# Flow Through Pipes

 Question 1
A high velocity water jet of cross section area $= 0.01 m^2$ and velocity $= 35 m/s$ enters a pipe filled with stagnant water. The diameter of the pipe is 0.32 m. This high velocity water jet entrains additional water from the pipe and the total water leaves the pipe with a velocity 6m/s as shown in the figure.

The flow rate of entrained water is _______litres/s (round off to two decimal places).
 A 152.45 B 125.36 C 132.55 D 148.45
GATE ME 2021 SET-2   Fluid Mechanics
Question 1 Explanation:
Applying continuity equation:
\begin{aligned} 35 \times 0.01+\dot{q} &=6 \times \frac{\pi}{4} \times 0.32^{2} \\ \dot{q} &=0.13255 \mathrm{~m}^{3} / \mathrm{s} \\ \dot{q} &=132.55 \; \mathrm{lit} / \mathrm{s} \end{aligned}
 Question 2
Water flows out from a large tank of cross-sectional area $A_t=1\; m^2$ through a small rounded orifice of cross-sectional area $A_0=1\; cm^2$, located at $y=0$. Initially the water level, measured from $y=0$, is $H=1m$. The acceleration due to gravity is $9.8 \; m/s^2$.

Neglecting any losses, the time taken by water in the tank to reach a level of $y=H/4$ is ________seconds (round off to one decimal place).
 A 2146.2 B 2258.8 C 1245.6 D 3251.2
GATE ME 2021 SET-2   Fluid Mechanics
Question 2 Explanation:

\begin{aligned} \dot{m}_{\text {in }}-\dot{m}_{\text {out }} &=\frac{d m_{w}}{d t} \;\;\;\; \left(\dot{m}_{\text {in }}=0\right)\\ -\rho A_{0} \sqrt{2 g H} &=\frac{d}{d t}[A \times H \times \rho] \\ d t &=-\frac{A}{A_{0}} \times \frac{1}{\sqrt{2 g}} \times \frac{d H}{\sqrt{H}} \\ t &=-\frac{A}{A_{0}} \times \frac{1}{\sqrt{2 g}} \int_{1}^{0.25} \frac{d H}{\sqrt{H}} \\ t &=\frac{A}{A_{0}} \times \frac{1}{\sqrt{2 g}}(2 \sqrt{H})_{0.25}^{1} \\ t &=\frac{1}{10^{-4}}(2-1) \times \frac{1}{\sqrt{2 \times 9.81}} \\ t &=2258.8 \mathrm{~s} \end{aligned}
 Question 3
Water flows through two different pipes A and B of the same circular cross-section but at different flow rates. The length of pipe A is 1.0 m and that of pipe B is 2.0 m. The flow in both the pipes is laminar and fully developed. If the frictional head loss across the length of the pipes is same, the ratio of volume flow rates $Q_B/Q_A$ is ______ (round off to two decimal places).
 A 0.24 B 0.5 C 0.96 D 0.68
GATE ME 2019 SET-2   Fluid Mechanics
Question 3 Explanation:
Given data,
Water is flowing in two different pipes A and B.

Given that, $A_{A} = A_{B} \Rightarrow d_{A} = d_{B}$ Flow is laminar in both pipes.
In laminar flow,
$\Delta \mathrm{P}=\frac{32 \mu \mathrm{VL}}{\mathrm{d}^{2}}=\frac{128 \mu \mathrm{LQ}}{\pi \mathrm{d}^{4}}$
and $\mathrm{h}_{\mathrm{f}}=\frac{\Delta \mathrm{P}}{\rho \mathrm{g}}=\frac{128 \mu \mathrm{LQ}}{\rho \pi \mathrm{gd}^{4}}$
since, $\mu, \pi, \mathrm{g} \text{and } \rho$ are all same for both pipes, we can write
$\mathrm{h}_{\mathrm{f}} \propto \mathrm{L} \mathrm{Q}(\text { as d is same })$
or $\quad \frac{\mathrm{h}_{\mathrm{fA}}}{\mathrm{h}_{\mathrm{fB}}}=\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}} \times \frac{\mathrm{Q}_{\mathrm{A}}}{\mathrm{Q}_{\mathrm{B}}}$
$\text{since,}\quad h_{\mathrm{fA}}=\mathrm{h}_{\mathrm{fB}}$
$\frac{Q_{B}}{Q_{A}}=\frac{L_{A}}{L_{B}}=\frac{1}{2}=0.50$
 Question 4
The wall of a constant diameter pipe of length 1 m is heated uniformly with flux q'' by wrapping a heater coil around it. The flow at the inlet to the pipe is hydrodynamically fully developed. The fluid is incompressible and the flow is assumed to be laminar and steady all through the pipe. The bulk temperature of the fluid is equal to $0^{\circ}C$ at the inlet and $50^{\circ}C$ at the exit. The wall temperatures are measured at three locations, P, Q and R, as shown in the figure. The flow thermally develops after some distance from the inlet. The following measurements are made:

Among the locations P, Q and R, the flow is thermally developed at
 A P, Q and R B P and Q only C Q and R only D R only
GATE ME 2019 SET-1   Fluid Mechanics
Question 4 Explanation:

Energy balance: between points 1 and 5:
$\dot{\mathbf{q}} \times \pi \mathrm{d} \times \mathrm{L}=\dot{\mathrm{m}} \mathrm{c}_{\mathrm{p}}\left(\mathrm{T}_{\mathrm{s}}-\mathrm{T}_{\mathrm{i}}\right)$
$\dot{\mathbf{q}} \times \pi \times \mathrm{d} \times 1=50 \mathrm{m} \mathrm{c}_{\mathrm{p}}$
$\dot{\mathrm{q}}=\frac{50 \mathrm{mc}_{\mathrm{p}}}{\pi \mathrm{d}}$
Similarly,
Energy balance between points 1 and 2:
$\mathrm{\dot{m}c}_{\mathrm{p}}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)=\dot{\mathrm{q}} \times \pi \times \mathrm{d} \times 0.4$
$\dot{\mathrm{m}} \times \mathrm{c}_{\mathrm{p}} \times\left(\mathrm{T}_{2}-0\right)=\frac{50 \mathrm{m} \mathrm{c}_{\mathrm{p}}}{\pi \mathrm{d}} \times \pi \mathrm{d} \times 0.4$
$\mathrm{T}_{2}=20^{\circ} \mathrm{C}$
Energy balance between points 1 and 3:
$\dot{\mathrm{m}} \mathrm{c}_{\mathrm{p}}\left(\mathrm{T}_{3}-\mathrm{T}_{1}\right)=\dot{\mathrm{q}} \times \pi \times \mathrm{d} \times 0.6$
$\dot{\mathrm{m}} \times \mathrm{c}_{p} \times\left(\mathrm{T}_{3}-0\right)=\frac{50 \mathrm{mc}_{\mathrm{p}}}{\pi \mathrm{d}} \times \pi \mathrm{d} \times 0.6 \\ \mathrm{T}_{3}=30^{\circ} \mathrm{C}$
Energy balance between points 1 and 4
$\mathrm{mc}_{p}\left(\mathrm{T}_{4}-\mathrm{T}_{1}\right)=\dot{\mathrm{q}} \times \pi \times \mathrm{d} \times 0.8$
$\dot{\mathrm{m}} \times \mathrm{c}_{\mathrm{p}} \times\left(\mathrm{T}_{4}-0\right)=\frac{50 \mathrm{mc}_{\mathrm{p}}}{\pi \mathrm{d}} \times \pi \mathrm{d} \times 0.8$
$\mathrm{T}_{4}=40^{\circ} \mathrm{C}$
In constant heat flux condition:
$\mathrm{Nu}=4.36(\text { constant })$
$\frac{\mathrm{hd}}{\mathrm{k}}=4.36$
$\mathrm{h}=4.36 \frac{\mathrm{k}}{\mathrm{d}}=\text { constant (because k and d are constant )}$
Heat flux $\mathrm{q}=\mathrm{h} \Delta \mathrm{T}$
q = constant
h = constant
that means, $\Delta \mathrm{T}$= constant (for steady and fully developed flow).
$(\Delta \mathrm{T})_{\mathrm{P}}=\mathrm{T}_{\mathrm{p}}-\mathrm{T}_{2}=50-20=30^{\circ} \mathrm{C}$
$(\Delta \mathrm{T})_{\mathrm{Q}}=\mathrm{T}_{\mathrm{Q}}-\mathrm{T}_{3}=80-30=50^{\circ} \mathrm{C}$
$(\Delta \mathrm{T})_{\mathrm{R}}=\mathrm{T}_{\mathrm{R}}-\mathrm{T}_{4}=90-40=50^{\circ} \mathrm{C}$
$(\Delta \mathrm{T})_{\mathrm{Q}}=(\Delta \mathrm{T})_{\mathrm{R}}=50^{\circ} \mathrm{C}$
The flow is fully developed at locations Q and R

 Question 5
Water flows through a pipe with a velocity given by $\vec{V}=\left ( \frac{4}{t}+x+y \right )\hat{j}$ m/s, where $\hat{j}$ is the unit vector in the y direction, $t(\gt 0)$ is in seconds, and x and y are in meters. The magnitude of total acceleration at the point (x,y)=(1,1) at t=2s is ________ $m/s^2$.
 A 1 B 2 C 3 D 4
GATE ME 2019 SET-1   Fluid Mechanics
Question 5 Explanation:
$\begin{array}{l} \overrightarrow{\mathrm{V}}=\left(\frac{4}{\mathrm{t}}+\mathrm{x}+\mathrm{y}\right) \hat{\mathrm{j}} \\ \therefore \mathrm{u}=0 \\ \mathrm{v}=\left(\frac{4}{\mathrm{t}}+\mathrm{x}+\mathrm{y}\right) \\ \mathrm{w}=0 \\ \end{array}$
\begin{aligned} \mathrm{a}_{\mathrm{x}}&=\mathrm{u} \frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\mathrm{v} \frac{\partial \mathrm{u}}{\partial \mathrm{y}}+\mathrm{w} \frac{\partial \mathrm{u}}{\partial \mathrm{z}}+\frac{\partial \mathrm{u}}{\partial \mathrm{t}}=0+0+0+0=0 \\ \mathrm{a}_{\mathrm{z}}&=\mathrm{u} \frac{\partial \mathrm{w}}{\partial \mathrm{x}}+\mathrm{v} \frac{\partial \mathrm{w}}{\partial \mathrm{y}}+\mathrm{w} \frac{\partial \mathrm{w}}{\partial \mathrm{z}}+\frac{\partial \mathrm{w}}{\partial \mathrm{t}}=0+0+0+0=0 \\ \mathrm{a}_{\mathrm{y}}&=\mathrm{a}_{\mathrm{y}}= \mathrm{u} \frac{\partial \mathrm{v}}{\partial \mathrm{x}}+\mathrm{v} \frac{\partial \mathrm{v}}{\partial \mathrm{y}}+\mathrm{w} \frac{\partial \mathrm{v}}{\partial \mathrm{z}}+\frac{\partial \mathrm{v}}{\partial \mathrm{t}} \\ & =\mathrm{v} \frac{\partial \mathrm{v}}{\partial \mathrm{y}}+\frac{\partial \mathrm{v}}{\partial \mathrm{t}} \\ & =\left(\frac{4}{\mathrm{t}}+\mathrm{x}+\mathrm{y}\right) \cdot(1)+\left(\frac{-4}{\mathrm{t}^{2}}\right) \end{aligned}
$\begin{array}{l} =x+y+\frac{4}{t}-\frac{4}{t^{2}} \\ =1+1+\frac{4}{2}-\frac{4}{2^{2}} \\ =3 \\ |\vec{a}|=\sqrt{a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}=\sqrt{0^{2}+3^{2}+0^{2}}=3 \mathrm{m} / \mathrm{s}^{2} \end{array}$
 Question 6
Air flows at the rate of 1.5 $m^{3}/s$ through a horizontal pipe with a gradually reducing crosssection as shown in the figure. The two cross-sections of the pipe have diameters of 400 mm and 200 mm. Take the air density as 1.2 kg/$m^{3}$ and assume inviscid incompressible flow. The change in pressure$p_{2}-p_{1}$ (in kPa) between sections 1 and 2 is
 A -1.28 B 2.56 C -2.13 D 1.28
GATE ME 2018 SET-2   Fluid Mechanics
Question 6 Explanation:
\begin{aligned} \frac{P_{1}}{\rho g}+\frac{V_{1}^{2}}{2 g}+Z_{1} &=\frac{P_{2}}{\rho g}+\frac{V_{2}^{2}}{2 g}+Z_{2} \qquad \left\{Z= comstant\right\}\\ A_{1} &=\frac{\pi}{4} \times 0.4^{2} \\ A_{1} &=0.1256 \mathrm{m}^{2} \\ A_{2} &=\frac{\pi}{4} \times 0.2^{2} \\ \mathrm{A} 2 &=0.0314 \mathrm{m}^{2} \\ \frac{P_{2}-P_{1}}{\rho g} &=\frac{V_{1}^{2}-V_{2}^{2}}{2 g}\\ \frac{P_{2}-P_{1}}{1.2} &=\frac{1}{2}\left[\frac{Q^{2}}{A_{1}^{2}}-\frac{Q^{2}}{A_{2}^{2}}\right] \\ P_{2}-P_{1} &=0.6 \times(1.5)^{2}\left[\frac{1}{A_{1}^{2}}-\frac{1}{A_{2}^{2}}\right] \\ P_{2}-P_{1} &=1.35\left[\frac{1}{0.01577}-\frac{1}{0.00985}\right] \\ &=1.35[63.41-1015.22] \\ &=-1284.94=-1.28 \mathrm{kPa} \end{aligned}
 Question 7
Consider steady flow of an incompressible fluid through two long and straight pipes of diameters $d_{1}$ and $d_{2}$ arranged in series. Both pipes are of equal length and the flow is turbulent in both pipes. The friction factor for turbulent flow though pipes is of the form. $f=k(Re)^{-n}$, where K and n are known positive constants and Re is the Reynolds number. Neglecting minor losses, the ratio of the frictional pressure drop in pipe 1 to that in pipe 2,$(\frac{\Delta P_{1}}{\Delta P_{2}})$, is given by
 A $(\frac{d_{2}}{d_{1}})^{(5-n)}$ B $(\frac{d_{2}}{d_{1}})^{5}$ C $(\frac{d_{2}}{d_{1}})^{(3-n)}$ D $(\frac{d_{2}}{d_{1}})^{(5+n)}$
GATE ME 2017 SET-1   Fluid Mechanics
Question 7 Explanation:

For series connection,
$Q_{1}=Q_{2}=Q$
As given $\quad l_{1}=l_{2}=l$
For horizontal pipe,
\begin{aligned} h_{f}&=\Delta p \\ h_{f_{1}}&=\frac{f_{1} L Q^{2}}{12.1 d_{1}^{5}} ; \quad h_{f_{2}}=\frac{f_{2} L Q^{2}}{12.1 d_{2}^{5}} \\ \frac{\Delta P_{1}}{\Delta P_{2}}&=\frac{f_{1}}{d_{1}^{5}} \times \frac{d_{2}^{5}}{f_{2}} \\ f&=\frac{k}{\operatorname{Re}^{n}}=\frac{k}{\left(\frac{V D}{v}\right)^{n}} \\ Q&=\frac{\pi}{4} D^{2} \times V \\ V &\propto \frac{1}{D^{2}} \\ f &\propto \frac{1}{\left(\frac{1}{D^{2}} D\right)^{n}} \end{aligned}
By equation (i) $f \propto D^{n}$
$\frac{\Delta P_{1}}{\Delta P_{2}}=\frac{d_{1}^{n}}{d_{1}^{5}} \times \frac{d_{2}^{5}}{d_{2}^{n}}$
$\frac{\Delta P_{1}}{\Delta P_{2}}=\frac{d_{2}^{5-n}}{d_{1}^{5-n}}$
 Question 8
Water (density=1000 kg/$k^{3}$ ) at ambient temperature flows through a horizontral pipe of uniform cross section at the rate of kg/s. If the pressure drop across the pipe is 100 kPa, the minimum power required to pump the water across the pipe, in watts. is_________
 A 10 B 100 C 1000 D 10000
GATE ME 2017 SET-1   Fluid Mechanics
Question 8 Explanation:
\begin{aligned} \rho &=1000 \mathrm{kg} / \mathrm{m}^{3} \\ m &=1 \mathrm{kg} / \mathrm{s} \\ \Delta p &=100 \mathrm{kPa}=100 \times 10^{3} \mathrm{Pa} \\ P &=\rho Q g h_{f} \\ &=Q \Delta p \\ &=\frac{m}{\rho} \Delta p=\frac{1}{1000} \times 100 \times 10^{3} \\ &=100 \mathrm{W} \end{aligned}
 Question 9
A channel of width 450 mm branches into two sub-channels having width 300 mm and 200 mm as shown in figure. If the volumetric flow rate (taking unit depth) of an incompressible flow through the main channel is 0.9 $m^{3}$/s and the velocity in the sub-channel of width 200 mm is 3 m/s, the velocity in the sub-channel of width 300 mm is _____________ m/s.
Assume both inlet and outlet to be at the same elevation.

 A 2m/s B 1m/s C 4m/s D 5m/s
GATE ME 2016 SET-3   Fluid Mechanics
Question 9 Explanation:

Given data:
\begin{aligned} b_{1}&=450 \mathrm{mm}=0.45 \mathrm{m} \\ b_{2}&=300 \mathrm{mm}=0.30 \mathrm{m} \\ b_{3}&=200 \mathrm{mm}=0.20 \mathrm{m} Depth: \quad h&=1 \mathrm{m} \\ Q_{1}&=0.9 \mathrm{m}^{3} / \mathrm{s} \\ V_{3}&=3 \mathrm{m} / \mathrm{s} \\ V_{2}&=? \end{aligned}
Apply equation of continuity, we get
\begin{aligned} Q_{1}&=Q_{2}+Q_{3}\\ Q_{1}&=A_{2} V_{2}+A_{3} V_{3} \\ Q_{1}&=b_{2} h V_{2}+b_{3} h V_{3} \\ 0.9&=0.30 \times 1 \times V_{2}+0.20 \times 1 \times 3 \\ 0.9&=0.30 \mathrm{V}_{2}+0.60\\ or \quad 0.30 \mathrm{V}_{2}&=0.9-0.6=0.30 \\ or\qquad v_{2}=\frac{0.30}{0.30}&=1 \mathrm{m} / \mathrm{s} \end{aligned}
 Question 10
Three parallel pipes connected at the two ends have flow-rates $Q_{1}$ , $Q_{2}$ and $Q_{3}$ respectively, and the corresponding frictional head losses are $h_{L1},h_{L2} \; and \; h_{L3}$ respectively. The correct expressions for total flow rate (Q) and frictional head loss across the two ends ($h_{L}$ ) are
 A $Q=Q_{1}+Q_{2}+Q_{3};$ $h_{L}=h_{L1}+h_{L2}+h_{L3}$ B $Q=Q_{1}+Q_{2}+Q_{3};$ $h_{L}=h_{L1}=h_{L2}=h_{L3}$ C $Q=Q_{1}=Q_{2}=Q_{3};$ $h_{L}=h_{L1}+h_{L2}+h_{L3}$ D $Q=Q_{1}=Q_{2}=Q_{3};$ $h_{L}=h_{L1}=h_{L2}=h_{L3}$
GATE ME 2015 SET-3   Fluid Mechanics
Question 10 Explanation:
Total discharge,
$Q=Q_{1}+Q_{2}+Q_{2}$
Head loss: $h_{L}=h_{L_{1}}=h_{L_{2}}=h_{L_{3}}$

There are 10 questions to complete.