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Fluid Mechanics Test 1

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Fluid Mechanics Test 1
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  • Question 1
    2 / -0.33
    Oil in a hydraulic cylinder is compressed from an initial volume of 2 m3 to 1.96 m3. If the pressure of oil in the cylinder changes from 40 MPa to 80 MPa during compression, the bulk modulus of elasticity of oil is
    Solution

    Bulk modulus is defined as the ratio of compressive stress to Volumetric strain. It is a measure of the compressibility of a liquid.

    k=dP(dVV)

    k=8040(1.9622)=2000MPa
  • Question 2
    2 / -0.33

    For a steady two-dimensional flow, the scalar components of the velocity field are Vx = -2x, vy = 2y, vz = 0.The corresponding components of acceleration ax and ay, respectively are;

    Solution

    ax=ududx+vdudy+wdudz+dudt

    = (-2x) × (2) = 4x

    ay=udvdx+vdvdy+wdvdz+dvdt

    = 0 + 2y × 2 + 0 + 0 = 4y

  • Question 3
    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 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 on the  fluid is zero.

    Newton’s law of viscosity:

    τ=μdudy   

    Where

    τ is the shear stress on 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 4
    2 / -0.33
    The velocity in a wind tunnel is being measured using a Pitot-static tube connected to a vertical U-tube manometer. The density of air is 1.2 kg/m3 and the deflection of the manometer is 24 mm. The manometer fluid is water. The velocity measured by the Pitot-static tube is
    Solution

    Concept:

    V=2gha

    Calculation:

    Given:

    ρa = 1.2 kg/m3, deflection of the manometer = 24 mm

    Now,

    ρaha = ρwhw

    1.2 × ha = 1000 × 24 × 10-3

    ha = 20

    =2×10×20

    ∴ V = 20 m/s
  • Question 5
    2 / -0.33
    A centrifugal pump lifts water through a height h and delivers it at a velocity vd. The loss of head through piping is hf. The gross lift is
    Solution

    Manometric head/ gross lift is defined as the head against which a centrifugal pump has to work.

    It can be expressed as:

    Hm = Suction Head (hs) + Delivery Head (hd) + Friction head loss in the suction pipe (hfs) + Friction head loss in the Delivery pipe (hfd) + Velocity head of water in delivery pipe (V2/2g)

    Here Gross lift = h+hf+vd22g

  • Question 6
    2 / -0.33

    Select the correct option for the velocity profile

    uV=52(yδ)12(yδ)2+23(yδ)3

    Solution

    The separation point S is determined from the condition

    (uy)y=o=0

    So, If

    (uy)y=0<0flowhasseparated 

    (uy)y=0Theflowisonvergeofseparation(uy)y=0>0Flowwillnotseparate

    So, uv=52(yδ)12(yδ)2+23(yδ)3

    uy=5v2δyvδ2+2y2δ3(uy)y=0=52δv>0

    ⇒ The flow will not separate.

  • Question 7
    2 / -0.33
    The velocity field for a 2 D, steady and incompressible flow is represented as V = (5x) i – (5y) j. What would be the equation of stream line passing through point (1, 1).
    Solution

    Concept:

    A Stream line is an imaginary curve drawn through a flowing fluid in such a way that tangent to it any point gives the direction of instantaneous velocity at that point.  The equation of stream line for 2 D flow is given as:

    dxu=dyv 

    Where, u and v are the x and y components of the velocity.

    Calculation:

    Given:

    V = (5x) î - (5y) ĵ

    u = 5x and v = -5y

    The equation of streamline is given as

    dx5x=dy5ydyy=dxx

    ⇒ In y = -In x + C

    ⇒ In (xy) = C’ ⇒ xy = λ                                   [C, C’, λ are constants]

    If the stream line pases through point (1,1)  i.e. x = 1 and y = 1 ⇒ λ = 1

    ∴ The equation of sream line is xy = 1

  • Question 8
    2 / -0.33
    Oil (kinematic viscosity, νoil = 1.0 × 10-5 m2/s) flows through a pipe of 0.5 m diameter with a velocity of 10 m/s. Water (kinematic viscosity, vw = 0.89 × 10-6 m2/s) is flowing through a model pipe of diameter 20 mm. For satisfying the dynamic similarity, the velocity of water (in m/s) is _______.
    Solution

    Concept:

    For pipe flow, the dynamic similarity will be obtained if the Reynolds number in the model and prototype are equal.

    (Re)Model = (Re)Prototype

    ρmVmDmμm=ρpVpDpμp

    VmDmνm=VpDpνp

    Calculation:

    Given: νoil = 1.0 × 10-5 m2/s, vw = 0.89 × 10-6 m2/s, Doil = 0.5 m, Dw = 20 mm = 0.02 m, Voil = 10 m/s, Vw = ?

    (VDv)P=(VDv)m

    10×0.51.0×105=Vw×0.020.89×106

    VWater = 22.25 m/s
  • Question 9
    2 / -0.33
    The x and y - components of the velocity field for a two dimensional fluid flow is given by u = 3y – y2 and v = 3x - x2. What would be the rate of shear deformation and angular velocity of fluid flow in z direction respectively at point (2, 3)?
    Solution

    Concept

    If the velocity field represents the flow, it must satisfy the continuity equation, given as:

    ux+vy=0

    Where u and v are the x and y components of the velocity.

    If it satisfies, the rate of shear deformation is given as:

    θ=12(vx+uy)

    And the angular velocity in z-direction is given as,

    ω=12(vxuy)

    Calculation:

    u = 3y – y2 and v = 3x – x2

    ux=0andvy=0

    ux+vy=0+0=0

    i.e. equation of continuity is satisfied, so, it represents a flow field.

    Now,

    vx=32xanduy=32y

    θ=12(32x+32y)=12[62(x+y)]

    θ(x,y)=12(62(x+y))

    θ (2, 3) =12[62(2+3)]

    θ(2, 3) = -2 units

    Again,

    ω=12((32x)(32y))

    ⇒ ω=12(2y2x)=yx

    ∴  ω(2, 3) = 3 – 2 = 1 units

    θ = -2 units and ω = 1 units.  

  • Question 10
    2 / -0.33

    A smooth flat plate of length 5 m and width 2 m is moving with a velocity of 4 m/s in stationary air of density 1.25 kg/m3 and kinematic viscosity 1.5 × 10-5 m2/s. The total drag force (N) on one side of the plate assuming that the boundary layer is turbulent from the very beginning is _________.

    (Correct up to 2 decimals)

    Assume for turbulent flow over a flat plate, local drag coefficient is:

    CD,x=0.059Rex1/5
    Solution

    Given data; L = 5 m, B = 2 m, V = 4 m/s, ρ = 1.25 kg/m3

    ν = 1.5 × 10-5 m2/s

    ReL=ρVLμ=VLν=(4)(5)1.5×105=13.33×105

    ReL = 13.33 × 105 > 5 × 105

    ⇒ Flow is turbulent.

    Average drag coefficient (CD=1L0LCD,xdx

    CD=1L0L0.059Rex1/5dx=0.07375ReL1/5

    CD=0.07375(13.33×105)1/5=0.004393

    CD=FD12ρAV2FD=(0.004393).12.(1.25)(10)(16)

    FD = 0.4393 N ≈ 0.44 N
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