Question 1 |
A vertical shaft Francis turbine rotates at 300 rpm. The available head at the inlet to the turbine is 200 m. The tip speed of the rotor is 40 m/s. Water leaves the runner of the turbine without whirl. Velocity at the exit of the draft tube is
3.5 m/s. The head losses in different components of the turbine are: (i) stator and guide vanes: 5.0 m, (ii) rotor: 10 m, and (iii) draft tube: 2 m. Flow rate through the turbine is 20 m^3/s. Take g=9.8 m/s^2. The hydraulic efficiency of the turbine is ________% (round off to one decimal place).
91.2 | |
88.4 | |
122.4 | |
68.2 |
Question 1 Explanation:
\begin{aligned} H_{\text {inlet }} &=H_{\text {turbine }}+H_{L, \text { stator }+\text { Guide }}+H_{\text {rotor }}+H_{\text {draftuen }}+\frac{V_{3}^{2}}{2 g} \\ 200 &=H_{\text {turbine }}+5+10+2+\frac{3.5^{2}}{2 \times 9.8} \\ H_{\text {turbine }} &=182.375 \mathrm{~m} \\ \eta_{H} &=\frac{H_{\text {turbine }}}{H_{\text {inlet }}}=\frac{182.375}{200}=0.9118=91.187 \% \end{aligned}
Question 2 |
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 2 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 3 |
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 3 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 4 |
An object is moving with a Mach number of 0.6 in an ideal gas environment, which is at a temperature of 350 K. The gas constant is 320 J/kg.K and ratio of specific heats is 1.3. The speed of object is _______m/s (round off to the nearest integer).
147 | |
187 | |
229 | |
312 |
Question 4 Explanation:
\begin{aligned} M &=\frac{V}{C}=\frac{V}{\sqrt{\gamma R T}} \\ 0.6 &=\frac{V}{\sqrt{1.3 \times 350 \times 320}} \\ V &=228.94 \mathrm{~m} / \mathrm{s} \simeq 229 \mathrm{~m} / \mathrm{s} \end{aligned}
Question 5 |
A two dimensional flow has velocities in x and y directions given by u=2xyt and v=-y^2 t, where t denotes time. The equation for streamline passing through x=1, \; y=1 is
x^2y=1 | |
xy^2=1 | |
x^2y^2=1 | |
x/y^2=1 |
Question 5 Explanation:
\begin{aligned} u &=2 x y t \\ v &=-y^{2} t \\ \frac{d x}{u} &=\frac{d y}{v}=\frac{d z}{w} \\ \frac{d x}{2 x y t} &=\frac{d y}{-y^{2} t} \\ -y d x &=2 x d y \\ \ln x y^{2} &=c \\ x y^{2} &=1 \end{aligned}
Question 6 |
Which of the following is responsible for eddy viscosity (or turbulent viscosity) in a turbulent boundary layer on a flat plate?
Nikuradse stresses | |
Reynolds stresses | |
Boussinesq stresses | |
Prandtl stresses |
Question 6 Explanation:
Reynolds stresses are responsible for eddy viscosity
\tau _{\text{Reynolds}}=\eta \frac{\partial \bar{u}}{\partial y}
where \eta = Eddy viscosity.
\tau _{\text{Reynolds}}=\eta \frac{\partial \bar{u}}{\partial y}
where \eta = Eddy viscosity.
Question 7 |
For a two-dimensional, incompressible flow having velocity components u and v in the x and y directions, respectively, the expression
\frac{\partial (u^2)}{\partial x}+\frac{\partial (uv)}{\partial y}
can be simplified to
\frac{\partial (u^2)}{\partial x}+\frac{\partial (uv)}{\partial y}
can be simplified to
u\frac{\partial u}{\partial x}+u\frac{\partial v}{\partial y} | |
2u\frac{\partial u}{\partial x}+u\frac{\partial v}{\partial y} | |
2u\frac{\partial u}{\partial x}+v\frac{\partial v}{\partial y} | |
u\frac{\partial u}{\partial x}+v\frac{\partial v}{\partial y} |
Question 7 Explanation:
\frac{\partial\left(u^{2}\right)}{\partial x}+\frac{\partial(u v)}{\partial y}
By differentiating:
\Rightarrow 2 u\left[\frac{\partial u}{\partial x}\right]+u \frac{\partial v}{\partial y}+v \frac{\partial u}{\partial y}
\Rightarrow u \frac{\partial u}{\partial x}+\left[u \frac{\partial u}{\partial x}+u \frac{\partial v}{\partial y}\right]+v \frac{\partial u}{\partial y}
According to continity eq. : \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0
So, u\frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}
By differentiating:
\Rightarrow 2 u\left[\frac{\partial u}{\partial x}\right]+u \frac{\partial v}{\partial y}+v \frac{\partial u}{\partial y}
\Rightarrow u \frac{\partial u}{\partial x}+\left[u \frac{\partial u}{\partial x}+u \frac{\partial v}{\partial y}\right]+v \frac{\partial u}{\partial y}
According to continity eq. : \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0
So, u\frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}
Question 8 |
A single jet Pelton wheel operates at 300 rpm. The mean diameter of the wheel is 2 m. Operating head and dimensions of jet are such that water comes out of the jet with a velocity of 40 m/s and flow rate of 5 m^3/s. The jet is deflected by the bucket at an angle of 165^{\circ}. Neglecting all losses, the power developed by the Pelton wheel is _________MW (round off to two decimal places).
1.54 | |
6.25 | |
2.65 | |
4.22 |
Question 8 Explanation:
\begin{aligned} P &=? &P=\dot{m}\left[V_{w_{1}}-V_{w_{2}}\right] u \\ N &=300 \mathrm{rpm} \\ D &=2 \mathrm{~m} \\ V_{1} &=40 \mathrm{~m} / \mathrm{s} \\ Q &=5 \mathrm{~m}^{3} / \mathrm{s} \\ \beta &=180-165=15^{\circ} \end{aligned}
\begin{aligned} u &=\frac{\pi D N}{60}=\frac{\pi \times 2 \times 300}{60} \\ u_{1} &=u_{2}=u=31.42 \mathrm{~m} / \mathrm{s} \\ V_{r 1} &=v_{1}-u \\ V_{r 1} &=40-31.42 \\ V_{r 1} &=8.58\\ V_{r 1}&=V_{r 2}=8.58 \mathrm{~m} / \mathrm{s} \text { [neglecting blade friction] } \\ V_{r 2} \cos \beta_{2} &=u_{2}-V_{w 2} \\ P &=\rho Q\left[V_{w-1}-V_{w 2}\right] \\ &=2.65 \\ P &=2.65 \mathrm{MW} \end{aligned}
\begin{aligned} u &=\frac{\pi D N}{60}=\frac{\pi \times 2 \times 300}{60} \\ u_{1} &=u_{2}=u=31.42 \mathrm{~m} / \mathrm{s} \\ V_{r 1} &=v_{1}-u \\ V_{r 1} &=40-31.42 \\ V_{r 1} &=8.58\\ V_{r 1}&=V_{r 2}=8.58 \mathrm{~m} / \mathrm{s} \text { [neglecting blade friction] } \\ V_{r 2} \cos \beta_{2} &=u_{2}-V_{w 2} \\ P &=\rho Q\left[V_{w-1}-V_{w 2}\right] \\ &=2.65 \\ P &=2.65 \mathrm{MW} \end{aligned}
Question 9 |
A cylindrical jet of water (density = 1000 kg/m^3) impinges at the center of a flat, circular plate and spreads radially outwards, as shown in the figure. The plate is resting on a linear spring with a spring constant k=1 \; kN/m. The incoming jet diameter is D=1 \; cm.
If the spring shows a steady deflection of 1 cm upon impingement of jet, then the velocity of the incoming jet is _______m/s (round off to one decimal place).
If the spring shows a steady deflection of 1 cm upon impingement of jet, then the velocity of the incoming jet is _______m/s (round off to one decimal place).
11.3 | |
18.5 | |
8.6 | |
14.2 |
Question 9 Explanation:
\begin{aligned} \delta &=1 \mathrm{~cm} \\ D &=1 \mathrm{~cm} \\ \rho &=1000 \mathrm{~kg} / \mathrm{m}^{3} \\ K &=1 \mathrm{kN}-\mathrm{m}\\ \text{Force due to jet }&=\text{ Spring force}\\ \rho A V^{2} &=k x \\ 10^{3} \times \frac{\pi}{4} \times(0.01)^{2} \times V^{2} &=1 \times 10^{3} \times 0.01 \\ V &=11.28 \mathrm{~m} / \mathrm{s} \simeq 11.3 \mathrm{~m} / \mathrm{s} \end{aligned}
Question 10 |
Consider fully developed, steady state incompressible laminar flow of a viscous fluid between two large parallel horizontal plates. The bottom plate is fixed and the top plate moves with a constant velocity of U = 4 m/s. Separation between the plates is 5 mm. There is no pressure gradient in the direction of flow. The density of fluid is 800 kg/m^3, and the kinematic viscosity is 1.25 \times 10^{-4} \; m^2/s. The average shear stress in the fluid is ________Pa (round off to the nearest integer).
58 | |
64 | |
92 | |
80 |
Question 10 Explanation:
\begin{array}{l} V=4 \mathrm{~m} / \mathrm{s} \\ \rho=800 \mathrm{~kg} / \mathrm{m}^{3} \\ \mathrm{v}=1.25 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s} \\ h=5 \mathrm{~mm} \\ \tau=\mu \cdot \frac{d u}{d y} \\ \tau=\left[800 \times 1.25 \times 10^{-4}\right] \times \frac{4}{5 \times 10^{-3}} \\ \tau=80 \mathrm{~N} / \mathrm{m}^{2} \\ =80 \mathrm{~Pa} \end{array}
There are 10 questions to complete.
