Fluid Mechanics IOE Question Solution

Subject: Fluid Mechanics B.E. (Civil) Year/Part: II/I

Solution of Board exam question of TU, IOE (New Course) - Dr. K. N. Dulal

2068, Baishakh (Regular exam) 1. Define stream line with its drawing equation. A steady, three dimensional velocity field is given by V = (0.657+1.73x+0.984y+az)i+(2.61+cx+1.91y+bz)j+(-2.73x-3.66y-3.64z)k. Calculate constants a, b, c such that the flow field is irrotational. Solution: Stream line is an imaginary curve drawn through the flow field in such a way that the tangent to it at any point indicates the direction of velocity at that point. As stream lines join points of equal velocity, these are velocitycontours.It is useful to visualize the flow pattern. Equation of streamline is

u = (0.657+1.73x+0.984y+az), v= (2.61+cx+1.91y+bz) and w= (-2.73x-3.66y-3.64z) Rotational components: (Curl of V)/2

(

),

(

),

(

)

For irrotational flow , , ( ( (

)

(

)

(

)

(

) (

)) )

a = 2.73, b = 3.66, c = 0.984 2. Define the term pressure and skin friction drag with neat sketch. Describe briefly, with neat sketches, the changes in flow pattern and the drag coefficient with the variation of Reynold’s number when a thin circular cylinder of infinite length is placed in a fluid stream. Solution: Consider an arbitrary shaped solid body placed in a real fluid, which is flowing with a uniform velocity V in a horizontal direction. Consider a small elemental area dA on the surface of the body. The forces

acting on surface dA are: pressure force acting perpendicular surface, shear force acting tangential to the surface. Let θ be the angle made by the pressure force with horizontal direction.

FL PdA τ0dA

FR

PdAcosθ

PdA FD

V

θ

τ0dA

θ PdAsinθ

τ0dAsinθ θ τ0dAcosθ

stationary body

Drag force Drag force on elemental area = Total force in the direction of motion Total drag is ∫

∫

The first term is known as pressure drag and the second term is known as friction drag. Variation of flow around a cylinder with different Re

Re = 2 to 30

Re = 40 to 70

Re = 90

For RePA) Equating the pressure at XX

-20x1000+13.6x9810x2a = 0 a = 0.075 m 2a = (26.8/100)Sinθ 2x0.075 = 0.268Sinθ θ = 340

7. A tank of area A is provided with an orifice 40mm in diameter at its bottom. Water flows into the tank at a uniform rate and is discharged through the orifice. It is found that when the head of water over the orifice is 0.68m, the water surface rose at a rate of 0.0014m/Sec. But, when the head of water is 1.24m, the water surface rose at 0.00062m/sec. Find the rate of flow and the cross-sectional area of the tank. Take Cd = 0.62. Solution: Diameter of orifice (d) = 40mm = 0.04m = 0.001257 m2

Area of orifice (a) =

Coeff.of discharge (Cd) = 0.62 =0.00345 √ √ Inflow rate (Qi) =? Cross-sectional area of tank (A) =? First case: Head (h) = 0.68m, dh/dt = 0.0014m/s Second case: Head (h) = 1.24m, dh/dt = 0.00062m/s General equation for tank with inflow (Qi) and outflow (Qo) ( ) √

Substituting values for both cases √

Qi=0.0014A+0.002845

(a)

√

Qi=0.00062A+0.003842 Solving a and b A =1.278 m2 Qi = 0.00463 m3/s

(b)

OR A tank is in the form of frustum of a cone having top diameter of 2m, a bottom diameter of 0.8m and height 2m and is full of water. Find the time of emptying the tank through an orifice 100mm in diameter provided at the bottom. Take Cd = 0.625. 1m

2m

0.4m

H0

Solution: Cd = 0.625 Area of orifice (a) =

= 0.00785 m2

H1 = 2m, H2 = 0m, R1 = 1m, R0 = 0.4m From similar triangles,

H0 = 1.33m √

(

)

(

√

)

= 13.03

Time of emptying the tank is [ ( [ (

) )

( (

) )

( (

)] )] = 160 Sec

8. For the two orifices shown in the figure below, determine Y2 such that

2m

.

orifice1

10m orifice2

Y2 X2

X1

Solution: H1 = 2m, Y1 = 10-2 = 8m, H2 = 10-Y2 Y2= ? Coefficient of velocity for orifice 1 (

)

Coefficient of velocity for orifice 2 (

)

√ √

Since the two orifices are identical Cv1 = Cv2 √

√

(

) (

)

Solving for Y2 Y2 = 1, 9 As Y1 = 8m, Y2 = 9 (>Y1) is not feasible. Hence Y2 = 1m

9. The diameter of a pipe bend is 30cm at inlet and 15cm at outlet and the flow is turned through 1200 in a vertical plane. The axis at inlet is horizontal and the center of the outlet section is 1.5m below the center of the inlet section. Total volume of water in the bend is 0.9m 3. Neglecting friction, calculate the magnitude and direction of the force exerted on the bend by water flowing through it at 250lps and when the inlet pressure is 0.15N/mm2.

Y

W

Z1

P1 V1

X 120 R

1

0

Ry

Rx

F Z2 P2 V2

2

Solution: Diameter at section 1 (d1) = 30cm = 0.3m = 0.07068m2

Area at section 1 (A1) =

Diameter at section 2(d2) = 15cm = 0.15m = 0.01767m2

Area at section 2 (A2) =

Discharge (Q) = 250 lps = 0.25 m3/s Volume of water within control volume (Vol) = 0.9m3 Weight of water within control volume (W) = = 9810x0.9 = 8829N Velocity at section 1 (V1) =Q/A1 = 0.25/0.07068 = 3.54 m/s Velocity at section 2 (V2) =Q/A2 = 0.25/0.01767=14.15 m/s Z2 = 0, Z1 = 1.5m θ= 180-120 = 600 Pressure at 1 (P1) = 0.15N/mm2 = 0.15x106N/m2 Resultant force exerted by the water on the bend = ? Applying Bernoulli’s equation between 1 and 2

P2 = 70870 N/m2 ∑ ( ( ( ( = 13882 N

) ) )

( ( (

) ) ) )

(

)

∑ ( (

) )

= 4681N Resultant force ( )

√

=14650N

Resultant force exerted by the water on the bend = 14650N (to the right and downward) Direction of resultant force =

=18.60

=

10. The pipe flow in the figure is driven by the pump. What gauge pressure is needed to be supplied by the pump to provide water flow rate of Q = 60m3/h? Neglect head loss from A to B. Head loss from C to D=

; Head loss from D to E =

; dAB (diameter of pipe AB) = dCD = 5cm; dDE = 2cm. where VCD =

velocity in pipe CD and VDE = velocity in pipe DE. D

1

80m

10m B A

C

Pump

Solution: Discharge (Q) = 60m3/h = 60/3600 m3/s = 0.0167 m3/s Diameter of pipe AB and CD = 5cm = 0.05m C/s Area of pipe AB and CD (AAB, ACD) =

= 0.001963m2

Diameter of pipe DE =2cm = 0.02m C/s Area of pipe DE (ADE) =

= 0.000314m2

E

Q

Velocity of flow through AB and CD (VAB = VCD) =Q/AAB = 0.0167/0.001963 = 8.5m/s Velocity of flow through DE (VDE) = Q/ADE = 0.0167/0.000314 = 53.2m/s Head loss between A and B (hLAB) = 0 Head loss from C to D (hLCD) =

= 110.5m

Head loss from D to E (hLDE) =

= 73.65m

Total head loss (hL) = 0+110.5+73.65 = 184.15m Hp = Head supplied by the pump At point 1, V1 = 0, P1 = 0 (atm) At point E, PE = 0 (atm) Applying Bernoulli’s equation between 1 and E (Taking datum through A)

hp =398.4m Applying Bernoulli’s equation between 1 and C

Pc = 3970107 Pa = 3970 Kpa

11. Express the kinematic viscosity in stokes for a liquid with specific gravity 0.95 and dynamic viscosity 0.011 poise. A U-tube is made up of two capillarities of diameters 1.0mm and 1.5mm respectively. The U-tube is kept vertically and partially filled with water of surface tension 0.0075kg/m and zero contact angle. Calculate the difference in the level of the miniscii caused by capillarity.

h

d1 d2

Solution: Specific gravity = 0.95 Density of fluid (ρ) = 0.95x1000 = 950 kg/m3 Dynamic viscosity ( ) = 0.011 poise = 0.1x0.011 = 0.0011 NS/m2

Kinematic viscosity ( ) =? = 1.158x10-6 m2/s = 1.158x10-6x104 Stokes = 0.01158 Stokes Diameter of tube 1 (d1) = 1mm = 1x10-3 m Diameter of tube 2 (d2) = 1.5mm = 1.5x10-3 m Surface tension (σ) = 0.0075kg/m = 0.073575 N/m Contact angle (θ) = 0 Capillary rise in tube 1 (h1)

= 0.0316m

Capillary rise in tube 2 (h2)

= 0.021m

Difference in level (h) = 0.0316-0.021 = 0.0106m 12. The pressure drop in an air duct depends on the length and diameter of the duct, the mass density, viscosity of the fluid and the velocity of the flow. Obtain an expression for the pressure drop in dimensionless form using Buckingham –π theorem. Estimate the pressure drop in a 20m long air duct if a model of the duct operating with water produces a pressure drop of 10 KN/m2 over 10m length. The scale ratio is 1:50. Given, = 1000 kg/m3, = 1.2kg/m3 = 0.001NS/m2, = 0.0002NS/m2 Solution: ( Total number of variables = 6 No. of fundamental dimensions = 3 No. of π terms = 6-3= 3 ( ) (I) Choose D, V and ρ as repeating variables. First π term (II) Writing dimensions ( ) ( ) ( ) Equating the powers of M, L and T c1+1 =0 a1+b1-3c1-1 = 0 -b1- 2 = 0 c1 = -1, b1 = -2 a1-2+3-1=0

)

a1 = 0 Substituting the values of a1, b1 and c1 in II

Second π term (III) Writing dimensions ( ) ( ) ( ) Equating the powers of M, L and T c2 = 0 a2+b2-3c2+1 = 0 b2 = 0 a2 = -1 Substituting the values of a2, b2 and c2 in III

Third π term (IV) Writing dimensions ( ) ( ) ( ) Equating the powers of M, L and T c3+1 =0 a3+b3-3c3-1 =0 -b3-1 = 0 c3 = -1 b3 = -1 a3-1+3-1 = 0 a3 =-1 Substituting the values of a3, b3 and c3 in (IV)

Substituting the values of π1, π2 and π3 in (I) (

) (

)

Multiplying first πterm by 1/π2 and 1/π3 and expressing the product as a function of 1/π3

( (

)

)

Numerical Scale ratio (Lr) = 1/50 Pressure drop in model (Pm) = 10 KN/m2= 10x103 N/m2 Density of air (ρp) = 1.24 kg/m3 Density of water (ρm) = 1000 kg/m3 Viscosity of water (μm) = 0.001NS/m2 Viscosity of air (μp) = 0.0002NS/m2 Pressure drop in prototype (Pp) = ? As the problem involves both viscous and pressure force, we have to use both Reynolds and Euler model law. From Reynolds’ model law Re)model = Re) Prototype

Lm/Lp = 1/50

Vm = 0.31Vp From Euler model law Eu) model = Eu) prototype √

√

√ √

Pp = 129N/m2 Pressure drop in prototype corresponding to 10 KPa over 10m length in model = 129N/m2 Pressure drop for 20m length = 2x129 = 258N/m2 OR How are the repeating variables selected for dimensional analysis? Prove that the scale ratio for discharge for a distorted model is given as

( )

( )

where Qp = discharge through

prototype, Qm= discharge through model, (Lr)H = horizontal scale ratio, (Lr)V = Vertical scale ratio. Solution:

Repeating variables appear in most of the π groups.They have a large influence on the problem.There is great freedom in choosing these. Some rules which should be followed for selecting repeating variables are as follows:  There are n ( = 3) repeating variables.  In combination they must contain all of dimensions (M, L, T)  The repeating variables must not form a dimensionless group.  Dependent variable should not be selected as repeating variable.  No two repeating variables should have the same dimensions.  They should be measurable in an experiment.  They should be of major interest to the designer. Qp = discharge through prototype, Qm= discharge through model, (Lr)H = horizontal scale ratio, (Lr)V = Vertical scale ratio A = C/s Area of flow, V = Velocity, B = width, h = depth of flow (a) ( ) ( ) ( ) and

where

( )

From Froude’s model law √

√

( )

(c)

From a, b and c ( )

( )

(b)

2068, Shrawan (Back exam) 1. State and prove Newton’s law of viscosity. Prove that P is pressure and

, where K is Bulk Modulus of elasticity,

is density.

Solution: Newton’s law of viscosity states that the shear stress is proportional to the rate of deformation or velocity gradient.

Where = shear stress, du/dy = velocity gradient andthe constant of proportionality ( ) = coefficient of viscosity.

A

dx A’

D

D’

τ dφ

dy

τ B

C

Let us consider a fluid confined between two plates, where the bottom plate is stationary and the upper plate is moving. Let ABCD is the fluid at any time t. Due to the application of shear force τ, the fluid deforms to A’BCD’ at time t+dt. Let dy = distance between two layers, AA’ = dx and shear strain = dφ. For small angle, dx = dφ. dy Also dx = du. dt Equating (a) Shear stress is proportional to rate of shear strain. (b) From a and b

Bulk modulus is (c) Considering unit mass of substance, Differentiating w.r.t. ρ (

)

(d) From c and d,

2. An 8cm diameter piston compresses manometer oil into an inclined 7mm diameter tube, as shown in figure below. When a weight W is added to the top of the piston, the oil rises an additional distance of 10cm up the tube. How large is the weight, in N?

W 10cm

D = 8cm Piston

h

Initial level

X

X

dh Y

Y

Final level

Oil, sp gr = 0.827

D = 7mm

15

0

Solution: Diameter of piston (D) = 8cm = 0.08m Diameter of tube (d) = 7mm = 0.007m h = 0.10Sin15 = 0.0258m Due to compression, the fluid in the container moves down by dh and the fluid in the tube moves up by 10cm. Volume of fluid fallen = Volume of fluid risen dh = 0.000766m

Equating the pressure at new level (YY) (

)

(b)

(

)

W=1N

3. A cylinder, 2m in diameter and 3m long weighing 3KN rests on the floor of the tank. It has water to a depth of 0.6m on one side and liquid of sp gr 0.7 to a depth of 1.25m on the other side. Determine the magnitude and direction of the horizontal and vertical components of the force required to hold the cylinder in position.

C 0.6m

A

E O D

Water

Oil

1.25m

B

Solution: OA= OB = OC = 1m, BD = 0.6m OD = 1-0.6 = 0.4m ) = 0.9165m CD = ( (

)= 66.40

OE = 1.25-1 = 0.25m (

)= 75.50

=0 =-3 x y x y 1 1.0 1 2.0 2 2.0 2 2.6 3 3.0 3 3.5 4 4.0 4 4.4 5 5.0 5 5.3

Plotting of streamlines (si) and equipotential line (phi)

line)

6. Write down and explain the terms of Euler’s equation and Navier-Stoke’s equation of motion of fluid. Solution: Euler’s equation

= density of fluid = change in pressure with space = change in velocity with space = change in elevation with space Navier-Stoke’s equation in 1D First term= local acceleration, second term = convective acceleration, third term = pressure force per unit mass, fourth term= gravity force per unit mass, fifth term = shear force per unit mass

7. Derive an expression for the time required to empty a conical tank without inflow. Solution: R1 x

dh H1

h

H2

H0

O R0

Let A = cross-sectional area of conical tank, a = cross-sectional area of orifice, H1 = Initial level of liquid, H2 = Final level of liquid, R = radius of tank at H1, R0 = radius of tank at the bottom, T = time to fall from H1 to H2. Let at any instant of time, the liquid surface is at height h above the orifice and let dh is the decrease of liquid surface in an interval of time dt.

From similar triangles (

)

Time of emptying tank from H1 to H2 ∫

√

∫

√

(A also varies with h)

∫

√ √

√

[|

(

∫

[

(

)

)

| √

]

∫ (

|

where

)

|

(

[ (

|

| ]

)

)

(

)

(

)]

This is the expression to compute T. H0 is found from similar triangles.

8. The inlet pipe of a pump rises at an angle of 450 to the horizontal and is 10cm in diameter. For design reasons, it is undesirable that the pressure at the inlet to the pump should fall below 25 KN/m2. The inlet pipe is 6m long, the lower end being just below the surface of the water in the sump. Find the maximum discharge that the pump may deliver. 10cm dia 2

6m5 m

Inlet 0

45 sump

Solution:

1

Pump

Taking datum at 1 Z1 = 0m Z2 = 6sin45 = 4.24m V1 =0m/s P1 = Patm = 0 (in terms of gauge) Gauge pressure at the inlet is negative. The given pressure 25Kpa is absolute pressure. Pabs=Patm+Pgauge Pgauge = Pabs-Patm = 25-101.3 = -76.3 Kpa = P2 Applying Bernoulli’s equation at section 1 and 2

V2 = 8.33 m/s Q = A2V2 =

=0.065m3/s = 65 lps

9. When a jet of fluid strikes a moving vertical plate, show that the maximum efficiency of the system is 8/27. Solution:

u

V

Jet of water

Consider a jet moving with velocity V strikes a flat vertical plate, which is moving with velocity u. The velocity with which the jet strikes the plate will be the relative velocity V-u. Force exerted by the jet in X-direction (Fx) = mass of water striking/sec x (V1x – V2x) ] ( )[( ) ( ) Work done per sec by the jet on the plate = ( Efficiency ( ) = For maximum efficiency,

(

) )

(

)

(

(

)

)

(V-u)(v-3u) = 0 V=u (work done =0) V=3u (

)

( (

) )

10. Explain with neat sketches Boundary layer concept along a thin plate. Solution: Turbulent Boundary layer

U Laminar Boundary layer

Laminar zone

Laminar sublayer

Transition Turbulent zone zone

Consider the flow of fluid, having free stream velocity U, over a smooth thin plate which is flat and parallel to the direction of free stream of fluid. At the leading edge of the fluid, the thickness of the boundary layer is zero. On the downstream, for the fluid in contact with the boundary, the velocity of flow is zero and at some distance from the boundary the velocity is u. Hence a velocity gradient is set up which retards the motion of fluid due to the shear resistance. Near the leading edge, the fluid is retarded in thin layer. At subsequent point downstream the leading edge, the boundary layer region increases because the retarded fluid is further retarded. Laminar boundary layer As the boundary layer develops, up to some distance from the leading edge, the flow in the boundary layer is laminar irrespective of whether the incoming flow is laminar or turbulent. This is known as laminar boundary layer. When the boundary layer is thin, the velocity gradient is large. As the boundary layer thickens, velocity gradient reduces and shear stress decreases. Eventually it is too small to drag the slow fluid along. Up to this point the flow is laminar. Viscous force is dominant in this case. Transition zone

As the boundary layer grows to a certain thickness, instability occurs, leading to transition from laminar to turbulent boundary layer. This length is known as transition zone. Turbulent boundary layer After the transition zone, the boundary layer is turbulent and continues to grow in thickness. This layer is known as turbulent boundary layer. If the viscous forces were the only action, the fluid would come to a rest. Viscous shear stresses have held the fluid particles in a constant motion within layers. Eventually they become too small to hold the flow in layers and the fluid starts to rotate. The fluid motion rapidly becomes turbulent. Momentum transfer occurs between fast moving main flow and slow moving near wall flow. Thus the fluid by the wall is kept in motion. The net effect is an increase in momentum in the boundary layer.

11. A jet plane having a wing area of 30 m2 and weighing 28 KN flies at 900 km/hr. The engine develops 8000 KW and has a mechanical efficiency of 70%. Determine the lift and drag coefficient for the wind taking density of air as 1.2 kg/m3. Solution: Weight of plane (W) = 28KN Wing area (A) = 30m2 Speed (V) = 900 km/hr =

= 250m/s

Power = 8000KW Efficiency = 70% Power used to overcome drag resistance (P) = 0.7x8000 =5600KW Density of air (ρ) = 1.2 kg/m3 Coefficient of drag (CD) = ? Coefficient of lift (CL) = ? P = FD x V 5600x1000 = FD x 250 FD = 22400 N

CD = 0.019 Lift force (FL) = Weight of plane = 28000N

CL = 0.025

12. Show that the resistance F to the motion of a sphere of diameter D moving with a uniform velocity V through a real fluid of density and viscosity is given by (

)

Solution: ( ) Total number of variables = 5 No. of fundamental dimensions = 3 No. of π terms = 5-3= 2 ( ) (I) Choose V, D and ρ as repeating variables. First π term (II) Writing dimensions ( ) ( ) ( ) Equating the powers of M, L and T c1+1 =0> c1 = -1 a1+b1-3c1+1 = 0 -a1-2 = 0> a1 = -2 -2+b1+3+1=0 b1 = -2 Substituting the values of a1, b1 and c1 in II

Second π term (III) Writing dimensions ( ) ( ) ( ) Equating the powers of M, L and T c2+1 =0> c2 = -1 a2+b2-3c2-1 = 0 -a2-1 = 0> a2 = -1 -1+b2+3-1 =0 b2 =-1 Substituting the values of a2, b2 and c2 in III

Substituting values of π1 andπ2 in I (

) (

) (

)

2070 Ashadh (Back exam) 1. What is continuum concept in Fluid Mechanics? Explain the cavitation phenomena. Solution: In Fluid Mechanics, a fluid is considered as a continuous substance. This concept is called continuum concept. In this concept, molecular structure of the fluid is not considered and the separation between molecules is neglected. The fluid properties such as velocity and pressure are a continuous function of space and time. The fluid properties can be considered to be constant at any point in space, which is average of large number of molecules surrounding that point within a characteristic distance. Using continuum concept, the mathematical equations relating the physical laws can be derived easily as we don’t need to consider the motion of individual molecule. This concept is not valid if the mean free path of molecules is greater than the characteristic dimension of fluid considered for analysis. The ratio of mean free path length to the characteristic length is known as Knudesn number (Kn). The continuum hypothesis is valid for Kn