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
Question 6 |
For an ideal gas, a constant pressure line and a constant volume line intersect at a point,
in the Temperature (T) versus specific entropy (s) diagram. C_P is the specific heat at
constant pressure and C_V is the specific heat at constant volume. The ratio of the slopes
of the constant pressure and constant volume lines at the point of intersection is
\frac{C_P-C_V}{C_P} | |
\frac{C_P}{C_V} | |
\frac{C_P-C_V}{C_V} | |
\frac{C_V}{C_P} |
Question 6 Explanation:
\begin{aligned} \text { If } \qquad v & =C \\ T d s & =d u+p d v \\ T d s & =d u\\ d u &=C_{V} d T &[\therefore V=C ]\\ T d s &=C_{V} d T \\ \left(\frac{d T}{d s}\right)_{V} &=\frac{T}{C_{V}} \\ \text{If}\qquad P &=C\\ T d s&=d h-v d P \\ d h &=C_{P} d T \qquad [\therefore P=C]\\ T d s &=C_{P} d T \\ \left(\frac{d T}{d s}\right)_{P} &=\frac{T}{C_{P}} \\ \text{Ratio, }\quad\frac{(d T / d s)_{P}}{(d T / d s)_{V}} &=\frac{\left(T / C_{P}\right)_{P}}{\left(T / C_{V}\right)_{V}}=\frac{C_{V}}{C_{P}} \end{aligned}
Question 7 |
For an ideal gas, the value of the Joule-Thomson coefficient is
Positive | |
Negative | |
Zero | |
Indeterminate |
Question 7 Explanation:
Value of Joule Thomson coefficient for ideal gas \mu=\left(\frac{\partial T}{\partial P}\right)_{h}=0
Question 8 |
The INCORRECT statement about the characteristics of critical point of a pure substance is that
there is no constant temperature vaporization process | |
it has point of inflection with zero slope | |
the ice directly converts from solid phase to vapor phase | |
saturated liquid and saturated vapor states are identical |
Question 8 Explanation:
At critical point liquid directly converts vapour phase.
Question 9 |
For water at 25^{\circ}C , dp_{s}/dT_{s}=0.189 KPa/K (p_{s} is the saturation pressure in kPa and T_{s} is the saturation temperature in K) and the specific volume of dry saturated vapour is 43.38 m^{3} /kg. Assume that the specific volume of liquid is negligible in comparison with that of vapour. Using the Clausius-Clapeyron equation, an estimate of the enthalpy of evaporation of water at 25^{\circ}C (in kJ/kg) is __________
2444 | |
1000 | |
1050 | |
2000 |
Question 9 Explanation:
At 25^{\circ}C, \frac{dp_s}{dT_s}=0.189kPa/K
Specific volume of saturated vapour,
v_s=43.38 m^3/kg
From clausis-clapeyron equation neglecting specific volume of liquid
\begin{aligned} \frac{dp_s}{dT_s}&=\frac{h_{fg}}{T_s \times v_s}\\ h_{fg}&=\text{Enthalpy of vapourisation}\\ 0.189&=\frac{h_{fg}}{(25+273) \times 43.38}\\ \therefore \;\; h_{fg}&=2443.248\; kJ/kg \end{aligned}
Specific volume of saturated vapour,
v_s=43.38 m^3/kg
From clausis-clapeyron equation neglecting specific volume of liquid
\begin{aligned} \frac{dp_s}{dT_s}&=\frac{h_{fg}}{T_s \times v_s}\\ h_{fg}&=\text{Enthalpy of vapourisation}\\ 0.189&=\frac{h_{fg}}{(25+273) \times 43.38}\\ \therefore \;\; h_{fg}&=2443.248\; kJ/kg \end{aligned}
Question 10 |
A rigid container of volume 0.5 m^{3} contains 1.0 kg of water at 120^{\circ}C (v_{f} = 0.00106 m^{3}/kg, v_{g}= 0.8908 m^{3}/kg). The state of water is
compressed liquid | |
saturated liquid | |
a mixture of saturated liquid and saturated vapor | |
superheated vapor |
Question 10 Explanation:
v=\frac{V}{m}=\frac{0.5}{1}=0.5 \mathrm{m}^{3} / \mathrm{kg}
\therefore It is lying between v_{f} and v_{g}. The state of water is a mixture of saturated liquid and saturated vapour.
\therefore It is lying between v_{f} and v_{g}. The state of water is a mixture of saturated liquid and saturated vapour.
There are 10 questions to complete.