  # Fluid Mechanics Test 2

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• ###### Question 1
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When a dolphin glides through air, it experiences an external pressure of 0.75 m of mercury. The absolute pressure on dolphin when it is 5 m below the free surface of the water is
###### Solution
Concept:

Pabs = Pgauge + Patm

Calculation:

Given:

h = 5 m

Patm = 0.75 m of mercury = ρHg × g × 0.75 = 13600 × 9.81 × 0.75 = 100062 Pa

Pgauge = ρwater × g × h

Pgauge = 1000 × 9.81 × 5 = 49050 Pa

Pabs = Pgauge + Patm

Pabs = 49050 + 100062

Pabs = 149112 Pa = 149112 N/m2 = 0.149 N/mm20.15 N/mm2

• ###### Question 2
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Calculate the capillary rise or fall in a glass tube of 2.5 mm diameter when immersed vertically in mercury. Take surface tension σ = 0.52 N/m for mercury in constant with glass and angle of contact = 130°
###### Solution
Concept:

Expression for capillary rise or fall

$$h = \frac{{4\sigma \cos \theta }}{{\rho gd}}$$

Calculation:

Given: θ = 130°, σ = 0.52 N/m, d = 2.5 × 10-3m

$$h = \frac{{4\; \times \;0.52\; \times\; \cos 130^\circ }}{{13600\; \times \;9.81\; \times \;2.5\; \times \;{{10}^{ - 3}}}}$$

h = - 0.4 cm

• ###### Question 3
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Velocity profile over a curved surface is given as : $$\frac{u}{U} = 2{\left( {\frac{y}{\delta }} \right)^2} - {\left( {\frac{y}{\delta }} \right)^3}$$

The flow ______

###### Solution
Concept:

$${\left( {\frac{{\partial u}}{{\partial y}}} \right)_{y = 0}} \Rightarrow + {\rm{ve\;}} \Rightarrow {\rm{\;stick\;to\;surface}}$$

$${\left( {\frac{{\partial u}}{{\partial y}}} \right)_{y = 0}} \Rightarrow 0{\rm{\;}} \Rightarrow {\rm{\;on\;verge\;of\;separation}}$$

$${\left( {\frac{{\partial u}}{{\partial u}}} \right)_{y = 0}} \Rightarrow - {\rm{ve\;}} \Rightarrow flow~is~seperated$$

Calculation:

$$\frac{u}{U} = 2{\left( {\frac{y}{\delta }} \right)^2} - {\left( {\frac{y}{\delta }} \right)^3}$$

$$\frac{{\partial u}}{{\partial y}} = U\left[ {4\left( {\frac{4}{\delta }} \right) - 3{{\left( {\frac{y}{\delta }} \right)}^2}} \right]$$

$${\left( {\frac{{\partial u}}{{\partial y}}} \right)_{y = 0}} = 0$$

∴ Flow is on the verge of separation.

• ###### Question 4
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The value of friction factor is misjudged by 25% in using Darcy-Weisbach equation. The resulting error in the discharge will be?
###### Solution
Concept:

$${h_L} = \frac{{fL{Q^2}}}{{12{D^5}}}$$

$$Q\; \propto {\rm{\;}}\frac{1}{{\sqrt f }}\;\;\; - - - \left( 1 \right)$$

Calculation:

Here f is misjudged by 25 %, it means f can be 1.25f or 0.75f

Case: 1

Let it be 1.25f

$${Q_1} = \frac{k}{{\sqrt f }}$$

$${Q_2} = \frac{k}{{\sqrt {1.25f} }}$$

$$Error = \frac{{{Q_2} - {Q_1}}}{{{Q_1}}} \times 100 = - 11\;\%$$

Case: 2

Let it be 0.75f

$${Q_1} = \frac{k}{{\sqrt f }}$$

$${Q_2} = \frac{k}{{\sqrt {0.75f} }}$$

$$Error = \frac{{{Q_2} - {Q_1}}}{{{Q_1}}} \times 100$$

Error = 15.47%

• ###### Question 5
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A centrifugal pump lifts 0.0125 m3 / s of water from a well with a static lift of 30 m. If the brake power of the driving electric motor is 5 kW, what is the overall efficiency of the pump-set?
###### Solution
Explanation:

Pump is defined as a device which transfer the input mechanical energy of a motor or a of an engine into pressure energy or kinetic energy or both of a liquid.

Centrifugal pump is type of dynamic pressure pump.it also known as velocity pump.

$$Q = 0.0125\frac{{{m^3}}}{s}$$

Shaft power or rating of motor (PS) = 5 kW

$$density\;\left( \rho \right) = 1000\frac{{kg}}{{{m^3}}}$$

$$\eta = \;\frac{{Power\;output\;}}{{shaft\;power}}$$

$$\eta = \;\frac{{\rho \times g \times Q \times H}}{{{P_S}}}$$

$$\eta = \;\frac{{1000 \times 9.81 \times 0.0125 \times 30}}{{5000}}$$

η = 0.73575 ≈ 0.7360∴ η = 73.60 %

• ###### Question 6
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Which of the following conditions is true regarding the validity of Pascal’s Law?

I- It is valid for ideal fluid in motion.

II- It is valid when real fluids with rigid body motion.
###### Solution
Concept:

Pascal’s Law states that at any point in a fluid the intensity of pressure is equal in all the directions if and only if shear stresses acting fluid are zero.

Newton’s law of viscosity:

$$\tau = \frac{{\mu du}}{{dy}}$$

Where,

τ is the shear stress on the plate.

μ is dynamic viscosity of fluid.

Explanation:

Statement- I -­ It is valid for ideal fluid in motion.

Since fluid is in motion, u ≠ 0 but

For ideal fluid, μ = 0 ⇒ τ = 0

Therefore, Pascal’s law is valid for ideal fluids even in state of motion.

Statement II- It is valid when real fluids with rigid body motion

Rigid body motion means every layer of fluid has uniform velocity.

For real fluid in motion mean that μ ≠ 0 and u ≠ 0

But, du = 0 τ = 0∴ For Real fluids Pascal’s law is also valid with rigid body motion.

• ###### Question 7
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A head loss in 100 m length of a 0.1 m diameter pipe (f=0.02) carrying water is 10 m. The boundary shear stress in Pa is ____
###### Solution
Concept:

For any type of flow, the shear stress at wall,

$$\tau = - \left( {\frac{{dp}}{{dx}}} \right)\frac{R}{2}\;$$

$$\therefore \tau = \frac{{\rho g{h_L}}}{L} \times \frac{R}{2}$$

$$\tau = \frac{{1000 \times 9.81 \times 10}}{{100}} \times \frac{{0.1}}{4}$$

∴ τ = 24.525 Pa

• ###### Question 8
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Which of the following represents generalised continuity equation?
###### Solution
Concept:

In generalised equation, density (ρ) is not constant and the flow is three dimensional

$$\frac{{D\rho }}{{Dt}} + \;\rho \left( {\nabla .\vec V} \right) = 0$$

After expansion,

$$\frac{{\partial \rho }}{{\partial t}} + u\frac{{\partial \left( \rho \right)}}{{\partial x}} + \rho \frac{{\partial \left( u \right)}}{{\partial y}} + v\frac{{\partial \left( \rho \right)}}{{\partial y}} + \rho \frac{{\partial \left( v \right)}}{{\partial y}} + w\frac{{\partial \left( \rho \right)}}{{\partial z}} + \rho \frac{{\partial \left( w \right)}}{{\partial z}} = 0$$

$$\frac{{\partial \rho }}{{\partial t}} + \frac{{\partial \left( {\rho u} \right)}}{{\partial x}} + \frac{{\partial \left( {\rho v} \right)}}{{\partial y}} + \frac{{\partial \left( {\rho w} \right)}}{{\partial z}} = 0$$

$$\frac{{\partial \rho }}{{\partial t}} + \;\nabla .\left( {\rho \vec V} \right) = 0$$So, options a, b and c are true.

• ###### Question 9
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In an orifice meter, coefficient of discharge is 0.58 and coefficient of vena contracta is 0.61, what is coefficient of velocity?
###### Solution
Concept:

Cd = Cc × Cv

Where, Cd = Coefficient of discharge, Cc = Coefficient of vena contracta, Cv = Coefficient of velocity

Calculation:

Cd = Cc × Cv

0.58 = 0.61 × Cv

∴ Cv = 0.95

• ###### Question 10
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In a flow over a flat plate, laminar boundary layer exists where velocity distribution is

$$\frac{u}{U} = 0.5\frac{y}{\delta }$$

Ratio of momentum thickness to boundary layer thickness is____
###### Solution
Concept:

θ = momentum thickness

$${\rm{\theta \;}} = \mathop \smallint \limits_0^\delta \frac{u}{U}\left[ {1 - \frac{u}{U}} \right]dy$$

Calculation:

$$\theta = \mathop \smallint \limits_0^\delta \frac{u}{U}\left[ {1 - \frac{u}{U}} \right]dy$$

$$\theta = \;\mathop \smallint \limits_0^\delta \frac{y}{{2\delta }}\left[ {1 - \frac{y}{{2\delta }}} \right]dy$$

$$\theta = \;\mathop \smallint \limits_0^\delta \left[ {\frac{y}{{2\delta }} - \frac{{{y^2}}}{{4{\delta ^2}}}} \right]dy$$

$$\theta = \left[ {\frac{{{y^2}}}{{4\delta }} - \frac{{{y^3}}}{{12{\delta ^2}}}} \right]_0^\delta$$

$$\theta = \frac{\delta }{4} - \frac{\delta }{{12}}$$

$$\therefore \frac{\theta }{\delta } = \frac{1}{6}$$

• ###### Question 11
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Match the following:

 P Reynold’s model i. Cavitation Q Froude model ii. Capillary rise R Euler model iii. Pipe flow S Weber model iv. Flow over spillways
###### Solution
Explanation:

Reynold’s model law is applicable for:

(i) Pipe flow

(ii) Resistance experienced by sub-marines, airplanes, etc.

Froude model law is applicable for:

(i) free surface flows such as spillways, channels, etc.

(ii) Sea waves

Euler model law is applicable for:

Where pressure forces are dominant like flow taking place in closed pipe, cavitation phenomenon etc.

Weber model is applicable for:

(i) Capillary rise in narrow passage

(ii) Capillary movement of water in soil

• ###### Question 12
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A 1 : 6 scale model of a passenger car is tested in a wind tunnel. The prototype velocity is 60 km/hr. The model drag is 250 N. The air in the model and prototype can be assumed to have the same properties. Which of the following statements are true?
###### Solution
Concept:

Reynolds number will be same for both model and prototype

Calculation:

Given:

$$\frac{{{L_M}}}{{{L_P}}} = \frac{1}{6}$$

$${V_p} = 60 \times \frac{5}{{18}}\;\frac{m}{s}$$

FM = 250 N

$$Re = \;\frac{{\rho {V_M}{L_M}}}{\mu } = \frac{{\rho {V_P}{L_P}}}{\mu }\; - - - \left( 1 \right)$$

Now,

From (1)

$${V_M} = \frac{{{L_P}}}{{{L_M}}}{V_P} = 100\;\frac{m}{s}$$

Inertial force = ρL2V2

$$\frac{{{F_P}}}{{{F_M}}} = \frac{{{\rho _P}}}{{{\rho _M}}}{\left( {\frac{{{L_P}}}{{{L_M}}}} \right)^2}{\left( {\frac{{{V_P}}}{{{V_M}}}} \right)^2}\;\; - - - \left( 2 \right)$$

Density is same for model and prototype (Given)

From (2),

FP = 250 N

Power on the prototype

P = FPVP

P = 4.167 kW

• ###### Question 13
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A smooth flat plate of total length 2 m is in parallel flow stream of water. The water has uniform free stream velocity of 1 m/s parallel to plate. The critical Reynolds number for a flat plate is 500000. Assume density of water is 1000 kg/m3 and dynamic viscosity of water 1 centipoises. What is the maximum distance measured (in m, up to two decimal places) from leading edge of plate, up-to which the boundary layer formed to be laminar?
###### Solution
Concept:

The Reynolds number for flow over plate can be given as:

$${R_e} = \frac{{\rho UX}}{\mu }$$

Where

X is the distance from leading edge.

U is the free mean velocity.

μ is the dynamic viscosity of fluid flowing.

ρ is the density of fluid flowing.

If Reynolds number is less than or equal to critical Reynold’s number than boundary layer formed over flat plate is laminar i.e.Re ≤ ReCr­

Calculation:

Given:

U = 1 m/s ; ρ = 1000 kg/m3 ; μ = 1 centipoise = 10-3 N s/m2

Re cr = 5 × 105

Let ‘x’ is the distance from leading edge up to which Boundary layer formed is laminar,

Now,

Rex ≤ Re cr

$$\frac{{\rho Ux}}{\mu } \le 5 \times {10^5}$$

$$\frac{{{{10}^3} \times 1 \times x}}{{{{10}^{ - 3}}}} \le 5 \times {10^5}$$

x ≤ 0.5 m

∴ The maximum distance from leading edge of the plate up to which boundary layer is laminar is 0.50 m.

• ###### Question 14
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In a small pipe of 0.5mm diameter and length 2m, water flows at 48cm3/minute. Viscosity is 1.92 centipoise at low temperature flow condition. Pressure drop in MPa is­­­­­­______
###### Solution
Concept:

$$u = \left( {\frac{{ - \partial P}}{{\partial x}}} \right)\left[ {\frac{{{R^2} - {r^2}}}{{4\mu }}} \right]$$

Calculation:

Given: D = 0.5 mm ⇒ R = 0.25 mm, L = 2m, Q = 48 cm3/min = 8 × 10-7 m3/sec,

μ = 1.92 centipoise = 1.92 × 10-3 Ns/m2

Now,

Q = A × Vavg

$$8 \times {10^{ - 7}} = \frac{\pi }{4} \times {\left( {0.5 \times {{10}^{ - 3}}} \right)^2} \times {V_{avg}}$$

∴ Vavg = 4.074 m/s

$$\therefore {u_{max}} = \left( {\frac{{ - \partial P}}{{\partial x}}} \right)\left[ {\frac{{{R^2}}}{{4\mu }}} \right]$$

And umean = Vavg umax

$$\therefore {V_{avg}} = \left( {\frac{{ - \partial P}}{{\partial x}}} \right)\left[ {\frac{{{R^2}}}{{8\mu }}} \right]$$

$$\therefore {\rm{\Delta }}P = \frac{{{V_{avg}} \times L \times 8\mu }}{{{R^2}}}$$

∴ ΔP = 2 MPa

Points to Remember:

$${\rm{\Delta }}P = \frac{{32\;\mu \;{V_{avg}}\;L}}{{{D^2}}}$$∴ ΔP = 2 MPa

• ###### Question 15
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The velocity potential function in a two-dimensional flow field is given by Ø = x2 – y2. Which of the following statements are true?
###### Solution
Concept:

$$U = \; - \frac{{\partial \emptyset }}{{\partial x}} = \; - \frac{{\partial \varphi }}{{\partial y}}\;\; - - - \left( 1 \right)$$

$$V = \; - \frac{{\partial \emptyset }}{{\partial y}} = \;\frac{{\partial \varphi }}{{\partial x}}\;\;\; - - - \left( 2 \right)$$

$$Q = \;{\varphi _2} - {\varphi _1}\;\;\; - - - \left( 3 \right)$$

Q is flow per unit width

Now,

From (1),

$$U = \; - 2x = \; - \frac{{\partial \varphi }}{{\partial y}}\; \Rightarrow \;\varphi \left( {x,y} \right) = 2xy + f\left( x \right)\; - - - \;\left( 4 \right)$$

$$V = \;2y = \;\frac{{\partial \varphi }}{{\partial x}}\; \Rightarrow \;\varphi \left( {x,y} \right) = 2xy + f\left( y \right)\;\; - - - \;\left( 5 \right)$$

At (1, 2)

U = -2 m/s

V = 4 m/s

$$Velocity = \;\sqrt {{{\left( { - 2} \right)}^2} + {4^2}}$$

Velocity = 4.47 m/s

Now,

From (4) and (5)

φ(x, y) = 2xy

φ(1, 3) = 6

φ(3, 4) = 24

Q = φ(3, 4) - φ(1, 3) = 24 - 6

Q = 18 units

• ###### Question 16
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For a two dimensional, unsteady flow of an incompressible fluid, the velocity field is given by V(2x2 t + y) î - 4xyt ĵ  Then which of the following statements are true?

###### Solution
Concept: For a fluid flow to be possible, it should obey the continuity equation which is given by $$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0~$$ (Incompressible, 2D) The convective acceleration is given by $${{\alpha }_{xC}}=u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}~~;~~{{\alpha }_{yC}}=u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}$$ The local acceleration is given by $${{\alpha }_{xL}}=\frac{\partial u}{\partial t}~~;~~{{\alpha }_{yL}}=\frac{\partial v}{\partial t}$$ The acceleration in a direction is given by α = αC + αL   Calculation: Given V = (2x2t + y) î - 4xyt ĵ ⇒ u = 2 x2 t + y ; v = - 4 x y t ; $$\frac{\partial \left( 2{{x}^{2}}t+y \right)}{\partial x}+\frac{\partial \left( -4xyt \right)}{\partial y}=4xt-4xt=0~~$$  (Option 1 is wrong)  Convective acceleration in the x-direction is $${{\alpha }_{xC}}=\left( 2{{x}^{2}}t+y \right)\frac{\partial \left( 2{{x}^{2}}t+y \right)}{\partial x}+\left( -4xyt \right)\frac{\partial \left( 2{{x}^{2}}t+y \right)}{\partial y}$$ ⇒ αxc = [(2 x2 t + y) × 4 x t] + [(- 4 x y t) × 1] = 8 x3 t2 ∴ αxc at (1, 2) and t = 3 sec is =  8 × 13 × 9 = 72 m/s2 (Option 2) Now, Local acceleration in y-direction is $${{\alpha }_{yL}}=\frac{\partial \left( -4xyt \right)}{\partial t}=-4xy$$ ∴ αyL = -4 × 1 × 2 = - 8 m/s2 (Option 3 is wrongNow, Local acceleration in x-direction is $${{\alpha }_{x}}=\frac{\partial \left( 2{{x}^{2}}t+y \right)}{\partial t}=2{{x}^{2}}=2\times {{1}^{2}}=2$$ ∴ Acceleration in x-direction is αx = 72 + 2 = 74 m/s2 ; Now, Convective acceleration in y-direction is $${{\alpha }_{yC}}=\left( 2{{x}^{2}}t+y \right)\frac{\partial \left( -4xyt \right)}{\partial x}+\left( -4xyt \right)\frac{\partial \left( -4xyt \right)}{\partial y}$$ ⇒ αyc = [(2 x2 t + y) × - 4 y t] + [(- 4 x y t) × - 4 x t] = 16 x2 y t2 – 4 y2 t – 8 x2 y t2 ; ⇒ αyc = 8 x2 y t2 – 4 y2 t αyc at (1, 2) and t = 3 sec is = 96 m/s2 ; ∴ Acceleration in y-direction is αy = 96 – 8 = 88 m/s2; Total acceleration = (262 + 882)0.5 ∴ αtotal = (262 + 882)0.5 = 91.76 m/s2 (Option 4 wrong)
• ###### Question 17
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A Fluid flow is given by

$$V = \left( {2x + \frac{{{y^3}}}{3} - {x^2}y} \right)\hat i + \left( {x{y^2} - \frac{{{x^3}}}{3} - 2y} \right)\hat j$$

Check whether the given flow is possible case of fluid flow or not. If possible, determine angular velocity about z-axis.
###### Solution
Concept:

To check possibility of fluid flow, continuity equilibrium should be verified, i.e.

$$\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} = 0$$

$${\omega _z} = \frac{1}{2}\left( {\frac{{\partial v}}{{\partial x}} - \frac{{\partial u}}{{\partial y}}} \right)$$

Calculation:

$$u = 2x + \frac{{{y^3}}}{3} - {x^2}y$$

$$v = x{y^2} - 2y - \frac{{{x^3}}}{3}$$

$$\frac{{\partial u}}{{\partial x}} = 2 - 2xy\;\;\;;\;\frac{{\partial u}}{{\partial y}} = {y^2} - {x^2}$$

$$\frac{{\partial v}}{{\partial x}} = {y^2} - \frac{{3{x^2}}}{3} = {y^2} - {x^2}$$

$$\frac{{\partial v}}{{\partial y}} = 2xy - 2$$

For 2-D flow, continuity equation,

$$\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} = 0$$

(2 – 2xy) + (2xy - 2) = 0

Hence, it represents possible case of fluid flow.

Now,

$${\rm{Angular\;velocity\;}}\left( {{\omega _z}} \right) = \frac{1}{2}\left( {\frac{{\partial v}}{{\partial x}} - \frac{{\partial u}}{{\partial y}}} \right)$$

$${\omega _z} = \frac{1}{2}\left[ {\left( {{y^2} - {x^2}} \right) - \left( {{y^2} - {x^2}} \right)} \right] = 0$$∴ ωz = 0 • Correct -

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