GATE CE · Fluid Mechanics And Hydraulics

Fluid Kinematics – GATE CE Fluid Mechanics And HydraulicsPractice Questions & PYQs

Generate GATE-level questions on Fluid Kinematics in Fluid Mechanics And Hydraulics. Focus on core concepts, previous year patterns, and numerical problem-solving techniques.

24 questions · 20 PYQs · 0 AI practice · GATE CE 2027

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Q1GATE 2025NAT

A 5.0 m wide rectangular channel carries a discharge of $10 \mathrm{~m}^3 / \mathrm{s}$ at a depth of 1.5 m under uniform flow. To produce critical flow conditions without aflecting the upstream conditions, the channel bottom elevation should be raised (in m) by _________ (rounded oft to 2 decimal places). Assume that there is no loss of head at the raise, kinetic energy correction factor is 1.0 , and acceleration due to gravity is $9.81 \mathrm{~m} / \mathrm{s}^2$ .

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📖 Explanation

Given: Channel width, $b = 5.0\ \text{m}$ Discharge, $Q = 10\ \text{m}^3/\text{s}$ Initial flow depth, $y = 1.5\ \text{m}$ $g = 9.81\ \text{m/s}^2$ Step 1: Calculate flow per unit width: $q = \frac{Q}{b} = \frac{10}{5} = 2.0\ \text{m}^2/\text{s}$ Step 2: Critical depth for rectangular channel: $y_c = \left( \frac{q^2}{g} \right)^{1/3} = \left( \frac{2.0^2}{9.81} \right)^{1/3} = (0.4077)^{1/3} = 0.740\ \text{m}$ Step 3: Specific energy at initial condition: $E_1 = y + \frac{v^2}{2g}$ $v = \frac{Q}{A} = \frac{10}{5 \times 1.5} = 1.33\ \text{m/s}$ $E_1 = 1.5 + \frac{(1.33)^2}{2 \cdot 9.81} = 1.5 + \frac{1.7689}{19.62} = 1.5 + 0.0902 = 1.5902\ \text{m}$ Step 4: Specific energy at critical condition: $E_c = \frac{3}{2} y_c = \frac{3}{2} \cdot 0.740 = 1.11\ \text{m}$ Step 5: Bottom raise required: $z = E_1 - E_c = 1.5902 - 1.11 = 0.48\ \text{m}$ Answer: $\boxed{0.48\ \text{m}}$

Q2GATE 2025MSQ

Consider a velocity vector, $\vec{V}$ in $(x, y, z)$ coordinates given below. Pick one or more CORRECT statement(s) from the choices given below. $\vec{V} = u \vec{x} + v \vec{y}$

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📖 Explanation

Given: $\vec{V} = u(x, y)\vec{x} + v(x, y)\vec{y}$ Step 1: Compute Curl of velocity Curl of a vector field in Cartesian coordinates is given by:

\nabla \times \vec{V} = \left| \begin{array}{ccc} \vec{x} & \vec{y} & \vec{z} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ u & v & 0 \\ \end{array} \right|

Expanding the determinant, we get: $\nabla \times \vec{V} = \left( \frac{\partial (0)}{\partial y} - \frac{\partial v}{\partial z} \right)\vec{x} - \left( \frac{\partial (0)}{\partial x} - \frac{\partial u}{\partial z} \right)\vec{y} + \left( \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right)\vec{z}$ So, z-component of curl is: $(\nabla \times \vec{V})_z = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}$ So, option (A) is correct. Option (B) is incorrect. Step 2: Compute Divergence of velocity Divergence is given by: $\nabla \cdot \vec{V} = \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}$ So, option (C) is correct. Option (D) is incorrect.

Q3GATE 2024MCQ

A flow velocity field $\vec{V}: \vec{V}(x, y)$ for a fluid is represented by $\vec{V}=3 \tilde{\hat{i}}+(5 x) \hat{j}$ In the context of the fluid and the flow, which one of the following statements is CORRECT?

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📖 Explanation

Continuity equation for incompressible flow :

\begin{aligned} \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y} & =0 \\ \frac{\partial}{\partial x}(3)+\frac{\partial}{\partial y}(5 x) & =0+0 \\ & =0 \text { (Incompressible Flow) } \\ \omega_{z} & =\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)=\frac{1}{2}\left[\frac{\partial}{\partial x}(5 x)-\frac{\partial}{\partial y}(3)\right] \\ & =\frac{1}{2}(5)=2.5 \neq 0 \text { (Flow is rotational) } \end{aligned}

Q4GATE 2020MCQ

The velocity components in the x and y directions for an incompressible flow are given as u=(-5+6x) and v=-(9+6y), respectively. The equation of the streamline is

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📖 Explanation

\begin{aligned} \text{Given } \; \; \; & u=-5+6x \\ & v=-(9+6y) \end{aligned}

Equation of streamline

\begin{aligned} \frac{dx}{u}&=\frac{dy}{u} \\ \frac{dx}{-5+6x}&=\frac{dy}{-(9+6y)} \\ \text{Integrating it,} \\ \ln(-5+6x)^{1/6}&= -\ln(9+6y)^{1/6}+ \ln C^{1/6}\\ \frac{1}{6} \ln (-5+6x)(9+6y)&=\frac{1}{6} \ln C \\ \text{Take antilog,} \\ (-5 + 6x)(9 + 6y) &= \text{constant} \\ \text{u.v} &= \text{constant} \end{aligned}

Q5GATE 2020MCQ

Uniform flow with velocity U makes an angle $\theta$ with the y-axis, as shown in the figure The velocity potential ( $\Phi$ ), is

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📖 Explanation

Velocity in x-depth, $u_x=u\sin \theta$ Velocity in y-depth, $u_y=u\cos \theta$

\begin{aligned} -\frac{\partial \phi }{\partial x}&=u_x \\ &{Integrating it} \\ \phi &= -u_xx+f(y)+c\\ &= -(u \sin \theta )x+f(y)+c...(i)\\ -\frac{\partial \phi }{\partial y}&=u_y \\ &\text{Integrating it} \\ \phi &= -u_y y+f(x)+c\\ &= -(u \cos \theta )y+f(x)+c...(ii)\\ & \text{By equation (i) and (ii),}\\ \phi &= -u(x \sin \theta+y \cos \theta )\\ &\text{If we take,} \\ \frac{\partial \phi }{\partial x}&= u_x \; and \; \frac{\partial \phi }{\partial y}= u_y\\ \text{then}, \; \phi &= u(x \sin \theta +y \cos \theta )\\ \text{So}, \; \phi &= \pm u(x \sin \theta +y \cos \theta ) \end{aligned}

Q6GATE 2019MCQ

The velocity field in a flow system is given by $v=2i+(x+y)j+(xyz)k$ . The acceleration of the fluid at (1,1,2) is

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📖 Explanation

\begin{aligned} \vec{v} &=2\hat{i}(x+y)\hat{j}+xyz\hat{k} \\ u&=2 \\ v&=x+y \\ w&=xyz \\ a_x&=u\frac{\partial u}{\partial x} +v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}+\frac{\partial u}{\partial t}=0\\ a_y&=x+y+2 \\ a_z&= u\frac{\partial w}{\partial x} +v\frac{\partial w}{\partial y}+w\frac{\partial w}{\partial z}+\frac{\partial w}{\partial t}\\ &= 2(yz)+(x+y)(xz)+xyz(xy)\\ a_z&=2yz+x^2z+xyz+x^2y^2z \\ &At\;(1,1,2) \\ a_y&=11+2=4\\ a_z&=2(1)(2)+(1)^2(2)+(1)(1)(2)+(1)^2(1)^2(2)\\ &=4+2+2+2=10\\ \vec{a}&=4\hat{j}+10\hat{k} \end{aligned}

Q7GATE 2019MCQ

A solid sphere of radius, r, and made of material with density, $\rho _s$ , is moving through the atmosphere (constant pressure, p)with a velocity, v. The net force ONLY due to atmospheric pressure ( $F_p$ ) acting on the sphere at any time, t, is

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Q8GATE 2018NAT

A flow field is given by $u=y^{2} ,v=-xy,w=0.$ Value of the component of the angular velocity(in radians per unit time,up to two decimal places)at the point (0,-1,1) is_____

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📖 Explanation

\begin{aligned} \omega_{z} &=\frac{1}{2}\left[\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right] \\ &=\frac{1}{2}\left[\frac{\partial}{\partial x}(-x y)-\frac{\partial}{\partial y}(y)^{2}\right]\$ \\ &=\frac{1}{2}[-y-2 y] \\ &=-\frac{3 y}{2} \\ \text { At point } &(0,-1,1) \\ \omega_{z} &=-\frac{3}{2} \times -1=1.50 \mathrm{rad} / \mathrm{s} \end{aligned}

Note: Since the density variation is not given continuity equation of incompressible flow can not be applied directly to check the possibility of flow.

Q9GATE 2015MCQ

The velocity components of a two dimensional plane motion of a fluid are: $u=\frac{y^{3}}{3}+2x-x^{2}y$ and $v=xy^{2}-2y-\frac{x^{3}}{3}$ . The correct statement is:

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📖 Explanation

Continuity equation

\begin{aligned} \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y} &=0 \\ \Rightarrow \quad(2-2 x y)+(2 x y-2) &=0 \\ 0 &=0 \end{aligned}

Hence, incompressible fluid.

\begin{aligned} \omega_{z} &=\frac{1}{2}\left[\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right] \\ &=\frac{1}{2}\left[y^{2}-\frac{3 x^{2}}{3}-\left(\frac{3 y^{2}}{3}-x^{2}\right)\right]\$ \\ &=\frac{1}{2}\left[y^{2}-x^{2}-y^{2}+x^{2}\right]\$ \\ &=0 \end{aligned}

Hence, flow is irrotational.

Q10GATE 2015NAT

A nozzle is so shaped that the average flow velocity changes linearly from 1.5 m/s at the beginning to 15 m/s at its end in a distance of 0.375 m. The magnitude of the convective acceleration (in m/ $s^{2}$ ) at the end of the nozzle is _________.

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📖 Explanation

Convective acceleration $=\frac{u \partial u}{\partial x}$ $=15 \times \frac{15-1.5}{0.375}=540 \mathrm{m} / \mathrm{s}^{2}$

Q11GATE 2015MCQ

In a two-dimensional steady flow field, in a certain region of the x-y plane, the velocity component in the x-direction is given by $v_{x}=x^{2}$ and the density varies as $\rho =\frac{1}{x}$ . Which of the following is a valid expression for the velocity component in the y- direction, $v_{y}$ ?

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📖 Explanation

Continuity equation in steady flow.

\begin{aligned} \frac{\partial(\rho u)}{\partial x}+\frac{\partial(\rho v)}{\partial y} &=0 \\ \frac{\partial}{\partial x}\left[\frac{1}{x} \cdot x^{2}\right]\$+\frac{\partial}{\partial y}\left(\frac{1}{x} \cdot v_{y}\right) &=0 \\ \frac{\partial}{\partial x}(x)+\frac{1}{x} \frac{\partial v_{y}}{x \partial y} &=0 \\ 1+\frac{1}{x} \frac{\partial v_{y}}{\partial y} &=0 \\ -x &=\frac{\partial v_{y}}{\partial y} \end{aligned}

On integrating both sides $v_{y}=-x y$

Q12GATE 2014NAT

A particle moves along a curve whose parametric equations are: $x=t^{3}+2t$ , $y=-3e^{-2t}$ and $z=2 \sin (5t)$ , where x, y and z show variations of the distance covered by the particle (in cm)with time t (in s). The magnitude of the acceleration of the particle (in $cm/s^{2}$ ) at t = 0 is ________

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📖 Explanation

\begin{aligned} x &=t^{3}+2 t \\ y &=-3 e^{-2 t} \\ z &=2 \sin (5 t) \\ \frac{d x}{d t} &=3 e^{2}+2 \\ \Rightarrow\quad a_{x} &=\frac{d^{2} x}{d t^{2}}=6 t \\ \frac{d y}{d t} &=-3 e^{-2} \times (-2)=6 e^{-2 t} \\ \Rightarrow\quad a_{y} &=\frac{d^{2} y}{d t^{3}}=-12 e^{-2 t}\\ \Rightarrow\quad \frac{d z}{d t} &=2 \times 5 \cos (5 t) \\ &=10 \cos (5 t) \\ \Rightarrow\quad a_{z} &=\frac{d^{2} z}{d t^{2}}=-50 \sin 5 t \\ \vec{a} &=a_{x} \hat{i}+a_{y} \hat{j}+a_{z} \hat{k} \\ \vec{a} \text { at } t=0 &=0 \hat{i}-12 \hat{j}+0 \hat{k} \\ \vec{a} &=-12 \hat{j} \end{aligned}

$\Rightarrow\quad$ Magnitude of acceleration at t=0 $=12 \mathrm{cm} / \mathrm{s}^{2}$

Q13GATE 2014NAT

An incompressible homogeneous fluid is flowing steadily in a variable diameter pipe having the large and small diameters as 15 cm and 5 cm, respectively. If the velocity at a section at the 15 cm diameter portion of the pipe is 2.5 m/s, the velocity of the fluid (in m/s) at a section falling in 5 cm portion of the pipe is________

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📖 Explanation

\begin{aligned} \frac{\pi}{4}(15)^{2} \times 25 &=\frac{\pi}{4}(5)^{2} \times v \\ \Rightarrow &=2.5 \times 9=22.5 \mathrm{m} / \mathrm{s} \end{aligned}

Q14GATE 2014MCQ

A plane flow has velocity components $u=\frac{x}{T_{1}}, \; v=-\frac{y}{T_{2}}$ and w=0 along x, y and z directions respectively, where $T_{1}(\neq 0)$ and $T_{2}(\neq 0)$ are constants having the dimensions of time. The given flow is incompressible if

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📖 Explanation

For a flow to exist

\begin{aligned} \Rightarrow \quad \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}&=0 \\ \Rightarrow \quad \frac{1}{T_{1}}-\frac{1}{T_{2}}&=0 \\ \Rightarrow \quad T_{1}&=T_{2} \end{aligned}

Q15GATE 2013MCQ

For a two dimensional flow field, the stream function $\Psi$ is given as $\Psi =\frac{3}{2}(y^{2}-x^{2})$ . The magnitude of discharge occurring between the stream lines passing through points (0,3) and (3,4) is:

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📖 Explanation

\begin{aligned} a &=\Psi_{1}-\Psi_{2} \\ &=\frac{3}{2}(9-0)-\frac{3}{2}(16-9) \\ &=\frac{27}{2}-24+\frac{27}{2}=3 \text { units } \end{aligned}

Q16GATE 2006MCQ

The velocity field for a flow is given by : $\vec{V}=(5x+6y+7z)\vec{i}$ + $(6x+5y+9z)\vec{j}$ + $(3x+2y+\lambda z)\vec{k}$ and the density varies as $\rho =\rho _{0}exp(-2t)$ . In order that the mass is conserved, the value of $\lambda$ should be

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📖 Explanation

The continuity equation is given by

\begin{aligned} \frac{\partial p}{\partial t}&+\frac{\partial(p u)}{\partial x}+\frac{\partial(\rho v)}{\partial y}+\frac{\partial(p w)}{\partial z}=0 \\ \text{Now,}\quad \rho&=\rho_{0} e^{-2 t} \\ \frac{\partial \rho}{\partial t} &=-2 \rho_{0} e^{-2 t}=-2 p \\ p \frac{\partial u}{\partial x}=5 p, p \frac{\partial v}{\partial y}&=5 p, p \frac{\partial w}{\partial z}=\lambda p \\ \therefore-2 p+5 p+5 p&+\lambda p=0 \\ \Rightarrow \quad 8 p+\lambda p&=0 \\ \text{or} \quad 8+\lambda&=0 \\ \text{or} \quad \lambda&=-8 \\ \end{aligned}

Q17GATE 2005MCQ

An inert tracer is injected continuously from a point in an unsteady flow field. The locus of locations of all tracer particles at an instance of time represents

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Q18GATE 2005MCQ

A stream function is given by $\Psi =2x^{2}y+(x+1)y^{2}$ The flow rate across a line joining points A(3,0) and B(0,2) is

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📖 Explanation

The rate of flow is the difference in value of stream function at points A and B

\begin{aligned} \therefore \quad Q &=\left|\psi_{A}-\psi_{B}\right| \\ \psi_{A} &=2 \times (3)^{2} \times 0+(3+1) \times (0)^{2}=0 \\ \psi_{B} &=2 \times (0)^{2} \times 2+(0+1) \times (2)^{2}=4 \\ \therefore & Q=|0-4|=4 \text { units } \end{aligned}

Q19GATE 2005MCQ

The circulation 'G' around a circle of radius 2 units for the velocity field u=2x+3y and v=-2y is

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📖 Explanation

\begin{aligned} \text { Circulation } &=\text { Vorticity } \times \text { Area } \\ &=\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right) \times \text { Area } \\ &=(0-3) \times \pi \times (2)^{2} \\ &=-12 \pi \text { units } \end{aligned}

Q20GATE 2004MCQ

The velocity in m/s at a point in a two-dimensional flow is given as $\vec{V}=2\vec{i}+3\vec{j}$ . The equation of the stream line passing through the point is

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📖 Explanation

The equation of stream line in two dimensions is given as

\begin{array}{rr} & v d x-u d y=0 \\ \Rightarrow \quad & 3 d x-2 d y=0 \end{array}

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