c7 Soln Manual
October 8, 2022 | Author: Anonymous | Category: N/A
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Problem 7.1 No solution provided
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Problem 7.2 No solution provided
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Problem 7.3 Answer the questions below.
a. Wh What at are com commo mon n for forms ms of ener energy? gy? Whi Which ch of tthes hesee aare re rrele elev vant ant to fluid mec mechanics hanics?? •
Mechanical Energy = Energy that matter has because of its motion (i.e. kinetic energy) or its position.(potential energy associated with a spring; gravitational potential energy). Highly releva relevant nt to fluid mechanics. Examples of KE include. — KE in a river that is flowing. KE in air that is being used to drive a wind turbine. turbi ne. KE in the mo moving ving ai airr in the throa throatt of a ven venturi. turi. — Elastic potential energy associated with compression of a gas. Elastic potential energy associated with compression of water during water hammer. — Gravitational potential energy associated with water stored behind a dam. Gravitational Gravi tational potenti potential al energy associated with w water ater in a stand pipe. Grav Grav-itational potential energy associated with water stored in a water tower for a municipal water supply.
•
Other forms of energy include electrical energy, nuclear energy, thermal energy, an and d chem chemic ical al ener energy gy.. Ea Each ch of thes thesee fo form rmss of ener energy gy can be re rela late ted d to fluid mechanics.
b. What is w work? ork? Giv Givee three exam examples ples tha thatt are relev relevant ant tto o fluid mechanics. •
In mechanics, work is done when a force acts on a body as the body moves throug thr ough h a dis distan tance. ce. Oft Often, en, the magn magnitu itude de of work is for force ce mu multi ltipli plied ed by the distance moved ( moved (W W = F d). water in a river tumbles rock, theawater is exerting a force on the — When rock because the rock is movinga through distance. — When the heart contracts, the walls of the heart exert a force on the blood within wit hin the ch cham ambers bers of the heart heart,, the thereb reby y doi doing ng wo work rk (fo (force rce of wa wall ll x displacement of wall). — When water flows through a turbine in a dam, the water exerts a force on the turbine blade and the turbine blade rotates in response to this force.
c. Common units of power: horsepower and watts. d. Diff erences erences between power and energy •
power is a rate (amount per time) whereas energy in an amount. 3
•
•
diff erent erent primary dimensions: power (M (M L2 /T 3 ) and Energy (M (M L2/T 2 ). diff erent erent units units:: pow power er has units of horsepow horsepower, er, watts watts,, etc.; energy has units of Joules, ft-lbf, Btu, cal, etc.
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Problem 7.4 Apply the grid method to each situati Apply situation on described below. Not Note: e: Unit cancel cancellati lations ons are not shown in this solution. a) Situation: A pump operates for 6 hours. Cost = C = C = $0 $0..15 15/ / kW · h. P = P = 1 hp, t = 6 h. Find: Amount of energy used (joules). Cost of this energy ($’s). Solution:
∆E
= P ∆t =
µ ¶µ ¶µ 1 hp 1.0
∆
E = = 1. 61
Cost = C ∆E = =
×
µ
6h 1.0
745.7 W 745. 1.0 hp
7
¶µ ¶µ ¶ J W · s
3600s h
10 J
0.15 kW · h
¶µ
1. 611 × 107 J 1.0
¶µ
kW · h 3.6 × 106 J
¶
Cost = Cost = $0 $0..67 b) Situation: A motor is turning the shaft of a centrifugal pump. T = T = 100 100 lbf lbf · in, in, ω = 850 rpm. 850 rpm. P P = T ω. Find: Power in watts. P = T ω 100 100 lbf lbf · in = 1.0
µ
¶µ
850 rev 850 rev min
P = P = 1010 1010 W
¶µ
2π rad 1.0 rev
c) Situation: A turbine produces power. P = P = 7500 7500 ft lbf lbf / s. Find: Covert the power to watts and hp. 5
¶µ
1.0min 60 s
¶µ
1.0 m 39 39..37in
¶µ
4.448N lbf
¶µ ¶ W · s J
Solution:
P =
P =
µ
¶µ
75 7500 00 ft · lbf s
P P = 10 10..2 kW 75 7500 00 ft · lbf s
µ
P P = 13 13..6 hp
6
¶µ
¶
s · W 0.73 7376 76 ft · lbf s · hp 550ft · lbf
¶
7.5: PROBLEM DEFINITION Situation: Water is being sprayed out a spray bottle. Find: Estimate the power (watts) required. PLAN 1. Acquire data from a spray bottle. 2. Find Find pow power er usi using ng P P = ∆W/∆t, where ∆W W is the amount of work in the time interval ∆t. SOLUTION 1. Data: •
Ten sprays took approximately 15 seconds.
•
Deflecting the trigger involves a motion of ∆x = 1.0 in = 00..0254m 0254m..
•
The force to deflect the trigger was about F about F = = 3 lb which is about 15 about 15 N.
2. Power: P = =
∆W
F (N ( N ∆x) = = ∆t ∆t
µ
(15N)(10 sprays (15N)(10 sprays)) (0 (0..0254m 0254m//spray spray)) 15 s
0.25 W (estimate)
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¶µ ¶ W · s N · m
7.6: PROBLEM DEFINITION Situation: A water pik is producing a high-speed jet. d = 3 mm, V 2 = 25 25 m/ s.
Find: Minimum electrical power (watts). Assumptions: Neglect energy losses in the mechanical system—e.g. motor, gears, and pump. Neglect all energy losses associated with viscosity. Neglect potential energy changes because these are very small. Properties: Water (10 Water (10 C) , Table A.5: ρ = 1000 1000 kg/ kg/ m3 . ◦
PLAN Balance electrical power with the rate at which water carries kinetic energy out of the nozzle. SOLUTION Amo Amount unt of kineti kineticc energy that lea leaves ves the nozzle Power
= =
Each interval of time
V 2 ∆m 22
∆t
where ∆m is the mass that has fl owed out of the nozzle for each interval of time ( time (∆t) . ˙ = ρA Since the mass per time is mass flow rate: (∆m/∆t = m = ρA2 V 2 ) mV ˙ 22 Power = 2 ρA2 V 23 = 2 Exit area π
×
¡
×
2
3
−
¢
A2 = 4 3.0 10 m = 7.07 × 10 6 m2 8
−
Thus:
(1000 (1000 kg/ kg/ m3 ) (7 (7..07 × 10 Power = Power = 2
6
−
m2 )(25m )(25m// s)3 = 55 55..2 W
P P = 5555.2 W REVIEW Based on Ohm’s law, the current on a U.S. household circuit is about: I I = P /V /V = 55. 55.2 W/115V = 0. 0.48 A.
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7.7: PROBLEM DEFINITION Situation: A small wind turbine is being developed. D = 1.25 m, V V = 1155 mph = 6 = 6..71 m/ s. Turbine efficiency: η=20%.
Find: Power (watts) produced by the turbine. Properties: Air (10 (10 C, 0 0..9 bar bar = 90 kP kPaa), R ), R = = 287 287 J/ kg · K. ◦
PLAN Find the densi Find density ty of air us using ing th thee idea gas la law. w. The Then, n, find the kinetic energy of the wind and use 20% of this value to find the power that is produced. SOLUTION Ideal gas law ρ =
p
RT 90 90,, 000 Pa = (287J// kg · K) (10 + 2273) (287J 73) K 3 = 1.108kg 108kg// m Kinetic energy of the wind Rate of KE
Amo Amount unt of ki kinetic netic ener energy gy Interval of time ∆mV 2 /2 = ∆t =
where ∆m is the mass of air that flows through a section of area A = πD 2 /4 for each unit of time time (∆t) . Since the mass for each interv interval al of time is mass flow rate: 10
(∆m/∆t = m ˙ = ρAV = ρAV )) mV ˙ 2 2 ρAV 3 = 2
Rate of KE =
The area is (1 (1..108kg 108kg// m3 ) π (1 (1..25m)2 /4 (6 (6..71 m/ s)3 Rate of KE = KE = 2
¡
¢
Rate of KE = KE = 205 205 W Since the output power is 20% of the input kinetic energy: energy: P P = (0 (0..2)(205W) P P = 4411.0 W REVIEW The amount of energy in the wind is di ff use use (i.e. sprea spread d out). For thi thiss situat situation, ion, th thee wind turbine provides enough power for approximately one 40 watt light bulb.
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Problem 7.8 No solution provided.
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7.9: PROBLEM DEFINITION Situation: This is a linear velocity distribution in a rectangular channel.
Find: Kinetic energy correction factor: α SOLUTION ¯ = V max V V and V V = V max max/2 and maxy/d Kinetic Kinet ic energy correctio correction n facto factorr 1 α = d = 1 d
d
V max maxy 3 dy [( [(V V max / 2) d ] max
Z Z µ ¶ 0 d
3
2y d
0
α = 2 = 2
13
dy
7.10: PROBL PROBLEM EM DEFINITION Situation: Velocity distributions (a) through (d) are shown in the sketch.
Find: Indicate whether α whether α is less than, equal to, or less than unity. SOLUTION a) α = 1.0 b) α > 1. 1.0 c) α > 1. 1.0 d) α > 1. 1.0
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7.11: PROBL PROBLEM EM DEFINITION Situation: There is a linear velocity distribution in a round pipe.
Find: Kinetic energy correction factor: α SOLUTION Kinetic Kinet ic energy correctio correction n facto factorr 1 α = A Flow rate equation
3
Z µ ¶ V ¯ V
A
V = V m − (
dA
r )V m r0
V = V m(1 − ( r )) r0 Q =
Z Z Z
V dA
r0
=
V V (2 (2πrdr πrdr))
0 r0
=
0
r )2 )2πrdr πrdr r0 r2 )]dr [r − ( )]dr r0
V m(1 − r0
= 2πV m
Z 0
Integrating yields
0
Q = 2πV m[( [(rr 2 /2) − (r3 /(3 (3rr0 ))]0r 2 Q = 2πV m[(1 [(1//6) 6)rr0 ] Q = (1 (1//3) 3)V V mA Thus V V =
Q V m = 3 A
Kinetic Kinet ic energy correctio correction n facto factorr 3 r 1 r V m(1 − r ) α = 2πrdr A 0 ( 13 V m) 54 54π π r r 3 = (1 − ( )) rdr πr 02 0 r0
Z ∙ 0
0
0
Z
α = 2.7 15
¸
7.12: PROBL PROBLEM EM DEFINITION Situation: There is a linear velocity distribution in a round pipe.
Find: Kinetic energy correction factor: α SOLUTION Flow rate equation V = kr
r0
Z
Q =
V (2 V (2πrdr πrdr))
0 r0
2πk πkrr 2dr
= = ¯ = V = =
Z
20πkr πk r03 3 Q A 2 kπr kπ r03 3 πr 02 2 kr0 3
Kinetic Kinet ic energy correctio correction n facto factorr 3
α = 1 A 1 α = A
Z µ ¶ Z µ ¶ Z µ ¶ µ ¶ A r0
0
V ¯ V
dA
kr 2 kr 0 3
(3 (3//2)3 2π r α = (πr 02 ) 0 5 27 27//4 r0 α = r02 5r03 α = 27 20
16
0
3
2πrdr
r r0
3
rdr
7.13: PROBL PROBLEM EM DEFINITION Situation: The velocity distribution in a pipe with turbulent flow is given by n
y r0
V = V max max
µ¶
Find: Derive a formula for α for α as a function of n. n . Find α Find α for for n n = = 1/7. SOLUTION Flow rate equation V V max max
=
n
y r0
r0
=
r
= 1
r0
r r0
n
V dA
Q =
Z Z µ ¶ − Z µ − ¶ A r0
=
n
µ ¶ µ −¶ µ− ¶ n
r r0
V max max 1
0
2πrdr
r0
= 2πV max max
r r0
1
0
n
rdr
Upon integration 2 Q = 2πV max maxr0
∙µ 1
n+1
¶ − µ ¶¸ 1 n+2
Then ¯ = Q/ V Q/A A = 2V max max =
∙µ 1
n+1
2V max max (n + 1)(n 1)(n + 2)
¶ − µ ¶¸ 1 n+2
Kinetic Kinet ic energy correctio correction n facto factorr 1 α = A
n
3
Z ⎡⎣ ³ − ´ ⎤⎦ r0
V max max 1
r r0
2V max max (n+1)( +1)(n n+2)
0
Upon integration one gets 1 a = 4
h
[(n [(n+2)( +2)(n n+1)]3 (3 (3n n+2)(3 +2)(3n n+1)
17
i
2πrdr
If n = n = 1/7, then 1 α = 4
3
" £ # ¤ ¡ ¡ ¢ ¢¡ ¡ ¢ ¢ ( 17 + 2)( 17 + 1) 3 17 + 2 3 17 + 1
α = 1.06
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7.14: PROBL PROBLEM EM DEFINITION Situation: The velocity distribution in a rectangular channel with turbulent flow is given by V /V max y/d))n max = (y/d Find: Derive a formula for the kinetic energy correction factor. Find α Find α for for n n = = 1/7. SOLUTION Solve for q for q first in terms of u u max and and d d d
q = =
Z
d
V dy = dy =
0
y d
Z ³ ´ V max max
0
n
V max max dy = dy = n d
Integrating: n+1
µ ¶∙ ¸ µ ¶
V max y n+1 dn n+1 V max d n maxd = n+1 V max maxd = n+1
q =
d 0
−
Then
max ¯ = q = V max V V d n+1
Kinetic Kinet ic energy correctio correction n facto factorr α = 1 A 1 = d
Z µ ¶ Z " ¡ ¢ # Z A
d
0
3
V ¯ V
dA
n
y V max max d V max max/(n + 1)
(n + 1)3 = d3n+1
3
d
y 3n dy
0
Integrating (n + 1)3 d3n+1 α = d3n+1 3n + 1
∙
3
a = (3nn+1) +1 19
¸
d
Z
dy
0
y ndy
When n When n = = 1/7 (1 + 1/7)3 α = 1 + 3/ 3/7 α = 1.05
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7.15: PROBL PROBLEM EM DEFINITION Situation: Turbulent flow in a circular pipe. r = 3.5 cm cm.. Find: Kinetic energy correction factor: α. SOLUTION Kinetic Kinet ic energy correctio correction n facto factorr 1 α = A The integral is evaluated using 3
Z µ ¶ A
V ¯ V
dA '
1 ¯3 V
3
Z µ ¶ A
X i
V ¯ V
dA
µ
vi + vi π(ri2 − ri2 1 ) 2 −
1
−
3
¶
Result Results. velocity y is 24.3 24.322 m/s and the kinetic energy corre correction ction facto factorr is 1.19. s. The mean velocit
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Problem 7.16 Answer the questions below.
a. What is the conceptual meaning of the first law of thermodynamics for a system? •
The law can be written for — energy (amount) with units such as joule, cal, or BTU. — rate of energy (amount/time) with units such as watts, J/s, or BTU/s.
•
•
Meaning (rate form):
⎧⎨ net rate of energy ⎫⎬ ⎧⎨ entering system − at time t time t ⎩ ⎭ ⎩
˙ = dE Q˙ − W dt net rate of work net rate of change in energy done by system of matter in system = at time t time t during time t time t
⎫⎬ ⎧⎨ ⎭ ⎩
Meaning (amount form):
⎫⎬ ⎭
Q − W = ⎫⎬ ⎧⎨E net change in energy ⎫⎬ ⎧⎨ net amount of energy ⎫⎬ ⎧⎨ net amount of work − ⎩ done by system ⎭ = ⎩ of matter in system ⎭ system ⎭ ⎩ entering during time t during time t during time t ∆
∆
∆
b. What is flow work? How is the equation for flow work derived? •
Flow work is work done by forces associated with pressure.
•
Eq. for flow work is derived by using the de finition of work: W W = F · d.
c. What is shaft work? How is shaft work diff erent erent than flow work? •
Shaft work is any work that is not flow work.
•
Shaft work diff ers ers from flow work by the physical origin of the work: — flow work is done by a force associated with pressure. — shaft work is done by any other force.
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∆
Problem 7.17 Answer the questions below.
a. What is head? How is head related to energy? To power? •
•
Head is a way of characterizing energy or power or work. Head is the ratio of (energy of a fluid particle)/(weight of a fluid particle) at a point in a fluid flow ow.. — examp example: le: KE of a fluid particle divided by its weight gives "velocity head": velocity head = ( = (mV mV 2 /2) / (mg mg)) = V 2 /(2 (2gg )
•
Head is related to power by P P = mgh ˙
b. What is head of a turbine? •
Head of a turbine is a way of describing the work (or power) that is done on the turbine blades by a flowing fluid: P turbine turbine =
∆W turbine turbine ∆t
= mgh ˙ t
c. How is head of a pump related to power? To energy? P pump pump =
∆W pump pump ∆t
˙ p = mgh
d. What is head loss? •
•
•
Conversion of mechanical energy in a flowing fluid to thermal energy via the action of viscous stresses. Mec Mechanic hanical al energy is gener generally ally useful. The therma thermall energ energy y that is generated is generally not useful because it cannot be recovered and used productively. The the therm rmal al ene energy rgy usu usuall ally y hea heats ts the fluid uid.. Thi Thiss is ana analog logous ous to fric frictio tional nal heating.
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Problem 7.18 Answer the questions below. a. What are the fi ve main terms in the energy equation (7.29)? What does each term mean? See the text and figure that follow Eq. (7.29). b. How are terms in the energy equation related to energy? To power? •
Each Each term in the energ energy y equation equation is a "head ter term" m" wit with h units of leng length. th. To relate head, power, energy, and work use: P P =
∆W
∆E
∆t
∆t
=
= mgh ˙
c. What assumptions are required to use Eq. (7.29)? •
The flow is steady.
•
The control volume has one inlet port and one exit port.
•
The fluid density is the same at all spatial locations.
How is the energy equation similar to the Bernoulli equation? List three similarities. How is the energy equation diff erent erent from the Bern Bernoulli oulli equat equation? ion? List thre threee di ff ererences. Similarities: •
Both equations involve head terms.
•
Both equations are applied from a "location 1" to a "location 2."
•
There are some simila similarr terms (all terms in the Bernou Bernoulli lli equatio equation n appear in the energy equation).
Diff erences: erences: •
•
•
The energy equation applies to viscous or inviscid flow whereas the Bernoulli equation applies only to inviscid flow ow.. The energy equation includes terms for w work ork of a pump and turbine; these terms do not appear in the Bernoulli equation. The energy equation includes head loss, whereas the Bernoulli equation lacks any term to account for energy loss.
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Problem 7.19 Situation: Flow in a pipe. Find: Prove that fluid in a constant diameter pipe will flow from a location with high piezometric head to a location with low piezometric head. Assumptions: No pumps or turbines in the pipeline. Steady flow ow.. SOLUTION The energy equation (7.29). p1 V 12 p2 V 22 + α1 + z1 + h p = + α2 + z2 + ht + hL γ γ 2g 2g Term-by-t erm-by-term erm analysis: •
In a constant diameter pipe, KE terms cancel out.
•
No machines means that h that h p = h = h t = 0.
Energy equation (simplified form):
µ
¶ µ
¶
p1 p2 + z1 = + z2 + hL γ γ h1 = h2 + hL
Since section 1 is upstream of section 2, and head loss is always positive, we conclude that h that h 1 > h2 . Th This is m mea eans ns ttha hatt fluid flows from high to low piezometric head.
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7.20: PROBL PROBLEM EM DEFINITION Situation: Water flows in a vertical pipe. L = 10 10 m, p A = 10 kP kPaa. pB = 98. 98.1kPa 1kPa..
Find: Direction of flow in a pipe: (a) Upward. (b) Downward. (c) No flow ow.. PLAN 1. Calculate the piezometric head h head h at A and B., 2. To determine flow direction, compare the piezometric head values. Whichever location has the larger value of of h is upstream. If the h the h values are the same, there is no flow ow.. SOLUTION 1. Piezometric head: pA + zA = hA = γ hB = pB + zB = γ
µ µ
10000 kP 10000 kPaa 9810N// m3 9810N 98100 kP kPaa 9810N// m3 9810N
2. Since h Since h A > hB , the correct correct se selec lectio tion n is (b) .
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¶ ¶
+ 10 10 m = 11 11..02 m + 0 = 10 10 m
7.21: PROBL PROBLEM EM DEFINITION Situation: Water flowing from a tank into a pipe connected to a nozzle. α = 1.0, D B = 40cm. 40cm. D0 = 20cm, 20cm, z 0 = 0 m. zB = 3.5 m, z r = 5 m.
Find: 3
(a) Discharge in pipe (m /s). (b) Pressure at point B (kPa). Assumptions: γ = = 9810 9810 N/ m3 . PLAN 1. Find velocity at nozzle by applying the energy equation. 2. Find discharge by applying Q applying Q = = A A oV o 3. Find the pressure by applying the energy equation. SOLUTION 1. Energy equation (point 1 on reservoir surface, point 2 at outlet) preser. V r2 poutlet V 02 + + zr = + + z0 γ γ 2g 2g 2 V 0+0+5 = 0+ 0 2g V 0 = 9.90 90 m/s m/s 2. Flow rate equation Q = V 0A0
π = 9.90 m/ s × ( ) × (0 (0..20m)2 4 Q = 0.311 311 m m 3 /s 27
3. Energy equation (point 1 on reservoir surface, point 2 at B) 0+0+5=
pB V B2 + + 33..5 γ 2g
where 3
48 m/s V B = V Q = 0.311m / s 2 = 22..48 (π/ π/4) 4) × (0. (0.4 m) B V B2 = 0.312m 2g pB − 5 − 3.5 = 0.312 γ pB = 86 86..4 kP kPa a
28
7.22: PROBL PROBLEM EM DEFINITION Situation: Water drains from a tank into a pipe. x = 10ft, 10ft, y = 4 ft.
Find: Pressure at point A point A (psf). Velocity at exit (ft/s). Assumptions: α2 = 1 PLAN To find pressure at point A, apply the energy equation between point A and the pipe exit. Then, then ap apply ply energy equ equatio ation n betwe between en top of tank and the exit. SOLUTION Energy equation (point A to pipe exit). V A2 V 22 pA p2 + zA + αA + h p = + z2 + α2 + ht + hL γ γ 2g 2g A 2 (continuity); p2 = 0-gage; Term -gage; (zA − zB ) = y; h p = 0, ht = 0by , hLterm = 0.analysis: Thus V = V (continuity);
pA = −γy = −62 62..4lbf / ft3 × 4 ft pA = −250 250 psf psf Energy equation (1= top of tank; 2 = pipe exit) p1 V 2 p2 V 2 + z1 + α1 1 + h p = + z2 + α2 2 + ht + hL γ γ 2g 2g 2 z1 = V 2 + z2
2g
29
Solve for velocity at exit V 2 = =
p p
2g (z1 − z2 )
2 × 32 32..2 ft/ ft/ s2 × 14ft
V 2 = 30 30..0 ft/ ft/ s
30
7.23: PROBL PROBLEM EM DEFINITION Situation: Water drains from a tank into a pipe. x = 10 10 m, y = 4 m.
Find: Pressure at point A point A (kPa). Velocity at exit (m/s). Assumptions: α1 = 1. PLAN 1. Fin Find d pre pressu ssure re at poin pointt A by apply applying ing the ener energy gy equat equation ion betw between een point A and the pipe exit. 2. Find velocity at the exit by applying the energy equation between top of tank and the exit. SOLUTION 1. Energy equation (section A to exit plane): pA V 2 p2 V 2 + zA + αA A + h p = + z2 + α2 2 + ht + hL γ γ 2g 2g Term by term analysis: V A = V 2 (continuity); (continuity); p2 = 0-gage; -gage; (zA − zB ) = y; h p = 0, ht = 0, h˙ L = 0. Thus pA = −γy (1..5 m) pA = − 9810N 9810N// m3 (1
¡
pA =
¢
−14 14..7kPa
31
2. Energy equation (top of tank to exit plane): V 12 V 22 p1 p2 + α1 + z1 + h p = + α2 + z2 + ht + hL γ γ 2g 2g 2 V z1 = 2 + z2 2g V 2 = =
p p
2g (z1 − z2 ) 2 × 9. 9.81 m/ s2 × 11 11..5 m
V 2 = 15 15..0 m/ s
32
7.24: PROBL PROBLEM EM DEFINITION Situation: Water is pushed out a nozzle by a pump. Q = 0.1 m3 / s, D 2 = 30 30 cm cm.. Dn = 10cm, 10cm, z n = 7 m. z1 = 1 m, z 2 = 2 m.
Find: Pressure head at point 2. SOLUTION Flow rate equation to find nd V V n (velocity at nozzle) V n =
0.10 m3 / s Q = = 12 12..73 73 m/s m/s An (π/4) π/4) × (0 (0..10m)2
V n2 = 8.26 26 m m 2g Flow rate equation to find nd V V 2 Q 0.10 m3 / s V 2 = = = 11..41 41 m/s m/s A2 (π/ π/4) 4) × (0 (0..3 m)2 V 22 = 0.102 102 m m 2g Energy equation p2 + 00..102 + 2 = 0 + 8. 8.26 + 7 γ p2 = γ
1133.2 m
33
7.25: PROBL PROBLEM EM DEFINITION Situation: Oil moves through a narrowing section of pipe. DA = 20 20 cm cm,, D B = 12cm. 12cm. L = 1 m, Q = Q = 0.06 m3/ s.
Find: Pressure diff erence erence between A between A and and B B.. Properties: S = 00..90. 90. SOLUTION Flow rate equation Q A1 = 1.910 910m/s m/s
V A =
2
µ¶
20 V B = × V A 12 = 5.31 31 m/s m/s Energy equation pA − pB = 1γ + + (ρ/ ( ρ/2)( 2)(V V B2 − V A2 ) (0..9) + pA − pB = (1) 9810N 9810N// m3 (0
¡
¢
µ
900kg/ m3 900kg/ 2
pA − pB = 19 19..9kPa
34
¶
((5..31 m/ s)2 − (1 ((5 (1..91 m/ s)2 )
7.26: PROBL PROBLEM EM DEFINITION Situation: Gasoline flows in a pipe that narrows into a smaller pipe. Q = 5 ft3 / s, p 1 = 18 18 psig. psig. hL = 6 ft, ∆z = 12 ft. ft. 2 A1 = 0.8 ft , A 2 = 0.2 ft2.
Find: Pressure at section 2 (psig). Properties: S = 00..90. 90. Water, Table A.5: γ = = 62 62..4lbf / ft3 . PLAN Apply flow rate equation and then the energy equation. SOLUTION Flow rate equation: V 1 =
5 Q = = 66..25ft 25ft// s 0.8 A1
V 12 = (6 (6..25ft 25ft// s)2 = 00..60 6066 66 ft 2g 2(32.2 ft/ 2(32. ft/ s2 ) 5 Q = V 2 = = 25ft/ 25ft/ s 0.2 A2 V 22 (25ft/ (25ft/ s)2 = 99..705ft = 2(32..2 ft/ 2(32 ft/ s2 ) 2g Energy equation: p1 V 12 p2 V 22 + + z1 = + + z2 + 6 γ 2g γ 2g 18lbf / in2 × 14 1444 in2 / ft2 p2 0.8 × 62 62..4lbf / ft3 + 00..6066ft 6066ft + 12ft = 0.8 × 62 62..4lbf / ft3 + 99..70 7055 ft + 0 ft + 6 ft p2 = 2437 psfg 35
p2 = 16 16..9 psig
36
7.27: PROBL PROBLEM EM DEFINITION Situation: Water flows from a pressurized tank, through a valve and out a pipe. p1 = 100 kP kPaa, z 1 = 8 m. p2 = 0 kP kPaa, z2 = 0 m. hL = K = K L V , V , V 2 = 10 m/ s. 2
2g
Find: The minor loss coefficient cient ( (K K L) . Assumptions: Steady flow ow.. Outlet flow is turbulent so that α that α 2 = 1.0. V 1 ≈ 0.
Properties: Water (15 Water (15 C) C),, Table A.5: γ = = 9800 9800 N/ m3 . ◦
PLAN Apply the energy equation and then solve the resulting equation to find the minor loss coefficient. SOLUTION Energy equation (section 1 on water surface in tank; section 2 at pipe outlet) V 12 V 22 p1 p2 + α1 + z1 + h p = + α2 + z2 + ht + hL γ γ 2g 2g Term by term analysis: •
At the inlet. p1 = 100 kP kPaa, V 1 ≈ 0, z 1 = 8 m
•
At the exit , p , p 2 = 0 kP kPaa, V 2 = 10 10 m/ s, α2 = 1.0.
•
Pumps and turbines. h p = h = h t = 0
•
V 2
Head loss. hL = K = K L 2g
37
(1)
Eq. (1) simplifies to V 22 V 22 p1 + z1 = α2 + K L 2g 2g γ (100,, 00 (100 0000 Pa) (10m// s)2 (10m (10m/ (10m/ s)2 + 8 m = + K L 2 3 2 2 (9 (9..81 m/ s ) (9 (9800 800 N/ m ) 18 (9. .81 m/+s K ) (5. 18..2 m = (25. 5(9 .09 097 7 m) L (5.097m) Thus K L = 2.57 REVIEW
1. The mi minor nor lo loss ss coefficient cient ( (K K L = 2.57) 57) is is typical of a valve (this information is presented in Chapter 10). 2. The hea head d at the inle inlett
p1 + z1 = γ
³ ³ ³
this energy goes to head loss
22 22..2 m repre represent sentss av availa ailable ble energy energy.. Mos Mostt of
V 2 K L 22g = V 2 2
´ ´ ´
17 17..1 m . The remai remainder nder is carrie carried d as
kinetic energy out of the pipe α2 2g = 5.1 m .
38
7.28: PROBL PROBLEM EM DEFINITION Situation: Water flows from a pressurized tank, through a valve and out a pipe. Q = 0.1 ft3 / s, ∆z = 10ft. 10ft. p2 = 0 psig psig,, D = 1 in. hL = K = K L V . 2
2g
Find: Pressure in tank (psig). PLAN Apply the energy equation and then solve the resulting equation to give pressure in the tank. SOLUTION Energy equation (from the water surface in the tank to the outlet) p1 V 12 p2 V 22 + + z1 = + + z2 + hL γ γ 2g 2g V 22 p1 6V 22 = + hL z1 = 10 γ 2g 2g − − 0.1 ft3/ s Q = V 2 = 188.33 33 ft/s ft/s 2 = 1 1 A2 (π/ π/4) 4) ft
¡ ¢ 12 2
6(18. 6(18.33ft 33ft// s) p1 = − 10ft = 21. 21.3 ft γ 64 64..4 ft2/ s2 p1 = 62 62..4lbf / ft3 × 21 21..3 ft = = 1329 1329 psfg p1 = 9.23 psig
39
7.29: PROBL PROBLEM EM DEFINITION Situation: Water flows from an open tank, through a valve and out a pipe. A = 9 cm2 , ∆z = 11 11 m. p1 = p = p2 = 0 kP kPaa. V hL = 5 . 2
2g
Find: Discharge in pipe. Assumptions: α = 1. PLAN Apply the energy equation from the water surface in the reservoir (pt. 1) to the outlet end of the pipe (pt. 2). SOLUTION Energy equation:
2
2
1 2 2 1 2 L pγ 1 + V 2g + z = pγ + V 2g + z + h
Term by term analysis: p1 = 0; p2 = 0 z2 = 0; V 1 ' 0 The energy equation becomes. V 22 z1 = + hL 2g V 22 V 22 V 22 11 m = + 5 = 6 2g 2g 2g 2g 2 V 2 = 6 (11) V 2 =
µs ¶ µ
2 × 9 9..81 m/ s2 6
V 2 = 5.998 998 m/s m/s
¶
(1 (111 m)
Flow rate equation Q = V 2 A2
¡ ¢µ
= (5 (5..998m 998m// s) 9 cm2 = 5. 3982 × 10
3 m
10 4 m2 cm2
3
−
Q = 5.40 × 10 40
s 3
−
m3 / s
−
¶
7.30: PROBL PROBLEM EM DEFINITION Situation: A microchannel is designed to transfer fluid in a MEMS application. D = 200 μm, L = L = 5 cm. Q = 0.1 μ μL L/ s. 32 32μLV μLV hL = . 2
γD
Find: Pressure in syringe pump (Pa). Assumptions: α = 2. Properties: Table A.4: ρ = 799 799 kg/ kg/ m3 . PLAN Apply the energy equation and the flow rate equation. SOLUTION Energy equation (locate section 1 inside the pumping chamber; section 2 at the outlet of the channel) p1 V 2 = hL + α2 γ 2g 32 32μLV μLV V 2 = γD 2 + 2 2g Flow rate The cross-sectional area of the channel is 3.14×10 10 7 l/s or 10 10 m3 /s. The flow velocity is
8
−
−
m2. A flow rate of 0.1 μl/s is
−
Q A 10 10 m3 / s = 3.14 × 10 8 m2 = 0.318 × 10 2 m/s
V =
−
−
−
= 3.18 18 mm/s mm/s Substituting the velocity and other parameters in Eq. (1) gives 41
(1)
0..05 × 0 0..318 × 10 32 × 1 1..2 × 10 3 × 0 p1 = γ 7, 850 × 4 × (10 4 )2 = 0.0194 0194 m m −
−
2
−
(0 (0..318 × 10 2 ) 2 × 9. 9.81 −
+ 2 ×
The pressure is 0.0194 0194 m m 9..81 81 m/s2 × 0. p1 = 799 kg/m 799 kg/m3 × 9 p1 = 152 Pa
42
2
7.31: PROBL PROBLEM EM DEFINITION Situation: Water flows out of a fire nozzle. V V = = 40 40 m/ s, ∆z = 50 50 m. Ae /Ahose = 1/4. hL = 10 V . 2
2g
Find: Pressure at hydrant. Assumptions: Hydrant supply pipe is much larger than the firehose. Properties: Water, Table A.5, γ A.5, γ = = 9810 9810 N/ m3. PLAN Apply the energy equation. SOLUTION Energy equation V 22 p1 + z1 = + z2 + hL γ 2g where the kinetic energy of the fluid feeding the hydrant is neglected. Because of the contraction at the exit, the outlet velocity is 4 times the velocity in the pipe, so the energy equation becomes V 2 p1 V 2 = 2 + z2 − z1 + 10 γ 2g 16 × 2 2gg
µ µ
¶
1.625 V 2 + 50 γ 2g 1.625 2 = (40 0 m / s) + 50 50 m 9810N/ 9810N/ m3 × (4 2 2 × 9 9..81 m/ s = 1. 791 × 106 Pa
p1 =
¶
p1 = 1790 kP kPaa
43
7.32: PROBL PROBLEM EM DEFINITION Situation: Water flows out of a syphon. L1 = L = L 2 = 3 ft. Q = 2.8 ft3 / s, D = D = 8 in.
Find: Pressure at point B point B.. Assumptions: α = 1. SOLUTION Flow rate equation V c = V c =
Q A2 2.8 ft3 / s
(π/ π/4) 4) × = 8.02 02 ft/s ft/s
2 8 ft 12
¡ ¢
Energy equation (from reservoir surface to C to C )) p1 V 12 pc V c2 + + z1 = + + zc + hL γ 2g γ 2g (8. (8 .02ft 02ft// s)2 + 0 + hL 0 + 0 + 3 ft = 0 + 64 64..4 ft2 / s2 hL = 2.00 00 ft ft Energy equation (from reservoir surface to B to B)). pB
V B2
×
0 + 0 + 3 ft = γ γ + 2g + 6 ft + ((33/4)
44
2 ft ; V B = V C = 8.02 02 ft/s ft/s
pB = 3 − 1 − 6 − 1.5 = −5.5 ft γ pB = −5.5 ft × 62 62..4lbf / ft3 = −343 343 psfg psfg pB = −2.38 38 psig psig
45
7.33: PROBL PROBLEM EM DEFINITION Situation: A siphon transports water from one reservoir to another. zA = 30 30 m, z B = 32 32 m. zC = 27 27 m, z D = 26 26 m. D = 25cm. 25cm. hp i p e = V 2pg . 2
Find: Discharge. Pressure at point B point B.. Assumptions: α = 2. PLAN Apply the energy equation from A to C, then from A to B. SOLUTION Head loss hp i p e
V p2 = 2g
htotal
V p2 = hp i p e + ho utlet = 2 2g
Energy equation (from A to C) V p2 0 + 0 + 30m = 0 + 0 + 27 m + 2 2g V p = 5.42 42 m/s m/s
46
Flow rate equation Q = V p A p = 5.42 m/ s × (π/ (π/4) 4) × (0. (0.25m)2 Q = 0.266 266 m m 3 /s Energy equation (from A to B) V p2 pB V p2 + + 3322 m + 00..75 30 m = 2g 2g γ pB = −2 − 1.75 × 1 1..497 497 m m γ pB = −45 45..3 kPa, gage
47
7.34: PROBL PROBLEM EM DEFINITION Situation: A siphon transports water from one reservoir to another. pv = 1.23 kP kPaa, p atm = 100kPa. 100kPa. Q = 8 × 10 4 m3 / s, A = A = 10 4 m2 . hL,A B = 11..8V 2 /2g. −
−
→
Find: Depth of water in upper reservoir for incipient cavitation. PLAN Apply the energy equation from point A to point B. SOLUTION Flow rate equation V = =
Q A 8 × 10
4
m3 /s
−
1 × 10 = 8 m/ s
4
−
m2
Calculations V 2 (8 m/ s)2 = 3.262m = 2 × 9 9..81 m/ s2 2g V 2 872 m m hL,A B = 1.8 = 5.872 2g →
Energy equation (from A to B; let z let z = 0 at bottom of reservoir) pB V B2 pA V A2 + + zB + hL + + zA = 100000γ Pa 2g 100000 γ 1230 12302Pga + 0 + z = + 33..26 2622 m + 10 m + 5.872m A 9810N// m3 9810N 9810N/ m3 9810N/ 48
zA = depth depth = = 9.07 m
49
7.35: PROBL PROBLEM EM DEFINITION Situation: A reservoir discharges water into a pipe with a machine. d = 6 in, D = D = 12in. 12in. 12ft. ∆z1 = 6 ft, ∆z2 = 12ft. 3 Q = 10ft / s.
Find: Is the machine a pump or a turbine? Pressures at points A points A and and B B (psig) (psig).. Assumptions: Machine is a pump. α = 1.0. PLAN Apply the energy equation between the top of the tank and the exit, then between point B and the exit, finally between point A and the exit. SOLUTION Energy equation V 22 z1 + h p = + z2 2g Assuming the machine is a pump. If the machine is a turbine, then h p will be negative. The velocity at the exit is 10ft3 / s Q = π V 2 = 50..93 93 ft/s ft/s 2 = 50 A2 (0. (0 . 5ft) 4 Solving for h for h p and taking the pipe exit as zero elevation we have h p =
(50 (50..93ft 93ft// s)2 ×
(6 + 12) ft = 22 22..3 ft
2
2 32 32..2 ft/ ft/ s Therefore the machine is a pump.
− 50
Applying the energy equation between point B and the exit gives pB + zB = z2 γ Solving for p for p B we have pB = γ (z2 − zB ) pB = −6 ft × 62 62..4lbf / ft3 = −374 374 psfg psfg pB = −2.6 psig
Velocity at A V A =
2
µ¶ 6 12
50..93ft 50 93ft// s = 12. 12.73 73 ft/s ft/s
×
Applying the energy equation between point A and the exit gives pA V A2 V 22 γ γ + zA + 2g = 2g so V 22 V A2 pA = γ − zA − 2g 2g
µ
= 62 62..4lbf / ft3 ×
Ã
¶
2
2
(50..93ft (50 93ft// s) − (12 (12..73ft 73ft// s) 2 × 32 32..2 ft/ ft/ s2
= 1233 psfg pA = 8.56 psig
51
!
− 18
7.36: PROBL PROBLEM EM DEFINITION Situation: Water flows through a pipe with a venturi meter. D = 30cm, 30cm, d = d = 15 cm cm.. patm = 100 kP kPaa, H H = 5 m.
Find: Maximum allowable discharge before cavitation. Assumptions: α = 1.0. Properties: ◦
Water (10 Water (10 C) , Table A.5: pv = 2340 Pa, 2340 Pa, abs. SOLUTION Energy equation (locate 1 on the surface of the tank; 2 at the throat of the venturi) p1 V 12 p2 V 22 + + z1 = + + z2 γ 2g γ 2g p2,vapor V 22 + + 0 0+0+5 = γ 2g p2,vapor = 2340 − 100 100,, 000 = −97 97,, 660 660 Pa Pa gage Then V 22 97660 97660 Pa = 5m + = 14 14..97 97 m m 2g 9790N// m3 9790N V 2 = 17 17..1 m/s Flow rate equation Q = V 2 A2 = 17 17..1 m/ s × π/4 π/ 4 × (0. (0.15m)2 Q = 0.302 302 m m 3 /s
52
7.37: PROBL PROBLEM EM DEFINITION Situation: A reservoir discharges to a pipe with a venturi meter before draining to atmosphere. D = 40cm, 40cm, d = d = 25 cm cm.. patm = 100 kP kPaa, h L = 0.9V 22 /2 /2g.
Find: Head at incipient cavitation ( cavitation ( m) m).. Discharge at incipient cavitation ( cavitation ( m3 / s) s).. Assumptions: α = 1.0. Properties: From Table A.5 p A.5 p v = 2340 Pa, 2340 Pa, abs. PLAN First apply the energy equa First equation tion from th thee V Vent enturi uri sectio section n to the end of the pipe. Then apply the energy equation from reservoir water surface to outlet: SOLUTION Energy equation from Venturi section to end of pipe: p1 V 12 p2 V 22 + + z1 = + + z2 + hL γ 2g γ 2g V 2 V 2 pvapor V 12 = 0 + 2 + 00..9 2 + 2g 2g 2g γ pvapor = 2, 340 Pa 340 Pa abs. = abs. = −97 97,, 660 660 Pa Pa gage Continuity Contin uity principle V 1 A1 = V 2 A2 V 2 A2 V 1 = A1 = 2.56 56V V 2 Then
V 22 V 12 2g = 6.55 2g
Substituting into energy equation 53
V 22 V 22 −97 97,, 660 660//9, 790 + 6. 6.55 = 1.9 2g 2g V 2 = 6.49 49 m/s m/s Flow rate equation Q = V 2 A2 = 6.49 m/ s × π/ π/44 × (0. (0.4 m)2 Q = 0.815 815 m m 3 /s Energy equation from reservoir water surface to outlet: V 22 z1 = + hL 2g V 22 H = 1.9 2g H = 4.08 08 m m
54
7.38: PROBL PROBLEM EM DEFINITION Situation: A pump fills a tank with water from a river. Dtank = 5 m, D pipe = 5 cm. hL = 10V 10V 22 /2 /2g , h p = 20 − 4 × 104Q2 .
Find: Time required to fill tank to depth of 10 m. Assumptions: α = 1.0. SOLUTION Energy equation (locate 1 on the surface of the river, locate 2 on the surface of the water in the tank). p1 V 12 p2 V 22 + + z1 = + + z2 + hL γ γ 2g 2g but p but p 1 = p = p2 = 0, z1 = 0, V 1 = 0, V 2 ' 0 0.. The energy equation reduces to 0 + 0 + 0 + h p = 0 + 0 + (2 (2 m m + h) + hL where h where h = =depth depth of water in the tank 20 − (4)(104 )Q2 = h + 2 +
V 2 V 2 + 10 2g 2g
where V where V 2/2g is the head loss due to the abrupt expansion. Then V 2 18 = (4)(10 )Q + 11 + h 2g 4
2
Q A 2 V 11 Q2 11 = = (1 (1..45)(105 )Q2 2 2g 2g A 18 = 1.85 × 105 Q2 + h 18 − h Q2 = 1.85 × 105 (18 − h)0.5 Q = 430 V =
µ ¶ 55
But Q But Q = = A A T dh/dt dh/dt where where A A T = tank area, so
dh/(18 dh/ (18 − h)0.5 Integrate:
dh (18 − h)0.5 (18 − h)0.5 ∴ = = dt (430)(π/ (430)( π/4)(5) 4)(5)2 8, 443 = dt/ dt/88, 443
t −2(18 − h)0.5 = 8 , 443 + const.
But t But t = = 0 when when h h = = 0 so const. = −2(18)0.5 . Then t = (180.5 − (18 − h)0.5 )(16 )(16,, 886) For or h h = = 10 m 10 m t = (180.5 − 80.5)(16 )(16,, 886) = 23 23,, 880 880 s s t = 6.63 h
56
7.39: PROBL PROBLEM EM DEFINITION Situation: Water is flowing in a horizontal pipe. D = 0.15 m, L = L = 60 60 m. V V = 2 m/ s. hL = 2 m.
Find: Pressure drop (Pa). Pumping power (W). Properties: Water (15 (15 C), Table A.5: γ = = 9800 9800 N/ m3 . ◦
PLAN 1. Find pressure drop using the energy equation. 2. Find power using the power equation. SOLUTION 1. Energy equation: p1 V 12 p2 V 22 + α1 + z1 + h p = + α2 + z2 + ht + hL γ γ 2g 2g •
let ∆ p = p = p p1 − p2
•
KE terms cancel.
•
Elevation terms cancel.
•
h p = h = h t = 0 ∆ p
= γh L = 9800N/ 9800N/ m3 (2m)
¡
∆ p = p =
¢
19 19..6 kPa
2. Power equation: ˙ p = γ W = γQh Qh p = mgh ˙ p •
The head of the pump must equal the head loss. 57
•
Q = V = V A = (2 (2 m/ s) (π/ π/4)(0 4)(0..15m)2 = 0. 0 .0353m3 / s. ˙ p = γQh L W 9800N = m3
µ
¶¡
0.0353m3 / s (2m)
˙ W p = 692 692 W
¢
REVIEW The pump would need to supply about 0.9 hp to the water.
58
7.40: PROBL PROBLEM EM DEFINITION Situation: A fan is moving air through a hair dryer. ∆ p = p = 6 mm-H20 = 58 = 58..8 Pa Pa,, V V = = 10 10 m/ s. D = 0.044m 044m,, η = 60%. 60%.
Find: Electrical power (watts) to operate the fan. Assumptions: Air is at a constant temperature. Constant diameter: D1 = D = D2 = 0.044m 044m.. Properties: Air (60 (60 C), Table A.3, γ A.3, γ = 10 10..4 N/ m3 . ◦
PLAN 1. Find head loss using the energy equation. 2. Find power supplied to the air using the power equation. 3. Find electrical power using the efficiency equation. SOLUTION 1. Energy equation: p1 V 12 p2 V 22 + α1 + z1 + h p = + α2 + z2 + ht + hL γ γ 2g 2g •
locate section 1 just downstream of fan fan ( p1 = 58 58..8Pa) 8Pa);; section 2 at exit plane ( p2 = 0) .
•
KE terms cancel.
•
Elevation terms cancel.
•
h p = h = h t = 0. hL = pγ 1 = 10 58 58. 8 Pa 10. .4.N / m3 = 5.65 m
59
59
2. Power equation: ˙ p = γ ˙ p W = γQh Qh p = mgh •
Assume that head supplied by the fan equals the head loss.
•
Q = V = V A = (10 (10 m/ s) (π/ π/4)(0 4)(0..044m)2 = 0. 0 .0152m3 / s. fan = γQh L W ˙ fan 10 10..4 N = m3 = 0.893W
µ ¶¡
3. Efficiency equation: P electrical electrical =
0.0152m3 / s (5 (5..65m)
¢
0.893W P fan fan = η 0.6
P electrical electrical = 1.49 W REVIEW The electrical power to operate the fan ( fan ( ≈ 1.5 W) W) is is small compared to the electrical power to heat the air.
60
7.41: PROBL PROBLEM EM DEFINITION Situation: A pump supplies energy to a flowing fluid Q = 5 ft3 / s, α = 1.0. DA = 1.0 ft, ft, pA = 5 psig. DB = 00..5 ft ft,, pB = 90 90 psig. psig.
Find: Horsepower delivered by pump ( pump ( hp) hp). PLAN Apply the flow rate rate equatio equation, n, then the energy equati equation on from A to B. Then appl apply y the power equation. SOLUTION Flow rate equation: Q 5 ft3 / s V A = = = 6.366ft 366ft// s AA (π/ π/4) 4) × (1 ft) ft)2 V A2 (6 (6..366ft 366ft// s)2 = 0.6293 6293 ft = 2 × (32 (32..2 ft/ ft/ s2 ) 2g Q 5 ft3 / s V B = = = 25 25..47 47 ft/s ft/s AB (π/ π/4) 4) × (0 (0..5ft)2 2 B V
2g
2
10..07ft = (25 (25..47ft 47ft// s) 2 = 10 2 × (32 (32..2 ft/ ft/ s )
Energy equation:
5 ×
Power equation:
pA V A2 pB V 2 + + zA + h p = + B + zB γ γ 2g 2g 144 144 + (0. (0 .6293) + 0 + h p = 90 × + 10 10..07 + 0 62 62..4 62 62..4 h p = 205. 205.6 ft
61
Qγh p 550 3 205..6 ft 5 ft / s × 62 62..4lbf / ft3 × 205 = 550 P = P = 117 117 hp P ( P (hp) hp) = =
62
7.42: PROBL PROBLEM EM DEFINITION Situation: A pump moves water from a tank through a pipe. Q = 8 m3 / s, h L = 7V 2 /2g . D = 1 m.
Find: Power supplied to flow ow (MW) (MW).. Assumptions: α = 1.0. PLAN Find power using the power equation. The steps are 1. Find velocity in the pipe using the flow rate equation. 2. Find head of the pump using the energy equation. 3. Calculate power. SOLUTION 1. Flow rate equation Q A 8 = π/4 π/ 4 × ( (11 m)2 = 10 10..2 m/s
V =
2. Energy equation (locate 1 on the reservoir surface; locate 2 at the out of the pipe). V 12 V 22 p1 p2 + α1 + z1 + h p = + α2 + z2 + ht + hL γ 2g γ 2g 2 V V 2 0 + 0 + 40 + h p = 0 + + 20 + 7 2g 2g 2 2 V (10. (10.2 m/ s) = 5.30 30 m m = 2 × 9 9..81 m/ s2 2g Then V 2 V 2 40 + h p = + 20 + 7 2g 2g h p = 8 × 5 5..30 + 20 − 40 = 22 22..4 m
63
3. Power equation P = Qγh p = 8 ft ft3 / s × 98 9810 10 N/ m3 × 22 22..4 m P P = 11..76MW
64
7.43: PROBL PROBLEM EM DEFINITION Situation: An engineer is estimating the power that can be produced by a small stream. Q = 1.4 cfs, cfs, T T = 4400 F, H =34 ft. ft. ◦
Find: Estimate the maximum power that can be generated ( generated ( kW) W) if: if: hL = 0 ft, η t = 100%, 100%, η g = 100%. 100%. hL = 5.5 ft ft,, η t = 70%, 70%, η g = 90%. 90%. PLAN To find the head of the turbine (h ), apply the energy equation from the t upper water water surface (sect (section ion 1) to the low lower er water sur surface face (secti (section on 2). To calcula calculation tion power, use P use P = η( η (mgh ˙ t ), where where η η accounts for the combined efficiency of the turbine and generator. SOLUTION Energy equation p1 V 12 p2 V 22 + α1 + z1 = + α2 + z2 + ht + hL γ γ 2g 2g Term by term analysis p1 = 0; V 1 ≈ 0 p2 = 0; V 2 ≈ 0 z1 − z2 = H Eq. (1) becomes H = ht + hL ht = H − hL Flow rate mg ˙ = γQ = 62 62..4lbf / ft3 = 87 87..4lbf / s
¡
1.4 ft3 / s
¢¡
¢
(1)
65
Power (case a) P = = = =
mgh ˙ t mgH ˙ (87.4lbf / s)(34ft)(1 s)(34ft)(1..356J 356J/ / ft · lbf) 4.02kW
Power (case b). ˙ (H − hL) P = η mg = (0 (0..7)(0 7)(0..9) (8 (877.4lbf / s)(34ft − 5.5ft)(1 5ft)(1..356J 356J/ / ft · lbf) = 2.128kW
Power (case a) = 4.02 kW Power (case b) = 2.13 kW
REVIEW 1. In th thee idea ideall ca case se (c (cas asee a) a),, all all of the the el elev evat atio ion n head head is used used to ma mak ke pow power er.. When typical head losses and machine efficiencies are accounted for, the power production is cut by nearly 50%. 2. From Ohm’s la law, w, a pow power er of 2.13 kW will produce a current of about 17.5 amps at a vo volta ltage ge of 12 120V 0V.. Th Thus, us, the tur turbin binee wil willl pro provid videe eno enough ugh pow power er for about 1 typical typical househ household old circui circuit. t. It is unlikel unlikely y the turbine syste system m will be practical (too expensive and not enough power for a homeowner).
66
7.44: PROBL PROBLEM EM DEFINITION Situation: A pump draws water out of a tank and moves this water to a higher elevation. DA = 8 in, D C = 4 in. V C 12ft/ s, P = P = 25hp. 25hp. C = 12ft/ η = 60%, 60%,
hL =
2 C 2V /2g.
Find: Height ( Height (h h)above water surface. Assumptions: α = 1.0. PLAN Apply the energy equation from the reservoir water surface to the outlet. SOLUTION Energy equation p1 V 12 p2 V 22 + α1 + z1 + h p = + α2 + z2 + hL γ γ 2g 2g 0 + 0 + 0 + h p =
2 0+ + V V c +
2 2 V c 2V
2g h + 2g V c2 h p = h + 3 2g
Velocity head
V c2 (12ft (12ft// s)2 = 22..236 236 ft = 64 64..4 ft/ ft/ s2 2g
Q = V C C AC
µ ¶Ã
= s = 1.047ft3/ s
π (4 (4//12ft)2 4
(1)
(2)
Flow rate equation
12ft
!
67
Power equation Qγh p 550ηη 550 P (550) (550) η h p = Qγ
P ( h hp p) =
25hp(550)0. 25hp(550)0.6 1.047ft3 / s × 62 62..4lbf / ft3 = 126.3 ft
=
Substitute Eqs. (2) and (3) into Eq. (1) V c2 h p = h + 3 2g 126..3 fftt = h + (3 × 2. 126 2. 236) ft h = 119.6 ft
h = 120 ft 120 ft
(3)
68
7.45: PROBL PROBLEM EM DEFINITION Situation: A pump draws water out of a tank and moves this water to a higher elevation. DA = 20 20 cm cm,, D C = 10cm. 10cm. V C P = 35kW. 35kW. C = 3 m/ s, P = η = 60%, 60%, hL =
2 2V C /2g.
Find: Height h Height h in meters. Assumptions: α = 1.0. Properties: Water (20 (20 C), Table A.5: γ = = 9790 9790 N/ m3 . ◦
SOLUTION Energy equation: pA V A2 pC V C 2 + αA + zA + h p = + αC + zC + hL γ γ 2g 2g V c2 V c2 0 + 0 + 0 + h p = 0 + 2g + h + 2 2g V c2 h p = h + 3 2g Velocity head:
V c2 (3 m/ s)2 = 0.4587m = 2 (9 (9..81 m/ s2 ) 2g
Flow rate equation: Q = V C C AC π (0 (0..1 m)2 4
= (3 m m// s)
Ã
= 0.023 02356 56 m3 / s
!
(1)
69
Power equation: Qγh p η Pη h p = = P =
Qγ
35000W × 0. 0.6 3
3
= 9911.05 m
(0 (0..02 0235 3566 m / s)(9790N s)(9790N// m )
Eq. (1): V c2 h = h p − 3 = (91. (91.05m) − 3 (0 (0..45 4587 87 m) 2g h = 89 89..7 m
70
7.46: PROBL PROBLEM EM DEFINITION Situation: A subsonic wind tunnel is being designed. A = 4 m2 , V = V = 60 60 m/ s. V T hL = 0.025 2g . 2
Find: Power required ( required ( kW) W).. Assumptions: α = 1.0. Properties: ρ = 1.2 kg kg// m3 . PLAN To find power, apply the power equation. The steps are 1. Fin Find d the head of the pump by app applyi lying ng the ene energy rgy equa equatio tion n and the con contin tinuit uity y equation together. 2. Calculate power. SOLUTION 1. Finding head of the pump •
Energy equation p1 + V 12 + z1 + h p = p2 + V 22 + z2 + hL γ 2g γ 2g V 22 V T 2 0 + 0 + 0 + h p = 0 + + 0 + 0. 0.025 2g 2g
•
Continuity principle V T T AT = V 2 A2 V T T AT V 2 = A2 = V T 0.4 T × 0. 2 V 2 V T 2 2g = 0.16 2g
71
•
Combining previous two equations V T 2 (0 (0..185) 2g (60m/ (60m/ s)2 (0 (0..185) = 2 × 9 9..81 m/ s2
h p =
h p = 33 33..95 95 m 2. Power equation P = Qγh p = (V A) (ρg ρg)) h p = 60 m/ s × 4 m2
¡
1.2 kg/ kg/ m3 × 9. 9.81 m/ s2 (33 (33..95m)
¢¡
P P = 9955.9 kW
¢
72
7.47: PROBL PROBLEM EM DEFINITION Situation: A pumping system delivers water. zA = 110ft, 110ft, z B = 200ft. 200ft. zC = 110 ft, ft, z D = 90 ft. ft. 2
A = 0.1 ft .
Find: Power delivered by pump ( pump ( hp) hp). SOLUTION
0 + 0 + 110 + h p = 0 + 0 + 200; 200; h p = 90 90 ft ft Qγh p P (hp) P (hp) = 550 Q = V j A j = 0.10 10 V V j V j = 2g × (200 − 110) 110) ft = 76 76..13 13 ft/s ft/s 3 Q = 7.613 613 ft /s Power equation
p P = Qγh p 7.613ft3 / s × 62 62..4lbf / ft3 × 90 ft P = 550 P P = 77 77..7 hp
73
7.48: PROBL PROBLEM EM DEFINITION Situation: A pumping system delivers water. zA = 40 40 m, z B = 65 65 m. zC = 35 35 m, z D = 30 30 m. 2
A = 25 cm .
Find: Power delivered by pump ( pump ( kW) kW). PLAN Apply the energ Apply energy y equation from the reserv reservoir oir wat water er surface to point B. Then apply the power equation. SOLUTION Energy equation p V 2 pB V B2 + + z + h p = + + zB γ 2g γ 2g 0 + 0 + 40 + h p = 0 + 0 + 65 65;; h p = 25 25 m m Flow rate equation Q = V j A j = 25 × 10
p
4
−
m2 × V j
where V where V j j = 2g × (65 − 35) = 24. 24.3 m/s 24..3 = 00..0607 0607 m m 3 /s Q = 25 × 10 4 × 24 −
Power equation P = Qγh p P = 0.0607m3 / s × 9 9,, 810N 810N// m3 × 25 m P P = 1144.9 kW
74
7.49: PROBL PROBLEM EM DEFINITION Situation: A pumping system pumps oil. L = 1 mi, D = 1 ft. ft. Q = 3500 gpm, 3500 gpm, ∆z = 200ft. 200ft. ∆
p = p = 60 psi 60 psi.. Find: Power required for pump ( pump ( hp) hp). Properties: γ = = 0.53lbf / ft3 . SOLUTION Energy equation h p = z = z2 − z1 + hL Expressing this equation in terms of pressure γh p = γ = γzz2 − γz 1 + ∆ ploss Thus pressure rise across the pump is γh p = 53 53 lbf/ft lbf/ft3 × 200 200 ft +60 × 144 144 lbf/ft lbf/ft2 = 19 19,, 240 240 psf Flow rate equation Q = V × A ft3 /s Q = 3500 gpm × 0 0..002228 = 77..80 80 cfs gpm Power equation ˙ = Qγh p W = 7.80 80 cfs cfs ×
19 19,, 240 240 psf psf 550
˙ = 273 hp W W = 273 hp
75
7.50: PROBL PROBLEM EM DEFINITION Situation: Water is pumped from a reservoir through a pipe. V Q = 0.35 m3 / s, h L = 2 2g . z1 = 6 m, z 2 = 10 10 m. 2 2
D1 = 40cm, 40cm, D 2 = 10 10 cm cm.. p2 = 100 kP kPaa.
Find: Power (kW) that the pump must supply. Assumptions: α = 1.0. Properties: Water (10 (10 C), Table A.5: γ = = 9810 9810 N/ m3 . ◦
PLAN Apply the flow rate equation, then the energy equation from reservoir surface to the 10 m elevatio elevation. n. Then app apply ly the pow power er equat equation. ion. SOLUTION Flow rate equation: Q A 0.35 = (π/4) π/4) × (0 (0..3 m)2 = 4.95 95 m/s m/s
V =
V 22 = 1.250 250 m m 2g
76
Energy equation (locate 1 on the reservoir surface; locate 2 at the pressure gage) 1000 100000 00 Pa + 11..25 m + 10 m + 2.0 (1 (1..25m) 9810N// m3 9810N h p = 17 17..94 m
0 + 0 + 6 m + h p =
Power equation: P = Qγh p = 0.35 m3 / s
¡
(17..94m) 9810N// m3 (17 9810N
¢¡
P P = 61 61..6 kW
¢
77
7.51: PROBL PROBLEM EM DEFINITION Situation: Oil is pumped through a pipe attached to a manometer filled with mercury. Q = 6 ft3 / s, ∆h = 46in. 46in. D1 = 12in, 12in, D 2 = 6 in.
Find: Horsepower pump supplies (hp) supplies (hp).. Assumptions: α = 1.0. Properties: Table A.5: S mercury 13..55 55.. mercury = 13 S oil oil = 0.88 PLAN Apply the flow rate rate equa equati tion on,, then then the en ener ergy gy equ equat atio ion. n.
Th Then en app apply ly the pow power er
equation. SOLUTION Flow rate equation V 1122 2 V 12 /2g V 6 V 62 /2 /2g
6 ft3 / s Q = = = 77..64ft 64ft// s A12 (π/ π/4) 4) × (1 ft) ft)2 = 0.906 906 ft ft = 4V 1122 = 30 30..56 56 ft/s ft/s = 14 14..50 50 ft ft
78
Energy equation
µ
¶−µ
¶
p6 p12 + z6 + z12 = γ γ 2 p12 V 12 + z12 + + h p = γ 2g
µ
¶
h p = h p =
(13 (13..55 − 0.88) 46 ft 12 0.88 p6 V 62 + z6 + γ 2g 13 13..55 3.833 ft + 14 14..50ft − 0.906ft − 1 × 3. 0.88 68 68..8 ft
¡ ¢
µ
¶¶
Power equation P (hp) P (hp) = Qγh p /550 62..4lbf / ft3 6 ft3 / s × 0.88 × 62 P = 550 P P = 41 41..2 hp
¡
¢
68..8 ft 68
×
79
7.52: PROBL PROBLEM EM DEFINITION Situation: Q = 500 cfs, 500 cfs, η η = 90%. 90%. V hL = 1.5 2g , D , D = = 7 ft. z1 = 35 35 ft, ft, z 2 = 0 ft. 2
Find: Power output from turbine. Assumptions: α = 1.0. PLAN Apply the energy equation from the upstream water surface to the downstream water surface. surfa ce. Then ap apply ply the po power wer eq equatio uation. n. SOLUTION Energy equation: p1 V 12 p2 V 22 + + z1 = + + z2 + hL + ht γ 2g γ 2g V 2 0 + 0 + 35 = 0 + 0 + 0 + 1.5 + ht 2g Q 500ft3 / s V = = = 12 12..99 99 ft/s ft/s A (π/4) π/4) × (7 ft) ft)2 V 2 = 2.621 621 ft ft 2g ht = 35 ft f t − 1.5 (2 (2..621 621 ft) ft) = 31 31..07 07 ft Power equation: equation: P (hp) P (hp) = =
Qγh t η 550 (500ft3 / s)(62 s)(62..4lbf / ft3 )(31 )(31..07ft × 0. 0.9) 550 P = P = 1590hp 1590 hp = 11..18MW
80
7.53: PROBL PROBLEM EM DEFINITION Situation: A hydraulic power system consists of a dam with an inlet connected to a turbine. H = = 15 15 m, V 2 = 5 m/ s. Q = 1 m3 / s.
Find: Power produced by turbine ( turbine ( kW) W).. Assumptions: All head loss is expansion loss. 100% efficiency. PLAN Apply the energy equation from the upstream water surface to the downstream water surface. surfa ce. Then ap apply ply the po power wer eq equatio uation. n. SOLUTION Energy equation p1 V 12 p2 V 22 + + z1 = + + z2 + hL + ht γ γ 2g 2g V 22 0 + 0 + 15 m 15 m = 0 + 0 + 0 + ht + 2g (5 m/ s)2 ht = 15 15 m − 2 × 9. 9.81 m/ s2 = 13 13..73 73 m Power equation P = Qγh t = (1 (1 m m3 /s)(9810 /s)(9810 N/m N/m3 )(13 )(13..73 73 m m)) P = P = 135 kW 135 kW
81
7.54: PROBL PROBLEM EM DEFINITION Situation: A pump transfers SAE-30 oil between two tanks. Dtank = 12 12 m, D pipe = 20 cm. V hL = 20 2g , h , h p = 60 60 m. 2
zA = 20 20 m, z B = 1 m.
Find: Time required to transfer oil ( oil ( h). h). PLAN Apply the energy equation between the top of the fluid in tank A to that in tank B. SOLUTION Energy equation h p + zA = z = zB + hL or h p + zA = z = zB + 20
V 2 2g
Solve for velocity
+
V 2 2g
2g (h p + zA − zB ) 21 2 × 9 9..81 V 2 = (60 + zA − zB ) 21 V = 0.9666(60 + zA − zB )1/2
V 2 =
The sum of the elevations of the liquid surfaces in the two tanks is zA + zB = 21 So the energy equation becomes
82
V V = 0.9666(81 − 2zB )1/2 Continuity Contin uity equation (0. (0 .2 m)2 Apipe dzB = V = V Atank dt (12m)2 = =
¡¡
2.778 × 10 2.778 × 10
4
−
4
−
¢¢
V 0.9666(81 − 2zB )1/2
= 2.685 × 10 4(81 − 2zB )1/2 −
Separate variables dzB = 2.685 × 10 4 dt 1 / 2 (81 − 2zB ) −
Integrate
20ft
2.685 × 10 4 dt
=
(81 − 2zB )1/2
1
81
∆t −
Z ¡−√ − ¢ ¡ p − ´ ¡ ¡ Z
³−p −
dzB
81
2 (2 (20) 0) +
81
20ft 2zB 1 ft
0
4
=
2.685 × 10
−
2(1) =
2.685 × 10
−
2. 485 1 = 2.685 × 10 ∆t = 9256s
−
t = 9260s = 2. 2.57 h
4
4
¢ ¢ ¢
∆t ∆t ∆t
83
7.55: PROBL PROBLEM EM DEFINITION Situation: A pump is used to pressurize a tank. Dtank = 2 m, D pipe = 4 cm. hL = 10 V 2g , h , h p = 50 50 m. 2
zA = 20 20 m, z B = 1 m. pT = 4 3zt p0 , p 0 = 0 kP kPaa gage gage = = 100 kP kPaa. −
Find: Write a computer program to show how the pressure varies with time. Time to pressurize tank to 300 kPa (s) . PLAN Apply the energy equation between the water surface at the intake and the water surface inside the tank. SOLUTION Energy equation p2 + z2 + hL γ Expressing the head loss in terms of the velocity allows one to solve for the velocity in the form h p + z1 =
V 2 =
2g pt (h p + z1 − zt − ) γ 10
Substituting in values 3 1/2 ) 4 − zt The equation for the water surface elevation in the tank is 10..19 V V = 11..401(46 − zt − 10
∆zt = V = V A A p ∆t =
At
2500 V ∆t
84
A computer program can be written taking time intervals and finding the fluid level and pressure in the tank at each time step. The time to reach a pressure of 300 kPa abs in the tank is 69 6988 sec second ondss or 111.6 1.6 min minute utes. s. A plot of how how the pres pressur suree vari aries es with time is provided. 350 ) a P k ( e r u s s e r P
300 250 200 150 100 50 0
200
400
Time (sec)
600
800
85
7.56: PROBL PROBLEM EM DEFINITION Situation: A pipe discharges water into a reservoir. Q = 10 cfs, 10 cfs, D D = = 12 in. in.
Find: Head loss at pipe outlet (ft) outlet (ft).. PLAN Apply the flow rate equation, then the sudden expansion head loss equation. SOLUTION Flow rate equation Q A 10ft3 / s = (π/ π/4) 4) × (1 ft) ft)2 = 12 12..73 73 ft/s ft/s
V =
Sudden expansion head loss equation hL = V 2 /2g hL = 2.52 ft
86
7.57: PROBL PROBLEM EM DEFINITION Situation: A pipe discharges water into a reservoir. Q = 0.5 m3 / s, D = D = 50cm. 50cm.
Find: Head loss at pipe outlet ( outlet ( m) m).. PLAN Apply the flow rate equation, then the sudden expansion head loss equation. SOLUTION Flow rate equation Q A 0.5 ft3 / s = (π/ π/4) 4) × (0. (0.5ft)2 = 2.546 546 m/s m/s
V =
Sudden expansion head loss equation V 2 hL = 2g (2 (2..546m 546m// s)2 = 2 ×
2 9 9..81 m/ s hL = 0.330 330 m m
87
7.58: PROBL PROBLEM EM DEFINITION Situation: Water flows through a sudden expansion. D8 = 8 cm, V 8 = 2 m/ s. D15 = 15 15 cm cm.. Find: Head loss caused by the sudden expansion ( m) m).. PLAN Apply the continuity principle, then the sudden expansion head loss equation. SOLUTION Continuity principle: V 8 A8 = V 1155A15 V 8 A8 V 1155 = = 2 × A15 Sudden expansion head loss:
8
2
= 0.569 569 m/s m/s
µ¶ 15
(V 8 − V 1155)2 hL = (2 (2gg ) (2 m/ s − 0.569m 569m// s)2 hL = (2 × 9 9..81 m/ s2 ) hL = 0.104m
88
7.59: PROBL PROBLEM EM DEFINITION Situation: Water flows through a sudden expansion. D6 = 6 in = 00..5 ft. ft. D12 = 12 in = 1 ft. Q = 5 cfs. Find: Head loss PLAN Apply the flow rate equation, then the sudden expansion head loss equation. SOLUTION Flow rate equation Q 5 = = 25 25..46 46 ft/s; ft/s; A6 (π/ π/4) 4) × (0. (0.5ft)2 1 V 1122 = V 6 = 6.37 37 ft/s ft/s 4 V 6 =
Sudden expansion head loss equation (V 6 − V 1122)2 hL = 2g (25 (25..46ft 46ft// s − 6.37ft 37ft// s)2 = 2 × 32 32..2 ft/ ft/ s2 hL = 5.66 66 ft ft
89
7.60: PROBL PROBLEM EM DEFINITION Situation: Two tanks are connected by a pipe with a sudden expansion. ∆z = 10 10 m, A 1 = 8 cm2 . A2 = 25 cm2 .
Find: Discharge between two tanks ( tanks ( m3 / s) PLAN Apply the energy equation from water surface in A to water surface in B. SOLUTION Energy equation (top of reservoir A to top of reservoir B) pB V B2 pA V A2 + + zB + ΣhL + + zA = γ γ 2g 2g 0 + 0 + 10 m = 0 + 0 + 0 + ΣhL
(1)
Let the pipe from A be called pipe 1. Let the pipe into B be called pipe 2 Then (V 1 − V 2 )2 V 22 + ΣhL = 2g 2g Continuity principle: V 1A1 = V 2 A2 (25cm2 ) V 2 A2 = V 2 V 1 = = 33..125 125V V 2 A1 (8cm2 ) Combine Eq. (1), (2), and (3): V 2 2 (3 (3..125 125V V 2 − V 2 )2 + 10 m = 2 (9 (9..81 m/ s2) 2 (9 (9..81 m/ s2 ) V 2 = 5.964m 964m// s Flow rate equation: Q = V 2 A2 1.0 m
2
= (5 (5..964m 964m// s)(25cm ) Q = 0.0149m3 / s
µ
100cm
2
¶
(2)
(3)
90
7.61: PROBL PROBLEM EM DEFINITION Situation: A horizontal pipe with an abrupt expansion. D40 = 40 40 cm cm,, D 60 = cm cm.. 3 Q = 1.0 m / s, p 1 = 70 kP kPaa gage. Find: Horizontal force required to hold transition in place ( place ( kN) N).. Head loss ( loss ( m) m).. Assumptions: α = 1.0. Properties: Water, ρ Water, ρ = = 1000 1000 kg kg// m3 . PLAN Apply the flow rate equation, the sudden expansion head loss equation, the energy equation, and the momentum principle. SOLUTION Flow rate equation 1.0 m3 / s Q V 4400 = = = 77..96 96 m/s A40 (π/ π/4) 4) × (0 (0..40m)2 2 V 40 = 3.23 23 m m 2g 2 4 V 6600 = V 4400 × = 3.54 54 m/s m/s 6 2 V 60
µ¶
2g = 0.639 639 m m Sudden expansion head loss equation 2 (V 4400 − V 60 60 ) hL = 2g = 0.99 0.996 6m
Energy equation 2 2 p40 V 40 p60 V 60 + + = + hL γ γ 2g 2g
p60 = 70 70,, 000 Pa + 9981 8100 N/ m3 (3 (3..23 − 0.639 − 0.99 996) 6) m = 85 85,, 647 647 Pa Pa Momentum principle
91
Fx p 2 A2
p 1 A1
Substituting values
X
F x = m ˙ oV xx,o ˙ i V xx,i,i ,o − m
70 70,, 000Pa × π/ π/44 × (0. (0.4 m)2 − 85 85,, 64 6477 Pa × π/4 π/4 × (0 (0..6 m)2 + F x = 1000 kg/ m3 × 1 1..0 m3/ s × (3. (3.54 m/ s − 7.96 m/ s) The result is F x = −8796 + 24, 24, 216 − 4, 420 = 11 11,, 000 000 N N F x = 11.0 kN
92
7.62: PROBL PROBLEM EM DEFINITION Situation: Water flows in a horizontal pipe before discharging to atmosphere. A = 9 in2 , V V = 15 ft/ s. hL = 3 ft.
Find: Force on pipe joint. Assumptions: α = 1.0. Properties: γ = = 62. 62.4lbf / ft3 . PLAN Apply the momentum principle, then the energy equation. SOLUTION F j
Vx
p1A1
p2=0
Momentum Momen tum Equation F x = mV ˙ o,x ˙ ii,x o,x − mV ,x 2
2
X
F j + p + p1 A1 = −ρV x A + ρV x A F j = − p1 A1 Energy equation
p1 V 12 p2 V 2 + + z1 = + 2 + z2 + hL γ γ 2g 2g = γh hL p1 − p2 = γ p1 = γ = γ (3) (3) = 187. 187.2 psfg 9 in2 ) F j = −187 187..2lbf / ft × ( 144in2 / ft2 F j = 11 11..7lbf 2
−
F j = 1111.7lbf acting acting to the left
x
93
7.63: PROBL PROBLEM EM DEFINITION Situation: An abrupt expansion dissipates high energy flow ow.. D1 = 5 ft, p 1 = 5 psig. V V = = 25ft/ 25ft/ s, D 2 = 10 ft.
Find: (a) Horsepower lost ( lost ( hp) hp). (b) Pressure at section 2 ( 2 (psig psig)). (c) Force needed to hold expansion ( expansion ( lbf) lbf).. Assumptions: α = 1.0. Properties: Water, γ Water, γ = 62 62..4lbf / ft3 . PLAN Find the head loss by applying the sudden expansion head loss equation, first solving for V 2 by applyin for applyingg the con contin tinuit uity y pri princip nciple. le. The Then n app apply ly the pow power er equat equation ion,, the energy equation, and finally the momentum principle. SOLUTION Contin Continuity uity equation V 2 = V 1
A1 A2
1 = 25 ft f t/ s 4 = 6.25 25 ft/s ft/s
µ¶
Sudden expansion head loss equation (2gg ) hL = (V 1 − V 2 )2 /(2 (25ft/ (25ft/ s − 6.25ft 25ft// s)2 hL = 64 64..4 ft/ ft/ s2 = 5.46 46 ft ft a) Power equation P (hp) P (hp) = Qγh/550 Qγh/550 Q = V A = 25 ft/ ft/ s( s(π/ π/4)(5ft) 4)(5ft)2 = 490. 490.9 ft 3/s P = (490.9 ft3 / s)(62 s)(62..4lbf / ft3 )(5 )(5..46) 46)//550 P = P = 304 hp
94
b) Energy equation p1 V 12 + + z1 γ 2g (5 × 144) lb lbf f / ft2 (2 (255 ft/ ft/ s)2 + 64 64..4 ft/ ft/ s2 62 62..4lbf / ft3 p2 /γ p2
= = = = =
p2 V 22 + + z2 + hL γ 2g (6..25ft 25ft// s)2 p2 (6 + 55..46ft + 64 64..4 ft/ ft/ s2 γ 15 15..18 18 ft ft 15 15..18ft × 62 62..4lbf / ft3 947 psfg
p2 = 6.58 psig c) Momentum equation m ˙ = 1.94 sl slug ug// ft3 × (π/ (π/4) 4) × (5 ft) ft)2 × 25ft 25ft// s = 952.3 kg/s ˙ oV xx,o F x = m ,o
X−
p1 A1
m ˙ i V xx,i,i
−
p2 A2 + F x = m ˙ (V 2 − V 1 )
(5lbf / in2 )( )(144 144 in2 / ft2 )( )(π/ π/4)(5ft) 4)(5ft)2 − (6 (6..58lbf / in2 )(1 )(144 44 in2 / ft2)( )(π/ π/4)(10ft) 4)(10ft)2 + F x = 952.3 kg kg// s × (6 (6..25ft 25ft// s − 25ft 25ft// s) F x = 42,400 lbf
95
7.64: PROBL PROBLEM EM DEFINITION Situation: Rough aluminum pipe discharges water. L = 50 50 ft ft,, D = D = 6 in. W = (1 W (1..5lbf) L, Q = Q = 6 cfs. hL = 10 10 ft. ft.
Find: Longitudinal force transmitted through pipe wall ( wall ( lb lbf) f).. Properties: Water, γ Water, γ = 62 62..4lbf / ft3 . PLAN Apply the energy equation, then the momentum principle. SOLUTION 1
c.s.
2
p1 V 12 p2 V 22 + + z1 = + + z2 + hL γ γ 2g 2g but V but V 1 = V = V 2 and and p p 2 = 0. Therefore p1 /γ = −50 ft + 1100 ft p1 = −2496 2496 lbf/ft lbf/ft2
96
Momentum principle
X
F y = mV ˙ y,o ˙ y,i = ρQ ρQ((V 2y − V 1y ) y,o − mV y,i =
− p1A1 − γAL − 2L + F wall wall = 0
F wall wall = 1.5L + γA 1 L − p1 A1 = 75 lbf + (π/4) π/4) × (0 (0..5ft)2 (62 (62..4lbf / ft3 × 50ft − 2, 496lbf / ft3) = 75lbf + 122 122.5lbf
£
F wall 198 lbf acting upward wall = 198 lbf
¤
97
7.65: PROBL PROBLEM EM DEFINITION Situation: Water flows in a bend. Q = 5 m3 / s, p = p = 650 kP kPaa. hL = 10 10 m, D = D = 80cm. 80cm. d = 50cm. 50cm.
Find: Pressure at outlet of bend ( bend ( kP kPa) a).. Force on anchor block in the x the x-direction -direction ( ( kN kN)). Assumptions: α = 1.0. PLAN Apply the energy equation, then the momentum principle. SOLUTION Energy equation
2 2 p50 V 50 p80 V 80 + + + z50 = + z80 + hL γ γ 2g 2g
where p 50 = 650, where p 650, 000 000 Pa Pa and z and z 50 = z = z80 Flow rate equation 5 m3 / s Q = = 9.947 947 m/s m/s A80 (π/4) π/4) × (0 (0..8 m)2 2 V 80 /2g = 5.04 04 m m V 8800 =
Continuity Contin uity equation V 5500 = V 8800 × (8 (8//5)2 = 25 25..46 46 m/s m/s 2 V 50 = 33 33..04 04 m m 2g hL = 10 10 m m
98
Then 650 650,, 000Pa p80 = + 33 33..04 − 5.04 − 10 γ γ (33..04 − 5.04 − 10 10)) m = 826 826,, 600 600 Pa Pa p80 = 650, 00 0000 Pa + 9, 810N 810N// m3 (33
¡
p80 = 827 kPa
¢
Momentum principle
X
F x = mV ˙ o − mV ˙ i = ρQ = ρQ((V 80,x 80,x − V 50,x 50,x )
947m// s − 0.5 m × 25 25..46 m/ s) p80A80 + p + p50 A50 × co coss 60 + F x = 1, 000kg 000kg// m3 × 5 m3 / s(−9.947m F x = −415 415,, 494 − 63 63,, 814 − 113 113,, 385 = −592 592,, 693 N F x = -593 kN ◦
99
7.66: PROBL PROBLEM EM DEFINITION Situation: Fluid in a pipe flows around an accelerated disk. U = = 10 10 m/ s, D = 5 cm. d = 4 cm.
Find: Develop an expression for the force required to hold the disk in place in terms of U, D, d, and ρ. and ρ. Force required under given conditions ( N) N).. Assumptions: α = 1.0. Properties: ρ = 1.2 kg kg// m3 . PLAN Apply the energy equation from section (1) to section (2), and apply the momentum principle. SOLUTION Control volume
U2
U1
Fdis disk k on fluid fluid Energy equation ρU 12 ρU 22 p1 + = p2 + 2 2 2 2 p1 − p2 = ρU 2 − ρU 1 2 2
100
but U 1 A1 = U 2(π/4)( π/4)(D D2 − d2) U 1 D2 U 2 = (D2 − d2 ) Then p1 − p2 = Momentum principle for the C.V.
X
(1)
ρ D4 2 − 1 U 1 2 (D2 − d2 )2
¸
³´ ∙
(2)
F x = m ˙ o U o − m ˙ i U i = ρQ = ρQ((U 2x − U 1x )
p1 A − p2 A + F disk ρQ((U 2 − U 1 ) disk on fl uid = ρQ ( p1 − p2 )A F fluid on disk = F d = ρQ = ρQ((U 1 − U 2 ) + ( p Eliminate p Eliminate p 1 − p2 by Eq. (2), and U and U 2 by Eq. (1):
µ
F d = ρU = ρU A U 1 −
U 1 D2 (D2 − d2 ) 2
¶ µ ¶∙ h i
F d = ρU 8πD
+
2
ρU 2 2
¸
D4 − 1 A (D2 − d2 )2
1 (D2 /d2 −1)2
When U When U = 10 m/s, m/s, D D = = 5 cm, cm, d d = = 4 cm and ρ and ρ = = 1.2 kg/m3 1.2 kg/ kg/ m3 × (1 (100 m/ s)2 × π × (0 (0..05m)2 1 F d = 8 ((0..05 m/0.04m)2 − 1)2 ((0
∙
F d = 0.372 N
¸
101
Problem 7.67 No solution provided.
102
Problem 7.68 Answer the following questions. a. What are three important reasons that engineers use the HGL and EGL? •
Identify where head loss is occurring. • Identify potential sites of cavitation. •
Visualize how the energy equation is being satis fied.
b. Wh What at fac factor torss influenc uencee the mag magnit nitude ude of the HG HGL? L? Wh What at fact factors ors influence the magnitude of the EGL? •
Since the HGL = HGL = ( p/γ ) + z , the factors are: — Pressure head head ( p/γ ), which is influenced by the pressure and the specific weight of the fluid. — Elevation head ( head (zz ) .
•
Since the EGL = EGL = ( p/γ ) + z + (V ( V 2 /(2 (2gg)), )), the factors are: — Pressure head and elevation head (as described above). — Velocity head (V 2 /(2 (2gg )) , which is influenced by the flow rate and pipe section area.
c. Ho How w are the EGL and HGL rel related ated to the piezo piezometer meter?? To the stagna stagnation tion tube? •
•
When liquid flows in a pipe, the HGL is coincident with the water level in a piezometer that is tapped into the pipe. When liquid flows in a pipe, the EGL is coincident with the water level in a stagnation tube that is tapped into the pipe.
d. How is the EGL related to the energy equation? • •
The EGL involves three of the terms that appear in the energy equation. The terms represent transport of energy across the control surface (PE + KE) plus flow work at the control surface.
e. How can you use an EGL or an HGL to determine the direction of flow ow?? •
•
In a pipe of constant diameter, the flow goes from locations of high EGL & HGL to locations of low HGL & EGL. In a pipe of variable diameter, fl ow goes from locations of high EGL to locations of low EGL.
103
7.69: PROBL PROBLEM EM DEFINITION Situation: A piping system with a black box shows a large EGL change at the box.
Find: What the black box could be. SOLUTION •
•
•
Because the EGL slopes downward to the left, the flow is from right to left. fl
fl
the In the ”black box” is a device that remo removes ves energy from the owing uid because EGL drop. Thus, there could either be a turbine, an abrupt expansion or a partially closed val alv ve. b, c, d.
7.70: PROBL PROBLEM EM DEFINITION Situation: A constant diameter pipe is shown with an HGL.
Find: Whether this system is possible, and if so under what conditions. SOLUTION Possible if the fluid is being accelerated to the left.
104
7.71: PROBL PROBLEM EM DEFINITION Situation: Two tanks are connected by a pipe with a machine.
Find: (a) Direction of flow ow.. (b) What kind of machine is at point A. (c) Compare the diameter of pipe sections. (d) Sketch the EGL. (e) If there is a vacuum at anywhere, if so where it is. SOLUTION (a) Flow is from right to left. (b) Machine is a pump. (c) Pipe CA is smaller because of steeper HGL (d)
EGL
(e) No vacuum in the system.
105
7.72: PROBL PROBLEM EM DEFINITION Situation: An HGL and EGL are shown for a flow system.
Find: (a) Direction of flow ow.. (b) Whether there is a reservoir. (c) Whether the diameter at E is uniform or variable. (d) Whether there is a pump. (e) Sketch a physical set up that could exist between C between C and and D D.. (f) Whether there is anything else revealed by the sketch. SOLUTION (a) Flow is from A from A to E to E because because EGL slopes downward in that direction. (b) Yes, at D at D,, because EGL and HGL are coincident there. (c) Uniform diameter because V because V 2 /2g is constant (EGL and HGL uniformly spaced). (d) No, because EGL is always dropping (no energy added). (e)
(f) Nothing else.
106
7.73: PROBL PROBLEM EM DEFINITION Situation: Two tanks are connected by a uniformly tapered pipe.
Find: Draw the HGL and EGL. SOLUTION
EGL HGL
107
7.74: PROBL PROBLEM EM DEFINITION Situation: A system with an HGL and EGL is described in the problem statement.
Find: (a) Which line is HGL and which is EGL. (b) If pipes are the same size and which is smallest. (c) If and where pipe pressure falls below zero. (d) Point of max pressure. (e) Point of minimum pressure. (f) What is at point E. (g) Air pressure in the tank above or below atmospheric. (h) What is at point B. SOLUTION (a) Solid line is EGL, dashed line is HGL. (b) No; AB is smallest. (c) From B to C. (d) pmax is at the bottom of the tank. (e) pmin is at the bend C. (f) A nozzle. (g) above atmospheric pressure. (h) abrupt expansion.
108
7.75: PROBL PROBLEM EM DEFINITION Situation: Water flows from a tank through a pipe system before discharging through a nozzle. z1 = 100 100 m, z 2 = 30 30 m. L1 = 100 100 m, L 2 = 400 400 m. D1 = D = D2 = 60 60 cm cm,, D jet = 30cm. 30cm.
Find: (a) Discharge. (b) Draw HGL and EGL. (c) location of maximum pressure. (d) location of minimum pressure. (e) values for maximum and minimum pressure. SOLUTION Energy equation (locate 1 on the reservoir water surface; locate 2 at outlet of the nozzle). p1 V 12 p2 V 22 + + z1 = + + z2 + hL γ 2g γ 2g V 22
L
V p2
µµ ¶¶µ ¶
2g 0 + 0 + 100 = 0 + 2g + 30 + 0.014 D 2 V 2 500 V 22 100 = 0 + + 30 + 0.014 2g 0.6 2g Continuity Contin uity equation V 2 A2 = V p A p A p V 2 = V p AL V 2 = 4V p
109
Then V p2 (16 + 11. 11.67) = 70 m 2g V p = 7.045 045 m/s m/s V p2 53 m m 2g = 2.53 Flow rate equation Q = V p A p = 7.045m 045m// s × (π/ (π/4) 4) × (0 (0..60m)2 Q = 1.99 m3 /s pmin
EGL HGL 40.5 m pmax max
pmin V p2 100m V p2 0.014 pmin : 100 = + + 100 + 0. 2g γ 0.6 m 2g pmin + 100 + 3.33 × 2 2..53 100 = γ
µ ¶
pmin = -82.6 kPa, gage pmin = 40 40..5 m − 2.53 53 m m γ pmax = 373 kPa
110
7.76: PROBL PROBLEM EM DEFINITION Situation: A reservoir discharges into a pipe. Find: Draw the HGL and EGL. SOLUTION
37.2 m 80 m 42.6 m
2000 m
111
7.77: PROBL PROBLEM EM DEFINITION Situation: Water discharges through a turbine. Q = 1000 cfs, 1000 cfs, η η = 85%. 85%. hL = 4 ft, H H = = 100ft. 100ft.
Find: Power generated by turbine ( turbine ( hp) hp). Sketch the EGL and HGL. PLAN Apply the energy equation from the upper water surface to the lower water surface. Then apply the power equation. SOLUTION Energy equation p1 V 12 p2 V 22 + + z1 = + + z2 + γ γ 2g 2g 0 + 0 + 100 ft 100 ft = 0 + 0 + 4 ft + ht ht = 96 96 ft ft
X
hL + ht
Power equation P = (Qγh t )(e )(eff .) Qγh t η 1, 000ft3 / s × 62 62..4lbf / ft3 × 96 ft × 0. 0.85 P P ((hp) = = 550 550 P = P = 9260 hp 9260 hp
EGL HGL
Turbine urbi ne
112
7.78: PROBL PROBLEM EM DEFINITION Situation: Water discharges from a reservoir through a pipe and out a nozzle. L V hL = 0.025 D , D , D = 1 ft. 2g L = 1000 1000 ft, ft, D jet = 6 in. 2
z1 = 100ft, 100ft, z 2 = 60 ft. ft.
Find: (a) Discharge ( Discharge (cfs cfs)) . (b) Draw the HGL and EGL. Assumptions: α = 1.0. PLAN Apply the energy equation from the reservoir surface to the exit plane of the jet. SOLUTION Energy Energ y equat equation ion.. Let the v veloc elocit ity y in the 6 inch pipe be be V V 6. Let the vel velocity ocity in the 12 inch pipe be V be V 1122. p1 V 12 p2 V 62 + + z1 = + + z2 + hL γ 2g γ 2g
µ ¶
2 2 0 + 0 + 100 = 0 + V 6 + 60 + 0.025 1000 V 12 2g 1 2g
Continuity Contin uity principle V 6 A6 = V 1122A12 A12 V 6 = V 1122 A6 122 V 6 = V 1122 2 = 4V 12 12 6 V 62 V 2 = 16 12 2g 2g
113
Substituting into energy equation 2 V 12 40 fftt = (16 + 25) 2g 40ft 2 V 12 = 2 × 32 32..2 ft/ ft/ s2 41 V 1122 = 7.927 927 ft/s ft/s
µ ¶
Flow rate equation
µ ¶¡
¢
Q = V 1122A12 = (7 (7..927ft 927ft// s)( s)(π/ π/4)(1ft) 4)(1ft)2 Q = 6.23 23 ft ft 3 /s
EGL HGL
114
7.79: PROBL PROBLEM EM DEFINITION Situation: Water moves between two reservoirs through a contracting pipe. L V hL = 0.02 D . 2g Dd = 15cm, 15cm, L d = 100 100 m. 2
Lu = 100 100 m, Du = 30cm. 30cm. z1 = 100 100 m, z 2 = 60 60 m. Find: Discharge of water in system ( system ( m3 / s) . PLAN Apply energy equation from upper to lower reservoir. SOLUTION Energy equation p2 V 22 p1 V 12 + + z2 + + + z1 = γ γ 2g 2g 0 + 0 + 100m = 0 + 0 + 70 m +
X
hL = 30 30 m m
V 2 ) 2g 200 V u2 0.3 2g
X X
hL
hL
hL = .02 × (L/D (L/D)( )( 30 = 0.02 ×
V d2 + 1. 1.0 2g
µ ¶µ ¶ µ µ ¶ ¶ 100m + 0.02 0.15 m
(1)
Flow rate equation Q Q = Au (π/4) π/ 4) × (0 (0..3 m)2 Q Q = V d = Ad (π/4) π/ 4) × (0 (0..15m)2 V u =
Substituting Eq. (2) and Eq. (3) into (1) and solving for Q for Q yields: Q = 0.110m3 / s
(2)
(3)
115
7.80: PROBL PROBLEM EM DEFINITION Situation: Water is pumped from a lower reservoir to an upper one. z1 = 90 90 ft, ft, z 2 = 140ft. 140ft. L1 = 1000 1000 ft, ft, L 2 = 2000 2000 ft. ft. D1 = 8 in, D 2 = 8 in. L V Q = 3 cfs, cfs, h h L = 0.018 D 2g . 2
Find: (a) Power supplied to the pump ( pump ( hp hp)). (b) Sketch the HGL and EGL. Properties: Water (68 (68 F), Table A.5: γ = = 62 62..4lbf / ft3. ◦
PLAN Apply the flow rate equation to find the veloc velocit ity y. The Then n cal calcula culate te hea head d los loss. s. Nex Nextt apply the energy equation from water surface to water surface to find the head the pump provides. provides. Fina Finally lly,, apply the power equa equation tion.. SOLUTION Flow rate equation Q A 3.0 ft3 / s = (π/ π/4) 4) × (8/ (8/12ft)2 = 8.594 594 ft/s ft/s
V =
Head loss hL =
L V 2 0.018 D 2g
µ
V 2 2g
¶µ ¶ µ ¶ +
2
2
3000 00 ft (8 (8..594ft 594ft// s) + (8 (8..594ft 594ft// s) = 0.018 30 8/12ft 2(32 2(32..2 ft/ ft/ s2 ) 2(32 2(32..2 ft/ ft/ s2)
= 94 94.. 04ft 116
Energy equation p1 V 12 p2 V 22 + α1 + z1 + h p = + α2 + z2 + hL γ γ 2g 2g 0 + 0 + 90 + h p = 0 + 0 + 140 + 94 94..04 h p = 144.0 ft Power equation P = Qγh p = 3.0 ft3 / s × 62 62..4lbf / ft3 × 14 1444 ft ft lb lbf f ft · lbf = 26 269957 s 550hp · s
µ
¶
P P = 4499.0 hp EGL
EGL HGL
HGL
117
7.81: PROBL PROBLEM EM DEFINITION Situation: Water flows between two reservoirs. D = 1 m, L = L = 300 300 m. H = = 16 16 m, h = h = 2 m. 2
hL =
L V 0.01 D 2g .
Find: (a) Discharge in pipe ( pipe ( m3/ s) s).. (b) Pressure halfway between two reservoirs ( reservoirs ( kPa) kPa).. PLAN To find the discharge, apply the energy equation from water surface A to water surface in in B. B. To find the pressure at location P, apply the energy equation from water surface A to location P location P .. SOLUTION Energy equation V 12 V 22 p1 p2 + α1 + z1 + h p = + α2 + z2 + hL γ γ 2g 2g 0 + 0 + H = 0 + 0 + 0 + 0.01 × 16 = 4V p2 /2 /2g V p =
µ ¶
300 V p2 + V p2 2g 1 2g
√
4 × 2 × 9 9..81 = 8. 8.86 86 m/s m/s
Flow rate equation Q = VA = 8.86 × (π/ (π/4) 4) × 12 Q = 6.96 m3 /s Energy equation between the water surface in A in A and point P point P ::
118
p p V p2 + − h + 0. 0.01 × 0 + 0 + H = γ 2g V p2 p p − 2 + 2. 16 = 2.5 2g γ
150m V p2 1m 2g
µ ¶
where V where V p2 /2 /2g = 4 m. Then p p =
9, 810N 810N// m3 (16 + 2 − 10)m
¢
¡
p p = 78 78..5 kPa 2
V /2g /2g=4 =4 m EGL HGL p/γ
119
7.82: PROBL PROBLEM EM DEFINITION Situation: Two reservoirs are connected by a pipe with an abrupt expansion. L V hL = 0.02 D , Q , Q = = 16 ft3 / s. 2g 2
Find: Elevation in left reservoir. Assumptions: α = 1.0. PLAN Apply the energy equation from the left reservoir to the right reservoir. SOLUTION Energy equation pR V R2 pL V L2 = + + zR + hL + + zL γ γ 2g 2g 200 V 12 0 + 0 + zL = 0 + 0 + 11 1100 + 0.02 1.128 2g 300 V 22 (V ( V 1 V 2 )2 V 22 −2g + 2g +0..02 1.596 2g + +0
µ ¶
µ ¶
Flow rate equation
Q A1 16ft3 / s = 16 16 ft/s ft/s = 1 ft2 V 2 = 8 ft/s V 1 =
Substituting into the energy equation 0.02 200 2 L z = 110 + 2 × 32 32..2 1.238 (16) + = 110 110 + 16. 16.58 + 0. 0.99 + 0. 0.99
µ
¶ ∙µ ¶
300 (8ft/ (8ft/ s)2 (16ft// s 8 ft/ (16ft ft/ s)2 2 1.596 (8) + ( 64..4 ft/ ft/ s2 64 64..4− ft/ ft/ s2 ) + 64
µ ¶ ¸
120
zL = 129 ft 129 ft 2
V 1 /2 g EGL HGL V1
V2
V22/2g
121
7.83: PROBL PROBLEM EM DEFINITION Situation: Water is pumped from a lower to an upper reservoir. L1 = 100 100 m, L 2 = 1000 1000 m. D1 = 1 m, D 2 = 50 cm cm.. η = 74%, 74%, Q = Q = 3 m3 / s. z1 = 150 150 m, z 2 = 250 250 m.
Find: (a) Pump power (MW) power (MW).. (b) Sketch the HGL and EGL. Assumptions: α = 1.0. PLAN Apply the energy equation from the upper reservoir surface to the lower reservoir surface. SOLUTION Energy equation p1 + V 12 + z1 + h p = p2 + V 22 + z2 + hL γ 2g γ 2g 0 + 0 + 150 + h p = 0 + 0 + 25 2500 + Flow rate equation
L V 2 V 2 0.018 + D 2g 2g
X µ¶
3 m3 / s Q = V 1 = = 3.82 82 m/s m/s A1 (π/ π/4) 4) × (1 m)2 V 12 = 0.744 744 m m 2g V 2 = Q/ Q/A A2 = 4V 1 = 15 15..28 28 m/s m/s V 22 = 11 11..9 m 2g
122
Substituting into the energy equation h p = 250 m − 150 150m m + 00..018 = 541.6 m
∙µ 100m ¶ 1m
0 0..744 744 m +
×
µ
1000m 5 cm
Power equation Qγh p η 3 m3 / s × 9 9,, 810N 810N// m3 × 541 541..6 m = 0.74 P = P = 21.5 MW
P =
11.9 m
h p=535 =535 m
0.794 m
EGL
HGL
¶
¸
11..9 m + 11. 11 11.9 m
×
123
7.84: PROBL PROBLEM EM DEFINITION Situation: Water flows out of a reservoir into a pipe that discharges to atmosphere. L V hL = 0.018 D , z , z 1 = 200 200 m. 2g z2 = 185 185 m, z pipe = 200 200 m. D = 30cm, 30cm, L = L = 200 200 m. Find: (a) Water discharge in pipe ( pipe ( m3 / s) . (b) Pressure at highest point in pipe ( pipe ( kPa) kPa).. 2
Assumptions: α = 1.0. PLAN First apply energy equation from reservoir water surface to end of pipe to find the V the V to calculate the flow rate. rate. Then to so solve lve for th thee pressure mi midwa dway y along pi pipe, pe, apply the energy equation to the midpoint: SOLUTION Energy equation p1 V 12 p2 V 22 + + z1 = + + z2 + hL γ γ 2g 2g V 2 V 2 0 + 0 + 200 = 0 + + 185 + 0. 0.02(200 02(200//0.30) 2g 2g V 2 = 15 14 14..33 2g V 2 = 1.047 2V g = 4.53 53 m/s m/s Flow rate equation Q = VA = 4.53 m/ s × (π/ (π/4) 4) × (0 (0..30m)2 Q = 0.320 m3 /s
124
Energy equation to the midpoint: p1 V 12 pm V m2 + + z1 = + + zm + hL γ 2g γ 2g pm V m2 100 V 2 + 0 + 0 + 20 2000 = + 200 + 0. 0.02 γ 2g 0.3 2g 2 V pm/γ = − (1 + 6. 6.667) 2g = (−1.047)(7 047)(7..667) = −8.027 027 m m pm = −8.027 027γ γ = −78 78,, 745 745 Pa Pa
µ ¶
pm =
−78 78..7 kPa
µ ¶
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