Fluid Mechanics PYQ Set (2000 to 2025)

Air (density = $1.2\ \text{kg/m}^3$, kinematic viscosity = $1.5 \times 10^{-5}\ \text{m}^2/\text{s}$) flows over a flat plate with a free-stream velocity of 2 m/s. The wall shear stress at a location 15 mm from the leading edge is $\tau_w$. What is the wall shear stress at a location 30 mm from the leading edge?

(A) $ \dfrac{\tau_w}{2} $
(B) 2$\tau_w$
(C) 1$\tau_w$
(D) $ \dfrac{\tau_w}{\sqrt{2}} $

Solution =

To determine the wall shear stress at a location 30 mm from the leading edge, we analyze the flow over a flat plate and how the wall shear stress varies with distance from the leading edge.

Given:
Density of air ($\rho$): $1.2\ \text{kg/m}^3$
Kinematic viscosity ($\nu$): $1.5 \times 10^{-5}\ \text{m}^2/\text{s}$
Free-stream velocity ($U$): $2\ \text{m/s}$
Wall shear stress at 15 mm ($\tau_{w1}$): $\tau_w$
Distance 1 ($x_1$): $15\ \text{mm} = 0.015\ \text{m}$
Distance 2 ($x_2$): $30\ \text{mm} = 0.030\ \text{m}$

Key Concepts:
For laminar flow over a flat plate, the wall shear stress ($\tau_w$) is given by:
\[ \tau_w = 0.332 \rho U^2 \left( \frac{U x}{\nu} \right)^{-1/2} \] This shows that the wall shear stress is inversely proportional to the square root of the distance from the leading edge:
\[ \tau_w \propto \frac{1}{\sqrt{x}} \] Calculation:
Let $\tau_{w2}$ be the wall shear stress at $x_2 = 30\ \text{mm}$.
Using the proportional relationship:
\[ \frac{\tau_{w2}}{\tau_{w1}} = \sqrt{\frac{x_1}{x_2}} = \sqrt{\frac{0.015}{0.030}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \] Thus:
\[ \tau_{w2} = \frac{\tau_{w1}}{\sqrt{2}} = \frac{\tau_w}{\sqrt{2}} \] Final Answer: \[ \boxed{\dfrac{\tau_w}{\sqrt{2}}} \]

For a fully-developed pipe flow, which of the following options is/are correct?

(A) For the same maximum velocity, the average velocity is higher in the turbulent regime than that of the laminar regime
(B) Compressibility effects are important if Mach number is less than 0.3
(C) For laminar flow, the friction factor is independent of surface roughness
(D) For laminar flow, friction factor decreases with decrease in Reynolds number

Solution =

Option A) “For the same maximum velocity, the average velocity is higher in the turbulent regime than that of the laminar regime.”
✓ Correct= In laminar flow, the velocity profile is parabolic:
\[ V_{avg} = \frac{1}{2}V_{max} \] In turbulent flow, the velocity profile is flatter (due to mixing):
\[ V_{avg} \approx 0.8V_{max} \text{ to } 0.9V_{max} \] Conclusion: For the same $V_{max}$, $V_{avg}$ is indeed higher in turbulent flow.

Option B) “Compressibility effects are important if Mach number is less than 0.3.”
✗ Incorrect = Compressibility effects become significant when:
\[ Ma > 0.3 \] For $Ma < 0.3$, flow is typically treated as incompressible.
Conclusion: The statement reverses the correct condition.

Option C) “For laminar flow, the friction factor is independent of surface roughness.”
✓ Correct = Laminar flow friction factor depends only on Reynolds number:
\[ f = \frac{64}{Re} \] Surface roughness affects only turbulent flow friction factors.
Conclusion: The statement is correct.

Option D) “For laminar flow, friction factor decreases with decrease in Reynolds number.”
✗ Incorrect = The laminar friction factor relationship:
\[ f = \frac{64}{Re} \] This shows that $f$ increases as $Re$ decreases.
Conclusion: The statement is exactly opposite of the truth.

Final Answer = The correct options are: \[ \boxed{A \text{ and } C} \]

Which of the following is responsible for eddy viscosity (or turbulent viscosity) in a turbulent boundary layer on a flat plate?

(A) Nikuradse stresses
(B) Reynolds stresses
(C) Boussinesq stresses
(D) Prandtl stresses

Solution =

Reynolds stresses arise from the time-averaged product of fluctuating velocity components in turbulent flow:

\[ \tau_{ij}^{\text{Reynolds}} = -\rho \overline{u_i’ u_j’} \]
These are responsible for the additional momentum transport in turbulent flows and are modeled using the concept of eddy viscosity.
Boussinesq introduced a hypothesis to relate Reynolds stresses to the mean rate of strain using eddy viscosity:

\[ \tau_{ij}^{\text{Reynolds}} \approx \mu_t \left( \frac{\partial \overline{u_i}}{\partial x_j} + \frac{\partial \overline{u_j}}{\partial x_i} \right) \]
But the physical origin is still the Reynolds stresses.
Correct Answer: B. Reynolds stresses

In the space above the mercury column in a barometer tube, the gauge pressure of the vapour is __.

(A) positive, but more than one atmosphere
(B) negative
(C) zero
(D) positive, but less than one atmosphere

Solution =

Explanation:
A barometer measures atmospheric pressure using a column of mercury in a closed-end glass tube.
The space above the mercury column is essentially a vacuum or near-vacuum (called the Torricellian vacuum).
In this space: There is very little mercury vapour.
The pressure is extremely low compared to atmospheric pressure.
Key Concept:
Gauge pressure is defined as:
Gauge Pressure = Absolute Pressure – Atmospheric Pressure

Since the absolute pressure in the vacuum space is near zero:
\[ \text{Gauge Pressure} = 0 – 1\, \text{atm} = -1\, \text{atm} \]
So, the gauge pressure is negative.
Final Answer: Option B — Negative

Which of the following conditions is used to determine the stable equilibrium of all partially submerged floating bodies?

(A) Centre of buoyancy must be above the centre of gravity
(B) Centre of buoyancy must be below the centre of gravity
(C) Metacentre must be at a higher level than the centre of gravity
(D) Metacentre must be at a lower level than the centre of gravity

Solution =

Explanation:
Centre of Gravity (G): Point at which the weight of the body acts.
Centre of Buoyancy (B): Centroid of the submerged volume (point of buoyant force).
Metacentre (M): Point where the line of action of buoyant force intersects the vertical axis when tilted.
For a floating body to be in stable equilibrium:
Metacentre (M) must be above Centre of Gravity (G)
So,
\[ MG > 0 \]
This ensures that a restoring moment is produced when the body is tilted.

Summary Table: (Condition => Stability)
$ M > G $ => Stable
$ M = G $ => Neutral
$ M < G $ => Unstable

Final Answer: Option C — Metacentre must be at a higher level than the centre of gravity

A pump handling a liquid raises its pressure from 1 bar to 30 bar. Take the density of the liquid as $ 990 \, \text{kg/m}^3 $. The isentropic specific work done by the pump in kJ/kg is

(A) 0.1
(B) 0.3
(C) 2.5
(D) 2.93

Solution =

Given:
Initial pressure: $ P_1 = 1 \, \text{bar} = 10^5 \, \text{Pa} $
Final pressure: $ P_2 = 30 \, \text{bar} = 30 \times 10^5 \, \text{Pa} $
Density: $ \rho = 990 \, \text{kg/m}^3 $
We are to calculate the isentropic specific work done by the pump:
\[ w = \frac{P_2 – P_1}{\rho} \]
Step-by-step:
\[ \Delta P = P_2 – P_1 = (30 – 1) \times 10^5 \]\[= 29 \times 10^5 \, \text{Pa} \]
\[ w = \frac{29 \times 10^5}{990} = \frac{2900000}{990} \]\[\approx 2929.29 \, \text{J/kg} = 2.93 \, \text{kJ/kg} \]
Final Answer: D) 2.93

For the stability of a floating body under the influence of gravity alone, which of the following is TRUE?

(A) Metacentre should be below centre of gravity
(B) Metacentre should be above centre of gravity
(C) Metacentre and centre of gravity must lie on the same horizontal line
(D) Metacentre and centre of gravity must lie on the same vertical line

Solution =

Explanation:
For a floating body to be stable under gravity, the metacentre (M) must lie above the centre of gravity (G). This ensures that when the body tilts, the buoyant force and gravity create a restoring moment that returns the body to equilibrium.

Key Definitions:
Metacentre (M): The intersection point of the buoyant force line (after tilting) and the original vertical axis.
Centre of Gravity (G): The point where the entire weight of the body acts.
Stability Conditions:
Stable Equilibrium: $ M $ above $ G $
→ Generates a restoring moment: $ \tau = W \cdot GM \cdot \sin\theta $
where $ W $ is weight, $ GM $ is metacentric height, and $ \theta $ is tilt angle.
Unstable Equilibrium: $ M $ below $ G $
→ Generates an overturning moment (may capsize).
Neutral Equilibrium: $ M $ coincides with $ G $
→ No moment is produced.
Thus, for stability, the condition $ \text{Metacentric height} = GM > 0 $ must be satisfied, meaning $ M $ must be above $ G $.

The correct answer is: B) Metacentre should be above centre of gravity

A two-dimensional flow field has velocities along the x and y directions given by $ u = x^2 t, \quad v = -2 x y t $ respectively, where t is time. The equation of streamlines is:

(A) $ x^2 y = \text{constant} $
(B) $ xy^2 = \text{constant} $
(C) $ xy = \text{constant} $
(D) not possible to determine

Solution =

Given: A 2D flow field with velocity components:
\[ u = x^2 t, \quad v = -2 x y t \]
To Find: Equation of streamlines.
Step 1: Use streamline equation
The streamline equation is: \[ \frac{dx}{u} = \frac{dy}{v} \] Substituting the values of $ u $ and $ v $: \[ \frac{dx}{x^2 t} = \frac{dy}{-2 x y t} \]
Cancel the common term $ t $ (assuming $ t \ne 0 $):
\[ \frac{dx}{x^2} = \frac{dy}{-2 x y} \]
Step 2: Cross multiply and simplify
Multiply both sides by $ x^2 \cdot (-2 x y) $: \[ (-2 x y) \, dx = x^2 \, dy \] \[ -2 y \, dx = x \, dy \]
Step 3: Separate variables and integrate
\[ \frac{dy}{y} = -2 \frac{dx}{x} \] Integrating both sides: \[ \ln y = -2 \ln x + \ln C \] Using log properties: \[ \ln y + 2 \ln x = \ln C \]\[\Rightarrow \quad \ln (x^2 y) = \ln C \] \[ x^2 y = \text{constant} \]
Final Answer:
\[ \boxed{x^2 y = \text{constant}} \quad \text{(Option A)} \]

The velocity components in the x and y directions of a two dimensional potential flow are u and v, respectively, then $ \frac{\partial u}{\partial x} $ is equal to

(A) $ \frac{\partial v}{\partial x} $
(B) $ -\frac{\partial v}{\partial x} $
(C) $ \frac{\partial v}{\partial y} $
(D) $ -\frac{\partial v}{\partial y} $

Solution =

Concept: For a two-dimensional potential flow, the flow is incompressible and irrotational.

We use two important conditions:
1. Continuity equation for incompressible flow:
\[ \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0 \Rightarrow \frac{\partial u}{\partial x} = -\frac{\partial v}{\partial y} \]
So the correct answer is:
\[ \boxed{D\ \left(-\frac{\partial v}{\partial y}\right)} \]

At a hydro electric power plant site, available head and flow rate are 24.5 m and $10.1 \, \text{m}^3/\text{s} $ respectively. If the turbine to be installed is required to run at 4.0 revolution per second (rps) with an overall efficiency of 90%, the suitable type of turbine for this site is

(A) Francis
(B) Kaplan
(C) Pelton
(D) Propeller

Solution =

Given:
$ H = 24.5 \, \text{m} $
$ Q = 10.1 \, \text{m}^3/\text{s} $
$ N = 4.0 \, \text{rps} = 4 \times 60 = 240 \, \text{rpm} $
$ \eta = 90\% = 0.9 $

Step 1: Compute Power Output
\[ P = \eta \cdot \rho g Q H \]\[= 0.9 \cdot 1000 \cdot 9.81 \cdot 10.1 \cdot 24.5 \] \[ P \approx 0.9 \cdot 1000 \cdot 9.81 \cdot 247.45 \]\[= 0.9 \cdot 1000 \cdot 2427.6 \approx 2184840 \, \text{W} \] \[ P \approx 2184.84 \, \text{kW} \]
Step 2: Use the Specific Speed Formula
\[ N_s = \frac{N \sqrt{P}}{H^{5/4}} = \frac{240 \cdot \sqrt{2184.84}}{(24.5)^{1.25}} \]\[= \frac{240 \cdot 46.74}{(24.5)^{1.25}} \approx \frac{11217.6}{61.43} \approx 182.6 \]
Step 3: Turbine Type Based on Specific Speed
Pelton: $ N_s = 10 $–$ 35 $
Francis: $ N_s = 60 $–$ 250 $
Kaplan: $ N_s = 250 $–$ 850 $
Propeller: $ N_s > 300 $

Final Answer:
Since $ N_s \approx 183 $, the suitable turbine is: \[ \boxed{\text{A — Francis}} \]

In a laboratory experiment using a scaled down model to measure scour at a bridge pier, the Froude number is important. The ratio of the prototype length to the model length is 100. If the velocity of the model is $1 \, \text{ms}^{-1}$, the velocity (in $\text{ms}^{-1}$) of the prototype is:

(A) 0.1
(B) 1
(C) 10
(D) 100

Solution =

Understanding the Problem
We have a scale model experiment where:
Length scale ratio ($L_r$) = $\frac{L_p}{L_m} = 100$
Model velocity ($V_m$) = $1 \, \text{ms}^{-1}$
We need to find the prototype velocity ($V_p$) while maintaining Froude number similarity.
Froude Number Similarity
The Froude number ($Fr$) is given by:
\[ Fr = \frac{V}{\sqrt{gL}} \] For dynamic similarity, the Froude number must be the same in both model and prototype:
\[ Fr_m = Fr_p \] \[ \frac{V_m}{\sqrt{gL_m}} = \frac{V_p}{\sqrt{gL_p}} \] Velocity Scale Relationship
Since gravitational acceleration ($g$) cancels out:
\[ \frac{V_m}{\sqrt{L_m}} = \frac{V_p}{\sqrt{L_p}} \] The velocity scale ratio ($V_r$) is therefore:
\[ V_r = \frac{V_p}{V_m} = \sqrt{\frac{L_p}{L_m}} = \sqrt{L_r} \] Given $L_r = 100$:
\[ V_r = \sqrt{100} = 10 \] Calculating Prototype Velocity
Using the velocity scale ratio:
\[ V_p = V_r \times V_m = 10 \times 1 \, \text{ms}^{-1} = 10 \, \text{ms}^{-1} \] Conclusion
The correct prototype velocity is $10 \, \text{ms}^{-1}$, which corresponds to option C.
Final Answer: $\boxed{C}$

Consider two identical tanks with a bottom hole of diameter $ d $. One tank is filled with water and the other tank is filled with engine oil. The height of the fluid column $ h $ is same in both the cases. The fluid exit velocity in the two tanks are $ V_1 $ and $ V_2 $. Neglecting all losses, which one of the following options is correct?

(A) $ V_2 > V_1 $
(B) $ V_2 = V_1 $
(C) $ V_2 < V_1 $
(D) Insufficient data to definitively conclude the relationship between $ V_1 $ and $ V_2 $

Solution =

Using Torricelli’s Law, the exit velocity of a fluid from a hole under the force of gravity is given by:
$$ V = \sqrt{2gh} $$
Where:
$ g $ = acceleration due to gravity
$ h $ = height of the fluid column above the hole
However, Torricelli’s Law assumes the fluid is ideal and incompressible. Practically, the pressure at the hole is due to hydrostatic pressure:
$$ P = \rho g h $$
Where:
$ P $ = hydrostatic pressure
$ \rho $ = fluid density
Thus, the velocity of efflux can also be expressed as:
$$ V = \sqrt{\frac{2P}{\rho}} = \sqrt{2gh} $$
Since both tanks have the same fluid height $ h $ and experience gravity equally, in ideal conditions, both would have the same exit velocity:
$$ V_1 = V_2 $$
However, considering real-world behavior:
Engine oil has a lower density than water.
This results in lower hydrostatic pressure at the same height.
Therefore:
$$ P_{\text{oil}} < P_{\text{water}} $$
Leading to:
$$ V_2 < V_1 $$
Hence, the fluid with higher density (water) will have a higher exit velocity.
Final Answer: Option C. $ V_2 < V_1 $

As per common design practice, the three types of hydraulic turbines, in descending order of flow rate, are

(A) Kaplan, Francis, Pelton
(B) Pelton, Francis, Kaplan
(C) Francis, Kaplan, Pelton
(D) Pelton, Kaplan, Francis

Solution =

In hydraulic turbines, the flow rate handling capacity generally depends on the type of turbine. Based on common design practices, the descending order of flow rate capacity is:
Kaplan > Francis > Pelton
This is because:
Kaplan turbine: Axial flow turbine used for low head and high flow applications.
Francis turbine: Mixed flow turbine suitable for medium head and moderate flow.
Pelton turbine: Impulse turbine used for high head and low flow applications.
So, the correct descending order of flow rate is:
Kaplan > Francis > Pelton

Select the correct statement for 50% reaction stage in a steam turbine.

(A) The rotor blade is symmetric.
(B) The stator blade is symmetric.
(C) The absolute inlet flow angle is equal to absolute exit flow angle.
(D) The absolute exit flow angle is equal to inlet angle of rotor blade.

Solution =

Explanation: In a 50% reaction stage, the steam pressure drop is equally divided between the stator and rotor blades, meaning the energy transfer to the fluid is the same in both stages. This symmetry leads to the absolute exit flow angle from the rotor being equal to the inlet angle of the rotor blade.
Why other options are incorrect:
The rotor blade is symmetric:
While the blade design might be symmetrical in some 50% reaction turbines, this is not the defining characteristic of the reaction stage itself.
The stator blade is symmetric:
Similar to the rotor blade, symmetry in the stator blade is not the defining feature of a 50% reaction stage.
The absolute inlet flow angle is equal to absolute exit flow angle:
This statement describes the condition for a fully symmetrical turbine, but in a 50% reaction turbine, only the absolute exit flow angle is equal to the rotor blade inlet angle.

In a Lagrangian system, the position of a fluid particle in a flow is described as: $ x = x_0 e^{-kt} \quad \text{and} \quad y = y_0 e^{kt} $ where $ t $ is the time while $ x_0, y_0, $ and $ k $ are constants. The flow is:

(A) unsteady and one-dimensional
(B) steady and two-dimensional
(C) steady and one-dimensional
(D) unsteady and two-dimensional

Solution =

Given Position Equations:
\[ x = x_0 e^{-kt}, \quad y = y_0 e^{kt} \] where:
$x_0$, $y_0$, and $k$ are constants,
$t$ is time.

Step 1: Check for Time Dependence (Steady vs. Unsteady)
The equations for $x$ and $y$ explicitly depend on time $t$ (through the terms $e^{-kt}$ and $e^{kt}$).
Conclusion: The flow is unsteady because the particle’s position changes with time.

Step 2: Check the Dimensionality (1D vs. 2D)
The flow is described by two independent coordinates: $x$ and $y$.
Both coordinates are functions of time, and neither can be eliminated or expressed purely in terms of the other.
Conclusion: The flow is two-dimensional because it requires both $x$ and $y$ to describe the motion.

Step 3: Given Options
Option A: Unsteady and one-dimensional → Incorrect (the flow is 2D).
Option B: Steady and two-dimensional → Incorrect (the flow is unsteady).
Option C: Steady and one-dimensional → Incorrect (both properties are wrong).
Option D: Unsteady and two-dimensional → Correct.

Final Answer: D) The flow is unsteady and two-dimensional.

An incompressible fluid flows over a flat plate with zero pressure gradient. The boundary layer thickness is 1mm at a location where the Reynolds number is 1000. If the velocity of the fluid alone is increased by a factor of 4, then boundary layer thickness at the same location, in mm will be

(A) 4
(B) 2
(C) 0.5
(D) 0.25

Solution =

To solve this, we use the relationship between boundary layer thickness $ \delta $, Reynolds number $ Re $, and flow velocity $ U $ for a flat plate with zero pressure gradient:
Formula:
\[ \delta \propto \frac{x}{\sqrt{Re_x}} \quad \text{where} \quad Re_x = \frac{Ux}{\nu} \]
So,
\[ \delta \propto \frac{1}{\sqrt{U}} \quad \text{(since $x$ and $ \nu $ are constant)} \]
Given:
Initial Reynolds number: $ Re_1 = 1000 $
Initial boundary layer thickness: $ \delta_1 = 1 \, \text{mm} $
Velocity increases by a factor of 4: $ U_2 = 4U_1 $
So,
\[ \delta_2 = \delta_1 \cdot \frac{1}{\sqrt{4}} = 1 \cdot \frac{1}{2} = 0.5 \, \text{mm} \]
Correct Answer: C) 0.5

Water at 25ºC is flowing through a 1.0km long G.I pipe of 200mm diameter at the rate of $ 0.07 \, \text{m}^3/\text{s} $ . If value of Darcy friction factor for this pipe is 0.02 and density of water is $ 1000 \, \text{kg/m}^3 $, the pumping power (in kW) required to maintain the flow is

(A) 1.8
(B) 17.4
(C) 20.5
(D) 41

Solution =

We are given:
Length of pipe: $ L = 1.0 \, \text{km} = 1000 \, \text{m} $
Diameter of pipe: $ D = 200 \, \text{mm} = 0.2 \, \text{m} $
Flow rate: $ Q = 0.07 \, \text{m}^3/\text{s} $
Darcy friction factor: $ f = 0.02 $
Density of water: $ \rho = 1000 \, \text{kg/m}^3 $
We need to calculate the pumping power required to overcome frictional head loss.
Step 1: Velocity of flow
\[ A = \frac{\pi D^2}{4} = \frac{\pi (0.2)^2}{4} = 0.0314 \, \text{m}^2 \] \[ V = \frac{Q}{A} = \frac{0.07}{0.0314} \approx 2.23 \, \text{m/s} \] Step 2: Head loss due to friction (Darcy-Weisbach equation)
\[ h_f = f \cdot \frac{L}{D} \cdot \frac{V^2}{2g} \] \[ h_f = 0.02 \cdot \frac{1000}{0.2} \cdot \frac{(2.23)^2}{2 \cdot 9.81} \]\[= 0.02 \cdot 5000 \cdot \frac{4.97}{19.62} \approx 100 \cdot 0.2534 \]\[= 25.34 \, \text{m} \] Step 3: Pumping power required
\[ P = \rho g Q h_f = 1000 \cdot 9.81 \cdot 0.07 \cdot 25.34 \]\[\approx 17390.4 \, \text{W} = 17.4 \, \text{kW} \] Final Answer: Option B: 17.4 kW

In a Pelton wheel, the bucket peripheral speed is 10 m/s, the water jet velocity is 25 m/s and volumetric flow rate of the jet is $ 0.1\ \text{m}^3/\text{s} $. If the jet deflection angle is120º and the flow is ideal, the owe developed is:

(A) 7.5 kW
(B) 15.0 kW
(C) 22.5 kW
(D) 37.5 kW

Solution =

To solve this Pelton wheel problem, we’ll use the power developed by the jet on the Pelton wheel formula for ideal flow:
Given:
Bucket peripheral speed, $ u = 10\ \text{m/s} $
Jet velocity, $ V = 25\ \text{m/s} $
Volumetric flow rate, $ Q = 0.1\ \text{m}^3/\text{s} $
Jet deflection angle, $ \theta = 120^\circ $
Ideal flow (no losses)
Formula for power developed:
\[ P = \rho Q (V – u) \left(1 + \cos\theta\right) u \]
Where:
$ \rho = 1000\ \text{kg/m}^3 $ (density of water)
$ \theta = 120^\circ \Rightarrow \cos(120^\circ) = -0.5 $
Substitute values:
\[ P = 1000 \times 0.1 \times (25 – 10) \]\[\times (1 – 0.5) \times 10 \]
\[ P = 1000 \times 0.1 \times 15 \times 0.5 \times 10 \]\[= 7500\ \text{W} = 7.5\ \text{kW} \]
Final Answer: (A) 7.5 kW

A correctly designed convergent-divergent nozzle working at a designed load is

(A) always isentropic
(B) always choked
(C) never choked
(D) never isentropic

Solution =

Explanation: A convergent-divergent (C-D) nozzle, also known as a De Laval nozzle, is used to accelerate fluids (typically gases) to supersonic speeds. When it is correctly designed and operating at design conditions, the flow behaves as follows:
The throat (the narrowest section) is where the flow becomes sonic, i.e., the Mach number $ M = 1 $. This is called the choked condition.
In the divergent section, the flow expands and accelerates to supersonic speeds, i.e., $ M > 1 $.
A “designed load” means the nozzle operates under specific pressure conditions such that the exit pressure equals the ambient (back) pressure.
Under these conditions, choking always occurs at the throat, ensuring proper acceleration through the nozzle.

Why not the other options?
A) always isentropic: Not necessarily. Although design conditions aim for isentropic flow, real conditions may introduce friction or shocks.
C) never choked: Incorrect. At design conditions, the flow must be choked at the throat.
D) never isentropic: Incorrect. The flow can be isentropic if there are no shocks or dissipative effects.

Flow separation in flow past a solid object is caused by

(A) a reduction of pressure to vapour pressure
(B) a negative pressure gradient
(C) a positive pressure gradient
(D) the boundary layer thickness reducing to zero

Solution =

Flow separation occurs when the fluid particles near a surface reverse direction and detach from the surface. This typically happens due to an adverse pressure gradient, which means the pressure is increasing in the direction of the flow.

Key Concepts:
A positive pressure gradient means:
\[ \frac{dp}{dx} > 0 \] i.e., pressure increases in the flow direction.
As the fluid moves from a region of low pressure to high pressure, the kinetic energy decreases since some of it is used to overcome the pressure rise.
In the boundary layer, where the fluid has already been slowed by viscosity, this loss of energy causes the flow to reverse direction, leading to separation.

Why the other options are incorrect:
A) a reduction of pressure to vapour pressure → This causes cavitation, not flow separation.
B) a negative pressure gradient → This helps the flow stay attached by accelerating it.
D) the boundary layer thickness reducing to zero → This does not lead to separation. In fact, separation is associated with the boundary layer thickening.

Correct Answer: C) a positive pressure gradient

The velocity field of a two-dimensional, incompressible flow is given by: \[ \vec{V} = 2 \sinh x \hat{i} + v(x,y)\hat{j} \] where $\hat{i}$ and $\hat{j}$ denote the unit vectors in $x$ and $y$ directions, respectively. If $v(x, 0) = \cosh x$, then $v(0, -1)$ is:

(A) 1
(B) 2
(C) 3
(D) 4

Solution =

Step 1: Apply Incompressibility Condition
For an incompressible flow, the divergence of velocity must be zero:
\[ \nabla \cdot \vec{V} = 0 \] \[ \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0 \] Given $u = 2\sinh x$, we compute its derivative:
\[ \frac{\partial u}{\partial x} = 2\cosh x \] Thus, the y-component must satisfy: \[ \frac{\partial v}{\partial y} = -2\cosh x \]
Step 2: Integrate to Find v(x,y)
Integrate with respect to y:
\[ v(x,y) = -2\cosh x \cdot y + f(x) \] Where $f(x)$ is an integration constant that may depend on x.
Step 3: Apply Boundary Condition
Given $v(x,0) = \cosh x$:
\[ \cosh x = -2\cosh x \cdot 0 + f(x) \Rightarrow f(x) \]\[= \cosh x \] Thus, the complete expression for v(x,y) is: \[ v(x,y) = -2y\cosh x + \cosh x \]\[= \cosh x (1 – 2y) \]
Step 4: Evaluate at (0,-1)
Compute $v(0,-1)$:
\[ v(0,-1) = \cosh 0 (1 – 2(-1)) \]\[= 1 \times (1 + 2) = 3 \] Note that $\cosh 0 = 1$.
Final Answer: $\boxed{C}$

Consider a steady flow through a horizontal divergent channel with supersonic flow at the inlet. The direction of flow is from left to right. Pressure at location B is observed to be higher than at upstream location A. Which among the following options can be the reason?

(A) Since volume flow rate is constant, velocity at B is lower than velocity at A
(B) Normal shock
(C) Viscous effect
(D) Boundary layer separation

Solution =

Solution: Key Concepts:
1. Supersonic flow in divergent channels behaves differently than subsonic flow
2. Pressure increases when supersonic flow decelerates (contrary to subsonic behavior)
3. Shock waves can cause sudden pressure increases

Option A Analysis:
“Volume flow rate constant → lower velocity at B”
– For supersonic flow in a divergent channel: – Area ↑ → Velocity ↑ (opposite to subsonic behavior) – Pressure actually increases when velocity decreases in supersonic flow – However, the argument about volume flow rate alone doesn’t explain the pressure rise mechanism. Not the best answer.

Option B Analysis:
“Normal shock”
– A normal shock causes: – Sudden transition from supersonic to subsonic flow – Large pressure increase (while decreasing velocity) – Exactly matches the observed phenomenon – This is the correct explanation.

Option C Analysis:Viscous effect
– Viscosity causes gradual pressure losses – Cannot explain sudden pressure increase – Incorrect.

Option D Analysis:
“Boundary layer separation”
– Separation causes pressure recovery, but: – Typically occurs downstream of shocks – Not the primary cause of pressure increase in supersonic flow – Secondary effect, not main reason.

Final Answer: The pressure increase is caused by a normal shock. $\boxed{B}$

Which one of the following statement is TRUE?

(A) Both Pelton and Francis turbines are impulse turbines.
(B) Francis turbine is a reaction turbine but Kaplan turbine is an impulse turbine.
(C) Francis turbine is an axial’flow reaction turbine.
(D) Kaplan turbine is an axial’flow reaction turbine.

Solution =

Let’s analyze each statement about turbine types:
A. Both Pelton and Francis turbines are impulse turbines.
This is FALSE because:
Pelton turbine is an impulse turbine (uses jet of water impacting buckets)
Francis turbine is a reaction turbine (water pressure changes through the turbine)
B. Francis turbine is a reaction turbine but Kaplan turbine is an impulse turbine.
This is FALSE because:
Francis turbine is correctly identified as reaction turbine
But Kaplan turbine is also a reaction turbine (not impulse)
C. Francis turbine is an axial-flow reaction turbine.
This is FALSE because:
Francis turbine is a radial/centripetal-flow reaction turbine
Axial-flow refers to turbines where water flows parallel to the axis (like Kaplan)
D. Kaplan turbine is an axial-flow reaction turbine.
This is TRUE because:
Kaplan turbine is indeed a reaction turbine (pressure changes through runner)
It has axial-flow design (water flows parallel to the axis of rotation)
Adjustable blades make it suitable for low-head, high-flow situations

Conclusion:
The TRUE statement is: D. Kaplan turbine is an axial-flow reaction turbine.

Which one of the following is Not a rotating machine?

(A) Centrifugal pump
(B) Gear pump
(C) Jet pump
(D) vane

Solution =

We need to identify which of the given options is not a rotating machine. Let’s analyze each option:
A. Centrifugal pump
This is a rotating machine. It uses an impeller (rotating component) to increase the pressure of the fluid.
B. Gear pump
This is a rotating machine. It uses meshing gears to pump fluid by displacement.
C. Jet pump
This is not a rotating machine. It works on the principle of fluid dynamics (Venturi effect) without any rotating parts.
D. Vane pump
This is a rotating machine. It uses rotating vanes to pump fluids.

Conclusion:
The correct answer is: \[ \boxed{C\ \text{Jet pump}} \]

For steady flow of a viscous incompressible fluid through a circular pipe of constant diameter, the average velocity in the fully developed region is constant. Which one of the following statements about the average velocity in the developing region is true?

(A) It increases until the flow is fully developed.
(B) It is constant and is eaual to the average velocity in the fully devloped region.
(C) It decreases until the flow is fully developed
(D) It is constant but is always lower than the average velocity in the fully developed region.

Solution =

Solution: For steady flow of a viscous incompressible fluid through a circular pipe:

Key Concepts:
In the developing region (entrance length), the velocity profile evolves from a uniform profile to a parabolic profile
In the fully developed region, the velocity profile becomes parabolic and remains unchanged
Due to continuity (conservation of mass), the average velocity remains constant throughout the pipe

Analysis of Options:
A. It increases until the flow is fully developed
Incorrect – The average velocity doesn’t increase
B. It is constant and equal to the average velocity in the fully developed region
Correct – Conservation of mass requires the average velocity to remain constant
C. It decreases until the flow is fully developed
Incorrect – The average velocity doesn’t decrease
D. It is constant but is always lower than the average velocity in the fully developed region
Incorrect – The average velocity is equal in both regions

Conclusion:
The correct statement is B: It is constant and is equal to the average velocity in the fully developed region

Oil flows through a 200 mm diameter horizontal cast iron pipe (friction factor, f = 0.0225) of length 500 m. The volumetric flow rate is $ 0.2 \, \text{m}^3/\text{s} $. The head loss (in m) due to friction is ( $ assume g = 9.81 \, \text{m/s}^2 $)

(A) 116.18
(B) 0.116
(C) 18.22
(D) 232.36

Solution =

Given:
Diameter $ D = 200 \, \text{mm} = 0.2 \, \text{m} $
Length $ L = 500 \, \text{m} $
Friction factor $ f = 0.0225 $
Flow rate $ Q = 0.2 \, \text{m}^3/\text{s} $
Acceleration due to gravity $ g = 9.81 \, \text{m/s}^2 $
We need to calculate head loss due to friction using the Darcy-Weisbach equation:
\[ h_f = \frac{f L v^2}{D \cdot 2g} \]
Step 1: Calculate cross-sectional area
\[ A = \frac{\pi D^2}{4} = \frac{\pi (0.2)^2}{4} = 0.0314 \, \text{m}^2 \]
Step 2: Calculate velocity
\[ v = \frac{Q}{A} = \frac{0.2}{0.0314} \approx 6.369 \, \text{m/s} \]
Step 3: Plug into the Darcy-Weisbach equation
\[ h_f = \frac{0.0225 \cdot 500 \cdot (6.369)^2}{0.2 \cdot 2 \cdot 9.81} \]
\[ h_f = \frac{0.0225 \cdot 500 \cdot 40.56}{3.924} \]
\[ h_f \approx \frac{456.3}{3.924} \approx 116.3 \, \text{m} \]
Final Answer: A) 116.18

You are asked to evaluate assorted fluid flows for their suitability in a given laboratory application. The following three flow choices, expressed in terms of the two-dimensional velocity fields in the xy-plane, are made available. P: $ u = 2y, \, v = -3x $, Q: $ u = 3xy, \, v = 0 $, R: $ u = -2x, \, v = 2y $ Which flow(s) should be recommended when the application requires the flow to be incompressible and irrotational?

(A) P and R
(B) Q
(C) Q and R
(D) R

Solution =

Incompressibility condition:
\[ \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0 \]
Irrotationality condition (2D):
\[ \frac{\partial v}{\partial x} – \frac{\partial u}{\partial y} = 0 \]
Option P: $ u = 2y, \, v = -3x $
Incompressibility: $ \frac{\partial u}{\partial x} = 0, \, \frac{\partial v}{\partial y} = 0 \Rightarrow 0 + 0 = 0 \quad \text{✅} $
Irrotationality: $ \frac{\partial v}{\partial x} = -3, \, \frac{\partial u}{\partial y} = 2 $$\Rightarrow -3 – 2 = -5 \quad \text{❌} $
Option Q: $ u = 3xy, \, v = 0 $
Incompressibility: $ \frac{\partial u}{\partial x} = 3y, \, \frac{\partial v}{\partial y} = 0 \Rightarrow 3y \neq 0 \quad \text{❌} $
Irrotationality: $ \frac{\partial v}{\partial x} = 0, \, \frac{\partial u}{\partial y} = 3x $$\Rightarrow -3x \neq 0 \quad \text{❌} $
Option R: $ u = -2x, \, v = 2y $
Incompressibility: $ \frac{\partial u}{\partial x} = -2, \, \frac{\partial v}{\partial y} = 2 $$\Rightarrow -2 + 2 = 0 \quad \text{✅} $
Irrotationality: $ \frac{\partial v}{\partial x} = 0, \, \frac{\partial u}{\partial y} = 0 $$\Rightarrow 0 – 0 = 0 \quad \text{✅} $
Correct Answer: Option D — R

A large hydraulic turbine is to generate 300 kW at 1000 rpm under a head of 40 m. For initial testing, a 1:4 scale model of the turbine operates under a head of 10 m. The power generated by the model (in kW) will be

(A) 2.34
(B) 4.68
(C) 9.38
(D) 18.75

Solution =

Given Parameters:
Prototype Turbine:
Power: $ P_{\text{prototype}} = 300 \, \text{kW} $
Head: $ H_{\text{prototype}} = 40 \, \text{m} $
Speed: $ N_{\text{prototype}} = 1000 \, \text{rpm} $
Model Turbine: Scale ratio: $ \frac{D_{\text{model}}}{D_{\text{prototype}}} = \frac{1}{4} $
Head: $ H_{\text{model}} = 10 \, \text{m} $

Solution:
The power similarity law for hydraulic turbines is: \[ \frac{P_{\text{model}}}{P_{\text{prototype}}} \]\[= \left( \frac{H_{\text{model}}}{H_{\text{prototype}}} \right)^{3/2} \times \left( \frac{D_{\text{model}}}{D_{\text{prototype}}} \right)^2 \]
Step 1: Substitute the given values \[ \frac{P_{\text{model}}}{300} = \left( \frac{10}{40} \right)^{3/2} \times \left( \frac{1}{4} \right)^2 \]
Step 2: Simplify the equation \[ \frac{P_{\text{model}}}{300} = \left( \frac{1}{4} \right)^{3/2} \times \left( \frac{1}{16} \right) \] \[ \frac{P_{\text{model}}}{300} = \left( \frac{1}{8} \right) \times \left( \frac{1}{16} \right) = \frac{1}{128} \]
Step 3: Solve for model power \[ P_{\text{model}} = \frac{300}{128} = 2.34375 \, \text{kW} \]
Step 4: Round the result \[ P_{\text{model}} \approx 2.34 \, \text{kW} \]
Final Answer:
The power generated by the model turbine is 2.34 kW, which corresponds to option A.

For a fluid flow through a divergent pipe of length ( L ) having inlet and outlet radii and $ R_1 $ and $ R_2 $ respectively and a constant flow rate of Q, assuming the velocity to be axial and uniform at any cross section, the acceleration at the exit is

(A) $ \frac{2Q(R_1 – R_2)}{\pi L R_2^3} $
(B) $ \frac{2Q^2(R_1 – R_2)}{\pi L R_2^3} $
(C) $ \frac{2Q^2(R_1 – R_2)}{\pi^2 L R_2^5} $
(D) $ \frac{2Q^2(R_2 – R_1)}{\pi^2 L R_2^5} $

Solution =

Given:
Length of pipe: $ L $
Inlet radius: $ R_1 $, Outlet radius: $ R_2 $
Constant flow rate: $ Q $
Axial and uniform velocity at all cross-sections

Step-by-Step Solution:
Velocity at any cross-section: \[ v = \frac{Q}{A} = \frac{Q}{\pi R^2} \] At the exit (radius $ R_2 $): \[ v = \frac{Q}{\pi R_2^2} \]
Acceleration Along the Axis:
Using material derivative: \[ a = \frac{dv}{dt} = v \frac{dv}{dx} \] Let the radius vary linearly from $ R_1 $ to $ R_2 $ over the length $ L $: \[ R(x) = R_1 + \left( \frac{R_2 – R_1}{L} \right) x \] Then, \[ v(x) = \frac{Q}{\pi R(x)^2} \]\[ \Rightarrow \frac{dv}{dx} = \frac{d}{dx} \left( \frac{Q}{\pi R(x)^2} \right) = \frac{-2Q}{\pi R^3} \cdot \frac{dR}{dx} \] Since \[ \frac{dR}{dx} = \frac{R_2 – R_1}{L} \] At the exit ($ R = R_2 $):

Acceleration at the Exit:
\[ a = v \cdot \frac{dv}{dx} \]\[ = \left( \frac{Q}{\pi R_2^2} \right) \cdot \left( \frac{-2Q}{\pi R_2^3} \cdot \frac{R_2 – R_1}{L} \right) \]\[ = \frac{-2Q^2 (R_2 – R_1)}{\pi^2 L R_2^5} \] Taking the magnitude (acceleration is opposite to flow direction): \[ a = \boxed{ \frac{2Q^2 (R_1 – R_2)}{\pi^2 L R_2^5} } \]
Correct Option: C

If there are m physical quantities and n fundamental dimensions in a particular process, the number of non-dimensional parameters is

(A) m + n
(B) m * n
(C) m – n
(D) m / n

Solution =

Explanation:
This question is based on the Buckingham Pi Theorem, a key principle in dimensional analysis.

Buckingham Pi Theorem:
If a physical problem involves:
$ m $ physical quantities
$ n $ fundamental dimensions (e.g., mass, length, time, temperature, etc.)
Then, the number of independent non-dimensional parameters (also called $ \pi $ terms) that describe the system is:
\[ \boxed{m – n} \] Example:
$ m = 5 $ physical quantities
$ n = 3 $ fundamental dimensions (e.g., M, L, T)
Then the number of dimensionless parameters is:
\[ 5 – 3 = 2 \]