Question 1 |
Consider a fully adiabatic piston-cylinder arrangement as shown in the figure. The piston is
massless and cross-sectional area of the cylinder is A. The fluid inside the cylinder is air
(considered as a perfect gas), with \gamma being the ratio of the specific heat at constant pressure to
the specific heat at constant volume for air. The piston is initially located at a position L_1 . The
initial pressure of the air inside the cylinder is P_1 \gt \gt P_0 , where P_0 is the atmospheric pressure.
The stop S_1 is instantaneously removed and the piston moves to the position L_2 , where the
equilibrium pressure of air inside the cylinder is P_2 \gt \gt P_0 .
What is the work done by the piston on the atmosphere during this process?
What is the work done by the piston on the atmosphere during this process?
0 | |
P_0A(L_2-L_1) | |
P_1AL_1 \ln \frac{L_1}{L_2} | |
\frac{(P_2L_2-P_1L_1)A}{(1-\gamma )} |
Question 1 Explanation:
Initial volume \mathrm{V}_{1}=\mathrm{L}_{1} \times \mathrm{A}
Final volume V_{2}=L_{2} \times A
Work done by atmospheric air =\int_{V_{1}}^{V_{2}} P d v
\begin{aligned} & =P_{0} \int_{V_{1}}^{V_{2}} d V \\ & =P_{0}\left(L_{2} A-L_{1} A\right) \\ & =P_{0} A\left(L_{2}-L_{1}\right) \end{aligned}
Question 2 |
A heat engine extracts heat (Q_H) from a thermal reservoir at a temperature of
1000 K and rejects heat (Q_L) to a thermal reservoir at a temperature of 100 K,
while producing work (W). Which one of the combinations of (Q_H,Q_L,W) given is allowed?
Q_H=2000J, Q_L=500J, W=1000J | |
Q_H=2000J, Q_L=750J, W=1250J | |
Q_H=6000J, Q_L=500J, W=5500J | |
Q_H=6000J, Q_L=600J, W=5500J |
Question 2 Explanation:
For a reversible engine, the rate of heat rejection
is minimum.
For process to be feasible
\oint \frac{dQ}{T}\leq 0
for option (b) \oint \frac{dQ}{T}=\frac{Q_H}{T_1}-\frac{Q_2}{T_2}=\frac{2000}{1000}-\frac{750}{100} \lt 0
So cyclic process is possible.
For process to be feasible
\oint \frac{dQ}{T}\leq 0
for option (b) \oint \frac{dQ}{T}=\frac{Q_H}{T_1}-\frac{Q_2}{T_2}=\frac{2000}{1000}-\frac{750}{100} \lt 0
So cyclic process is possible.
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 |
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 4 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 5 |
In a steam power plant, superheated steam at 10 MPa and 500^{\circ}C, is expanded isentropically
in a turbine until it becomes a saturated vapour. It is then reheated at constant pressure
to 500^{\circ}C. The steam is next expanded isentropically in another turbine until it reaches
the condenser pressure of 20 kPa. Relevant properties of steam are given in the following
two tables. The work done by both the turbines together is ______ kJ/kg (roundoff to
the nearest integer).
1513 | |
1245 | |
832 | |
1825 |
Question 5 Explanation:
Given data: h_{1}=3373.6 \mathrm{kJ} / \mathrm{kg}, h_{3}=3478.4 \mathrm{kJ} / \mathrm{kg}, h_{2}=2778.1 \mathrm{kJ} / \mathrm{kg}, s_{1}=s_{2} (as from table )
\begin{aligned} s_{3} &=s_{4} \\ s_{3} &=7.7621=0.8319+x+(7.9085-0.8319) \\ x_{4} &=0.9793 \\ h_{4} &=h_{f}+x_{4} \times\left(h_{g}-h_{f}\right)=2560.91 \mathrm{kJ} / \mathrm{kg} \\ W_{T} &=\left(h_{1}-h_{2}\right)+\left(h_{3}-h_{4}\right)=1512.95 \mathrm{kJ} / \mathrm{kg} \end{aligned}
\begin{aligned} s_{3} &=s_{4} \\ s_{3} &=7.7621=0.8319+x+(7.9085-0.8319) \\ x_{4} &=0.9793 \\ h_{4} &=h_{f}+x_{4} \times\left(h_{g}-h_{f}\right)=2560.91 \mathrm{kJ} / \mathrm{kg} \\ W_{T} &=\left(h_{1}-h_{2}\right)+\left(h_{3}-h_{4}\right)=1512.95 \mathrm{kJ} / \mathrm{kg} \end{aligned}
There are 5 questions to complete.