Question 1 |
In the vicinity of the triple point, the equation of liquid-vapour boundary in the P-T phase diagram for ammonia is \ln P=24.38-3063/T, where P is pressure (in Pa) and T is temperature (in K). Similarly, the solid-vapour boundary is given by \ln P=27.92-3754/T. The temperature at the triple point is ________K (round off to one decimal place).
195.2 | |
25.6 | |
254.6 | |
125.5 |
Question 1 Explanation:
Liquid vapour, \ln P=24.38-\frac{3063}{T}
Solid vapour,\ln P=27.92-\frac{3754}{T}
At triple point temperature of solid, liquid and vapour is same.
\therefore equating
24.38-\frac{3063}{T}=27.92-\frac{3754}{T}
Multiplying by T on both sides
\begin{aligned} 24.38 T-3063 &=27.92 T-3754 \\ 3.54 T &=691 \\ T &=195.197 \mathrm{~K} \end{aligned}
Question 2 |
A closed vessel contains pure water, in thermal equilibrium with its vapour at 25^{\circ}C (Stage
#1), as shown.
The vessel in this stage is then kept inside an isothermal oven which is having an atmosphere of hot air maintained at 80^{\circ}C. The vessel exchanges heat with the oven atmosphere and attains a new thermal equilibrium (Stage #2). If the Valve A is now opened inside the oven, what will happen immediately after opening the valve?
The vessel in this stage is then kept inside an isothermal oven which is having an atmosphere of hot air maintained at 80^{\circ}C. The vessel exchanges heat with the oven atmosphere and attains a new thermal equilibrium (Stage #2). If the Valve A is now opened inside the oven, what will happen immediately after opening the valve?
Water vapor inside the vessel will come out of the Valve A | |
Hot air will go inside the vessel through Valve A | |
Nothing will happen - the vessel will continue to remain in equilibrium | |
All the vapor inside the vessel will immediately condense |
Question 2 Explanation:
As inside pressure is less than outside pressure, hot
air will ?ow into the vessel.
Question 3 |
For an ideal Rankine cycle operating between pressures of 30 bar and 0.04 bar, the work
output from the turbine is 903 kJ/kg and the work input to the feed pump is 3 kJ/kg.
The specific steam consumption is _________________ kg/kW.h (round off to 2 decimal
places).
2 | |
3 | |
4 | |
5 |
Question 3 Explanation:
\begin{aligned} (WD)_{\text {turbine }} &=903 \mathrm{kJ} / \mathrm{kg} \\ (WD)_{\text {pump}} &=3 \mathrm{kJ} / \mathrm{kg} \\ \text { Specific steam consumption }=& ?(\mathrm{kg} / \mathrm{kW}-\mathrm{hr}) \\ \mathrm{SSC} &=\frac{3600}{W_{T}-W_{C}} \\ &=\frac{3600}{903-3}=4 \mathrm{kg} / \mathrm{kW}-\mathrm{hr} \end{aligned}
Question 4 |
Air (ideal gas) enters a perfectly insulated compressor at a temperature of 310 K. The
pressure ratio of the compressor is 6. Specific heat at constant pressure for air is
1005 J/kg.K and ratio of specific heats at constant pressure and constant volume is
1.4. Assume that specific heats of air are constant. If the isentropic efficiency of the
compressor is 85 percent, the difference in enthalpies of air between the exit and the
inlet of the compressor is ________ kJ/kg (round off to nearest integer).
245 | |
264 | |
312 | |
224 |
Question 4 Explanation:
\begin{aligned} T_{1} &=310 \mathrm{K} \\ r_{p} &=\frac{P_{2}}{P_{1}}=6 \quad \eta_{s}=0.85 \\ C_{p} &=1.005 \mathrm{kJ} / \mathrm{kgK} \\ \gamma &=1.4 \\ \therefore \quad \frac{T_{2}}{T_{1}} &=\left(\frac{P_{2}}{P_{1}}\right)^{\frac{\gamma-1}{\gamma}} \\ \Rightarrow \quad \frac{T_{2}}{310} &=(6)^{1.4-1 / 1.4} \\ \Rightarrow \quad T_{2} &=517.225 \mathrm{K} \\ \text{Now,}\quad \eta_{\text {isen }} &=\frac{W_{\text {isen }}}{W_{\text {actual }}}=\frac{h_{2}-h_{1}}{h_{2}-h_{1}} \\ 0.85 &=\frac{C_{p}\left(T_{2}-T_{1}\right)}{h_{2}-h_{1}} \\ \Rightarrow \quad h_{2}-h_{1} &=\frac{1.005(517.22-310)}{0.85}=245 \mathrm{kJ} / \mathrm{kg} \end{aligned}
Question 5 |
The compressor of a gas turbine plant, operating on an ideal intercooled Brayton cycle,
accomplishes an overall compression ratio of 6 in a two-stage compression process,
Intercooling is used to cool the air coming out from the first stage to the inlet temperature
of the first stage, before its entry to the second stage. Air enters the compressor at
300 K and 100 kPa. If the properties of gas are constant, the intercooling pressure for
minimum compressor work is________ kPa (round off to 2 decimal places).
600 | |
124.63 | |
312.26 | |
244.94 |
Question 5 Explanation:
\frac{P_{3}}{P_{1}}=6\, \, \Rightarrow P_{3}=600\, kPa
intermediate Pressure =\sqrt{P_{3}P_{1}}=\sqrt{600\times 100}=244.94kPa
intermediate Pressure =\sqrt{P_{3}P_{1}}=\sqrt{600\times 100}=244.94kPa
There are 5 questions to complete.