Thermodynamics

Question 1
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).
A
1.2
B
4.3
C
2.2
D
3.8
GATE ME 2021 SET-2      Second Law, Carnot Cycle and Entropy
Question 1 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 2
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).
A
25.2
B
45.6
C
62.3
D
18.4
GATE ME 2021 SET-2      Power System
Question 2 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}
Question 3
Consider adiabatic flow of air through a duct. At a given point in the duct, velocity of air is 300 m/s, temperature is 330 K and pressure is 180 kPa. Assume that the air behaves as a perfect gas with constant c_p=1.005 kJ/kg.K. The stagnation temperature at this point is ______K (round off to two decimal places).
A
374.71
B
352.24
C
874.65
D
458.32
GATE ME 2021 SET-2      Thermodynamic System and Processes
Question 3 Explanation: 
\begin{aligned} M_{1}&=\frac{V_{1}}{\sqrt{\gamma R T_{1}}}=\frac{300}{\sqrt{1.4 \times 287 \times 330}}=0.823 \\ \frac{T_{o1}}{T_{1}}&=1+\frac{\gamma-1}{2} M_{1}^{2}=1+\frac{1.4-1}{2}(0.823)^{2} \\ \frac{T_{o1}}{T_{1}}&=1.154 \\ T_{01}&=374.7037 \mathrm{~K} \end{aligned}
Question 4
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).
A
195.2
B
25.6
C
254.6
D
125.5
GATE ME 2021 SET-1      Pure Substances
Question 4 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 5
Consider a steam power plant operating on an ideal reheat Rankine cycle. The work input to the pump is 20 kJ/kg. The work output from the high pressure turbine is 750 kJ/kg. The work output from the low pressure turbine is 1500 kJ/kg. The thermal efficiency of the cycle is 50 %. The enthalpy of saturated liquid and saturated vapour at condenser pressure are 200 kJ/kg and 2600 kJ/kg, respectively. The quality of steam at the exit of the low pressure turbine is ________ % (round off to the nearest integer).
A
45
B
68
C
98
D
93
GATE ME 2021 SET-1      Power System
Question 5 Explanation: 


\begin{aligned} h_{f} &=200 \mathrm{~kJ} / \mathrm{kg} \\ h_{g} &=2600 \mathrm{~kJ} / \mathrm{kg} \\ w_{p} &=20 \mathrm{~kJ} / \mathrm{kg}=h_{6}-h_{5} \\ h_{1}-h_{2} &=750 \mathrm{~kJ} / \mathrm{kg} \\ h_{3}-h_{4} &=1500 \mathrm{~kJ} / \mathrm{kg} \\ \eta &=0.5=\frac{W_{\mathrm{NET}}}{Q_{s}}=\frac{W_{T}-W_{P}}{Q_{s}}\\ 0.5 &=\frac{750+1500-20}{Q_{S}} \\ Q_{S} &=4460 \mathrm{~kJ} / \mathrm{kg} \\ \eta &=1-\frac{Q_{R}}{Q_{S}} \\ \frac{Q_{R}}{Q_{S}} &=0.5 \\ Q_{R} &=2230 \mathrm{~kJ} / \mathrm{kg} \\ Q_{R} &=h_{4}-h_{5} \\ 2230 &=h_{4}-200 \\ h_{4} &=2430 \mathrm{~kJ} / \mathrm{kg} \\ h_{4} &=h_{f}+x\left(h_{g}-h_{f}\right) \\ 2430 &=200+x(2600-200) \\ x &=0.9291 \\ x &=93 \% \end{aligned}
Question 6
The fundamental thermodynamic relation for a rubber band is given by dU=TdS+\tau dL, where T is the absolute temperature, S is the entropy, \tau is the tension in the rubber band, and L is the length of the rubber band. Which one of the following relations is CORRECT:
A
\tau =\left ( \frac{\partial U}{\partial S}\right )_L
B
\left ( \frac{\partial T}{\partial L}\right )_S =\left ( \frac{\partial \tau}{\partial S}\right )_L
C
\left ( \frac{\partial T}{\partial S}\right )_L =\left ( \frac{\partial \tau}{\partial L}\right )_S
D
T=\left ( \frac{\partial U}{\partial S}\right )_ \tau
GATE ME 2021 SET-1      Second Law, Carnot Cycle and Entropy
Question 6 Explanation: 
\begin{aligned} d U&=T d s+\tau d L \\ \left(\frac{\partial T}{\partial L}\right)_{S}&=\left(\frac{\partial \tau}{\partial S}\right)_{L}\\ \text { Comparing with } M=\left(\frac{\partial z}{\partial x}\right)_{y}& \\ T&=\left(\frac{\partial U}{\partial S}\right)_{L}\\ \text { Comparing with } N=\left(\frac{\partial z}{\partial y}\right)_{x} \\ \tau&=\left(\frac{\partial U}{\partial L}\right)_{S} \\ M&=\left(\frac{\partial z}{\partial x}\right)_{y} \\ N&=\left(\frac{\partial z}{\partial y}\right)_{x} \\ d z&=M d x+N d y \end{aligned}
If z is exact different
\left(\frac{\partial M}{\partial y}\right)_{x}=\left(\frac{\partial N}{\partial x}\right)_{y}
Question 7
A rigid insulated tank is initially evacuated. It is connected through a valve to a supply line that carries air at a constant pressure and temperature of 250 kPa and 400 K respectively. Now the valve is opened and air is allowed to flow into the tank until the pressure inside the tank reaches to 250 kPa at which point the valve is closed. Assume that the air behaves as a perfect gas with constant properties (c_p=1.005\; kJ/kg.K, c_v=0.718\; kJ/kg.K, R=0.287 kJ/kg.K). Final temperature of the air inside the tank is _______K (round off to one decimal place).
A
512
B
248
C
688
D
560
GATE ME 2021 SET-1      Thermodynamic System and Processes
Question 7 Explanation: 


\begin{array}{l} T_{2}=\frac{C_{P}}{C_{V}} T_{1} \\ T_{2}=\left(\frac{1.005}{0.718}\right) \times 400 \\ T_{2}=559.888 \mathrm{~K} \approx 560 \mathrm{~K} \end{array}
Question 8
In which of the following pairs of cycles, both cycles have at least one isothermal process?
A
Diesel cycle and Otto cycle
B
Carnot cycle and Stirling cycle
C
Brayton cycle and Rankine cycle
D
Bell-Coleman cycle and Vapour compression refrigeration cycle
GATE ME 2021 SET-1      Second Law, Carnot Cycle and Entropy
Question 8 Explanation: 
1.Brayton

\begin{array}{ll} 1 \text { to } 2: & S=C \\ 2 \text { to } 3: & P=C \\ 3 \text { to } 4: & S=C \\ 4 \text { to } 1: & P=C \end{array}

2. VCRS

1 to 2 : Isentropic compression
2 to 3: constant pressure (P=C)
3 to 4: Isenthalpic expansion \left(h_{3}=h_{4}\right)
4 to 1: Constant pressure (P=C)

3. Carnot

1 to 2 : lsentropic compression
2 to 3 : Isothermal heat addition
3 to 4 : lsentropic expansion
4 to 1 : Isothermal heat rejection

4. Bell Coleman

1 to 2 : lsentropic compression
2 to 3 : Constant pressure
3 to 4 : lsentropic expansion
4 to 1 : Constant pressure

5. Rankine

1 to 2 : lsentropic compression
2 to 3 : Constant pressure
3 to 4 : lsentropic expansion
4 to 1 : Constant pressure

6. Diesel

1 to 2 : lsentropic compression
2 to 3 : Constant pressure (P = C)
3 to 4 : lsentropic expansion
4 to 1 : Constant volume

7. Otto

1 to 2 : lsentropic compression
2 to 3 : Constant pressure (V = C)
3 to 4 : lsentropic expansion
4 to 1 : Constant volume (V = C)
Question 9
Keeping all other parameters identical, the Compression Ratio (CR) of an air standard diesel cycle is increased from 15 to 21. Take ratio of specific heats = 1.3 and cut-off ratio of the cycle r_c = 2.

The difference between the new and the old efficiency values, in percentage,

(\eta _{new}|_{ CR = 21})-(\eta _{old}|_{CR = 15})= _______ %. (round off to one decimal place)
A
4.8
B
2.4
C
6.2
D
2.8
GATE ME 2020 SET-2      Availability and Irreversibility
Question 9 Explanation: 
\begin{aligned} \eta_{d, r=21} &=1-\left(\frac{1}{r}\right)^{\gamma-1} \times \frac{\left(\rho^{\gamma}-1\right)}{\gamma(\rho-1)}=54.87 \% \\ \eta_{d, r=15} &=1-\left(\frac{1}{r}\right)^{\gamma-1} \times \frac{\left(\rho^{\gamma}-1\right)}{\gamma(\rho-1)}=50.08 \% \\ \eta_{d, r=21}-\eta_{d, r=15} &=4.8 \% \end{aligned}
Question 10
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).
A
1513
B
1245
C
832
D
1825
GATE ME 2020 SET-2      First Law, Heat, Work and Energy
Question 10 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}


There are 10 questions to complete.

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