Question 1 |
At steady state, 500 kg/s of steam enters a
turbine with specific enthalpy equal to 3500 kJ/
kg and specific entropy equal to 6.5 kJ \dot kg^{-1}\dot K^{-1}. It
expands reversibly in the turbine to the condenser
pressure. Heat loss occurs reversibly in the turbine
at a temperature of 500 K. If the exit specific
enthalpy and specific entropy are 2500 kJ/kg and
6.3 kJ\dot kg^{-1} \dot K^{-1}, respectively, the work output from
the turbine is ________ MW (in integer).
320 | |
650 | |
275 | |
450 |
Question 1 Explanation:
Mass flow rate of stem \dot{m}=500kg/s
h_1=300 kJ/kg
S_1=6.5 kJ/kgL
h_2=2500kJ/kg
S_2=6.3 kJ/kgK

Surrounding temperature T_o=500 K
Work output from turbine
\begin{aligned} W_T&=(h_1-h_2)-T_o(S_1-S_2)\\ &=(3500-2500)-500(6.5-6.3)\\ &=1000-100\\ &=900 kJ/kg \end{aligned}
Power output of turbine
\begin{aligned} P&=\frac{W_T}{kg} \times \dot{m}\\ &=900 \times 500\\ &=450000 kW\\ &=450 MW \end{aligned}
h_1=300 kJ/kg
S_1=6.5 kJ/kgL
h_2=2500kJ/kg
S_2=6.3 kJ/kgK

Surrounding temperature T_o=500 K
Work output from turbine
\begin{aligned} W_T&=(h_1-h_2)-T_o(S_1-S_2)\\ &=(3500-2500)-500(6.5-6.3)\\ &=1000-100\\ &=900 kJ/kg \end{aligned}
Power output of turbine
\begin{aligned} P&=\frac{W_T}{kg} \times \dot{m}\\ &=900 \times 500\\ &=450000 kW\\ &=450 MW \end{aligned}
Question 2 |
A rigid tank of volume of 8 m^3 is being filled up
with air from a pipeline connected through a valve.
Initially the valve is closed and the tank is assumed
to be completely evacuated. The air pressure and
temperature inside the pipeline are maintained at
600 kPa and 306 K, respectively. The filling of the
tank begins by opening the valve and the process
ends when the tank pressure is equal to the pipeline
pressure. During the filling process, heat loss to
the surrounding is 1000 kJ. The specific heats of
air at constant pressure and at constant volume are
1.005 kJ/kg.K and 0.718 kJ/kg.K, respectively.
Neglect changes in kinetic energy and potential
energy.
The final temperature of the tank after the completion of the filling process is _________ K (round off to the nearest integer).
The final temperature of the tank after the completion of the filling process is _________ K (round off to the nearest integer).
395 | |
254 | |
355 | |
125 |
Question 2 Explanation:

Initially the tank is completely evacuated (m_1 = 0)
After the gas is filled with tank, the gas pressure in the tank is 600 kPa.
Heat lost to surroundings (Q) = 1000 kJ
c_p=1.005 KJ/kgK
c_v=0.718KJ/kgK
R=c_p-c_v=0.287 kJ/kgK
(Changes in KE and PE are negligible)
Final temperature of gas T_f=?
m = mass of gas entering to tank
The energy balance equation is energy carried by gas in the pipe = energy of gas in the rigid tank + Heat lost to surrounding
\begin{aligned} mh_i&=mu_f+Q\\ h_i&=u_f+\frac{Q}{m}\\ C+pT_i&=C_vT_f+\frac{1000}{\frac{16724.73868}{T_f}} \end{aligned}
Mass of gas in the tank
\begin{aligned} m=\frac{P_fV_f}{RT_f}\\ &=\frac{600(8)}{0.287(T_f)}\\ &=\frac{16724.73868}{T_f}\\ 1.005(306)&=0.718T_f+\frac{1000T_f}{16724.73868}\\ &=0.718T_f+0.05979T_f\\ &=0.77779T_f\\ \therefore \; T_f&=\frac{1.005 \times 306}{0.77779}\\ &=395.389K\approx 395K \end{aligned}
Question 3 |
Consider 1 kg of an ideal gas at 1 bar and 300 K
contained in a rigid and perfectly insulated container.
The specific heat of the gas at constant volume c_v
is equal to 750 \; Jkg^{-1}K^{-1}. A stirrer performs 225 kJ of
work on the gas. Assume that the container does not
participate in the thermodynamic interaction. The
final pressure of the gas will be ______ bar
(in integer).
1 | |
2 | |
3 | |
4 |
Question 3 Explanation:
m = 1 kg, P_1
= 1 \;bar, T_1
= 300 \;K

\begin{aligned} V &= \text{Constant} \\ W_{expansion}&= 0\\ C_V &=750\frac{J}{kgK}=0.75\frac{kJ}{kgK} \\ W_{stirrer}&= 225kJ\;\;\;(-ve \;\; work)\\ P_2&=? \\ \therefore W&= W_{expansion}+W_{stirrer}\\ &=0-225=-225kJ \end{aligned}
Using Ist law of thermodynamics
\begin{aligned} Q-W&=dU=mc_v(T_2-T_1)\\ 0-(-225)&=1 \times 0.75(T_2-300)\\ T_2&=600K\\ \therefore \frac{P_2}{P_1}&=\frac{T_2}{T_1}\\ P_2&=\frac{600}{300} \times 1 =2\; bar \end{aligned}

\begin{aligned} V &= \text{Constant} \\ W_{expansion}&= 0\\ C_V &=750\frac{J}{kgK}=0.75\frac{kJ}{kgK} \\ W_{stirrer}&= 225kJ\;\;\;(-ve \;\; work)\\ P_2&=? \\ \therefore W&= W_{expansion}+W_{stirrer}\\ &=0-225=-225kJ \end{aligned}
Using Ist law of thermodynamics
\begin{aligned} Q-W&=dU=mc_v(T_2-T_1)\\ 0-(-225)&=1 \times 0.75(T_2-300)\\ T_2&=600K\\ \therefore \frac{P_2}{P_1}&=\frac{T_2}{T_1}\\ P_2&=\frac{600}{300} \times 1 =2\; bar \end{aligned}
Question 4 |
Which one of the following is an intensive property
of a thermodynamic system?
Mass | |
Density | |
Energy | |
Volume |
Question 4 Explanation:
Intensive property -> Density
Mass, energy and volume are extensive properties.
Mass, energy and volume are extensive properties.
Question 5 |
An engine running on an air standard Otto cycle
has a displacement volume 250 cm^3 and a clearance volume 35.7 cm^3. The pressure and temperature at
the beginning of the compression process are 100
kPa and 300 K, respectively. Heat transfer during
constant-volume heat addition process is 800 kJ/kg. The specific heat at constant volume is 0.718 kJ/kg.K and the ratio of specific heats at constant
pressure and constant volume is 1.4. Assume the
specific heats to remain constant during the cycle. The maximum pressure in the cycle is ______ kPa
(round off to the nearest integer).
4811 | |
1254 | |
2589 | |
2547 |
Question 5 Explanation:

\begin{aligned} V_S&=250cm^3\\ V_C&=35.7 cm^3\\ T_1&=300K\\ P_1&=100kPa\\ Q_S&=800kJ/kg\\ C_v&=0.718kJ/kgK\\ \gamma &=1.4\\ P_3&=\_\_\_kPa\\ \frac{T_2}{T_1}&=\left ( \frac{P_2}{P_1} \right )^{\frac{\gamma -1}{\gamma }}=\left ( \frac{V_1}{V_2} \right )^{\gamma -1}=\left ( \frac{V_S+V_C}{V_C} \right )^{\gamma -1}\\ \frac{T_2}{300}&=\left ( \frac{P_2}{100} \right )^{\frac{1.4-1}{1.4 }}=\left ( \frac{250+35.7}{35.7} \right )^{1.4 -1}\\ T_2&=689.31K\\ P_2&=1838.82kPa\\ Q_S&=c_v \times (T_3-T_2)\\ 800&=0.718(T_3-689.31)\\ T_3&=1803.516K\\ &\text{For Process 2-3(Volume is constant)}\\ \frac{P_3}{P_2}&=\frac{T_3}{T_2}\\ P_3&=\frac{1803.516}{689.31} \times 1838.82\\ P_3&=4811kPa \end{aligned}
Question 6 |
In a steam power plant based on Rankine cycle,
steam is initially expanded in a high-pressure
turbine. The steam is then reheated in a reheater
and finally expanded in a low-pressure turbine. The
expansion work in the high-pressure turbine is 400
kJ/kg and in the low-pressure turbine is 850 kJ/kg,
whereas the pump work is 15 kJ/kg. If the cycle
efficiency is 32%, the heat rejected in the condenser
is ________ kJ/kg (round off to 2 decimal places).
2624.37 | |
1225.36 | |
3625.25 | |
1475.36 |
Question 6 Explanation:

\begin{aligned} W_{HPT}&=400kJ/kg\\ W_{LPT}&=850kJ/kg\\ W_{P}&=15kJ/kg\\ \eta &=0.35=\frac{W_{net}}{Q_S}\\ Q_S&=3859.375 kJ/kg\\ \therefore \; W_{net}&=Q_S-Q_R\\ Q_R&=3859.375-1235\\ Q_R&=2624.37 kJ/kg \end{aligned}
Question 7 |
A polytropic process is carried out from an initial
pressure of 110 kPa and volume of 5 m^3
to a final volume of 2.5 m^3. The polytropic index is given by
n = 1.2. The absolute value of the work done during
the process is _______ kJ (round off to 2 decimal
places).
408.92 | |
215.58 | |
852.36 | |
789.14 |
Question 7 Explanation:
Polytropic process]
\begin{aligned} P_1 &= 110 kPa,\\ V_1&= 5m^3,\\ V_2&=2.5 m^3,\\ n&=1.2\\ \Rightarrow \; P_1V_1^n&=P_2V_2^n\\ P_2&=P_1\left ( \frac{V_1}{V_2} \right )^n\\ &=110 \times \left ( \frac{5}{2.5} \right )^{1.2}\\ P_2&=252.71 kPa\\ W&=\frac{P_1V_1=P_2V_2}{n-1}\\ &=\frac{110 \times 5-252.71 \times 2.5}{1.2-1}\\ W&=-408.92kJ\\ |W|&=408.92kJ \end{aligned}
\begin{aligned} P_1 &= 110 kPa,\\ V_1&= 5m^3,\\ V_2&=2.5 m^3,\\ n&=1.2\\ \Rightarrow \; P_1V_1^n&=P_2V_2^n\\ P_2&=P_1\left ( \frac{V_1}{V_2} \right )^n\\ &=110 \times \left ( \frac{5}{2.5} \right )^{1.2}\\ P_2&=252.71 kPa\\ W&=\frac{P_1V_1=P_2V_2}{n-1}\\ &=\frac{110 \times 5-252.71 \times 2.5}{1.2-1}\\ W&=-408.92kJ\\ |W|&=408.92kJ \end{aligned}
Question 8 |
The Clausius inequality holds good for
any process | |
any cycle | |
only reversible process | |
only reversible cycle |
Question 8 Explanation:
The Clausius inequality holds good for any cycle.
\oint \frac{dQ}{T}=0\Rightarrow Reversible cycle
\oint \frac{dQ}{T} \lt 0\Rightarrow Irreversible cycle
\oint \frac{dQ}{T} \gt 0\Rightarrow Impossible cycle
\oint \frac{dQ}{T}=0\Rightarrow Reversible cycle
\oint \frac{dQ}{T} \lt 0\Rightarrow Irreversible cycle
\oint \frac{dQ}{T} \gt 0\Rightarrow Impossible cycle
Question 9 |
An adiabatic vortex tube, shown in the figure given below is supplied with 5 kg/s of air (inlet 1) at 500 kPa and 300 K. Two separate streams of air are leaving the device from outlets 2 and 3. Hot air leaves the device at a rate of 3 kg/s from outlet 2 at 100 kPa and 340 K, and 2 kg/s of cold air stream is leaving the device from outlet 3 at 100 kPa and 240 K.

Assume constant specific heat of air is 1005 J/kg.K and gas constant is 287 J/kg.K. There is no work transfer across the boundary of this device. The rate of entropy generation is ________kW/K (round off to one decimal place).

Assume constant specific heat of air is 1005 J/kg.K and gas constant is 287 J/kg.K. There is no work transfer across the boundary of this device. The rate of entropy generation is ________kW/K (round off to one decimal place).
1.2 | |
4.3 | |
2.2 | |
3.8 |
Question 9 Explanation:

\begin{aligned} \left(\frac{d S}{d t}\right)_{C . V} &=\dot{S}_{i}+\dot{S}_{g e n}-\dot{S}_{\theta} \\ \dot{S}_{\text {gen }} &=\dot{S}_{e}-\dot{S}_{i} \\ &=\dot{m}_{2} s_{2}+\dot{m}_{3} s_{3}-\dot{m}_{1} s_{1} \\ &=3\left(s_{2}-s_{1}\right)+2\left(s_{3}-s_{1}\right) \\ &=3 \times 0.587+2(0.237) \\ &=2.235 \mathrm{~kW} / \mathrm{K} \simeq 2.2 \mathrm{~kW} / \mathrm{K} \end{aligned}
Question 10 |
Consider the open feed water heater (FWH) shown in the figure given below:

Specific enthalpy of steam at location 2 is 2624 kJ/kg, specific enthalpy of water at location 5 is 226.7 kJ/kg and specific enthalpy of saturated water at location 6 is 708.6 kJ/kg. If the mass flow rate of water entering the open feed water heater (at location 5) is 100 kg/s then the mass flow rate of steam at location 2 will be kg/s (round off to one decimal place).

Specific enthalpy of steam at location 2 is 2624 kJ/kg, specific enthalpy of water at location 5 is 226.7 kJ/kg and specific enthalpy of saturated water at location 6 is 708.6 kJ/kg. If the mass flow rate of water entering the open feed water heater (at location 5) is 100 kg/s then the mass flow rate of steam at location 2 will be kg/s (round off to one decimal place).
25.2 | |
45.6 | |
62.3 | |
18.4 |
Question 10 Explanation:


\begin{aligned} 100 h_{5}+(x-100) h_{2} &=x h_{6} \\ 100 \times 226.7+(x-100) 2624&=708.6 x \\ 22670+2624 x-262400 &=708.6 x \\ 2624 x-708.6 x &=239730 \\ 1915.4 x &=239730 \\ x &=125.159 \simeq 125.2 \mathrm{~kg} / \mathrm{s} \end{aligned}
Mass flow rate at state 2(x-100)=25.2 \mathrm{~kg} / \mathrm{s}
There are 10 questions to complete.
Hello,
It would have been good if options weren’t provided for numerical type questions.
Can you provide the same for XE-E,and XE-D