendix A-6 at 10 ºC (assumed), µ = 12.9 × 10−4 Ns m 2 , ρ = 999.6 kg m3 . The Reynolds numbers are:

ρ VD ( 999.6 kg m ) ( 3.82 m s )(1m ) = = 2.96 × 106 Re = 12.9 × 10−4 Ns m 2 µ 3

Assuming a sharp edged entrance and exit, from Figures 9-13 and 9-14, K ent = 0.5 and K exit = 1.0 . The flow is turbulent and from Table 9-2, ε =0.045mm , so ⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0.045 1000 ⎞1.11 1 6.9 ⎤ f = 0.0113 = −1.8log ⎢⎜ → ⎥ = −1.8log ⎢⎜ ⎥ ⎟ + ⎟ + 3.7 Re ⎥⎦ 2.96 × 106 ⎥⎦ f ⎠ ⎢⎣⎝ 3.7 ⎠ ⎢⎣⎝ Substituting known values into the pressure drop equation 2 ⎡ ⎛ 245 ⎞ ⎤ ( 3.82m s ) hP = ( 730-673) m+ ⎢0.5 + 1 + ( 0.0113) ⎜ ⎟⎥ ⎝ 1 ⎠ ⎦ 2 ( 9.81m s ) ⎣ = 57.0m+3.2m=60.2m  P mgh W = =

ηP

Answer

( 999.6 kg m3 )( 3m3 s )( 9.81m s2 ) ( 60.2m ) (1kNs 2 1000kgm ) 0.80

9- 47

=2,213kW

Answer

9-44

Fresh air is distributed in a factory through a 250-ft long rectangular galvanized duct, which is 36-in. by 6in. For a flow rate of 5000 ft3/min of 60 °F air at 14.2 lbf/in.2, determine the fan pumping power required if the fan has a mechanical efficiency of 65% (in hp).

Approach: The required fan head can be obtained directly from the steady, incompressible flow energy equation. Once we determine the required head, the fan power can be calculated.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: The steady, incompressible flow energy equation is:

P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

There is no turbine, so hT = 0 . The duct has constant area and is horizontal V1 = V2 and z1 = z2 . Point 1 is just upstream of the fan and point 2 is outside of the duct, so both are at atmospheric pressure and P1 = P2 . Therefore, with only an exit minor loss and line loss: 2 ⎛ L ⎞V hP = ⎜ f + K exit ⎟ ⎝ D ⎠ 2g where K exit = 1 . Assuming air is an ideal gas:

2 2 2 lbm PM (14.2 lbf in. ) ( 28.97 lbm lbmol ) (144in. ft ) =0.0737 3 ρ= = RT ft (1545ft lbf lbmR )( 60+460 ) R

The friction factor is a function of Reynolds number and roughness. For the duct (from Table 9-2),

ε =0.0005 ft , and for air from Appendix B-7 at 60 ºF, µ = 1.214 × 10−5 lbm ft s . The velocity is:

V =

3 ft V ( 5000 ft min ) (1min 60s ) =55.6 = s A ( 0.5ft )( 3ft )

Because this is a non-circular duct, we need the hydraulic diameter: 4 ( 0.5ft )( 3ft ) 4 Ax 4 HW =0.857ft = = Dh = pwetted 2 ( H + W ) 2 ( 0.5+3) ft Re =

3 ρ VDh ( 0.0737 lbm ft ) ( 55.6 ft s )( 0.857ft ) = = 289, 000 1.214 × 10 −5 lbm ft s µ

The flow is turbulent, so ⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0.0005 0.857 ⎞1.11 1 6.9 ⎤ 1.8log = −1.8log ⎢⎜ + = − ⎥ ⎢ ⎥ ⎟ ⎜ ⎟ + 3.7 289, 000 ⎦⎥ Re ⎦⎥ f ⎠ ⎣⎢⎝ 3.7 ⎠ ⎣⎢⎝ Therefore, 2 ⎡ ⎛ 250 ⎞ ⎤ ( 55.6 ft s ) hP = ⎢( 0.0186 ) ⎜ =308ft ⎟ +1⎥ ⎝ 0.857 ⎠ ⎦ 2 ( 32.2 ft s 2 ) ⎣

→

f = 0.0186

Power is lbm ⎞ ⎛ ft 3 ⎞ ⎛ 1min ⎞ ⎛ ft ⎞ ⎛ ⎛ 1hp s ⎞ 0.0737 5000 ⎟⎜ ⎜ ⎟ ⎜ 32.2 2 ⎟ ( 308ft ) ⎜ ⎟ 3 ⎟⎜  ft min 60s s  mgh ρ Vgh ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ 550 ft lbf ⎠ ⎝ ⎠ P P W P = = = ηP ηP ⎛ 32.2ft lbm ⎞ ( 0.65 ) ⎜ ⎟ 2 ⎝ lbf s ⎠ = 5.3hp

Answer 9- 48

9-45

Water at 20 °C is to be siphoned from a large tank, as shown below. The siphon is a 2.5-cm diameter smooth tube and has a reentrant inlet. Determine: a. the volume flow rate if only the minor losses are taken into account (in m3/s) b. the volume flow rate if both the minor and line losses are taken into account (in m3/s).

Approach: The flow can be determined by applying the steady, incompressible flow energy equation between points 1 and 2. For part (b) an iterative solution is required.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: The steady, incompressible flow energy equation is:

P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

There is no pump or turbine, so hT = hp = 0 . We assume the area at 1 large, so V1 ≈ 0 , and the pressure at 1 and 2 are atmospheric, so P1 = P2. The losses, in general, include one entrance, one bend, and line loss: 2 ⎛ L ⎞V ∑ hL = ⎜⎝ f D + K ent + Kbend ⎟⎠ 2 g Combining and solving for the velocity: 2 g ( z2 − z1 ) ⎡ ⎤ V =⎢ ⎥ ⎣1 + K ent + K bend + f L D ⎦ From Figure 9-13a, for a reentrant entrance Kent = 0.8. From Table 9-3, for a 180º bend, Kbend = 1.5. a) Ignoring line loss, f = 0 0.5

⎡ 2 ( 9.81m s )(1.5m ) ⎤ m V =⎢ ⎥ =2.99 1+0.8+1.5+0 s ⎣ ⎦ 2 V = VA = ( 2.99 m s )( π 4 )( 0.025m ) =0.00147 m3 s 0.5

b) The line loss will decrease the flow, so the procedure we will follow is: assume a velocity, calculate Reynolds number, evaluate friction factor, and then calculate velocity from the complete velocity equation. For smooth tubing ε =0 , and for water from Appendix A-6 at 20 ºC, µ = 9.85 × 10−4 Ns m 2 , ρ = 998.2 kg m3 . The friction factor is a function of Reynolds number and roughness. 3 ρ VD ( 998.2 kg m ) ( V m s )( 0.025m ) Re = = = 25,340 V 9.85 × 10 −4 Ns m 2 µ For a velocity even one-fourth the velocity obtained without line loss, the flow still would be turbulent, so −2 ⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0 ⎞1.11 6.9 ⎤ ⎛ 1 ⎡ 6.9 ⎤ ⎞ f 1.8log 1.8log = −1.8log ⎢⎜ + = − + → = + ⎥ ⎢⎜ ⎥ ⎜ ⎟ ⎟ ⎢ ⎥⎟ Re ⎦⎥ Re ⎦⎥ f ⎣ Re ⎦ ⎠ ⎝ ⎣⎢⎝ 3.7 ⎠ ⎣⎢⎝ 3.7 ⎠ Guess V = 2 m/s → Re = 50,700 → f = 0.0207 → ⎡ ⎤ 2 ( 9.81m s )(1.5m ) m V =⎢ ⎥ =2.47 1+0.8+1.5+ 0.0207 0.1+0.1+1.5+0.1 /2 0.025 s π ( )( ) ⎥⎦ ⎣⎢ For a second iteration: V = 2.47 m/s → Re = 62,500 → f = 0.0197 → V = 2.49 m/s This is close enough so 2 V = VA = ( 2.49 m s )( π 4 )( 0.025m ) =0.00122 m3 s Answer 0.5

9- 49

9-46

A pump draws 40 °F water from a lake through 20 ft of commercial steel pipe; the line has a reentrant inlet and a 90° regular flanged elbow. The pump elevation is 12 ft above the lake surface. For a design flow rate of 100 gal/min, the head at the suction side of the pump must not be less than –20 ft of water. Determine the minimum pipe diameter (in in.).

Approach: The pressure drop in the inlet piping must be calculated, and the pressure at point 2 expressed in terms of head. The steady, incompressible flow energy equation must be used. Because friction factor depends on velocity and, hence, diameter, an iterative solution is required.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: The steady, incompressible flow energy equation is:

P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

There is no pump, so hp = 0 . We assume the area at 1 large, so V1 ≈ 0 , and the pressure at 1 is atmospheric, so using gage pressure P1 = 0. Let z1 = 0 . The losses, in general, include one entrance, one bend, and line loss: 0=

2 P2 V2 ⎛ L ⎞V + 2 + z2 + ⎜ f + K ent + K bend ⎟ 2 ρ g 2g ⎝ D ⎠ 2g

2 P2 L ⎛ ⎞V = h2 = − z2 − ⎜1 + f + K ent + K bend ⎟ 2 ρg D ⎝ ⎠ 2g The velocity is determined from conservation of mass for incompressible flow: 4 (100gal min ) ( 0.1337ft 3 gal ) (1min 60s ) 0.284 ft V 4V V2 = = = = 2 A π D2 D2 s π ( Dft )

Solving for P2 ρ g :

For commercial steel pipe ε =0.00015ft , and for water from Appendix B-6 at 40 ºF, µ = 104 × 10−5 lbm ft s , ρ = 62.4 lbm ft 3 . From Figure 9-13a, for a reentrant entrance Kent = 0.8. From Table 9-3, for a 90º bend, Kbend = 0.3. The friction factor is a function of Reynolds number and roughness. 3 2 ρ VD ( 62.4 lbm ft )( 0.284 D ft s ) ( Dft ) 17, 000 = = Re = µ 104 × 10−5 lbm ft s D For any pipe with a diameter of less than about 8 ft, the flow will be turbulent, so ⎡⎛ ε D ⎞1.11 6.9 ⎤ 1 = −1.8log ⎢⎜ ⎥ ⎟ + Re ⎦⎥ f ⎣⎢⎝ 3.7 ⎠ We know that h2 ≤ −20ft . The procedure is: guess diameter, calculate velocity, Reynolds number, and friction factor, and then calculate h2 from the energy equation and compare to the limit. D (ft) Re f h2 (ft) V (ft/s) 0.2 7.10 85,000 0.0214 -15.4 0.17 9.83 100,000 0.0215 -19.0 0.16 11.1 106,300 0.0216 -21.2 Based on these calculations, the standard pipe size about 0.17 ft = 2.04 in. in diameter or larger would be chosen for the suction pipe. Answer

9- 50

9-47

Large farm implements and road construction equipment use hydraulically actuated cylinders to position scoops, cutting blades, and other tools. High-pressure pumps are used to circulate the hydraulic fluid (ρ = 880 kg/m3 and µ = 0.033 N·s/m2). Consider a hydraulic system that has a pump outlet pressure of 20 MPa and which requires a minimum pressure at the hydraulic cylinder of 18 MPa at a flow rate of 0.0005 m3/s. If the hydraulic fluid flows through 25 m of smooth, drawn steel tubing, determine the minimum tubing diameter required (in cm).

Approach: The steady, incompressible flow energy equation must be used. For line friction losses, friction factor is a function of velocity, which depends on pipe diameter. Depending on the flow, an iterative solution may be required.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: The steady, incompressible flow energy equation is:

P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

There is no pump or turbine, so hp = hT = 0 . The pipe diameter and elevation are constant, so V1 = V2 and z1 = z2 . Ignoring minor losses: P1 − P2 L V2 = f ρg D 2g The friction factor is a function of Reynolds number and roughness. However, because of the low flow rate and large viscosity, we can speculate that the flow might be laminar. Therefore, we will assume (initially) that the flow is fully developed laminar flow. For this situation, f = 64 Re Substituting into the pressure drop relation, using V = V A = 4V π D 2 , and solving for diameter: 14 ⎡128 ( 0.033 Ns m 2 ) ( 25m ) ( 0.0005 m3 s ) ⎤ ⎡128µ LV ⎤ ⎢ ⎥ = D=⎢ ⎥ π ( 2000 kN m 2 ) (1000N 1kN ) ⎢⎣ ⎥⎦ ⎣ π∆P ⎦ Checking Reynolds number ρ VD m 4 ρV 4 ρV Re = → V = = → Re = 2 µ ρA πD πµ D

Re =

4 ( 880 kg m3 )( 0.0005 3 s )

π ( 0.033 N s m 2 ) ( 0.00957m )

14

=0.00957m=9.57mm

=1770

This is laminar, so our assumption was valid. The diameter would be the next standard diameter greater Answer than 9.57 mm.

Comments: If the flow had been turbulent, we would have needed to solve the pressure drop equation for diameter, and then iterated to find a solution. We would have assumed a friction factor, calculated a diameter, calculated Reynolds number, and then checked the friction factor for convergence.

9- 51

9-48

Water at 70 °F with a flow rate of 30 gal/min flows from a 1-in. diameter tube into a 2-in. diameter tube through a sudden expansion. Determine the pressure rise across the expansion (in lbf/in.2).

Approach: Applying the steady, incompressible flow energy equation between points 1 and 2, we can determine the pressure rise across the expansion.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: The steady, incompressible flow energy equation is:

P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

There is no pump or turbine, so hT = hp = 0 . The pipe is horizontal, so z1 = z2 . The only loss is due to the sudden expansion. Solving for the pressure rise: V2 ρ P2 − P1 = ( V1 2 − V22 ) − K exp ρ 1 2 2 2 From conservation of mass, m 1 = m 2 → V2 = V1 ( A1 A2 ) = V1 ( D1 D2 ) Substituting this into the pressure drop equation and simplifying: 4 ⎤ V1 2 ⎡ ⎛ D1 ⎞ ⎢ P2 − P1 = ρ 1− ⎜ ⎟ − K exp ⎥ 2 ⎢ ⎝ D2 ⎠ ⎥⎦ ⎣ 4 ( 30 gal min ) ( 0.1337 gal ft 3 ) (1min 60s ) V 4V ft The velocity is: V1 = = = =12.2 2 2 A π D1 s π (1 12 ft ) Using Figure 9-15b for a sudden expansion, with A1 A2 = ( D1 D2 ) = (1 2 ) = 0.25 2

( 62.2 lbm ft ) (12.2 ft s ) 3

Therefore,

P2 − P1 =

2 = 0.380

lbf in.2

2

2

→

⎡ ⎛ 1 ⎞4 ⎤ ⎛ lbf s 2 ⎞ ⎛ 1ft ⎞ 2 ⎢1+ ⎜ ⎟ -0.56 ⎥ ⎜ ⎟⎜ ⎟ ⎢⎣ ⎝ 2 ⎠ ⎥⎦ ⎝ 32.2ft lbm ⎠ ⎝ 12in. ⎠

Answer

9- 52

K exp ≈ 0.56

9-49

A Class 100 clean room is to be supplied with 15 m3/min of air, which enters the duct (shown below) at 100 kPa, 25 °C. All entrances and exits are all sharp edged. Determine: a. the pressure in the clean room (in kPa) b. the fan power required (in W).

Approach: The pressure in the clean room can be determined by applying the steady, incompressible flow energy equation between points 2 and 3. The fan power can be obtained by applying the energy equation between points 1 and 3.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: a) The steady, incompressible flow energy equation is:

P V2 P2 V2 + 2 + z2 + hP = 3 + 3 + z3 + hT + ∑ hL ρ g 2g ρ g 2g

There is no pump or turbine, so hT = hp = 0 . The pipe is horizontal, so z1 = z2 = z3 . The pressure loss is due to line loss, plus one expansion and one contraction, so L V2 V2 V2 hL = f + K ent + K exit D 2g 2g 2g Assuming the room at 2 and the outside at 3 are large, V2 ≈ V3 ≈ 0 . Solving the energy equation for P2 (note that P3 = P1 ): 2 ⎛ L ⎞ V P2 = P3 + ⎜ f + K ent + K exit ⎟ ρ 2 ⎝ D ⎠ The velocity is determined from conservation of mass for incompressible flow: 4 (15m3 min ) (1min 60s ) V 4V m V = = = =5.09 2 2 A πD s π ( 0.25m )

The friction factor is a function of Reynolds number and roughness. For smooth pipe ε =0 , and for air from Appendix A-7 at 100 kPa, 25 ºC, µ = 1.832 × 10−5 Ns m 2 , ρ = 1.169 kg m3 . 3 ρ VD (1.169 kg m ) ( 5.09 m s )( 0.25m ) Re = = = 81, 200 1.832 × 10 −5 Ns m 2 µ

The flow is turbulent, so ⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0 ⎞1.11 1 6.9 ⎤ = −1.8log ⎢⎜ ⎥ = −1.8log ⎢⎜ ⎥ → f = 0.0186 ⎟ + ⎟ + Re ⎥⎦ 81, 200 ⎥⎦ f ⎢⎣⎝ 3.7 ⎠ ⎢⎣⎝ 3.7 ⎠ Using Figure 9-15a for the entrance and Figure 9-15b for a sudden expansion, with A1 A2 = 0 → K ent ≈ 0.5 and K exit = 1.0 Therefore: 2 15 kg ⎞ ( 5.09 m s ) ⎛ kN s 2 ⎞ ⎡ ⎤⎛ P2 = 100kPa+ ⎢( 0.0186 ) +0.5+1⎥ ⎜1.169 ⎜ ⎟ ⎟ 0.25 m ⎠ 2 ⎣ ⎦⎝ ⎝ 1000 kg m ⎠ =100+0.0397=100.04kPa Answer b) For fan power, we recognize that the pressure drop from just after the fan to the clean room is the same as from the clean room to the outside (calculated above). Using the energy equation, again assuming incompressible flow, and taking into account the two identical lengths of pipe: ⎛ m3 ⎞ ⎛ kN ⎞ ⎛ 1min ⎞ ⎛ 1000W s ⎞ W = −V ∆Ptot = −2V ∆P23 = -2 ⎜ 15 Answer ⎟ ⎜ 0.0397 2 ⎟ ⎜ ⎟ =-19.9W ⎟⎜ min m ⎠ ⎝ 60s ⎠ ⎝ kN m ⎠ ⎝ ⎠⎝ 9- 53

9-50

In Problem P 9-49, the sharp-edged entrances and exits are replaced with well-rounded entrances and exits. For the same fan power as in the original installation, determine the new volumetric flow rate (in m3/min).

Approach: We will use the information from the solution to Problem P 9-49. The main difference is that Kent and Kexit will change because of the well-rounded entrances and exits.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: We have two equations to combine: W = −2V ∆P23 = −2 VA ( P2 − P3 ) 2 ⎛ L ⎞ V P2 − P3 = ⎜ f + K ent + K exit ⎟ ρ 2 ⎝ D ⎠ The friction factor is a function of velocity, but the new flow probably will not cause a significant change in the friction factor. Therefore, we will assume it is approximately constant. Combining the two equations and solving for velocity: 13

⎡ ⎤ −4W V =⎢ ⎥ 2 ⎣⎢ π D ρ ( fL D + K ent + K exit ) ⎦⎥

For well-rounded entrances and exits, from Figure 9-13d and Figure 9-14d, K ent ≈ 0.04 and K exit = 1.0 . 13

⎡ ⎤ ⎢ -4 ( -19.9W )(1J Ws )(1Nm 1J ) ( kgm Ns 2 ) ⎥ m ⎥ =5.43 =⎢ s ⎢ π 0.25m 2 1.169 kg m3 ⎛ 0.0186 15 +0.04+1⎞ ⎥ ) ( ) ⎜⎝ ⎟⎥ ⎢ ( 0.25 ⎠⎦ ⎣ The volume flow rate is: 2 V = VA = ( 5.43m s )( π 4 )( 0.25m ) ( 60s 1min ) =16.0 m3 s % increase =

16.0 − 15.0 × 100 = 6.7% 15.0

Answer

Comments: Note that the new Reynolds number is 86,600 and the friction factor is 0.0184, which would not change our answer very much.

9- 54

9-51

Frictional pressure loss in fluid flow is converted to unwanted thermal energy. Consider an 18 gal/min flow of 70 °F water through a 1.25-in. diameter smooth tube. The tube is sloped so that the pressure remains constant throughout the tube. Determine: a) the slope (in ft/100 ft), b) the heat transfer per 100 ft of tube if the temperature remains constant (Btu/hr), and c) the temperature rise if the tube is perfectly insulated (in °F).

Approach: For constant pressure flow, the steady, incompressible flow energy equation is used. Frictional losses can be calculated because the flow is given. With constant temperature, Eq. 9-29 can be used to calculate the heat transfer; the same equation can be used to calculate temperature rise if the pipe is insulated.

Assumptions: 1. 2. 3. 4.

The system is steady. The flow is fully developed. Properties are constant. Neglect minor losses.

Solution: a) The slope (in ft/100 ft of pipe) requires that the elevation difference, z2 − z1 , be determined. Using the steady, P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

incompressible flow energy equation:

The areas at 1 and 2 are equal, so V1 = V2 . There is no pump or turbine, so hP = hT = 0 , and minor losses are neglected. For constant pressure flow P1 = P2 . Therefore,

−hL = z2 − z1 = − f

L V2 D 2g

4 (15gal min ) ( 0.1337ft 3 gal ) (1min 60s ) V 4V ft = = =3.93 2 2 A πD s π ( 0.104ft ) The friction factor is a function of Reynolds number and roughness. Assuming a smooth pipe ε = 0 , and

Velocity is : V =

for water from Appendix B-6 at 70 ºF, µ = 65.8 × 10−5 lbm fts , ρ = 62.2 lbm ft 3 :

ρ VD ( 62.2 lbm ft ) ( 3.93ft s )( 0.104ft ) Re = = = 38, 700 65.8 × 10 −5 lbm fts µ 3

The flow is turbulent, so

⎡⎛ 0 D ⎞ = −1.8log ⎢⎜ ⎟ f ⎢⎣⎝ 3.7 ⎠

1.11

1

+

6.9 ⎤ ⎥ 38, 700 ⎥⎦

→

f = 0.022

L V2 ⎛ 100 ⎞ ( 3.93ft s ) = − ( 0.022 ) ⎜ = −5.06ft ⎟ D 2g ⎝ 0.104 ⎠ 2 ( 32.2 ft s 2 ) 2

−hL = z2 − z1 = − f

So the slope is -5.06 ft/100 ft or θ = sin −1 ( −5.06 100 ) = −2.9o .

Answer

b) For the heat transfer , using Eq. 9-29 hL = ⎡⎣( u2 − u1 ) − q ⎤⎦ g . For isothermal flow, u2 − u1 = 0 , so that  = − mh  g = − ρ VAh g q = − h g → Q = mq L

L

L

lbm ⎞ ⎛ ft ⎞⎛ π ⎞ ft ⎞ ⎛ 1Btu ⎞ ⎛ lbf s 2 ⎞ Btu 2 ⎛ ⎛ Q = − ⎜ 62.2 2 ⎟ ⎜ 3.93 ⎟⎜ ⎟ ( 0.104ft ) ( 5.06ft ) ⎜ 32.2 2 ⎟ ⎜ Answer ⎟ =-0.0135 ⎟⎜ ft ⎠ ⎝ s ⎠⎝ 4 ⎠ s ⎠ ⎝ 778ft lbf ⎠ ⎝ 32.2ft lbm ⎠ s ⎝ ⎝ c) For an insulated pipe, q = 0, so again using Eq. 9-28, and assuming an ideal liquid with constant specific heat: hL = ⎡⎣( u2 − u1 ) − q ⎤⎦ g → u2 − u1 = hL g → c (T2 − T1 ) = hL g T2 − T1 = hL g c =

( 5.06ft ) ( 32.2 ft s2 ) (1Btu 778ft lbf ) =0.0065o F (1.00 Btu lbmR ) ( 32.2 ft lbm lbf s 2 ) 9- 55

9-52

A gas turbine power plant consists of a compressor, a combustor in which the fuel and air are mixed and combusted, and a turbine that drives an electrical generator. The air compression process takes from 40 to 80% of the turbine output power, leaving only 20-60% to drive the electric generator. Some gas turbine plants store compressed air in salt domes or caverns for use during times when additional electric power is needed; the compressed air to be supplied to the power plant is taken from the stored air instead of just using the air compressor. Consider the system shown in the figure below. The air reservoir, which fills with 10 °C water when the air has been used, is connected to the outside by a 30-cm cast iron pipe. During charging of the reservoir with air, the air pressure, P, increases. Determine the gage pressure P required to produce a water flow rate of 0.15 m3/s (in kPa).

Approach: The required pressure can be determined by applying the steady, incompressible flow energy equation between points 1 and 2. The frictional losses must be calculated.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: The steady, incompressible flow energy equation is:

P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

There is no pump or turbine, so hT = hp = 0 . We assume the surface of the reservoir is large, so V1 ≈ 0 , and the pressure at 2 is atmospheric. Expressing P1 and P2 in terms of gage pressures, P2, gage = 0 Therefore: V22 + ρ g ∑ hL 2 The losses include line loss, one entrance, and two bends: 2 ⎛ L ⎞V hL = ⎜ f + K ent + K exit + 2 K bend ⎟ ⎝ D ⎠ 2g The entrance has an area ratio of zero, so from Figure 9-15a, Kent = 0.5. Assuming the bends are regular, 90º, flanged, from Table 9-3, Kbend = 0.3. The velocity is determined from conservation of mass for incompressible flow: 4 ( 0.15m3 s ) V 4V m V2 = = = =2.12 2 2 A πD s π ( 0.30m ) P1, gage = ρ g ( z1 − z2 ) + ρ

The friction factor is a function of Reynolds number and roughness. For cast iron pipe ε =0.26mm , and for water from Appendix A-6 at 10 ºC, µ = 12.9 × 10−4 Ns m 2 , ρ = 999.6 kg m3 .

ρ VD ( 999.6 kg m ) ( 2.12 m s )( 0.30m ) = = 493, 000 12.9 × 10−4 Ns m 2 µ 3

Re =

The flow is turbulent, so ⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0.26 300 ⎞1.11 1 6.9 ⎤ 1.8log = −1.8log ⎢⎜ + = − ⎥ ⎢⎜ ⎥ ⎟ ⎟ + Re ⎦⎥ 493, 000 ⎦⎥ f ⎣⎢⎝ 3.7 ⎠ ⎣⎢⎝ 3.7 ⎠

→

f = 0.0196

2 ⎧⎪⎛ kg ⎞⎛ m⎞ 850 kg ⎞ ( 2.12 m s ) ⎫⎪ ⎛ kN s 2 ⎞ ⎡ ⎤⎛ 9.81 50m + 0.0196 +0.5+1+2 0.3 999.6 P1, gage = ⎨⎜ 999.6 ( ) ( ) ( ) ⎬⎜ ⎟ ⎟⎜ ⎟ ⎟ ⎢ ⎥⎜ m ⎠⎝ s2 ⎠ 0.30 m ⎠ 2 ⎣ ⎦⎝ ⎪⎭ ⎝ 1000 kg m ⎠ ⎩⎪⎝ =620kPa (gage) Answer

9- 56

9-53

Water at 20 °C is pumped from a reservoir through a 20-cm commercial steel pipe for 5 km from the pump outlet to a reservoir whose surface is 150-m above the pump. The flow rate is 0.10 m3/s. Determine: a. the pressure at the pump outlet (in kPa) b. the pumping power required (in W).

Approach: We can determine the pressure at 2 by using the steady, incompressible flow energy equation between points 2 and 3. The pump power can be calculated with the same equation but applied between points 1 and 2.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: a) The steady, incompressible flow energy equation between points 2 and 3 is: P V2 P2 V2 + 2 + z2 + hP = 3 + 3 + z3 + hT + ∑ hL ρ g 2g ρ g 2g The reservoir is large, so V3 ≈ 0 . There is no pump or turbine, so hP = hT = 0 . Expressing P2 and P3 in terms of gage pressure, so P3, g = 0 . The losses include line loss and exit loss ( K exit = 1 ). Solving for P2, g ⎡ ⎡ V2 V2 L V22 ⎤ L V22 ⎤ P2, g = ρ g ⎢( z3 − z2 ) − 2 + K exit 2 + f g z z f ρ = − + ( ) ⎥ ⎢ ⎥ 3 2 D 2g ⎦ D 2 ⎦ 2g 2g ⎣ ⎣ The friction factor is a function of Reynolds number and roughness. For water from Appendix A-6 at 20 ºC, µ = 9.85 × 10−4 Ns m 2 , ρ = 998.2 kg m3 . For commercial steel pipe, ε = 0.045 mm, ε D = 0.045 200 = 0.000225 .

Velocity is :

4 ( 0.10 m3 s ) V 4V m V = = = =3.18 2 2 A πD s π ( 0.20m )

ρ VD ( 998.2 kg m ) ( 3.18 m s )( 0.20m ) Re = = = 645, 000 9.85×10-4 Ns m 2 µ 3

This is turbulent flow so ⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0.000225 ⎞1.11 1 6.9 ⎤ = −1.8log ⎢⎜ ⎥ = −1.8log ⎢⎜ ⎥ → f = 0.0152 ⎟ + ⎟ + Re ⎥⎦ 645, 000 ⎥⎦ f ⎢⎣⎝ 3.7 ⎠ ⎢⎣⎝ 3.7 ⎠ Substituting known values into the pressure drop equation: 2 kg ⎞ ⎡⎛ m⎞ ⎛ ⎛ 5000 ⎞ ( 3.18 m s ) ⎤ ⎛ kN ⎞ ⎢ ⎥⎜ P2, g = ⎜ 998.2 3 ⎟ ⎜ 9.81 2 ⎟ (150m ) + ( 0.0152 ) ⎜ Answer ⎟ ⎟ =3390kPa (gage) m ⎠ ⎢⎝ s ⎠ 0.2 ⎠ 2 1000N ⎠ ⎝ ⎝ ⎥ ⎣ ⎦⎝ b) Applying the same energy equation between points 1 and 2. Assume pressure 1 is atmospheric, and the pipe diameters at 1 and 2 are the same. Therefore: P2, g − P1, g P2, g hP = = ρg ρg The pumping power is P 3 2  ⎛⎜ 2, g ⎞⎟ = VP   P = ρVg W = mgh Answer 2, g = ( 0.10 m s )( 3390 kN m ) (1kWs 1kNm ) =339kW ⎝ ρg ⎠

9- 57

9-54

Two reservoirs are connected by three galvanized iron pipes in series. The first pipe is 600-m long, 20-cm diameter; the second pipe is 800-m long, 30-cm diameter; and the third pipe is 1200-m long, 40-cm in diameter. For a flow of 0.15 m3/s of water at 10 °C, determine the elevation difference between the reservoirs (in m).

Approach: The elevation difference drives the flow. That pressure head is consumed by the frictional losses in the line. The steady, incompressible flow energy equation is used to calculate the elevation difference.

Assumptions: 1. 2. 3. 4.

The system is steady. The flow is fully developed. Properties are constant. Neglect minor losses.

Solution: The steady, incompressible flow energy equation is: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g At points 1 and 2 the areas are large and the pressure is atmospheric, so V1 = V2 = 0 and P1 = P2 . There is no pump or turbine, so hP = hT = 0 , and minor losses are neglected. Therefore, z1 − z2 = f A

L V2 LA VA2 L V2 + f B B B + fC C C DA 2 g DB 2 g DC 2 g

The velocities are:

4 ( 0.15 m3 s ) V 4V m VA = = = =4.77 2 2 s AA π DA π ( 0.20m ) VB = VA ( DA DB ) = ( 4.77 m s )( 0.20 0.30 ) =2.12 m s 2

Using conservation of mass:

2

VC = VA ( DA DC ) = ( 4.77 m s )( 0.20 0.40 ) =1.19 m s The friction factor is a function of Reynolds number and roughness. For galvanized iron pipe ε =0.15mm , and for water from Appendix A-6 at 10 ºC, µ = 12.9 × 10−4 Ns m 2 , ρ = 999.6 kg m3 . The Reynolds numbers are: 2

2

ρ VD ( 999.6 kg m ) ( 4.77 m s )( 0.20m ) = = 739, 000 µ 12.9 × 10−4 Ns m 2 3

ReA =

ReB = ReA ( DA DB ) = 493, 000 ReC = ReA ( DA DC ) = 369, 000

T the flow is turbulent, so ⎡⎛ 0.15 200 ⎞1.11 1 6.9 ⎤ = −1.8log ⎢⎜ ⎥ ⎟ + 739, 000 ⎦⎥ fA ⎣⎢⎝ 3.7 ⎠

→

f A = 0.0188

Similarly, f B = 0.0176 and f C = 0.0170 . Therefore, 2 2 2 ⎛ 600 ⎞ ( 4.77 m s ) ⎛ 800 ⎞ ( 2.12 m s ) ⎛ 1200 ⎞ (1.19 m s ) z1 − z = ( 0.0188 ) ⎜ + 0.0176 + 0.0176 ( ) ( ) ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 0.20 ⎠ 2 ( 9.81m s 2 ) ⎝ 0.30 ⎠ 2 ( 9.81m s 2 ) ⎝ 0.40 ⎠ 2 ( 9.81m s 2 ) =65.4m+10.7m+3.8m=79.9m

Answer

Comments: Note that pipe A losses dominate the total pressure loss. For a given elevation difference, if more flow were desired through this system, increasing the diameter of pipe A would be the first option to try.

9- 58

9-55

In a water system, a reservoir is connected to a canal with an 8-in. cast iron pipe. The system has three regular 90° threaded elbows, a half closed gate valve, and the exit from the reservoir is sharp edged. With an elevation difference of 55 ft between the reservoir surface and the pipe outlet, 3 ft3/s of water at 50 °F flows through the pipe. Determine the total length of straight pipe in the system (in ft).

Approach: The length can be determined by applying the steady, incompressible flow energy equation between points 1 and 2. The frictional losses must be calculated.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: The steady, incompressible flow energy equation is:

P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

There is no pump or turbine, so hT = hp = 0 . We assume the surface of the reservoir and canal are large, so V1 ≈ V2 ≈ 0 , and the pressure at 1 and 2 are atmospheric, so P1 = P2. Therefore, taking into account all the minor losses: 2 ⎛ L ⎞V 0 = ( z2 − z1 ) + ⎜ f + K valve + K exit + 3K bend ⎟ ⎝ D ⎠ 2g Solving for the length: ⎤ D ⎡ 2 g ( z2 − z1 ) − ( K valve + K exit + 3K bend ) ⎥ L = ⎢− 2 f ⎣ V ⎦ From Figure 9-15b, Kexit = 1.0. From Table 9-3, Kbend = 1.5 and Kvalve = 2.1. For cast iron pipe ε =0.00085ft , and for water from Appendix B-6 at 50 ºF, µ = 88 × 10−5 lbm ft s , ρ = 62.4 lbm ft 3 . The velocity is determined from conservation of mass for incompressible flow: 4 ( 3ft 3 s ) V 4V ft V = = = =8.59 2 2 A πD s π ( 0.667ft )

The friction factor is a function of Reynolds number and roughness. 3 ρ VD ( 62.4 lbm ft ) ( 8.59 ft s )( 0.667ft ) = = 406,300 Re = 88 × 10−5 lbm ft s µ The flow is turbulent, so ⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0.00085 0.667 ⎞1.11 1 6.9 ⎤ 1.8log = −1.8log ⎢⎜ + = − ⎥ ⎢⎜ ⎥ ⎟ ⎟ + Re ⎥⎦ 3.7 406,300 ⎥⎦ f ⎠ ⎢⎣⎝ 3.7 ⎠ ⎢⎣⎝ L=

2 ⎤ 0.667ft ⎡ 2 ( 32.2 ft s ) ( -55ft ) ⎢- ( 2.1+1+3 [1.5]) ⎥ =1260ft 2 0.0214 ⎢ ⎥⎦ (8.59 ft s ) ⎣

9- 59

→

Answer

f = 0.0214

9-56

On your land high in the Rocky Mountains, you decide to produce your own electric power for your vacation home using a hydroturbine. The surface of the small lake from where you will get the 50 °F water is 500 ft above where you will locate the turbine. You connect the lake and turbine with 1000 ft of 6-in. cast iron pipe. The turbine discharge is the same diameter as the inlet and is open to the atmosphere. Determine the maximum power that can be produced (in W).

Approach: The elevation difference drives the flow. Pressure head is consumed by the frictional losses in the line. Turbine power can be calculated with the steady, incompressible flow energy equation.

Assumptions: 1. The system is steady with constant properties. 2. The flow is fully developed. 3. Neglect minor losses.

Solution: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g

The steady, incompressible flow energy equation is:

At points 1 and 2 the areas are large and the pressure is atmospheric, so V1 = V2 = 0 and P1 = P2 . There is no pump, ( hP = 0 ), and the losses are line and exit losses. Therefore, ⎛ L V2 V2 ⎞ (1) hT = z1 − z2 − ⎜ f + K exit ⎟ 2g ⎠ ⎝ D 2g  T . The friction factor is a function of Reynolds number and roughness, so to find where K exit = 1 and WT = mgh the maximum possible power, an iterative solution is required.. For cast iron pipe ε =0.00085 ft , and for water from Appendix B-6 at 50 ºF, µ = 88 × 10−5 lbm ft s , ρ = 62.4 lbm ft 3 . 4 (V ft 3 s ) V 4V ft The velocity is: V = = = =5.09V A π D 2 π ( 0.5ft )2 s

ρ VD ( 62.4 lbm ft )( 5.09V ft s ) ( 0.5ft ) = = 180, 600V 88 × 10−5 lbm ft s µ 3

Re =

For any reasonable flowrate, the flow is turbulent, so ⎡ hT = 500ft − ⎢ f ⎣

Therefore,

⎡⎛ ε D ⎞1.11 6.9 ⎤ = −1.8log ⎢⎜ ⎥ ⎟ + Re ⎦⎥ f ⎣⎢⎝ 3.7 ⎠

1

2  ⎛ 1000 ⎞ ⎤ ( 5.09V ft s ) ⎜ ⎟ +1⎥ ⎝ 0.5 ⎠ ⎦ 2 ( 32.2 ft s 2 )

(2) (3)

(4)

⎛ lbfs 2 ⎞ ⎛ 1hp s ⎞ ⎛ π lbm ⎞ ⎛ ft ⎞ ⎛ π ⎞ ft ⎞ W⎞ 2⎛ ⎛ WT = ρ V D 2 ghT = ⎜ 62.4 3 ⎟ ⎜ 5.09V ⎟ ⎜ ⎟ ( 0.5 ft ) ⎜ 32.2 2 ⎟ ( hT ft ) ⎜ ⎟⎜ ⎟ ⎜ 746 ⎟ 4 ft ⎠ ⎝ s ⎠⎝ 4 ⎠ s ⎠ hp ⎠ ⎝ ⎝ ⎝ 32.2ft lbm ⎠ ⎝ 550ft lbf ⎠ ⎝   W = 84.6h V W (5) T

T

The procedure to follow is to use the above five equations to calculate the maximum power: guess a volume flow rate, calculate velocity, Reynolds number, friction factor, turbine head, and turbine power. Increase the volume flow and recalculate until a maximum power is obtained. Proceeding with the calculations: V ( ft 3 s ) V ( ft s ) WT ( W ) 2.6 13.24 76,590 2.7 13.75 76,830 2.8 14.26 76,760 At maximum power, f = 0.0274 and Re = 487,500.

Answer

Comments:

For zero power output from the turbine, the volume flow rate would be 4.73 ft3/s and V = 24.1 ft/s. 9- 60

9-57

Fire trucks have pumps to boost the pressure of the water supplied by a fire hydrant. Consider a fire truck that has a 250-ft long, 2-in. diameter smooth fire hose. Water must reach the nozzle at the hose exit at 100 lbf/in.2 (gage). Water from the hydrant reaches the pump inlet at 60 lbf/in.2 (gage). If the design pressure drop specification for the hose is 25 psi/100 ft of length, determine: a. the design flow rate (in gal/min) b. the nozzle exit velocity (in ft/s) c. the pump power required if the pump has a mechanical efficiency of 75% (in hp).

Approach: Flow rate, nozzle exit velocity, and pump power are calculated with the steady, incompressible flow energy equation. However, the equation must be applied across three different segments of the system.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: a) To find the design flow rate, apply the steady, incompressible flow energy equation between points 2 and 3: P V2 P2 V2 + 2 + z2 + hP = 3 + 3 + z3 + hT + ∑ hL ρ g 2g ρ g 2g The duct is constant area and horizontal, so V2 = V3 and z2 = z3 There is no pump or turbine, so hP = hT = 0 , and there are no minor losses: ⎡ 2 ( P2 − P3 ) D ⎤ V2 = ⎢ ⎥ ρ fL ⎣ ⎦ From the design pressure drop specification: ⎡ ( 25lbf in.2 ) ⎤ ⎥ ( 250ft ) =62.5lbf in.2 P2 − P3 = ⎢ ⎢⎣ 100ft ⎥⎦ The friction factor is a function of Reynolds number and roughness, so an iterative solution is required. For water from Appendix B-6 at 70 ºF, µ = 65.8 × 10−5 lbm fts , ρ = 62.2 lbm ft 3 . 4 (V ft 3 s ) V ft Velocity is : V = = =45.84V 2 A π ( 2 12 ft ) s 0.5

P2 − P3 L V22 = f D 2g ρg

→

ρ VD ( 62.2 lbm ft )( 45.84V ft s ) ( 2 12 ft ) = = 722, 200V Re = 65.8 × 10−5 lbm fts µ 3

For any reasonable flow rate, this is turbulent flow, so assuming a smooth duct: ⎡⎛ ε D ⎞1.11 6.9 ⎤ 1 = −1.8log ⎢⎜ ⎥ ⎟ + Re ⎥⎦ f ⎢⎣⎝ 3.7 ⎠ ⎡ 2 ( P2 − P3 ) D ⎤ V2 = ⎢ ⎥ ρ fL ⎣ ⎦ ft 3 V = V 45.84 s Now perform the iteration: V ( ft 3 s )

0.5

⎡ 2 ( 62.5lbf in.2 ) ( 2 12 ft ) ( 32.2 ft lbm lbf s 2 ) ⎤ ⎥ =⎢ 2 3 ⎢ ⎥ 62.2 lbm ft f 250ft 1ft 12in. ( )( ) ( ) ⎣ ⎦

V (ft/s)

Re

0.5 22.9 361,000 0.462 21.2 333,500 This is satisfactory convergence, so the volume flow rate is: 9- 61

0.5

=

2.492 ft f 0.5 s

f

V (ft/s)

V ( ft 3 s )

0.0139 0.0141

21.2 21.0

0.462 0.458

V = ( 0.458ft 2 s ) ( 60s min ) (1gal 0.1337ft 3 ) =206 gal min

Answer

b) To find the nozzle exit velocity, we apply the energy equation between points 3 and 4. Assuming no losses: V2 P3 V2 P + 3 = 4 + 4 ρ g 2g ρ g 2g ⎡ 2 ( P3 − P4 ) ⎤ + V32 ⎥ V4 = ⎢ ρ ⎣ ⎦

0.5

2 ⎡ 2 (100 lbf in.2 )( 32.2 ft lbm lbf s 2 ) ft ⎞ ⎤ ⎛ 21.0 =⎢ + ⎜ ⎟ ⎥ 2 3 s⎠ ⎥ ⎢ ⎝ 62.2 lbm ft 1ft 12in. ( ) ( ) ⎣ ⎦

0.5

= 124

ft s

Answer

c) To find the pumping power, applying the energy equation from point 1 to2: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g The duct is constant area and horizontal, so V1 = V2 and z1 = z2 There is no turbine ( hT = 0 ) and no minor losses, and incorporating other losses in terms of pump efficiency: P −P  P1 − P2 = V ( P − P )  p = ρVg hp = 1 2 and W P = mgh 1 2 ρg ρg This is the power into the fluid. So taking into account pump efficiency: 3 2 2 2 V ( P1 − P2 ) ( 0.458ft s ) ⎡⎣60- (100+62.5 ) ⎤⎦ lbf in. (1hp s 550ft lbf ) (144in. ft ) W P = = ηP 0.75 = -16.4 hp

Answer

9- 62

9-58

Water is pumped from a lake to a pond that is 50 m above the lake. A suction pipe runs from the lake to a pump, and a connecting pipe runs from the pump to the pond. The suction pipe is constructed of 10-cm diameter cast-iron pipe (assume no minor losses). The connecting pipe also is 10-cm diameter cast iron and has five long radius 90° threaded elbows. The pump can be located in one of three places: 1) level with the lake surface, and the suction pipe would be 6-m long and the connecting pipe would total 150-m long, 2) 10-m below the lake surface, and the suction pipe would be 11-m long and the connecting pipe would total 160-m long, and 3) 5-m above the lake surface, and the suction pipe would be 8-m long and the connecting pipe would total 145-m long. For a flow of 0.025 m3/s, determine which installation requires the smallest required pumping power (in W).

Approach: The pumping power is calculated with the steady, incompressible flow energy equation applied between points 1 and 2. The preferred pump location can be determined by comparing losses for each system.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: The steady, incompressible flow energy equation between points 1 and 2 is: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g The lake and pond surface areas are large, so V1 ≈ V2 ≈ 0 and P1 = P2 There is no turbine, so hT = 0 . Therefore: hP = z2 − z1 + ∑ hL All piping is the same diameter. Minor losses in all pump locations include an entrance loss, an exit loss, and 5 elbow losses. The only difference in these locations is the length of straight pipe. The location level at the lake and 5 m above the lake have total pipe lengths of 156 m; the location 10 m below the surface has a length of 163 m. So based solely on the pumping power, the preferred location is either level or above the lake. Pumping power is: 2 ⎡ ⎛ L ⎞V ⎤  p = mg  ⎢ z2 − z1 + ⎜ f + K ent + K exit + 5 K bend ⎟ W P = mgh ⎥ ⎝ D ⎠ 2g ⎦ ⎣ The friction factor is a function of Reynolds number and roughness. For water from Appendix B-6 at 10 ºC, µ = 12.9 × 10−4 Ns m 2 , ρ = 999.6 kg m3 . Velocity is :

4 ( 0.025 m3 s ) V 4V m V = = = =3.18 2 2 A πD s π ( 0.10m )

ρ VD ( 999.6 kg m ) ( 3.18 m s )( 0.10m ) = = 246, 400 Re = 12.9 × 10−4 Ns m 2 µ This is turbulent flow so for cast iron pipe, ε = 0.26mm, ε D = 0.26 100 = 0.0026 : 3

⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0.0026 ⎞1.11 6.9 ⎤ f = 0.0257 = −1.8log ⎢⎜ ⎥ = −1.8log ⎢⎜ ⎥ → ⎟ + ⎟ + Re ⎥⎦ 246, 400 ⎥⎦ f ⎢⎣⎝ 3.7 ⎠ ⎢⎣⎝ 3.7 ⎠ From Table 9-3, Kbend = 0.7, and from Figure 9-15, Kent = 0.5 and Kexit = 1.0 2 The mass flow rate is: m = ρ VA = ( 999.6 kg m3 ) ( 3.18 m s )( π 4 )( 0.10m ) =25.0 kg s 1

The power is: 2 2 kg ⎞⎛ m ⎞⎡ 156 ⎛ ⎛ ⎞ ( 3.18 m s ) ⎤ ⎛ N s ⎞ ⎛ W s ⎞  ⎥ +0.5+1.0+5 [ 0.7 ] ⎟ WP = ⎜ 25.0 ⎟⎜ 9.81 2 ⎟ ⎢50m+ ⎜ [ 0.0257 ] ⎜ ⎟⎜ ⎟ s ⎠⎝ s ⎠⎢ 0.10 ⎝ ⎝ ⎠ 2 ( 9.81m s 2 ) ⎥⎦ ⎝ kg m ⎠ ⎝ N m ⎠ ⎣

= 17,960 W

Answer 9- 63

9-59

Many universities have a central facility that produces chilled water for use in cooling all the buildings on campus. The water at 10 °C is continuously circulated through a closed flow loop and used as needed. Consider a system that consists of 5 km of 30-cm commercial steel pipe with a flow rate of 0.15 m3/s. The pump has a mechanical efficiency of 75% and is driven by a motor that has an efficiency of 92%. Determine: a. the pressure drop (in kPa) b. the pumping power required (in kW) c. the annual cost if electricity costs $0.10/kWhr and the system runs 7,500 hr/yr.

Approach: Pressure drop and pumping power both can be calculated with the steady, incompressible flow energy equation. The equation must be applied across different sections of the system.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: a) The steady, incompressible flow energy equation between points 2 and 1 is: P2 V2 P V2 + 2 + z2 + hP = 1 + 1 + z1 + hT + ∑ hL ρ g 2g ρ g 2g The pipe diameter is constant, so V1 = V2 and horizontal z1 = z2 There is no pump or turbine, so hP = hT = 0 . Ignoring minor losses, ⎛ L V2⎞ L V2 P2 − P1 = ρ g ∑ hL = ρ g ⎜ f ⎟= f ρ D 2 ⎝ D 2g ⎠ The friction factor is a function of Reynolds number and roughness. For water from Appendix A-6 at 10 ºC, µ = 12.9 × 10−4 Ns m 2 , ρ = 999.6 kg m3 . Velocity is:

4 ( 0.15 m3 s ) 4V m V =2.12 V = = = 2 s A π D2 π ( 0.30m )

ρ VD ( 999.6 kg m ) ( 2.12 m s )( 0.30m ) = = 493,300 Re = µ 12.9 × 10−4 Ns m 2 This is turbulent flow so for commercial steel pipe, ε = 0.045mm, ε D = 0.045 300 = 0.00015 : 3

⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0.00015 ⎞1.11 6.9 ⎤ 1.8log = −1.8log ⎢⎜ + = − ⎥ ⎢⎜ ⎥ → ⎟ ⎟ + Re ⎦⎥ 493,300 ⎦⎥ f ⎣⎢⎝ 3.7 ⎠ ⎣⎢⎝ 3.7 ⎠

1

kg ⎞ ( 2.12 m s ) ⎛ 1kN ⎞ ⎛ 5000 ⎞ ⎛ P2 − P1 = ( 0.0148 ) ⎜ ⎟ ⎜ 999.6 3 ⎟ ⎜ ⎟ =555kPa 2 m ⎠ ⎝ 0.30 ⎠ ⎝ ⎝ 1000N ⎠ b) To find the pumping power, applying the energy equation from point 1 to2: P −P  P1 − P2 = V ( P − P )  p = ρVg and W P = mgh hp = 1 2 1 2 ρg ρg

f = 0.0148

2

kN ⎞ ⎛ 1kW s ⎞ ⎛ W P = ( 0.15 m3 s ) ⎜ -555 2 ⎟ ⎜ Answer ⎟ =-83.2kW m ⎠ ⎝ 1kN m ⎠ ⎝ The minus sign indicates power is input to the pump. c) For the cost of running the pump:  ( $0.10 kWhr )(83.2kW )( 7500 hr yr ) CWt = $90,500 yr Cost = = ηPηm ( 0.75)( 0.92 )

9- 64

Answer

Answer

9-60

The Alaskan oil pipeline is 48-in. in diameter, with a wall roughness of approximately 0.0005 ft. The design flow rate is 1.6 × 106 barrels per day (1 barrel = 42 gal). To limit the required pipe wall thickness, the maximum allowable oil pressure is 1200 psig. To keep dissolved gases in solution in the crude oil, the minimum oil pressure is 50 psig. The oil has ρ = 58 lbm/ft3 and µ = 0.0113 lbm/ft·s. Determine: a. the maximum spacing between pumping stations (in km) b. the pumping power at each station if the pump mechanical efficiency is 85% (in kW).

Approach: The maximum length and pumping power are calculated with the steady, incompressible flow energy equation. However, the equation must be applied across different segments of the system.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: a) To find the maximum length, apply the steady, incompressible flow energy equation between points 2 and 1: P2 V2 P V2 + 2 + z2 + hP = 1 + 1 + z1 + hT + ∑ hL ρ g 2g ρ g 2g The pipe is constant area and horizontal, so V2 = V1 and z2 = z1 There is no pump or turbine, so hP = hT = 0 , and there are no minor losses: L=

Solving for length

P2 − P3 L V22 = f ρg D 2g 2 ( P2 − P3 ) D

ρf V2

Velocity is :

4 (1.6×106 barrel day ) ( 42gal barrel ) ( 0.1337ft 3 gal ) (1day 24hr )(1hr 3600s ) 4V ft V =8.28 = = 2 s A π D2 π ( 4ft ) The friction factor is a function of Reynolds number and roughness. With the given properties 3 ρ VD ( 58lbm ft ) ( 8.28ft s )( 4 ft ) = = 170, 000 Re = µ 0.0113lbm fts V =

This is turbulent flow so with ε = 0.0005ft, ε D = 0.0005 4 = 0.000125 ⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0.000125 ⎞1.11 6.9 ⎤ 1.8log = −1.8log ⎢⎜ + = − ⎥ ⎢⎜ ⎥ ⎟ ⎟ + Re ⎦⎥ 170, 000 ⎥⎦ f ⎣⎢⎝ 3.7 ⎠ ⎣⎢⎝ 3.7 ⎠

1

L=

2 (1200-50 ) lbf in.2 ( 4 ft ) ( 32.2 ft lbm lbf s 2 )

( 58lbm ft ) ( 0.0168)(8.28ft s ) (1ft 12in.) 3

2

2

=638,600ft =121mi

→

Answer

b) To find the pumping power, applying the energy equation from point 1 to 2: P −P  P1 − P2 = V ( P − P )  p = ρVg and W P = mgh hp = 1 2 1 2 ρg ρg This is the power into the fluid. So taking into account pump efficiency: 2 W = ( 8.28ft s ) ( -1150 lbf in.2 ) (π 4 )( 4ft ) (1hp s 550ft lbf ) (144in.2 ft 2 ) =-31,300hp P

This does not take into account pump efficiency, so -31,300hp W P = = -36,800hp Answer 0.85

Comments: The minus sign indicates that power must be supplied to the pump. 9- 65

f = 0.0168

9-61

For air at 300 K and one atmosphere, a fan performance curve can be approximated with h = 70 − 3 × 10−4 V 2 , where h is the pressure rise across the fan in cm of water and V is the air flow rate in m3/min. The fan discharges into a smooth rectangular duct 20-cm by 40-cm. Determine: a. the flow rate if the duct is 30-m long (in m3/min) b. the flow rate if the duct is 75-m long (in m3/min).

Approach: We must balance the pressure loss in the duct against the pressure head produced by the fan. Pressure drop is calculated directly with the steady, incompressible flow energy equation. Because the friction factor depends on flow rate, an iterative solution is required.

Assumptions: 1. 2. 3. 4.

The system is steady. The flow is fully developed. Properties are constant. The duct is smooth.

Solution: The steady, incompressible flow energy equation is: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g Let location 1 be just upstream from the fan and station 2 at the end of the duct, so P1 = P2 . The duct is constant area and horizontal, so V1 = V2 and z1 = z2 There is no turbine, so hT = 0 , and there are no minor losses: L V2 (1) D 2g where hP is produced by the fan (expressed in consistent units). The friction factor is a function of Reynolds number and roughness. For air from Appendix A-7at 100 kPa, 300 K, µ = 1.846 × 10−5 Ns m 2 , ρ = 1.1774 kg m3 . 3 V (V m min ) (1min 60s ) m =0.208V Velocity is : V = = (2) A s ( 0.20m )( 0.40m ) hP = f

Because this is a non-circular duct, the hydraulic diameter must be used: 4 ( 0.20m )( 0.40m ) 4 Ax Dh = = =0.2667m pwetted 2 ( 0.20m+0.40m )

ρ VDh (1.1774 kg m )( 0.208V m s ) ( 0.267m ) = = 3540V µ 1.846 × 10−5 Ns m 2 3

Re =

(3)

For any volume flow greater than 1 m3/min, this is turbulent flow so with a smooth duct: ⎡⎛ ε D ⎞1.11 6.9 ⎤ ⎡⎛ 0 ⎞1.11 6.9 ⎤ 1 1.8log (4) = −1.8log ⎢⎜ + = − ⎥ ⎢ ⎥ ⎟ ⎜ ⎟ + Re ⎦⎥ Re ⎥⎦ f ⎣⎢⎝ 3.7 ⎠ ⎣⎢⎝ 3.7 ⎠ The solution procedure is to solve the above four equations simultaneously using an iterative procedure: guess a volume flow rate V ; calculate velocity V, Reynolds number Re, and friction factor f. Then calculate the pressure drop across the duct and the pressure rise across the fan. When these two quantities are equal, then the flow has been determined. Performing the iterations, we obtain For L = 30 m  V = 285 m3 min V = 59.3m s f = 0.0116 h fan = 387m of air = 45.5cm of water Answer For L = 75 m  V = 198 m3 min V = 41.1m s

f = 0.0123

h fan = 495m of air = 58.3cm of water

9- 66

Answer

9-62

A town water system is constructed to supply water at a flow rate of 0.04 m3/s as shown below. Available cast iron pipe is to be used and the gate valve is fully open. The water is at 20 °C. Determine the height to which the upper reservoir dam (reservoir surface elevation) must be built (in m).

Approach: The elevation difference drives the flow. That pressure head is consumed by the frictional losses in the line. The steady, incompressible flow energy equation is used to calculate the elevation of the upper reservoir.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: The steady, incompressible flow energy equation is: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g At points 1 and 2 the areas are large and the pressure is atmospheric, so V1 = V2 = 0 and P1 = P2 . There is no pump or turbine, so hP = hT = 0 . Therefore, ⎡ L ⎤ V2 ⎡ L ⎤ V2 z1 = z2 + ⎢ K ent + K valve + K contract + f A A ⎥ A + ⎢ K exit + f B B ⎥ B DA ⎦ 2 g ⎣ DB ⎦ 2 g ⎣

The velocities are: Using conservation of mass:

VA =

4 ( 0.04 m3 s ) 4V m V =1.27 = = 2 2 s AA π DA π ( 0.20m )

VB = VA ( DA DB ) = (1.27 m s )( 0.20 0.15 ) =2.26 m s 2

2

The friction factor is a function of Reynolds number and roughness. For cast iron pipe ε =0.26mm , and for water from Appendix A-6 at 10 ºC, µ = 12.9 × 10−4 Ns m 2 , ρ = 999.6 kg m3 . The Reynolds numbers are:

ρ VD ( 999.6 kg m ) (1.27 m s )( 0.20m ) ReA = = = 197, 000 µ 12.9 × 10−4 Ns m 2 3

ReB = ReA ( DA DB ) = 263, 000

The flow is turbulent, so ⎡⎛ 0.26 200 ⎞1.11 1 6.9 ⎤ = −1.8log ⎢⎜ ⎥ → f A = 0.0221 ⎟ + 197, 000 ⎦⎥ fA ⎣⎢⎝ 3.7 ⎠ Similarly, f B = 0.0233 From Table 9-3, Figures 9-13, 9-14, and 9-15, Kent = 0.5, Kvalve = 0.15 (assumed fully open), Kexit = 1.0

For AB AA = ( DB DA ) = ( 0.15 0.20 ) = 0.56 , and Kcontract ≈ 0.15. 2

2

Therefore, ⎡ ⎡ ⎛ 200 ⎞ ⎤ (1.27 m s ) ⎛ 150 ⎞ ⎤ ( 2.26 m s ) z1 = 175m + ⎢ 0.5 + 0.15 + 0.15 + ( 0.0221) ⎜ + ⎢1 + ( 0.0233) ⎜ ⎟⎥ ⎟⎥ 2 ⎝ 0.20 ⎠ ⎦ 2 ( 9.81m s ) ⎣ ⎝ 0.15 ⎠ ⎦ 2 ( 9.81m s 2 ) ⎣ 2

=175m+1.88m+6.32m=183.2m

Answer

9- 67

2

9-63

The pump in an existing water system (shown below) fails and must be replaced. A duplicate is not available, so a manufacturer proposes a pump with a pump curve: hP = −4 × 10−6 V 2 + 0.0038V + 86 , where V is in gal/min and hp is in ft. The gate valve is fully open and a friction factor of 0.018 can be used for both pipes. Determine the flow rate in the system (in gal/min).

Approach: The pump must supply enough head to overcome frictional losses in the suction and discharge pipes, as well as the elevation head. The steady, incompressible flow energy equation is used to calculate the required pump head. The given pump head curve is equated to the head loss calculation to determine the operating point.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: From point 1 to point 2, the steady, incompressible flow energy equation is: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + + hT + ∑ hL ρ g 2g ρ g 2g There is no turbine, so hT = 0 . The areas at 1 and 2 are large, so V1 = V2 = 0 , and pressure is atmospheric, so P1 = P2 . The losses include entrance, exit, valve, and line losses. Therefore, ⎛ L ⎞V2 ⎛ L ⎞V2 hP = ( z2 − z1 ) + ⎜ K ent + f A A ⎟ A + ⎜ K valve + K exit + f B B ⎟ B DA ⎠ 2 g ⎝ DB ⎠ 2 g ⎝

From conservation of mass: m A = m B

→

VB = VA ( DA DB )

2

Substituting this into the above pump head equation: 4 ⎡ LA ⎛ D A ⎞ ⎛ LB ⎞ ⎤ VA2 hP = ( z2 − z1 ) + ⎢ K ent + f A +⎜ ⎟ ⎜ K valve + K exit + f B ⎟⎥ DA ⎝ DB ⎠ ⎝ DB ⎠ ⎥ 2 g ⎢⎣ ⎦ 3  4 (V gal min )( 0.1337 ft gal ) (1min 60s ) V 4V ft =0.00409V The velocity is: VA = = = 2 2 AA π DA s π (10 12 ft ) The head loss must be balance against the head added by the pump: hP = −4 × 10−6 V 2 + 0.0038V + 86 where hp is in ft and V is in gal/min.

(1)

(2)

From Table 9-3 and Figures 9-13 and 9-14, K ent = 0.5, K exit = 1.0, and K valve = 0.15 , assuming the valve is fully open. Substituting the known information into the head loss equation: 4 ⎡ VA2 40 ⎛ 0.833 ⎞ ⎛ 800 ⎞ ⎤ +⎜ hP = (1204-1152 ) ft+ ⎢ 0.5+ ( 0.018 ) ⎟ ⎜ 0.15+1+ ( 0.018 ) ⎟⎥ 0.833 ⎝ 0.667 ⎠ ⎝ 0.667 ⎠ ⎥⎦ 2 ( 32.2 ft s 2 ) ⎢⎣ (3) Simplifying: hP = 52.0ft+0.882VA2 The above three equations are solved simultaneously to find the operating point of the pump. Solving them, we obtain: V = 1450 gal min VA = 5.93ft s

Answers

hP = 83.1ft

9- 68

9-64

In drier regions, large central pivot sprinkler systems are used to irrigate large areas. Consider the simplified schematic of a portion of such a sprinkler (shown below). Water at 10 °C is pumped through the spray arm that is constructed of 2.5-cm diameter galvanized iron. The flow area of each nozzle is 1.5 cm2. The pressure at the first nozzle is 250 kPa (gage). Ignoring friction in each nozzle but not in the connecting lengths of pipe, determine the flow rate through the sprinkler (in m3/s).

Approach: The total flow is such that all the pressure head at location 1 is used to balance line losses in the galvanized pipe and the creation of kinetic energy at the nozzle exits. The flow through each nozzle is different and can be estimated using the Bernoulli equation. Because the line losses are a function of flow rate and friction factor, an iterative solution is required. The steady, incompressible flow energy equation must be used.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: The frictional losses between each nozzle can be calculated with the steady, incompressible flow energy equation. P1 V2 P V2 For flow between location 1 and 2: + 1 + z1 + hP = 2 + 2 + z2 + hT + ∑ hL ρ g 2g ρ g 2g There is no pump or turbine, so hp = hT = 0 . The pipe diameter and elevation are constant, so V1 = V2 and z1 = z2 . Ignoring minor losses: L V122 ρ D 2 Note that velocity and friction factor are different for each segment of pipe. The velocity is: V12 = V12 A = 4V12 π D 2 The flow between points 1 and 2 is the difference between what enters the system and what is leaving through the nozzle: V = V − V P2 = P1 − f12

12

1

j1

Assuming no losses in the nozzle, we use the Bernoulli equation to calculate the flow through the nozzle: V j12 Pj V2 P1 + 1 + z1 = + + z j1 ρ g 2g ρ g 2g Neglecting the inlet velocity and the elevation difference, V1 ≈ 0 and z1 = z j , and expressing the pressures in gage pressure, Pj = 0 , we obtain: V j1 = ( 2 P1 ρ )

0.5

and the volume flow rate through a jet is Vj1 = V j1 Aj . Assuming turbulent flow in each pipe segment, with µ = 12.9 × 10−4 Ns m 2 , ρ = 999.6 kg m3 from Appendix A-6 and from Table 9-2 for galvanized pipe, ε = 0.15mm so that ε D = 0.15 25 = 0.006 . The flow is turbulent, so ⎡⎛ ε D ⎞1.11 6.9 ⎤ 1 ρ VD = −1.8log ⎢⎜ ⎥ where Re = ⎟ + 3.7 Re µ f ⎠ ⎣⎢⎝ ⎦⎥ To solve for the total flow, we generalize the above equations for each pipe segment and nozzle: V1 = Vj1 + Vj 2 + Vj 3 + Vj 4 (1) P2 = P1 − f12

L V122 ρ D 2

(2) 9- 69

L ρ D L P4 = P3 − f 34 ρ D V12 = V1 − Vj1 V = V − V P3 = P2 − f 23

V232 2 V342 2

(3) (4) (5)

j2

(6)

V34 = V23 − Vj 3

(7)

23

12

V j1 = ( 2 P1 ρ )

0.5

V j 2 = ( 2 P2 ρ ) V j 3 = ( 2 P3 ρ )

(8)

0.5

(9)

0.5

(10)

V j 4 = ( 2 P4 ρ )

0.5

(11) The eleven unknowns in this system of equations are: Vj1 , Vj 2 , Vj 3 , Vj 4 , P2 , P3 , P4 , V12 , V23 , V34 , V1 . Solving the system of equations with appropriate software: V1 = 0.00699 m3 s V = 0.00336 m3 s P = 250kPa (gage) j1

1

Vj 2 = 0.00179 m s

P2 = 70.9kPa (gage)

Vj 3 = 0.00104 m 3 s

P3 = 23.9kPa (gage)

Vj 4 = 0.00081m3 s

P4 = 14.6kPa (gage)

3

Answers

V12 = 0.00363m3 s V23 = 0.00185 m3 s V34 = 0.00081m3 s

Comments: Note that with this design different flows exit at each nozzle. This would not be a good design, since a farmer would want uniform coverage on a field. Hence, different size nozzles or some other modification would be used to ensure equal flow through all nozzles.

9- 70

9-65

Two pipes are connected in parallel. The first pipe is 2-cm in diameter and 100-m long with a friction factor of 0.012. The second pipe is 5-cm in diameter 50-m long with a friction factor of 0.010. Determine the ratio of the flow rates in the two pipes.

Approach: We assume the same pressure drop across these two pipes and ignore minor losses. A ratio of the steady, incompressible flow energy equations written for both pipes will give the ratio of the velocities, which can be used to obtain a ratio of the flow rates.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: From point 1 to point 2, the steady, incompressible flow energy equation is: P1 V2 P V2 + 1 + z1 + hP = 2 + 2 + z2 + + hT + ∑ hL ρ g 2g ρ g 2g There is no pump or turbine, so hP = hT = 0 . The pipe is constant area, so V1 = V2 , and the pipe is horizontal, so z1 = z2 . Therefore, P1 − P2 L V2 = f ρg D 2g Applying this equation to pipe A and pipe B, and taking the ratio L V2 fA A A 0.5 ⎡⎣( P1 − P2 ) ρ g ⎤⎦ A DA 2 g VA ⎛ f B LB DA ⎞ =1= → =⎜ ⎟ VB ⎝ f A LA DB ⎠ L V2 ⎡⎣( P1 − P2 ) ρ g ⎤⎦ B fB B B DB 2 g Substituting in known information: VA ⎡ ( 0.010 )( 50 )( 0.02 ) ⎤ =⎢ ⎥ = 0.408 VB ⎢⎣ ( 0.012 )(100 )( 0.05 ) ⎥⎦ Because V = VA = Vπ D 2 4 , from conservation of mass: 0.5

2 2 VA VA ⎛ DA ⎞ ⎛ 0.02 ⎞ = = ( 0.408 ) ⎜ = 0.065 ⎜ ⎟ ⎟ VB VB ⎝ DB ⎠ ⎝ 0.05 ⎠

Answer

Comments: This answer is reasonable using the given friction factors. In actual practice, with a known total flow, the friction factors would need to be evaluated, and the solution would be iterative.

9- 71

9-66

For a storm sewer modification project, a 24-in. pipe and a 30-in. pipe both open at their ends to the atmosphere are to be joined using three existing (but under-utilized pipes) as shown on the figure. All the pipes are concrete, and the friction factors are shown on the figure. The branches are horizontal For a total flow rate of 20 ft3/s of 60 °F water, determine the flow rate in each of the three connecting pipes (in ft3/s) and the elevation difference from the entrance to the exit.

Approach: We know the total flow through the system and that the head loss across the middle three branches (B, C, and D) is the same. There are four unknowns in the problem: the head loss from 2 to 3, and the volume flows through the three branches. We use conservation of mass and the steady, incompressible flow energy equations to obtain the required four equations.

Assumptions: 1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution: Conservation of mass with constant density water for the whole system is: m tot = m B + m C + m D where m = ρV . Therefore, V = V + V + V

(1) From point 2 to point 3, the steady, incompressible flow energy equation is: P V2 P2 V2 + 2 + z2 + hP = 3 + 3 + z3 + + hT + ∑ hL ρ g 2g ρ g 2g tot

B

C

D

There is no pump or turbine, so hP = hT = 0 , and we ignore minor losses. Each of the pipes is constant area, so the velocity is constant. The pipe is horizontal, so z2 = z3 . Therefore, P2 − P3 L V2 = f ρg D 2g

Let hbr = ( P2 − P3 ) ρ g , which is the same for all three branches. Applying the pressure drop equation to each branch: hbr = f B

LB VB2 DB 2 g

(2)

hbr = fC

LC VC2 DC 2 g

(3)

hbr = f D

LD VD2 DD 2 g

(4)

V 4V = A π D2 We have four equations and four unknowns: VB , VC , VD , hbr . With the given information, the equations can be solved simultaneously (iteratively) for the unknowns. Doing so, we obtain: VB = 9.23ft 3 s VC = 4.20 ft 3 s VD = 6.57 ft 3 s hbr = 26.7ft Answers Now we need to evaluate the elevation difference between points 1and 4. Using the energy equation P1 V2 P V2 + 1 + z1 + hP = 4 + 4 + z4 + hT + ∑ hL ρ g 2g ρ g 2g

For each of the velocities:

V =

With no pump or turbine, so hP = hT = 0 , no minor losses, and P1 = P4 , and noting that there are line losses in pipe 1-2 and 3-4 and we know the line loss between 2-3, since we calculated that above: 9- 72

z1 − z4 = −

L V2 ⎛ L ⎞V2 V122 V342 L V2 L ⎞V2 ⎛ + + hbr + f12 12 12 + f 34 34 34 = ⎜ −1 + f12 12 ⎟ 12 + ⎜ 1 + f34 34 ⎟ 34 + hbr D12 2 g D34 2 g ⎝ D12 ⎠ 2 g ⎝ D34 ⎠ 2 g 2g 2g

For the velocities 4 ( 20 ft 3 s ) V 4V ft =2.83 V12 = = = 2 A π D122 s π ( 3ft )

V34 =

4 ( 20 ft 3 s ) π ( 2.5ft )

2

=4.07

ft s

⎡ ⎡ ⎛ 2300 ⎞ ⎤ ( 2.83ft s ) ⎛ 4000 ⎞ ⎤ ( 4.07 ft s ) z1 − z4 = ⎢-1+ ( 0.022 ) ⎜ + ⎢1+ ( 0.020 ) ⎜ +26.7ft ⎟⎥ ⎟⎥ 2 ⎝ 3 ⎠ ⎦ 2 ( 32.2 ft s ) ⎣ ⎝ 2.5 ⎠ ⎦ 2 ( 32.2 ft s 2 ) ⎣ 2

= 1.97ft+8.48ft+26.7ft=37.2ft

2

Answer

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