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).
The flow rate of entrained water is _______litres/s (round off to two decimal places).
152.45 | |
125.36 | |
132.55 | |
148.45 |
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}
\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).
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).
2146.2 | |
2258.8 | |
1245.6 | |
3251.2 |
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).
0.24 | |
0.50 | |
0.96 | |
0.68 |
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
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
Among the locations P, Q and R, the flow is thermally developed at
P, Q and R | |
P and Q only | |
Q and R only | |
R only |
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.
1 | |
2 | |
3 | |
4 |
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}
\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}
There are 5 questions to complete.