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A

Group 'A' (Multiple choice questions)

Rewrite the correct option of each question in your answer sheet.

9 questions·1 marks each
1mcq1 marks

Bernoulli's equation relates:

  • A

    Various force with change in momentum

  • B

    Torque with change in angular momentum

  • C

    Various forces involved in a fluid.

  • D

    Various energy forms

Correct answer: D

Various energy forms

Bernoulli's equation relates various energy forms (pressure, kinetic, potential) of a fluid.

bernoulli-equation
2mcq1 marks

For a given value of specific force, there are two possible depths which are known as:

  • A

    sequent depths

  • B

    Alternate depths

  • C

    Critical depths

  • D

    unsteady uniform

Correct answer: A

sequent depths

Two depths for a given specific force are conjugate (sequent) depths.

open-channelspecific-force
3mcq1 marks

The flow of liquid through a tapering pipe at constant rate is

  • A

    steady uniform

  • B

    Steady non-uniform

  • C

    Unsteady non-uniform

  • D

    Unsteady uniform

Correct answer: B

Steady non-uniform

Constant rate but velocity changes with the tapering section: steady non-uniform flow.

flow-classification
4mcq1 marks

The energy heads are expressed in term of energy per unit.

  • A

    mass

  • B

    weight

  • C

    volume

  • D

    density

Correct answer: B

weight

Energy heads are expressed as energy per unit weight.

energy-head
5mcq1 marks

Poise is a unit of :

  • A

    Density

  • B

    velocity gradient

  • C

    dynamic viscosity

  • D

    kinematic viscosity

Correct answer: C

dynamic viscosity

Poise is the CGS unit of dynamic viscosity.

viscosityunits
6mcq1 marks

The contraction loss coefficient K in term of coefficient of contraction in a pipe with a sudden contraction is:

  • A

    (1Cc1)\left(\frac{1}{C_c} - 1\right)

  • B

    (1Cc21)\left(\frac{1}{C_c^{2}} - 1\right)

  • C

    (1Cc21)2\left(\frac{1}{C_c^{2}} - 1\right)^{2}

  • D

    (1Cc1)2\left(\frac{1}{C_c} - 1\right)^{2}

Correct answer: D

(1Cc1)2\left(\frac{1}{C_c} - 1\right)^{2}

The sudden-contraction loss coefficient is K=(1Cc1)2K = \left(\frac{1}{C_c} - 1\right)^2.

minor-lossescontraction
7mcq1 marks

Continuity equation is valid for

  • A

    incompressible fluid with unsteady flow.

  • B

    incompressible fluid with steady flow.

  • C

    Compressible fluid with unsteady flow.

  • D

    Compressible fluid with steady flow.

Correct answer: B

incompressible fluid with steady flow.

The simple (one-dimensional) continuity equation Q = A·v applies to incompressible fluid with steady flow.

continuity-equation
8mcq1 marks

The relation between Manning's roughness coefficient 'n' and Chezy's constant C is given by:

  • A

    C=1nR1/6C = \frac{1}{n} R^{1/6}

  • B

    C=1nR2/3C = \frac{1}{n} R^{2/3}

  • C

    C=nR1/6C = nR^{1/6}

  • D

    C=nR2/3C = nR^{2/3}

Correct answer: A

C=1nR1/6C = \frac{1}{n} R^{1/6}

Chezy's C=1nR1/6C = \frac{1}{n}R^{1/6}.

manningchezy
9mcq1 marks

The error in measurement of discharge is ... times the error in measurement of height of fluid over rectangular notch.

  • A

    11

  • B

    1.251.25

  • C

    1.151.15

  • D

    1.51.5

Correct answer: D

1.51.5

For a rectangular notch, QH3/2Q \propto H^{3/2}, so the percentage error in Q is 1.5 times the percentage error in H.

weirerror-analysis
B

Group 'B' (Short answer questions)

Attempt all the questions.

5 questions·5 marks each
10short5 marks

Define surface tension, viscosity compressibility, Newtonian fluid and Real fluid.

Definitions

1. Surface tension (σ\sigma): The property of a liquid by virtue of which its free surface behaves like a stretched elastic membrane. It is the force acting per unit length along a line drawn on the free surface, acting perpendicular to that line and tangential to the surface. SI unit: N/m\text{N/m}.

2. Viscosity (μ\mu): The property of a fluid by which it offers resistance to the relative (shear) motion between adjacent fluid layers. By Newton's law of viscosity, the shear stress is

τ=μdudy\tau = \mu \frac{du}{dy}

where dudy\dfrac{du}{dy} is the velocity gradient. SI unit of dynamic viscosity: Pa\cdotps\text{Pa·s} (or poise in CGS).

3. Compressibility: The measure of the change in volume (or density) of a fluid under the action of applied pressure. It is the reciprocal of the bulk modulus of elasticity KK:

Compressibility=1K=dV/Vdp\text{Compressibility} = \frac{1}{K} = -\frac{dV/V}{dp}

It indicates how easily a fluid can be compressed.

4. Newtonian fluid: A fluid in which the shear stress is directly proportional to the velocity gradient (rate of shear strain), i.e. it obeys Newton's law of viscosity τ=μdudy\tau = \mu \dfrac{du}{dy} with constant μ\mu. Examples: water, air, thin oils.

5. Real fluid: A fluid that possesses viscosity, surface tension and compressibility, and therefore offers resistance to flow (experiences shear stress) when in motion. All actual fluids existing in nature (water, oil, air) are real fluids, as opposed to the idealized non-viscous "ideal fluid."

fluid-properties
11numeric5 marks

A 25cm25\,\text{cm} diameter pipe carries oil of sp. gr 0.9 at a velocity of 3m/s3\,\text{m/s}. At another section the diameter is 20cm20\,\text{cm}. Find the velocity at this section and also mass rate of flow of oil.

Numeric answer

continuitymass-flow-rate
12short5 marks

Derive Darcy- Weisbach equation for major head loss.

Derivation of the Darcy–Weisbach Equation

Consider steady, uniform flow of a fluid through a horizontal pipe of diameter DD (cross-sectional area AA) and length LL between sections 1 and 2. Let hfh_f be the head loss due to friction, ff' the frictional resistance per unit wetted area per unit velocity-squared, and vv the mean velocity.

Step 1 — Forces acting on the fluid element.

  • Pressure force at section 1: p1Ap_1 A (driving the flow)
  • Pressure force at section 2: p2Ap_2 A (opposing)
  • Frictional resistance at the pipe wall: F=f×(wetted surface area)×v2=f(πDL)v2F = f' \times (\text{wetted surface area}) \times v^2 = f' \,(\pi D L)\, v^2

Step 2 — Apply equilibrium (steady flow, no acceleration).

p1Ap2A=Fp_1 A - p_2 A = F (p1p2)π4D2=f(πDL)v2(p_1 - p_2)\,\frac{\pi}{4}D^2 = f'(\pi D L)v^2 p1p2=f(πDL)v2π4D2=4fLv2Dp_1 - p_2 = \frac{f'(\pi D L)v^2}{\frac{\pi}{4}D^2} = \frac{4 f' L v^2}{D}

Step 3 — Relate pressure difference to head loss.

Applying Bernoulli's equation between 1 and 2 for the horizontal pipe of constant diameter (z1=z2z_1=z_2, v1=v2v_1=v_2):

p1p2ρg=hfp1p2=ρghf\frac{p_1 - p_2}{\rho g} = h_f \quad\Rightarrow\quad p_1 - p_2 = \rho g\, h_f

Step 4 — Combine.

ρghf=4fLv2D\rho g\, h_f = \frac{4 f' L v^2}{D} hf=4fLv2ρgDh_f = \frac{4 f' L v^2}{\rho g D}

Putting f=fρ/2=2fρf = \dfrac{f'}{\rho/2} = \dfrac{2f'}{\rho} (the dimensionless Darcy friction coefficient), we obtain the Darcy–Weisbach equation:

hf=4fLv22gD\boxed{\,h_f = \frac{4 f L v^2}{2 g D}\,}

where ff is the coefficient of friction (and 4f4f is the Darcy friction factor). This gives the major (frictional) head loss in a pipe of length LL and diameter DD carrying a mean velocity vv.

darcy-weisbachhead-loss
13numeric5 marks

Water flows at a uniform depth of 1.5m1.5\,\text{m} in a trapezoidal channel having a bottom width 5m5\,\text{m} and side slope 2H:1V. It has to carry a discharge of 60m3/s60\,\text{m}^3/\text{s}. Compute the bottom slope required to be provided. Assume the bed material is gravel.

Numeric answer

open-channeltrapezoidal-channel
14numeric5 marks

A horizontal venturimeter with inlet and throat diameter 30cm30\,\text{cm} and 15cm15\,\text{cm} respectively is used to measure the flow of water. The reading of differential manometer connected to the inlet and throat is 20cm20\,\text{cm} of mercury. Determine the rate of flow take Cd=0.98C_d = 0.98.

Numeric answer

venturimeterdischarge
C

Group 'C' (Long answer questions)

Attempt all the questions.

2 questions·8 marks each
15numeric8 marks

a) If 1 liter container is immersed halfway in water, what is the volume of water displaced? What is buoyant force in the container?

b) A hinged gate in a dam is inclined at 3030^\circ from horizontal. The gate is 3.2m3.2\,\text{m} long 2.0m2.0\,\text{m} wide and hinged at A, at a depth of 1.7m1.7\,\text{m}. Calculate the magnitude of force to hold the gate closed.

Inclined hinged dam gate: water at depth 1.7 m, gate 3.2 m long at 30° hinged at A, force F_gate shown

Numeric answer

buoyancyhydrostatic-forceinclined-gate
16short8 marks

Mention general applications of Bernoulli's principle. Show, how discharge is measured through a pipe and using this principle.

General Applications of Bernoulli's Principle

Bernoulli's equation, pρg+v22g+z=constant\dfrac{p}{\rho g} + \dfrac{v^2}{2g} + z = \text{constant}, states that for steady, incompressible, non-viscous flow the total energy head (pressure + kinetic + potential) is conserved along a streamline. Common applications include:

  • Venturimeter – measuring discharge (flow rate) through a pipe.
  • Orifice meter and nozzle meter – discharge measurement.
  • Pitot tube – measuring flow velocity at a point.
  • Aerofoil / aeroplane wing lift – higher velocity over the curved upper surface lowers pressure, producing lift.
  • Spinning ball (Magnus effect), atomizer/carburettor (spray), and chimney draught.

Measurement of Discharge Through a Pipe (Venturimeter)

A venturimeter consists of a converging cone, a throat, and a diverging cone fitted in a pipe. Let:

  • Section 1 = inlet, area A1A_1, velocity v1v_1, pressure p1p_1
  • Section 2 = throat, area A2A_2, velocity v2v_2, pressure p2p_2

Step 1 — Bernoulli's equation (horizontal pipe, z1=z2z_1=z_2):

p1ρg+v122g=p2ρg+v222g\frac{p_1}{\rho g} + \frac{v_1^2}{2g} = \frac{p_2}{\rho g} + \frac{v_2^2}{2g} p1p2ρg=v22v122g=h\frac{p_1 - p_2}{\rho g} = \frac{v_2^2 - v_1^2}{2g} = h

where hh is the pressure-head difference (read from the differential manometer).

Step 2 — Continuity equation:

A1v1=A2v2v1=A2A1v2A_1 v_1 = A_2 v_2 \quad\Rightarrow\quad v_1 = \frac{A_2}{A_1} v_2

Step 3 — Substitute and solve for v2v_2:

h=v222g(1A22A12)h = \frac{v_2^2}{2g}\left(1 - \frac{A_2^2}{A_1^2}\right) v2=2gh1(A2/A1)2=A1A12A222ghv_2 = \sqrt{\frac{2gh}{1 - (A_2/A_1)^2}} = \frac{A_1}{\sqrt{A_1^2 - A_2^2}}\sqrt{2gh}

Step 4 — Theoretical discharge:

Qth=A2v2=A1A2A12A222ghQ_{th} = A_2 v_2 = \frac{A_1 A_2}{\sqrt{A_1^2 - A_2^2}}\sqrt{2gh}

Step 5 — Actual discharge (introducing the coefficient of discharge CdC_d to account for real losses):

Qact=CdA1A2A12A222gh\boxed{\,Q_{act} = C_d\,\frac{A_1 A_2}{\sqrt{A_1^2 - A_2^2}}\sqrt{2gh}\,}

Here, for a mercury–water differential manometer, the equivalent head of flowing liquid is

h=x(SmSw1)h = x\left(\frac{S_m}{S_w} - 1\right)

with xx the manometer deflection. Thus, by measuring the manometer deflection, the discharge through the pipe is determined using Bernoulli's principle.

bernoulliventurimeterdischarge-measurement

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