Fluid Mechanics IOE Question Solution

8/12/2019 Fluid Mechanics IOE Question Solution

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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 suchthat 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 thincircular 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.

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 Re<0.2, the inertia force is negligible and the flow pattern is symmetrical. With the increase of Re,the

flow pattern becomes unsymmetrical with respect to the axis perpendicular to the direction of flow. At

Re from 2 to 30, very weak vortices are formed on the downstream of the cylinder. It is the initial stage

for the development of the wake. At Re from 40 to 70, the wake as well as a pair of vortices become

quite distinct. With further increase in the value of Re, the vortices become more and more elongated in

τ0dA

PdAcosθ

FD

PdAsinθ

θ

θ

τ0dAsinθ

τ0dAcosθ

τ0dA

θ

stationary body

FR

V

PdA

FL

PdA



the direction of flow. At Re = 90, these vortices become cylindrical, they leave the cylinder and slowly

move in the downstream direction.

Coefficient of drag (CD) of cylinder for different Reynold no. (Re)

a. For Re<1,

b. For Re between 1-2000: CD decreases and reaches a minimum value of 0.95 at Re = 2000.

c. For Re = 2000-30000: CD increases and attains a maximum value of 1.2 at Re = 30000.

d. For Re = 30000-300000: CD decreases. At Re = 300000, CD = 0.3

e. For Re>300000: CD increases and it becomes equal to 0.7 in the end.

3. Derive the expression for center of pressure on a plane inclined surface immersed in liquid. Also show

pressure diagram for such surface.

Consider a plane surface of area A totally immersed in a liquid of specific weight and inclined an angle

θ to the free surface. Consider an element of area dA at a vertical distance y from the free surface.

P = pressure on each element, = vertical distance of CG from free surface, yp = vertical distance of CP

from the free surface, F = resultant force, z = distance of element from O-O, = distance of CG from O-O

and zp = distance of CP from O-O.

Finding resultant force

Force on element (dF) =

Summing the forces,

Assuming to be constant

CG

zp

z

O

θ

F

CP

y

yp

O



Here, ∑ = first moment of area about an axis through O-O =

Hence,

Finding the position center of pressure

Taking moment of force on element (dM) about O

dM= dF x z = Total moment of all forces (M) = ∑ ∑ ∑

∑ = second moment of area about an axis through O-O = moment of inertia = I0

(a)

Moment of resultant force F about O = (b)

Equating a and b

From parallel axis theorem

where IG = M.I. about an axis through CG. Substituting for I0

Pressure diagram for inclined surface



OR

The swing check valve in figure covers a 22.86cm diameter circular opening in a slanted wall. The hinge

is 15cm from the center line. The valve will open when the hinge moment is 50Nm. Find the value of h

for the water to cause this condition.

Solution:

Hinge moment = 50 Nm

Area (A) = = 0.041m2

Location of CG () = h

Resultant force on the valve is

9810x.041h = 402.21h

Vertical distance of Center of pressure from the free surface

ycp

Position of CP from CG (BO) =

Location of F from A (AB) =

Taking moment about A,

FxAB = 50

h = 0.82m

4. Consider a homogeneous right circular cylinder of length L, radius R, and specific gravity SG, floating in

water (SG = 1) with its axis vertical. Show that the body is stable is .

ycp

AO

B

F

22.86cmdia

30cmdia

600

Hinge

15cm

h

Water

Air



Solution:

Length of cylinder = L

Radius of cylinder = R

Specific gravity of cylinder = SG

h = depth of immersion

Weight of cylinder = Weight of water displaced

h = SGxL

Position of center of buoyancy (OB) = SGxL/2

Position of CG (OG) = L/2

BG = OG-OB =

GM = MB-BG =

For stable equilibrium, GM 0

OR

A U-tube shown in figure is filled with a liquid of specific gravity 1.25 to a height of 15cm in both the

limbs. It is rotated about a vertical axis 15cm from one limb and 30cm from the other. If the speed of

rotation is 60rpm, find the difference in the liquid levels in the two limbs. Also find the pressure at

points M and N at the base of U-tube.

Z0

0

M

G

B

N M

S

30cm 15cm

60rpm

Z1

Z2



Solution:

Distance from the axis of rotation to left side (r1) = 0.3m

Distance from the axis of rotation to right side (r2) = 0.15m

Speed of rotation (N) = 60 rpm

Angular velocity

= 6.28 rad/s

= 0.181m

= 0.045m

Difference in level = 0.181-0.045 = 0.136m

Z1+ Z2 = 0.3 (the total height should be same until no spilling)

Z1 - Z2 = 0.136

Solving

Z1 = 0.218m, Z1 = 0.082m

Pressure at N = = 1.25x9810x0.218 = 2673 PaPressure at M = = 1.25x9810x0.082 = 1005.5 Pa

5. Write precisely how the boundary layer theory helped in simplifying the complex motion of fluid. Why

does the boundary layer increase with distance from the upstream edge? A simplistic laminar boundary

layer model is: √

for

( √ )

(√ )for

Does this expression satisfy boundary conditions applicable to the laminar boundary layer velocityprofile?

Solution:

Boundary layer is a thin layer of the flow in the vicinity of the boundary within which the velocity varies

from zero at the solid boundary to the free stream velocity in the direction normal to the boundary. In

the boundary layer velocity gradient (du/dy) is large and the shear stress acts. Outside the boundary,

velocity is constant and velocity gradient is zero and hence shear stress is zero. Hence, there are two

regions of flows: one is the boundary layer zone close to the boundary where the effect of viscosity is

mostly confined and another region outside theboundary layer zone where the flow is inviscid. Thus, the

boundary layer theoryhelped in simplifying the complex motion of fluid by considering the region wherethe velocity gradient exists.

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.

Boundary conditions for laminar flow: (a) At y = 0, u = 0, (b) At , (c) At and in

between y = 0 to , velocity gradient exists. The general equation of velocity profile for laminarboundary layer is u = ay+by

2+cy

3+dy

4.

First equation

√

for

√

For this equation,

At y = 0, u = 0

At

,

√ . Velocity gradient is constant i.e. the velocity profile is linear.

Hence the equation satisfies the boundary conditions for laminar boundary layer velocity profile.

Second equation ( √ )

(√ )for

( √ )

At , √ .Velocity gradient is constant i.e. the velocity profile is linear.

At and

Hence the equation satisfies the boundary conditions for laminar boundary layer velocity profile.

6. Two water tanks are connected to each other through a mercury manometer with inclined tubes, as

shown in the figure below. The pressure difference between two tanks is 20Kpa. Calculate “a” and “θ”.

Solution:

26.8cm

θ

XX

a

2aa

A

Tank A

B Tank B

Mercury

SG = 13.6



PB-PA = 20 Kpa (As Hg is deflected towards A, PB>PA)

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

Area of orifice (a) = = 0.001257 m2

Coeff.of discharge (Cd) = 0.62

√ =0.00345

Inflow rate (Q i) =?

Cross-sectional area of tank (A) =?

First case: Head (h) = 0.68m, dh/dt = 0.0014m/sSecond case: Head (h) = 1.24m, dh/dt = 0.00062m/s

General equation for tank with inflow (Q i) and outflow (Q o)

√

Substituting values for both cases

√

Q i=0.0014A+0.002845 (a)

√

Q i=0.00062A+0.003842 (b)

Solving a and b

A =1.278 m2

Q i = 0.00463 m3/s



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.

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

0.4m

2m

1m

H0



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

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 0.9m3. 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.

orifice2

orifice12m

X1 X2

Y2

10m



Solution:

Diameter at section 1 (d1) = 30cm = 0.3m

Area at section 1 (A1) = = 0.07068m2

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

Area at section 2 (A2) = = 0.01767m

2

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

R

F

Ry

Rx

P2

V2

Z1 P1

V1

Y

W

1

2

1200

X

Z2



∑

( )

= 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.

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) = = 0.000314m

2

1

D E

B C

10m

80m

A

Pump

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 contactangle. Calculate the difference in the level of the miniscii caused by capillarity.

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

d2 d1

h



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

mSurface 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 ifa model of the duct operating with water produces a pressure drop of 10 KN/m

2 over 10m length. The

scale ratio is 1:50. Given,

= 1000 kg/m3, = 1.2kg/m

3

= 0.001NS/m2, = 0.0002NS/m

2

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 Tc1+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 = -1Substituting the values of a2, b2 and c2 in III

Third π term

(IV)

Writing dimensions


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= 10x10

3 N/m

2

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 Q p = discharge through

prototype, Q m= 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.

Q p = discharge through prototype, Q m= 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)

(b)

where and

From Froude’s model law

(c)

From a, b and c



2068, Shrawan (Back exam)

1. State and prove Newton’s law of viscosity. Prove that , where K is Bulk Modulus of elasticity,

P is pressure and

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.

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

dx

dφ

τ

CB

A’ D D’ A

τ

dy



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 infigure 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?

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

h

YY

X

X

150

Initial level

W

Piston

D = 8cm

D = 7mm

10cm

Oil, sp gr

= 0.827

Final level



Equating the pressure at new level (YY)

(b)

W = 1 N

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.

Solution:

OA= OB = OC = 1m, BD = 0.6m

OD = 1-0.6 = 0.4m

CD = = 0.9165m

= 66.40

OE = 1.25-1 = 0.25m

= 75.50

<AOB = 180-75.5 = 104.50

Weight of cylinder = 3KN = 3000N

Net horizontal force

= 0.7x9810x1.25x3x1.25/2-9810x0.6x3x0.6/2 = 10797N (left)

Net vertical force (FV)= Weight of volume of water vertically above AB + Weight of volume of watervertically above BC = FvAB (up) + FvBC (up)

=

=

= 32917N (up)

OilWater

C

A

B

0.6m

E

O

D 1.25m



The components to hold the cylinder in place are 10797 N to the right and 32917-3000 = 29917N down

.

4. An open cylinder tank 0.5m in diameter and 1m height is completely filled with water and rotated

about its axis at 240 rpm. Determine the radius up to which the bottom will be exposed and the volume

of water spilled out of the tank.

Solution:

Radius (r) = 0.25m

N = 240rpm

Angular velocity

= 25.13 rad/s

= 2.01m

Z1 = 2.01-1= 1.01m

r1 = ?

Volume of water spilled = ?

r1 = 0.18m

Volume of water spilled = Volume of paraboloid (APB-CPD)

= 0.146 m3

r1

P

r =0.25m

AB

C D

1m

z

z1



OR

Develop an expression for the position of the Metacenter relative to the center of buoyancy.

Solution:

Consider a vessel with water line AC, B as the CB and G as the CGin original position. When the vessel is

tilted through a small angle θ, the CB changes from B to B’, the position of water line changes to ED and

two wedges AOE and COD are formed. M is the metacenter, W is the weight of vessel and FB is the

buoyant force.

Consider an element of area dA at a distance x from the CG of plan of vessel on either side.

Volume of element (dV) = dA.xθ

(xθ = height for small angle)

Buoyant force on the element of left side (dFB) = Weight of element

=

where is the sp wt of water.

Similarly,

Buoyant force on the element of right side (dFB) =

Buoyant force on the two elements produces a couple.

Moment of couple = =

Total moment due to altered displacement (M) = ∫

= ∫

∫ = I = Second moment of plan of the vessel about an axis through CG (Moment of inertia)

(a)

Moment due to the movement of CB from B to B’ (M’) = FB BB’ = WBB’

(b)

Where V is the volume of water displaced.

For equilibrium, M = M’

θ

FB =W

G

B B’

O

C

D

A

E

M

xx



GM = BM-BG

5. Derive the continuity equation for cylindrical coordinates.

Consider any point P(r, θ, Z) in space. Let dr, dθ and dZ be small increments in r, θ and Z direction

respectively. Let Vr, Vθ and Vz be the velocity in r, θ and Z direction respectively, and ρ be the density of

fluid.

Mass of fluid entering face PQNM per unit time =

Mass of fluid leaving face RSTO per unit time =

Gain of mass in z-direction =

(a)

Similarly,

Gain of mass in r-direction (b)

Gain of mass in θ-direction (c)

Net gain of mass

(d)

Rate of increase of mass per unit time (e)

Equating d and e

Vθ S

T

Q

M

NP

O

R

dθ

dZ

θ

r

dr

Y

X

VZ

Vr



This is the continuity equation in 3D cylindrical polar co-ordinate.

For steady flow,

. The continuity equation is

For incompressible flow, . The continuity equation is

6. Derive Bernoulli’s equation from Euler’s equation of motion. A venturimeter with a throat diameter of

100mm is fitted in a vertical pipeline of 200mm diameter with oil of sp.gr. 0.88 flowing upwards at a rate

of 0.06m/s. The venturimeter coefficient is 0.96. Two pressure gauges calibrated in KN/m2 are fitted at

tapping points, one at the throat and the other at the inlet pipe 320mm below the throat. The difference

between two pressure gauge readings is 28 KN/m2. Working from Bernoulli’s equation, determine (I) the

volume rate of oil through the pipe, (II) the difference in level in the two limbs of mercury manometer if

it is connected to the tapping points.

Solution:

Euler’s equation in 1D is

Integrating ∫ ∫ ∫

Dividing by g

or,

This is the Bernoulli’s equation.

Diameter at inlet (d1) = 20cm = 0.2m

C/s area of inlet (A1) = = 0.0314m

2

Diameter at throat (d2) = 10cm = 0.1m

C/s area of throat (A2) =

= 0.00785m2

Sp.gr. of oil (S0) =0.88

Density of oil (ρ) =0.88x1000 = 880 kg/m8

Difference in elevation (Z2-Z1) = 320cm =0.32m

Difference in pressure (P1-P2)= 28KN/m2 = 28000N/m2

Sp.gr. of mercury (S) =13.6

Cd =0.961

2Throat

Inlet



Discharge of oil (Q) =?

Manometer reading (x) =?

(I) Applying Bernoulli’s equation between inlet (1) and throat (2)

(a)

According to continuity equation

A1 V1=A2V2

=

=4V1 (b)

Solving a and b

V1 = 1.95m/s

Discharge (Q) =A1V1 = 0.0314x1.95 = 0.612 m3/s

Actual discharge = Cd Q = 0.96x0.612 = 0.5875 m3/s

II. =

=2.92

x = 0.2m

7. Derive an expression for estimating discharge for partially submerged orifice.

Solution:

In partially sub-merged orifice, the outlet side of the orifice is partially submerged under liquid. This type

of orifice has two portions: free discharging orifice at the upper part and submerged orifice at the lower

part.

HH2 H1



Let H1 = height of liquid above the upper edge of orifice, H2 =height of liquid above the lower edge of

orifice, H= Difference in liquid level between two tanks, b = width of orifice, and Cd = coefficient of

discharge

Discharge through orifice (Q) = Discharge through free portion (Q1) + Discharge through submerged

portion (Q2)

Free portion is treated as large rectangular orifice. Hence

For submerged portion,

Q = Q1 + Q2

8. A 450 reducing bend is connected in a pipe line carrying water. The diameter at inlet andoutlet of the

bend is 400mm and 200mm respectively. Find the force exerted by water on the bend if the intensity of

pressure at inlet of the bend is 215.8KN/m2. The rate of flow of water is 0.5m

3/s. The loss of head in the

bend is 1.25m of oil of sp.gr. 0.85.

Solution:

Solution:

Diameter at section 1 (d1) = 400mm = 0.6m

Area at section 1 (A1) = = 0.1256m

2

Diameter at section 2(d2) = 200mm = 0.2m


Discharge (Q) = 0.5 m3/s

P2

2

1

450

Fy

P1 Fx +X

Reactions

+Y



Velocity at section 1 (V1) =Q/A1 = 3.98 m/s


Pressure at section 1 (P1) = 215.8KN/m2 = 215800N/m2

Loss of head = 1.25m of oil of sp gr 0.85, P= 0.85x9810x1.25 N/m2

Loss of head in terms of water (hL) =

=1.0625m

Angle of bend = 450

Resultant force (FR) =?

Direction of resultant force =?

Applying Bernoulli’s equation between 1 and 2 (Z1 = Z2)

P2 = 86574N/m

2

∑

= 21544N

∑

( )

= 7551N




=19.30

9. Define boundary layer with appropriate sketches. How laminar sub-layer is developed in boundary

layer. Differentiate between hydraulically smooth and rough boundary.

Solution:

When a real fluid flows past a solid boundary, the fluid particle on the surface will have the same

velocity as that of the surface because of viscosity. This is called no-slip condition. If the boundary is

U



stationary, the velocity of the fluid at the boundary is zero. Further away from the boundary, the

velocity gradually increases.

The term boundary layer is defined as the thin layer of the flow on the boundary within which the

velocity varies from zero at the solid boundary to the free stream velocity in the direction normal to the

boundary.

Boundary layer along a thin flat plate

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. As the fluid moves along the plate, the thickness goes on increasing.The zones

formed during boundary layer development are: Laminar boundary layer, Transition zone, turbulent

boundary layer. If the plate is very smooth, even in the region of turbulent boundary layer, there is a

very thin layer just adjacent to the boundary, in which the flow is laminar. This layer is known as laminar

sub-layer.

Let k is the average height of irregularities projecting from the surface of a boundary and = thickness

of laminar sub layer. If k is much less than , the boundary is called hydrodynamically smooth boundary.

In this case eddies cannot reach the surface irregularities. If is much less than k, the boundary is called

hydrodynamically rough boundary. In this case eddies come in contact with the surface irregularities and

lot of energy will be lost.

From Nikuradse’s experiment:

stream

velocity (U)

Boundary layer

Solid body

Turbulent

Boundary layer

Transition

zone

Turbulent zoneLaminar zone

Laminar

Boundary layer Laminar sub-

layer

U



<0.25: Smooth boundary

>6: rough boundary

0.25<<6: transition

10. Define aerofoil with accepted terminology with neat sketch. A kite, which may be assumed to be a

flat plate and mass 1kg, soars at an angle to the horizontal. The tension in the string holding the kite is

60N when the wind velocity is 50 km/h horizontally and the angle of string to the horizontal direction is

350. The density of air is 1.2 kg/m

3. Calculate the drag coefficient for the kite in the given position if the

lift coefficient in the same position is 0.45. Both coefficients have been based on the full area of the kite.

Solution:

An airfoil is a streamlined body which may be either symmetrical or unsymmetrical. Some of the

terminologies used to characterize airfoil are as follows.

Chord line: It is the line joining the front and rear edge.

Angle of attack: It is the angle between the direction of flowing fluid and chord line.

Camber: It is the curvature of an airfoil.

Span: The overall length of airfoil is called its span (L).

Aspect ratio: The ratio of span (L) to mean chord (C) is called aspect ratio.

Stall: An airfoil is said to be in stall condition when the angle of attack of an airfoil is greater than the

angle of attack at maximum lift.

Solution:

Lift

String

α

chord line

L

350

W

Kite

T

Drag



Mass of kite = 1kg

Weight of kite (W) = 1x9.81 N = 9.81N

Area of kite = A

Velocity (V) = 50 km/hr =

=13.88/s

Density of air (

) = 1.2kg/m3

Tension (T) = 60N

Coefficient of lift (CL) = 0.45

Coefficient of drag (CD) = ?

Forces in X-dir (Drag)

FD = Tcos35 = 60cos35 =49.14N

Force in Y-dir (Lift)

FL = Tsin35+9.81 = 60sin35+9.81=44.22N

A = 0.85 m2

CD = 0.5

11. Determine the relationship between model and prototype kinematic viscosity if both Reynold’s

number and Froude number are to be satisfied.

Solution:

From Reynolds model law.

(Re) model = (Re) prototype

(a)

From Froude model law

(Fr) model = (Fr) prototype

(b)

From a and b



12. Show by dimensional analysis that the power P required to operate a test tunnel is given by

where is density of fluid, is viscosity, V is fluid mean velocity, P is the power required and L is the

characteristics tunnel length.

Solution:



No. of π terms = 5-3= 2

(I)

Choose ρ, L and V as repeating variables.

First π term

(II)

Writing dimensions


a1+1 =0

-3a1+b1+c1+2 = 0

-c1-3 = 0

Solving

a1 = -1, c1 = -3

-3x-1+b1-3+2 = 0

b1 = -2


Second π term

(III)

Writing dimensions


a2+1 = 0

-3a2+b2+c2-1=0-c2-1 = 0

Solving

a2 = -1, c2 = -1

-3x-1+b2-1-1 = 0

b2 = -1

Substituting the values of a2, b2 and c2 in III



Substituting the values of π1 and π2 in I



2068, Chaitra (Regular exam)

1. Define kinematic and dynamic viscosity. A small thin plane surface is pulled through the liquid filled

space between two large horizontal planes in the parallel direction. Show that the force required will be

minimum if the plate is located midway between the planes.

Solution:

According to Newton’s law of viscosity, 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. The constant is also called dynamic viscosity or absolute viscosity.Kinematic viscosity is defined as the ratio of dynamic viscosity to density.

Where = Kinematic viscosity and ρ = density

= shear stress in the upper part

= shear stress in the upper part

F = pull required to drag the plate

For F to be minimum, dF/dy = 0

Thus, the plate is located midway between the planes for the force to be minimum.

yV

h



2. Define atmospheric, gauge and pressure. A U-tube shown in figure is 10mm in diameter and contains

mercury. If 12ml of water is poured into the right hand leg, what are the ultimate heights in the two

legs?

Solution:

The pressure exerted by the atmospheric air on the surface with which it is in contact is known as

atmospheric pressure. If the pressure is quoted by taking atmospheric pressure as a datum, it is called

gauge pressure. The pressure measured above a perfect vacuum is called absolute pressure.

Solution:

With the addition of 12ml of water,

Drop in mercury level in right leg = y

Rise in mercury level in left leg = y

Depth of water column = h

h= Vol of water added/Cross sectional area = = 0.153m = 153mm

PE = PF

13.6x9810x2y = 9810x0.153

y = 0.0056m = 5.6mm

Height of free mercury level in the left leg = 120+5.6 = 125.6mm

Height of free water level in the right leg = 120-5.6+153 = 267.4mm

3. Prove that in case of force vortex, rise of liquid level at the end is equal to the fall of liquid level at the

axis of rotation. The hemispherical dome shown in figure is filled with water and is attached to the floor

by two diametrically opposite bolts. What force in either bolt is required to hold the dome down, if the

dome weighs 25KN?

Solution:

Water

y

y

FE

120mm

h

Mercury



R = Radius of cylinder

MN = Water level at absolute equilibrium (original water level)

After rotation, AOB is the profile of the liquid surface.

hr = Rise of liquid at end

hf = Fall of liquid at the end

Volume of liquid before rotation = ( )

Volume of liquid rotation = Volume of cylinder ABQP- Volume of paraboloid AOB

( ) ( )

Volume of liquid before rotation = Volume of liquid after rotation

( ) ( ) ( )

Hence, rise of liquid level at the end = fall of liquid level at the axis of rotation

Water

4cm

1.5m

5m

h

P

hr

hf N

R

A B

C D

MO

Q

Bolt



Vertical force on the dome due to water (Fv) = Weight of volume of water (imaginary) vertically above

dome

= [ ]

= 381325N = 381.325KN (up)

Weight of dome = 25 KN (down)

Net Vertical force = 318.325-25 = 356.325 (up)

Force on either bolt = 356.325/2 = 178.163KN (up)

4. Show that if ‘B’ is the center of buoyancy and ‘M’ is the metacenter of a partially immersed floating

body, then BM=I/V, where I is the second moment of the area of the surface of floatation about the

longitudinal axis and V is the immersed volume.

Solution:

Consider a vessel with water line AC, B as the CB and G as the CGin original position. When the vessel is

tilted through a small angle θ, the CB changes from B to B’, the position of water line changes to ED and

two wedges AOE and COD are formed. M is the metacenter, W is the weight of vessel and FB is the

buoyant force.

Consider an element of area dA at a distance x from the CG of plan of vessel on either side.

Volume of element (dV) = dA.xθ

(xθ = height for small angle)

Buoyant force on the element of left side (dFB) = Weight of element

=

where is the sp wt of water.

Similarly,

Buoyant force on the element of right side (dFB) =

Buoyant force on the two elements produces a couple.

dFB

dFB

xx

θ

FB =W

G

B B’

O

C

D

A

E

M



Moment of couple = =

Total moment due to altered displacement (M) = ∫

= ∫

∫ = I = Second moment of plan of the vessel about an axis through CG (Moment of inertia)

(a)

Moment due to the movement of CB from B to B’ (M’) = FB BB’ = WBB’

(b)

Where V is the volume of water displaced.

For equilibrium, M = M’

5. Explain Lagrangian and Eulerian methods of describing fluid flow. [2]

Solution:Lagrangian method deals with the motion of single fluid particle. This approach is more complex and the

equations of motion are difficult to solve. Eulerian method deals with the motion of fluid particles at a

point in space. This method is simple and widely used as the resulting equations can be solved easily.

6. A flow is described by the stream function √ . Locate the point at which the velocity vector

has a magnitude of 6 unit and makes an angle 1450 with the X-axis.

Solution:

√

√

√

Resultant velocity √ = 6

(a)

√ √

y = 0.7x (b)

Solving a and bx = 1.16, y = 0.81

7. Prove that the Cippoletti weir is a trapezoidal weir having side slopes 14.20 with vertical. Water flows

over a rectangular sharp crested weir 2m long (divided into two bays of 1m each by pier). The head over

the sill is 0.75m. The weir is fitted at the end of the long rectangular channel 3m wide and flow depth



1.5m. Starting from the first principles, determine the rate of discharge over the weir. Consider the

velocity of approach and the effect of end contraction. Coefficient of discharge for the weir is 0.7.

Solution:

Discharge through Cippoletti weir is

For a rectangular weir with two end contractions, the discharge is given by

From the above expression, it is clear that discharge of a rectangular weir is reduced by

due to end contractions, which is compensated by discharge through triangular notch.

Discharge through triangular notch = loss of dischargedue to end contraction

Length (L) = 2m

1 bay = 2 end contractions

n = 2x2 = 4

Head over sill (H) = 0.75m

C/s of channel (A) = 3mx1m = 3m2

Coefficient of discharge (CD) = 0.7

Discharge over weir (Q) = ?

Discharge over a rectangular weir without velocity of approach and with end contraction is

= 2.28 m3/s

Velocity of approach (Va) = Q/A = 2.28/3 = 0.76 m/s

θ/2 θ/2

L

H



Velocity head

= 0.03m

Discharge by considering velocity of approach

( )

[ ]=2.38 m3/s

Next trial

Velocity of approach (Va) = Q/A = 2.38/3 = 0.79 m/s

Velocity head

= 0.032m

Discharge by considering velocity of approach

( )

[ ] = 2.39 m3/s

As the difference in Q is negligible, the iteration is stopped.Q=2.39 m

3/s

8. A pipe bend placed in a horizontal plane tapers from 50cm diameter at inlet to 25cm diameter at

outlet. An oil of density 850kg/m3 enters the reducing bend horizontal and gets turned by 450 clockwise

direction. The discharge is 0.45 m3/s and pressure at inlet of 40 KN/m2 drops to 23 KN/m2 at the outlet

due to frictional effect. Find out the magnitude and direction of resultant force on the bend.

Solution:

Diameter at section 1 (d1) = 0.5m


Diameter at section 2(d2) = 0.25m




+Y

2

1

450

Fy

Fx

+XReactions

P1

P2



Velocity at section 2 (V2) =Q/A2 = 9.16m/s

Pressure at section 1 (P1) = 40KN/m2 = 40000N/m

2

Pressure at section 2 (P2) =23KN/m2 =23000N/m2

Angle of bend = 450

Resultant force (FR) =?

Direction of resultant force =?

∑

= 5458N

∑

( )

= -3275N


Resultant force exerted by the water on the bend = 6365N (to the right and upward)


=310

9. What is hydrodynamic boundary layer? Workout the shape factor

parameter for the velocity

distribution given by

.

Solution:

When a real fluid flows past a solid boundary, the fluid particle on the surface will have the same

velocity as that of the surface because of viscosity. This is called no-slip condition. If the boundary is

stationary, the velocity of the fluid at the boundary is zero. Further away from the boundary, the

velocity gradually increases. U

stream

velocity (U)

Boundary layer

Solid body



The term boundary layer is defined as the thin layer of the flow on the boundary within which the

velocity varies from zero at the solid boundary to the free stream velocity in the direction normal to the

boundary. In the boundary layer velocity gradient (du/dy) is large and the shear stress exerted by the

fluid is given by Newton’s law of viscosity as

. Outside the boundary, velocity is constant and

velocity gradient is zero and hence shear stress is zero.

Displacement thickness ()

∫

∫

Momentum thickness ()

∫

∫

∫

10. A 2mm diameter spherical ball of unit weight 117.5 KN/m3 is dropped in a mass of fluid of viscosity

15 poise and sp.gr. 0.95. Find out drag force, pressure drag, skin drag and terminal velocity of ball.

Solution:

Diameter of ball (D) = 2mm = 0.002m

Sp.wt. of fluid () = 0.95x9810 N/m3

Viscosity of fluid () = 15 poise = 15/10 = 1.5 PaS

Sp.wt. of ball () = 117.5 kN/m3

Drag force (FD) = ?



Pressure drag and friction drag = ?

Terminal velocity (V) = ?

Weight of ball (W) = Drag force (FD) + Buoyant force on ball (FB)

FD = W – FB

= 0.000453N

Pressure drag =

= 0.000151 N

Friction drag =

= 0.000302N

From Stoke’s law,

V = 0.016 m/s

Checking the Reynold’s no.

= 0.02

As Re<0.2, above expression for FD is valid.

11. Discuss types of model studies. If gravity, viscosity and surface tension are equally important in a

model, show that for dynamic similarity the relationship between viscosity ratio , surface tension ratio

and model scale ratio Lr is given by

.

Solution:

There are two types of models.

a. Undistorted model: An undistorted model is that model which is geometrically similar to its

prototype. Result from such model can be directly applied to the prototype.

b. Distorted model: A distorted model is that model in which one or more terms of the model are not

identical with their counter-parts in the prototype. A distorted model may have either geometrical

distortion or material distortion or distortion of hydraulic quantities or a combination of these. Results

obtained from the distorted models cannot be directly applied.

(Re) model = (Re) prototype

(a)

(Fr) model = (Fr) prototype



(b)

(Wb) model = (Wb) prototype

(c)

Equating b and c, and squaring

(d)

Substituting Vr from b and from d into a

12. Derive an expression representing Euler’s equation of motion.

Solution:

Consider a streamline AB in which flow is taking place in S direction. Consider a cylindrical element of

length ds and area dA. Let P be the pressure on left face and on the right face. Forces acting:

pressure force and gravity forceResultant force in S- direction = m as where as = acceleration in s direction

B

A

dzθ

ds

PdA

dW



For steady flow

Dividing by ρ

As all are function of S,

Or,



2069 Ashadh (Back exam)

1. Give two examples of non-Newtonian fluid. If the velocity distribution of fluid over a plate is given by

where u is velocity in m/s at a distance y from the plate, determine the shear stress at

y=0.2m. Take dynamic viscosity as 8.45x10-4

NS/m2.

Solution:

Examples of non-Newtonian fluid: Blood, slurry

Dynamic viscosity () = 8.45x10-4

NS/m2

Shear stress () at y=0.2m = ?

At y = 0.2m, shear stress is

= 0 N/m2

2. Find the pressure difference between the container A and B as shown in the figure.

Solution:

Pressure at E = pressure at F

450

FE

550mm

A

B

300mm

230mm

230mm

Water

Liquid of

sp.gr = 0.8

Mercury

130mm



= -90901 N/m2 = -90.9 Kpa

3. When a 22N weight is placed at the end of the uniform floating wooden beam as shown in figure

below, the beam tilts at an angle θ with its upper right corner at the surface, as shown. Determine (a)

the angle θ and (b) the specific gravity of wood. Assume that the point of application of 22N weight cuts

at point B.

Water

Solution:

h = 2.75tanθ

Weight of wood

Buoyant force (FB)= Weight of water displaced =

=

=

E

θ

A

C

D

B

22N

10cmx10cm

2.75m

θ

C

D

B

W= 22N10cmx10cmh

FB

Wwood

A



Wwood + W = FB

(a)

Taking moment about B

∑ = 0

(b)

Solving a and b

N/m3

Sp. Gr. Of wood= 6656/9810 = 0.7

4. Find the resultant force on ABCDEF of 10m width and its location due to the liquid as shown in figure.

Liquid of sp. gr. = 1.2

Solution:

Equivalent head of liquid due to -117.72 Kpa pressure =

= -10m

3m

A

B

C

D

E

G

F

r= 1.6m

4m

2m

2m

2m

5m

2m2m

-117.72 Kpa



This negative pressure will reduce the fluid surface by 10m. Make calculation by taking imaginary fluid

surface (IFL) as free surface.

Liquid of sp. gr. = 1.2

IFL

Net Horizontal force on curved surface (FH) = = 1.2x9810x(8.2x10)x4.1 = 3957746N = 3957.7KN

= 5.47m from C’

Vertical force on D’D

= 706320N (down)

Vertical force on EF = 473140N (up)

Net vertical force (FV) = 473140-706320 = -233180N = -233.18KN

Location of FV

Taking moment about C’

x = 2.65m

FV

yp

FH

C’

D’

10m

3m

A

B

C

D

E

G

F

r= 1.6m

4m

2m

2m

2m

5m

2m2m

-117.72 Kpa

x



Resultant force (FR) = = 3964.6KN

Resultant force acts at a distance of 5.47m down IFL and 2.65m left from C’D.

5. Consider the following steady, incompressible two dimensional velocity field:

m/s. Generate an analytical expression for the flowstreamline and draw several streamlines in the upper right quadrant from x = 0 to 3 and y = 0 to 3.

Determine if there are any stagnation points in this flow field and if so, where are they.

Solution:

u = 4.35+0.656x, v = -1.22-0.656y

Equation of streamline

Integrating

∫ ∫

(let -K1x0.656 = K2)

(Replacing eK2 by K3)

-5.307-2.85y+0.8x+0.43xy=K3

0.8(x+0.54xy-3.6y)=K3+5.307

Replacing (K3+5.307)/8 by c

X+0.54xy-3.6y=c. This is the equation of streamline.

Plotting of streamlines

C=0 C=1 C=2

x y x y x y

0 0

1 0.3 1 0.0

2 0.8 2 0.4 2 0.0

3 1.5 3 1.0 3 0.5



Streamlines

At stagnation point u =0 and v =0

4.35+0.656x = 0

x = -6.6

-1.22-0.656y = 0

y = -1.9

Stagnation occurs at (-6.6, -1.9).

OR

A flow has a potential function given by: . Derive the corresponding stream

function and show that some of the stream lines are straight lines passing through the origin of

coordinates. Find the inclination of these lines.

Solution:

(a)

(b)

Integrating eq. (a) w.r.t. x (c)

Where c= f(y)

Differentiating eq. (c) w.r.t. y

(d)

Equating eq. (b) and (d)



Integrating w.r.t. y

Substituting the value of c in eq.(c)

For

√

For , this equation (in the form of y=ax) shows that some of the stream lines are straight lines

passing through the origin of coordinates.

Inclination of the straight lines

√

6. Describe a cippoletti weir. How does it differ from a rectangular sharp crested weir.

A cippoletti weir is a trapezoidal weir having side slopes of 4 vertical to 1 horizontal.

Discharge through Cippoletti weir is

For a rectangular weir with two end contractions, the discharge is given by

From the above expression, it is clear that discharge of a rectangular weir is reduced by

due to end contractions, which is compensated by discharge through triangular notch.

Discharge through triangular notch = loss of dischargedue to end contraction

θ/2 θ/2

L

H



7. Find the velocity V1 of the water flowing in a vertical pipe as shown in figure below. The loss of head

is two times the velocity head of small pipe. The pipe diameter at section 1 is 15cm and that at section 2

is 6cm.

Solution:

Applying Bernoulli’s equation between 1 and 2 (Taking datum through 2)

(a)

From continuity equation,

A1V1= A2V2

(b)

Writing manometric equation

( = dynamic head due to stagnation, total head = static +dynamic)

(c)

0.5m

5m

V1

2

1

Hg



From a, b and c

0.02m/s

8. Water flows through a 900 reducer bend. The pressure at the inlet is 206 KN/m

2 (gauge) where the

cross-sectional area is 0.01m2. At the exit section, the area is 0.0025m2 and the velocity is 15m/s.

Assume that the bend does not discharge into the atmosphere but 0.0025m2 are conduit continues.

Further assume that the head loss in the reducer bend is, where u is the average velocity at the inlet

and exit of the reducer. Determine the force exerted by the flow on the reducer bend.

Solution:

Area at section 1 (A1) = 0.01m2

Area at section 2 (A2) = 0.0025m2

Velocity at section 2 (V2) =15 m/s

Discharge (Q) =A2V2= 0.0025x15 = 0.0375 m3/s


Pressure at section 1 (P1) = 206KN/m2 = 206000N/m

2

u = (3.75+15)/2 = 9.375m/s

Head loss (hL) =

= 4.5m

Applying Bernoulli’s equation between 1 and 2 (Z1 = Z2)

= 56386 N/m2

P2

Fy

+Y

P1

1

2

+X

Fx

Reactions



= 2200 N

∑

( )

= 703 N

Resultant force

=2310N



= 17.70

9. Air at 200c flows parallel to a smooth thin flat plate at 4.75m/s. The plate is 3.23m long. Determine

whether the boundary layer on the plate is most likely laminar, turbulent or somewhere in between

(transitional). Compute the boundary layer thickness at the end of the plate for two cases: (a) The

boundary layer is laminar everywhere, and (b) The boundary layer is turbulent everywhere.

Solution:

Velocity of air (V) = 4.75 m/s

Length of plate (L) = 3.23m

Assume kinematic viscosity of air () = 1.5x10-5

m2/s

Reynold no. (Re) =

= 10.2x105

As Re>5x105, the boundary layer is turbulent.

a. For laminar boundary layer, the upper value of Reynold no. is

Re = 3x105

Thickness of laminar boundary layer

√ = 0.029m = 29mm

b. Turbulent boundary layer

Re = 10.2x105

Thickness of turbulent boundary layer

= 0.076m = 76mm



10. Draw a neat sketch of aerofoil and define the terms chord line, camber, angle of attack and aspect

ratio associated with aerofoil. A wing of a small aircraft is rectangular in plan having a span of 10m and a

chord of 1,2m. In straight and level flight of 240km/hr, the total aerodynamic force acting on the wing is

20KN. If the lift/drag ratio is 10, calculate the coefficient of lift and the total weight the aircraft can

carry. Assume air density be 1.2 kg/m3.

Solution:

Chord line: It is the line joining the front and rear edge.

Angle of attack: It is the angle between the direction of flowing fluid and chord line.

Camber: It is the curvature of an airfoil.

Aspect ratio: The ratio of span (L) to mean chord (C) is called aspect ratio.

Speed of aircraft (V) = 240km/hr = (240x1000)/3600 = 66.67m/s

Plan area (A) = 10x1.2 = 12m2

Total aerodynamic force (FR) = 20KN

FD = 1.99KN

FL = 10x1.99 = 19.9KN

Weight the aircraft can carry = FL = 19.9KN

= 0.62

α

chord line

L



11. Using Buckingham’s π theorem, show that the velocity through a circular orifice is given by

where H = head causing flow, D= diameter of orifice, = specific mass, =

coefficient of viscosity, g = acceleration due to gravity.

Solution:

Total no. of variables (n) = 6

No. of fundamental dimensions (m) = 3

No. of π terms = 6-3 =3

(a)

Take , g, H as repeating variables.

First π term


a1 = 0

-3a1+b1+c1+1 = 0

-2b1-1 = 0

b1 = -1/2

-3x0+(-1/2)+c1+1 = 0

c1 = -1/2

Substituting the value of a1, b1 and c1

Second π term


a2+1 = 0

a2 = -1

-3a2+b2+c2-1 = 0

-2b2-1 = 0

b2 = -1/2

-3x-1+(-1/2)+c2-1 = 0

c2 = 3/2


Third π term




a3 = 0

-3a3+b3+c3+1 = 0

-2b3 = 0

b3 = 0

-3x-0+0+c3+1 = 0

c3 = -1


Substituting the values of π1, π2, π3 in eq. a

12. Write down Euler’s equation, Navier-Stoke’s equation and Bernoulli’s equation with appropriateexpression of each component.

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

Bernoulli’s equation

= pressure head

= velocity head

Z = elevation head



2069 Chaitra (regular exam)

1. A solid cone of maximum radius R and vertex angle is to rotate at an angular velocity . An oil of

viscosity and thickness t fills the gap between the cone and the housing. Derive an expression for the

torque required and the rate of heat dissipation in the bearing.

Solution:

Maximum radius of cone = R

Thickness of oil = t

Angular velocity = () =

Viscosity of oil =

Consider an elementary area dA at radius r of the cone.

Torque on the element (dT) = Shear force x r

Total torque (T) = ∫

Rate of heat dissipation = power utilized in overcoming resistance =

Alternative expression in terms of H

θ

ds

r

H

Oil

2θ

dr

R

thickness = t



2. The pipe and connection B as shown in figure below are full of oil of specific gravity 0.9 under

pressure. If the U-tube contains mercury, find the elevation of point A in meters.

Solution:

sp gr of oil = 0.9

Let y = elevation difference between A and B

Starting from right side and ending to A

y=10.65mElevation of A = 100+2.1+10.65 = 112.75m

Alternative method

Starting from right side and finding PB

= 394087 N/m2

Applying Bernoulli’s equation at section A and B (V = 0: no motion)

= 112.75m

3x10 N/m

y

1m

B

Oil, SG = 0.9

El. 100m

A

1.1m

4.12m

Hg

Zero of gauge scale



3. The tank on figure has a 4cm diameter plug which will pop out if the hydrostatic force on it reaches

25N. What will be the reading h on the manometer when this happens?

Solution:

Hydrostatic force

= 2.028m = 2.043m

Writing Manometric equation at XX

13.6x9810h=9810(2.043+0.02)h = 0.15m

4. Prove that the rise of water equals the fall in case of forced vortex motion in cylindrical tank at side

and center of tank.

Solution:

X X

h

2cm

F =25N

H

h

P

hr

hf N

R

A B

C D

M

O

Q

50

Water

Plug

D=4cm

Hg



R = Radius of cylinder

MN = Water level at absolute equilibrium (original water level)

After rotation, AOB is the profile of the liquid surface.

hr = Rise of liquid at end

hf = Fall of liquid at the end

Volume of liquid before rotation = ( )

Volume of liquid rotation = Volume of cylinder ABQP- Volume of paraboloid AOB

( ) ( )

Volume of liquid before rotation = Volume of liquid after rotation

( ) ( ) ( )

Hence, rise of liquid level at the end = fall of liquid level at the axis of rotation

5. The stream function for the 2-D flow of a liquid is given by . In the range of values of x and y

between 0 and 5 plot the stream lines and equipotential lines passing through co-ordinates (1,1), (1,2),

(2,2). Also determine the velocity in magnitude and direction at the point (1,2).

Solution:

(a)

(b)

Integrating eq. (a) w.r.t. x

(c)

Differentiating eq. (c) w.r.t. y

(d)

Equating eq. (b) and (d)

Integrating w.r.t. y

Hence

Velocity at (1,2)

u = -2x = -2x1 =-2

v = 2y =2x2 =4



Velocity = √ = 4.47 unit

Direction

= -63.40

Plotting of streamlines (constant line) and equipotential line (constant line)

For point (1,1), (1,2), (2,2), is 2, 4, 8 >

=2 =4 =8

x y x y x y

1 1.0 1 2.0 1 4.0

2 0.5 2 1.0 2 2.0

3 0.3 3 0.7 3 1.3

4 0.3 4 0.5 4 1.0

5 0.2 5 0.4 5 0.8

For point (1,1), (1,2), (2,2), is 0, -3, 0

> =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)



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:

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 H 1, 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.

H0

OR0

x

R1

dh

h

H2

H1



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.

Solution:

Inlet Pump

2

1

450

sump

10cm dia

6m5



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:

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,

uV

Jet of water



()

(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:

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

Turbulent

Boundary layer

Transition

zone

Turbulent zoneLaminar zone

Laminar

Boundary layer Laminar sub-layer

U





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:




(I)

Choose V, D and ρ as repeating variables.

First π term (II)

Writing dimensions


c1+1 =0> c1 = -1

a1+b1-3c1+1 = 0

-a1-2 = 0> a1 = -2

-2+b1+3+1=0

b1 = -2


Second π term

(III)

Writing dimensions


c2+1 =0> c2 = -1

a2+b2-3c2-1 = 0

-a2-1 = 0> a2 = -1

-1+b2+3-1 =0

b2 =-1




Substituting values of π1 andπ2 in I





Vertical height for fluid pressure in right limb = (10/16)x8 = 5cm

Vertical height for water pressure in right limb = (10/16)x8 = 5cm

Starting from left

Ppipe = 36218 pa = 36.218 Kpa

3. Gate AB in figure below is a homogeneous mass of 180kg, 1.2m wide into the paper, resting on

smooth bottom B. For what depth h will the force at point B be zero?

Solution:

Wt. of gate AB (W) = 180x9.81 = 1765.8N

Width of gate = 1.2m

AC = 1-1sin60 = 1.13m

AD =h-1sin60 =h-0.87

Hydrostatic force

= 12360x(1.2X1)x(1.13+0.87/2) = 23212 N

= 9810x(1.2x1)x(h-0.87+0.87/2) = 11772(h-0.435) N

Center of pressure

= 1.605m

ycp ycp(gly)

D

C

F

Glycerin

Sp wt= 12360 N/m3

A

h

Fgly

A

B

G

B

60

Patm

Water

W

2m1m

Fwater

Zero



AG = (1.605-1.13)/sin60 = 0.55m

AF = [ycp-(h-0.87)]/sin60 =

=

Taking moment about A

h =2.4m

4. A very tall 10cm diameter vase contains 1178 cm3 of water. When spun steadily to achieve rigid body

rotation, a 4cm diameter dry spot appears at the bottom of the vase. What is the rotation rate, in

rev/min, for this condition?

Solution:

Radius of vase (r1) = 0.05m

Radius of dry spot at bottom (r2) = 0.02m

Volume of water = 1178 cm3

= 1178x10-6

m3

N = ?

h = z1-z2 =

r2 = 0.02m

P

r1 =0.05m

AB

C D

h

Z1

Z2



Volume of water left=Volume of cylinder- Volume of ABDC

=

Volume of water left=1178x10-6 m3

Equating

= 57.2 rad/s

= 546rpm

5. The x and y components of fluid velocity in a 2-D flow field are u = x and v = -y respectively. Determine

the stream function and plot the streamline for .

Solution:

u = x, v = -y

(a)

(b)

Integrating a with respect to x

(c)

c is a constant which is independent of x, but may be a function of y.

Differentiating c with respect to y

(d)

From b and d

c = 0

Substituting c in eq. c

Si = 1 si =2 si = 3

x y x y x y

1 -1 1 -2 1 -3

2 -0.5 2 -1 2 -1.5

3 -0.3 3 -0.67 3 -1

4 -0.25 4 -0.5 4 -0.75

5 -0.2 5 -0.4 5 -0.6



Plot of stream lines

6. From Euler’s equation, derive the Bernoulli’s equation.

Solution:

Integrating

∫ ∫ ∫

Dividing by g

or,





For completely emptying the tank, H2 = 0

In this condition H1 =R

8. A turbine is set 43m below water level of a reservoir and is fed by a 60cm diameter pipe as shown in

the figure below. A short pipe of 45cm diameter discharges the water at a rate of 0.9m3/s from the

turbine to the atmosphere. If the total loss of head is 3m of oil of sp. gr. 0.9 and the turbine efficiency is

85%, find the power output of the turbine. [6]

Solution:

Diameter of bigger pipe (d1) = 60cm

Diameter of smaller pipe (d2) = 45cm = 0.45m


Velocity at point 1 (V1) = 0m/s

Velocity at point 2 (V2) =Q/A2 = 0.3/( )= 5.66 m/s

Pressure at 1 (P1) = 0 pa (atm)

Pressure at 2 (P2) = 0 Pa

Datum head at 1 (Z2) = 43m

Datum head at 2 (Z1) = 0m

Efficiency () = 0.85Head loss = 3m of oil of sg 0.9

Head loss in m of water (hL) = 3x0.9 = 2.7m

Energy output of machine = ?

ht = Head extracted by turbine

Applying Bernoulli’s equation at 1 (water surface) and 2 (exit point)

Turbine 45cm dia.

60cm dia.

43m



ht = 38.m

Energy output =

= 0.85x(9.81x0.9x38.7) = 290KW

9. A flat plate is struck normally by a jet of water 50mm diameter with a velocity of 18 m/s. Calculate (a)

the force on the plate when it is stationary (b) the force on the plate when it moves in the same

direction as the jet with a velocity of 6 m/s

Solution:

Diameter of jet (d) = 50mm =0.05m

Area of jet (A) = = 0.001963m2

Velocity of jet (V) = 18 m/s

a. Force exerted by the jet on the plate

= 636N

b. Velocity of plate (u) = 6m/s

Force exerted by the jet on the plate when the plate is moving

= 283N

10. When a jet of fluid strikes series of flat vanes, show that the maximum efficiency of the system is

1/2.

Solution:

V = absolute velocity of jet, u = absolute velocity of plate

Force exerted by the jet in X-direction (Fx) = mass of water striking/sec x (V1x – V2x)

(Since entire fluid mass issuing from the jet strikes the plate, mass of water striking/sec = )

Work done per sec by the jet on the plate = Fxu =

V



Efficiency () =

For maximum efficiency,

V=2u

11. The laminar boundary layer profile in a case is approximated by a cubic parabola as,

, where u is the velocity at a distance y from the surface and .

Calculate the displacement thickness and momentum thickness and workout the shear stress at the

surface.

Solution:

Displacement thickness () = ?

Momentum thickness () = ?

Shear stress at surface () = ?

∫

∫

∫

∫

∫

∫

Shear stress =

At y =

Shear stress =



12. A kite has an effective area of 0.6 m2 and mass 0.4 kg. It experiences a drag of 15 N in a wind speed

of 40 km/hr. Determine (a) the tension in the chord if it makes an angle 450 with the horizontal (b) lift

coefficient for the kite. Consider the density of air 1.2kg/m3.

Solution:

Solution:

Weight of kite (W) = 0.4x9.81=3.924N

Area of kite (A) = 0.6 m2

Velocity (V) = 40 km/hr = =11.11m/sDrag (FD) = 15N

Density of air () = 1.2kg/m3

Tension (T) = ?

Coefficient of lift (CL) = ?

Forces in X-dir

FD = Tcos45

15=Tcos45

T = 21.2N

Force in Y-dirFL = Tsin45+W = 21.2sin45+3.924=18.88N

CL = 0.425

13. The force F on a circular cylinder depends on the free stream velocity V, the diameter of the cylinder

D, density of the fluid

, viscosity of the fluid

and time t. By using Buckingham’s π theorem, show that

. Take , V and D as repeating variables.

Solution:



Lift

String

450

W

Kite

T

Drag




(I)

, V and D are repeating variables.

First π term

(II)


a1+1 =0

-3a1+b1+c1+1 = 0

-b1- 2 = 0

a1 = -1, b1 = -2

3-2+c1+1=0

c1 = -2


Second π term

(III)


a2+1 =0-3a2+b2+c2-1 = 0

-b2- 1 = 0

a2 = -1, b2 = -1

3-1+c2-1=0

c1 = -1


Third π term

(IV)


a3 =0

-3a3+b3+c3 = 0



-b3+ 1 = 0

b3 =1

0+1+c3=0

c3 = -1

Substituting the values of a3, b3 and c3 in IV

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Fluid Mechanics IOE Question Solution