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 |
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 2 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 3 |
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 3 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 4 |
For a Kaplan (axial flow) turbine, the outlet blade velocity diagram at a section is shown
in figure.
The diameter at this section is 3m. The hub and tip diameters of the blade are 2m and 4m, respectively. The water volume flow rate is 100 m^3/s. The rotational speed of the turbine is 300 rpm. The blade outlet angle \beta is _________ degrees (round off to one decimal place).
The diameter at this section is 3m. The hub and tip diameters of the blade are 2m and 4m, respectively. The water volume flow rate is 100 m^3/s. The rotational speed of the turbine is 300 rpm. The blade outlet angle \beta is _________ degrees (round off to one decimal place).
8.6 | |
12.7 | |
18.2 | |
14.6 |
Question 4 Explanation:
\begin{array}{rl} D_{b}=2 \mathrm{m} & Q=100 \mathrm{m}^{3} / \mathrm{sec} \\ D_{o}=4 \mathrm{m} & N=300 \mathrm{rpm} \\ \tan \beta & =\frac{C_{F}}{C_{b}}=\frac{V_{F 2}}{u_{2}} \\ u_{2} & =C_{b}=\frac{\pi D N}{60}=\frac{\pi \times 3 \times 300}{60} \\ u_{2} & =C_{b}=47.12 \mathrm{m} / \mathrm{s} \\ Q & =\frac{\pi}{4}\left(4^{2}-2^{2}\right) \times V_{F 2} \\ V_{F 2} & =C_{F}=10.61 \mathrm{m} / \mathrm{s} \\ \tan \beta & =\frac{10.61}{47.12} \\ \therefore \quad \beta & =12.69^{\circ} \simeq 12.7^{\circ} \end{array}
Question 5 |
Air discharges steadily through a horizontal nozzle and impinges on a stationay vertical
plate as shown in figure.
The inlet and outlet areas of the nozzle are 0.1 m^2 and 0.02 m^2, respectively. Take air density as constant and equal to 1.2 kg/m^3. If the inlet gauge pressure of air is 0.36 kPa, the gauge pressure at point O on the plate is ________ kPa (round off to two decimal places).
The inlet and outlet areas of the nozzle are 0.1 m^2 and 0.02 m^2, respectively. Take air density as constant and equal to 1.2 kg/m^3. If the inlet gauge pressure of air is 0.36 kPa, the gauge pressure at point O on the plate is ________ kPa (round off to two decimal places).
375 | |
0.375 | |
3.75 | |
37.5 |
Question 5 Explanation:
On applying continuity equation,
\begin{aligned} \dot{m} &=\rho_{1} \cdot A_{1} \cdot v_{1}=\rho_{2} \cdot A_{2} \cdot v_{2} \\ \Rightarrow 0.1 V_{1} &=0.02 V_{2} \\ \Rightarrow V_{2} &=\frac{10}{2} V_{1}=5 V_{1} \end{aligned}
Now on applying Bernoulli between 1 and 2 section
\begin{aligned} \frac{P_{1}}{\rho g}+\frac{V_{1}^{2}}{2 g}+z_{1} &=\frac{P_{2} /^{0}}{\rho g}+\frac{V_{2}^{2}}{2 g}+z_{2} \\ \Rightarrow \quad \frac{0.36 \times 10^{3}}{1.21 \times 9.81}+\frac{V_{1}^{2}}{2 g} &=\frac{25 V_{1}^{2}}{2 g} \\ \Rightarrow \quad V_{1} &=4.98 \mathrm{m} / \mathrm{s}\\ \Rightarrow \quad V_{2}&=24.89 \mathrm{m} / \mathrm{s} \end{aligned}
On applying Bernoulli between 2 and 3 sections
\begin{aligned} \frac{P_{2} }{\rho g}+\frac{V_{2}^{2}}{2 g}+z_{2} &=\frac{P_{3}}{\rho g}+\frac{V_{3}^{2}}{z g}+z_{3} \\ P_{3} &=\frac{\rho g \cdot V_{2}^{2}}{2 g}=\frac{1.21 \times(24.89)^{2}}{2}=375 \mathrm{Pa} \\ &=0.375 \mathrm{kPa} \text { (gauge) } \end{aligned}
\begin{aligned} \dot{m} &=\rho_{1} \cdot A_{1} \cdot v_{1}=\rho_{2} \cdot A_{2} \cdot v_{2} \\ \Rightarrow 0.1 V_{1} &=0.02 V_{2} \\ \Rightarrow V_{2} &=\frac{10}{2} V_{1}=5 V_{1} \end{aligned}
Now on applying Bernoulli between 1 and 2 section
\begin{aligned} \frac{P_{1}}{\rho g}+\frac{V_{1}^{2}}{2 g}+z_{1} &=\frac{P_{2} /^{0}}{\rho g}+\frac{V_{2}^{2}}{2 g}+z_{2} \\ \Rightarrow \quad \frac{0.36 \times 10^{3}}{1.21 \times 9.81}+\frac{V_{1}^{2}}{2 g} &=\frac{25 V_{1}^{2}}{2 g} \\ \Rightarrow \quad V_{1} &=4.98 \mathrm{m} / \mathrm{s}\\ \Rightarrow \quad V_{2}&=24.89 \mathrm{m} / \mathrm{s} \end{aligned}
On applying Bernoulli between 2 and 3 sections
\begin{aligned} \frac{P_{2} }{\rho g}+\frac{V_{2}^{2}}{2 g}+z_{2} &=\frac{P_{3}}{\rho g}+\frac{V_{3}^{2}}{z g}+z_{3} \\ P_{3} &=\frac{\rho g \cdot V_{2}^{2}}{2 g}=\frac{1.21 \times(24.89)^{2}}{2}=375 \mathrm{Pa} \\ &=0.375 \mathrm{kPa} \text { (gauge) } \end{aligned}
There are 5 questions to complete.