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Fluid Mechanics Test 5
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Fluid Mechanics Test 5
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  • Question 1/10
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    What flow rate in m3/s is needed using a 20:1 scale model of a dam over which 4 m3/s of water flows?
    Solutions

    Concept:

    If gravitational and inertial force are me only important forces, then the Frouds number must be the same in the model and prototype. Thus

    VmgLm=VpgLp

    Scale Ratio for length: Lr=LpLm

    From Froude Model Law:

    VmgLm=VpgLpVmLm=VpLp=VpVm=LpLm=Lr

    Scale Ratio for Time: Time = Length/Velocity

    Tr=TpTm=(LV)P(LV)m=LpLm.VmVp=Lr.1Lr=Lr

    Scale ratio for Discharge:

    Q = A × V = L2 × L/T = L3/T

    Qr=QpQm=(L3T)P(L3T)m=(LpLm)3.TmTp=Lr3.1Lr=Lr2.5

    Or

    Qr=QpQm=(A×V)P(A×V)m=APAm.VPVm=(LpLm)2.LPLm=(LpLm)2.5=Lr2.5

    Calculation:

    As here only Length scale ratio and discharge is give so we will use Froude Model Law.

    Qr = Lr2.5 = 202.5 = 1788.85

    Qr=QpQmQm=QpQr=41788.85=0.0022m3/s

  • Question 2/10
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    A flat plate is kept in an infinite fluid medium, the fluid has a uniform free - stream velocity parallel to the plate. For the laminar boundary layer formed on the plate, pick the correct option matching

    List - I and List - ll.

    List - l

    A. Boundary layer thickness

    B. Shear stress at the plate

    C. Pressure along the plate

    List - ll

    1. Decreases in the flow direction

    2. Increases in the flow direction

    3. Remains unchanged

    Solutions

    δ is the distance from the centre - line to the edge of boundary layer where the change in speed is 99% of maximum change in speed from centreline of outer flow. δ is not constant but varies along x. δ(x) increases with x.

    But there are certain flows such as rapidly accelerating enter flow along a wall, in which δ(x) decreases with x.

    Pressure drop is the driving force to push fluid to flow through a pipe, that means as far as there is no pressure difference between two points along the pipe length there is no flow. Generally pressure drop takes place even radially, but the value is too smaller than that in length, therefore we just consider the pressure drop in pipe length. When we have a fully developed flow it doesn't mean that there is no pressure drop, it just means there is no velocity difference along the axis where we have fully developed flow.

     Shear stress also decreases along plate direction.

  • Question 3/10
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    If δ1 and δ2 are the laminar boundary layer thicknesses over a flat plate at a point, distant x from the leading edge when the Reynolds number of the flow are 100 and 400, respectively, then the ratio δ12 will be
    Solutions

    Concept:

    For Laminar Boundary layer:

    δX=KReX

    Where δ is the boundary layer thickness.

    Rex is Reynolds Number at a distance X from the leading edge.

    K is any constant; generally it is equal to 5.

    Now for Given X;

    δ1ReX

    Calculation:

    Given

    Re x1 = 200 and Re x2 = 400

    δ1δ2=ReX2ReX1

    δ1δ2=400100=2.0

  • Question 4/10
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    The velocity distribution in the boundary layer is given as uU=yδ. The dimensionless-profile shape factor is

    Solutions

    Dimensionless profile shape factor (H) = Displacement thickness/Momentum thickness = δ*/θ

    δ=0δ(1uU)dy,θ=0δuU(1uU)dy

    uU=yδ

    δ=0δ(1yδ)dy={yy22δ}0δ=δδ22δ=δ2

    θ=0δuU(1uU)dy=0δyδ(1yδ)dy

    θ=0δ(yδy2δ2)dy

    ={y22δy33δ2}0δ

    δ2δ3=δ6

    H=δθ=δ2δ6=3

    So, shape factor (H) = 3
  • Question 5/10
    1 / -0

    For laminar boundary layer over a flat plate, the velocity at a particular section over a flat plate varies as

    uUo=a+byδ+c(yδ)2+d(yδ)3

    Where, y is the distance measured normal to the plate and δ is the boundary layer thickness. U0 is the maximum velocity at y = δ or also called mean free stream velocity of flow. What would be the ratio of b/d?
    Solutions

    Concept:

    The given problem can be solved by applying boundary conditions on the laminar boundary layer. Since, there are 4 unknowns, 4 boundary equations are needed. These are:

    1. At y = 0; u = 0.

    2. At y = δ; u = Uo

    3. At y = δ; τ = 0 ⇒ du/dy = 0 {∵ τ = μ du/dy}

    4. At y = 0; shear stress at boundary is maximum and to get maximum shear stress at boundary dτ/dy at y must be zero.

    d2udy2=0

    Calculation

    Condition 1: At y = 0 ; u = 0

    ⇒ a = 0

    Condition 2: At y = δ; u = Uo

    ⇒ b + c + d = 0

    Condition 3: At y = δ; τ = 0 ⇒ du/dy = 0

    uU0=a+byδ+c(yδ)2+d(yδ)3

    dudy=U0{bδ+2cyδ2+3dy2δ3} = 0 ⇒ 2c + 3d = 0

    Condition 4: d2udy2=0

    d2udy2=U0{2cδ2+6dyδ3} = 0 ⇒ c = 0

    Based on above, we get

    a = 0 (First condition)

    a + b + c + d = 0 (second condition)

    2c + 3d = 0 (third condition)

    c = 0 (fourth condition)

    On solving, a = 0; b = 3/2; c = 0 and d= -1/2

    ∴ The desired ratio b/d is - 4.

  • Question 6/10
    1 / -0

    The velocity profile of a fully developed laminar flow in a straight circular pipe, as shown in the figure, is given by the expression u(r)=R24μ(dpdx)(1r2R2) , where dpdx  is a constant. The average velocity if fluid in the pipe is

    Solutions

    u(r)=R24μ(dpdx)(1r2R2)

    Velocity profile in fully developed laminar how in pipe is parabolic with a maximum at center line & minimum at pipe wall then average velocity

    Vavg=1πR20Ru(r)dr=1πR20RR24μ(dpdx)(1r2R2)2πrdr=12μ(dpdx)0R(rr3R2)dr=12μ(dpdx)[r22r44R2]0R=12μ(dpdx)[R22R44R2]=R28μ(dpdx)

  • Question 7/10
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    A pipe of diameter 1.8 m is required to transport an oil of relative density 0.8 and viscosity 0.04 poise at the rate of 4 m3/s. Tests were conducted on a 20 cm diameter pipe using water. The discharge (litres/s) through the model is _____. (Correct up to 2 decimals)

    Take viscosity of water = 0.01 poise, density = 1000 kg/m3
    Solutions

    Concept:

    Specific Gravity or Relative Density: The ratio of the density of the fluid to the density of water—usually 1000 kg/m3 at a standard condition—is defined as Specific Gravity or Relative Density of fluids

    ∴ The density of fluid = Relative Density × Density of water

    Poise = 0.1 Ns/m= 0.1 kg/ms

    Calculation:

    Given data,

    RD = 0.8 ⇒ ρ = 0.8 × 1000 = 800 kg/m3

    Prototype

    Model

    DP = 1.8 m

    Dm = 0.2 m

    ρP = 800 kg/m3

    ρm = 1000 kg/m3

    μP = 0.004 kg/ms

    μm = 0.001 kg/ms

    QP = 4 m3/s

    Qm = ?

     

    (Re)P = (Re)m

    (ρVDμ)P=(ρVDμ)m     ...(1)

    ρmρP.VmVP.DmDP.μPμm=1

    1000800×VmVP×0.21.8×0.0040.001=1

    VmVP=1.8

    QmQP=AmVmAPVP=(DmDP)2×VmVP

    QmQP=(0.21.8)2×1.8=0.0222

    Qm = 0.0222 × Q= 0.0222 × 4 = 0.08888 m3/s

    1 litre = 1000 cm3 = 10-3 m3

    ∴ 1 m3/s = 1000 litre/s

    Qm = 0.08889 m3/s

    ∴ Q= 88.89 litres/s

  • Question 8/10
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    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 _______.
    Solutions

    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 9/10
    1 / -0

    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?
    Solutions

    Concept:

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

    Re=ρUXμ

    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

    ρUxμ5×105

    103×1×x1035×105

    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 10/10
    1 / -0

    A smooth flat plate is in parallel flow stream. What is the ratio of drag force over the upstream half of the plate to that of downstream half of the plate? Assume that the average local skin friction coefficient is proportional to √X, where X is the distance measured from leading edge of plate.
    Solutions

    Concept:

    The drag force acting on a plate is given by:

    FD=CD12ρV2A

    Where,

    V is the stream velocity

    A is the project area over the plate

    CD is the drag coefficient or average skin friction coefficient

    Now, CD ∝ √X ⇒ CD = C√X, where C is constant.

    Calculation:

    Let the length of plate is L, then upstream length is L/2 and downstream length is also L/2 as shown in the figure:

    Consider a small strip of length ‘dx’ at a distance ‘x’ from leading edge of the plate. Let width of strip is B and drag force acting on this strip is dF.

    Projected area strip is dA = B⋅dx

    dF=(Cx)12ρV02(Bdx)

    dF=(C2ρV2B)xdx 

    Ρ, V, B, C are constant = K (say)

    dF=kxdx

    Let FAB and FBC are the drag forces on portion AB and BC respectively.

    For AB portion: x varies from 0 to ; So, total drag force on AB is:

    ABdF=FAB=0L2kxdx=2kx323|0L/2

    FAB=23k(L2)32

    For BC: x varies from L/2 to L,

    BCdF=FBC=L/2Lkxdx=23k[(L)32(L2)32]

    Now,

    FAB=23k(L2)3/2

    FBC=23k[L32(L2)32]

    The desired ratio is, R=RABFBC

    R=23k(L2)3/22k3(L)3/22k3(L2)3/2=121.51=0.547

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