Fluid Mechanics Test 2

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Fluid Mechanics Test 2
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  • Question 1
    2 / -0.33
    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
    2 / -0.33
    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

    Note: - ve sign indicates capillary depression (or full)

  • Question 3
    2 / -0.33

    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
    2 / -0.33
    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
    2 / -0.33
    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

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

    Head (H) = 30 m

    \(\eta = \;\frac{{Power\;output\;}}{{shaft\;power}}\)

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

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

    η = 0.73575 ≈ 0.7360∴ η = 73.60 %

  • Question 6
    2 / -0.33

    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.

    du/dy is velocity gradient.

    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
    2 / -0.33
    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
    2 / -0.33
    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
    2 / -0.33
    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
    2 / -0.33

    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
    2 / -0.33

    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
    2 / -0.33
    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
    2 / -0.33
    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
    2 / -0.33
    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
    2 / -0.33
    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
    2 / -0.33

    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
    2 / -0.33

    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.

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

    Angular velocity about z-axis,

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

    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

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