# First Law, Heat, Work and Energy

 Question 1
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   Thermodynamics
Question 1 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}
 Question 2
Moist air at 105 kPa, 30$^{\circ}C$ and 80% relative humidity flows over a cooling coil in an insulated air-conditioning duct. Saturated air exits the duct at 100 kPa and 15$^{\circ}C$. The saturation pressure of water at 30$^{\circ}C$ and 15$^{\circ}C$ are 4.24 kPa and 1.7 kPa respectively. Molecular weight of water is 18 g/mol and that of air is 28.94 g/mol. The mass of water condensing out from the duct is ______ g/kg of dry air (round off to 2 decimal places).
 A 8.21 B 15.24 C 10.01 D 12.24
GATE ME 2020 SET-2   Thermodynamics
Question 2 Explanation:
\begin{aligned} P_{t 1} &=105 \mathrm{kPa}, \mathrm{DBT}_{1}=30^{\circ} \mathrm{C}, \phi_{1}=0.8 \\ P_{t 2} &=100 \mathrm{kPa}, \mathrm{DBT}_{2}=15^{\circ} \mathrm{C}, \phi_{2}=1 \\ P_{v s 1} &=4.24 \mathrm{kPa} \\ P_{v s 2} &=1.7 \mathrm{kPa} \\ M_{\text {water }} &=18 \mathrm{g} / \mathrm{mol} \\ M_{\mathrm{air}} &=28.94 \mathrm{g} / \mathrm{mol} \\ \omega_{1} &=\frac{18}{28.94} \times \frac{P_{v 1}}{P_{t 1}-P_{v 1}} \\ \phi_{1} &=\frac{P_{v 1}}{P_{v s 1}} \end{aligned}
\begin{aligned} 0.8&=\frac{P_{\mathrm{v}}}{4.24} \\ \therefore \qquad P_{\mathrm{v} 1}&=3.392 \\ \therefore \qquad \omega_{1}&=\frac{18}{28.94} \times \frac{3.392}{105-3.392}\\ &=0.02076 \mathrm{kgV} / \mathrm{kg} \mathrm{d} . \mathrm{a} \\ &=20.76 \mathrm{gv} / \mathrm{kgd.a} \\ \omega_{2}&=\frac{18}{28.94} \times \frac{P_{\mathrm{v} 2}}{P_{\mathrm{t} 2}-P_{\mathrm{v} 2}} \\ \phi_{2}&=\frac{P_{\mathrm{v} 2}}{P_{\mathrm{vs} 2}} \\ 1&=\frac{P_{\mathrm{v} 2}}{1.7} \\ \therefore \qquad P_{\mathrm{v} 2}&=1.7 \\ \therefore \qquad \omega_{2}&=\frac{18}{28.94} \times \frac{1.7}{100-1.7}\\&=0.01075 \mathrm{Kgv} / \mathrm{kgda} \end{aligned}
Mass of water condensing $=\omega_{1}-\omega_{2}$
$\begin{array}{l} =20.76-10.75 \\ =10.01 \mathrm{g} / \mathrm{kgd} . \mathrm{a} \end{array}$
 Question 3
Air is contained in a frictionless piston-cylinder arrangement as shown in the figure.

The atmospheric pressure is 100 kPa and the initial pressure of air in the cylinder is 105 kPa. The area of piston is 300 $cm^2$. Heat is now added and the piston moves slowly from its initial position until it reaches the stops. The spring constant of the linear spring is 12.5 N/mm. Considering the air inside the cylinder as the system, the work interaction is ________ J. (round off to the nearest integer).
 A 1 B 544 C 254 D 623
GATE ME 2020 SET-2   Thermodynamics
Question 3 Explanation:
\begin{aligned} P_{0}&=100 \mathrm{kPa}, P_{1}=105 \mathrm{kPa}, \mathrm{K}=12.5 \mathrm{N} / \mathrm{mm}=12.5 \mathrm{kN} / \mathrm{m} \\ A&=300 \mathrm{cm}^{2}=300 \times 10^{-4} \mathrm{m}^{2} \\ x&=8 \mathrm{cm}=8 \times 10^{-2} \mathrm{m} \end{aligned}
1-2 constant pressure
\begin{aligned} W_{1-2} &=P_{1} \times A \times x=105 \times 300 \times 10^{-4} \times 8 \times 10^{-2} \\ &=0.252 \mathrm{kJ}=252 \mathrm{J} \\ P_{2} &=105 \mathrm{kPa} \end{aligned}

\begin{aligned} P_{3} \times A &=P_{2} A+K x \\ P_{3} &=P_{2}+\frac{K x}{A}\\ &=105+\frac{12.5 \times 8 \times 10^{-2}}{300 \times 10^{-4}} \\& =138.33 \mathrm{kPa} \\ W_{2-3} &=\frac{1}{2}\left(P_{2}+P_{3}\right) \times\left(V_{3}-V_{2}\right)\\ &=\frac{1}{2}(105+138.33) A \times x \\ &=\frac{1}{2}(243.33) \times 300 \times 10^{-4} \times 8 \times 10^{-2} \\ W_{2-3} &=0.2919 \mathrm{kJ}=291.9 \mathrm{J} \\ \therefore \quad W_{\text {total }} &=W_{1-2}+W_{2-3}\\&=0.5439 \mathrm{kJ}=543.91 \approx 544 \mathrm{J} \end{aligned}
Alternate Solution:
Total work = Workdone because of 105 kPa pressure + Workdone against spring which is equal to energy stored in spring
\begin{aligned} \text { Workdone } &=P_{1} \times A \times 2 x+\frac{1}{2} k \cdot x^{2} \\ &=105 \times 300 \times 10^{-4} \times 2 \times 8 \times 10^{-2}\\ & +\frac{1}{2} \times 12.5 \times\left(8 \times 10^{-2}\right)^{2} \\ &=0.504+0.04 \\ &=0.544 \mathrm{kJ}=544 \mathrm{J} \end{aligned}
 Question 4
One kg of air, initially at a temperature of 127$^{\circ}C$, expands reversibly at a constant pressure until the volume is doubled. If the gas constant of air is 287 J/kg.K, the magnitude of work transfer is __________ kJ (round off to 2 decimal places).
 A 156.6 B 114.8 C 89.8 D 124.4
GATE ME 2020 SET-1   Thermodynamics
Question 4 Explanation:
\begin{aligned} m=1 \mathrm{kg} ; T_{1}=127^{\circ} \mathrm{C} &=400 \mathrm{K} \\ P=C ; V_{2}=2 V_{1} ; \quad R=& 0.287 \mathrm{kJ} / \mathrm{kgK} \\ W &=P_{1}\left(V_{2}-V_{1}\right) \\ &=P_{1}\left(2 V_{1}-V_{1}\right) \\ &=P_{1} \cdot V_{1}=\mathrm{mRT}_{1} \\ &=114.8 \mathrm{kJ} \end{aligned}

 Question 5
For a simple compressible system, v, s, p and T are specific volume, specific entropy, pressure and temperature, respectively. As per Maxwell's relations, $\left (\frac{\partial v}{\partial s} \right )_p$ is equal to
 A $\left (\frac{\partial s}{\partial T} \right )_p$ B $\left (\frac{\partial p}{\partial v} \right )_T$ C $-\left (\frac{\partial T}{\partial v} \right )_p$ D $\left (\frac{\partial T}{\partial p} \right )_s$
GATE ME 2019 SET-2   Thermodynamics
Question 5 Explanation:
According to Maxwell's relation $\left(\frac{\partial \mathrm{v}}{\partial \mathrm{s}}\right)_{\mathrm{p}}=\left(\frac{\partial \mathrm{T}}{\partial \mathrm{p}}\right)$
 Question 6
A steam power cycle with regeneration as shown below on the T-S diagram employs a single open feedwater heater for efficiency improvement. The fluids mix with each other in an open feedwater heater. The turbine is isentropic and the input (bleed) to the feedwater heater from the turbine is at state 2 as shown in the figure. Process 3-4 occurs in the condenser. The pump work is negligible. The input to the boiler is at state 5. The following information is available from the steam tables:

The mass flow rate of steam bled from the turbine as a percentage of the total mass flow rate at the inlet to the turbineat state 1 is _______
 A 10 B 20 C 30 D 40
GATE ME 2019 SET-1   Thermodynamics
Question 6 Explanation:
Let '$m_{1}$' kg of steam enters the turbine, '$m_{2}$' kg of steam is bled from the turbine into the feed water heater and '$m_{1}-m_{2}$' kg of steam enters the condenser.

From energy balance of open feed water heater,
\begin{aligned} \mathrm{m}_{2} \mathrm{h}_{2}+\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) \mathrm{h}_{4}&=\mathrm{m}_{1} \mathrm{h}_{5} \\ \mathrm{m}_{2} \times 2800+\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) 175 &=\mathrm{m}_{1} \times 700 \\ 2625 \mathrm{m}_{2} &=525 \mathrm{m}_{1} \end{aligned}
$\therefore$ Percentage of steam bled from the turbine $=\frac{\mathrm{m}_{2}}{\mathrm{m}_{1}} \times 100=0.2 \times 100=20 \%$
 Question 7
If one mole of $H_2$ gas occupies a rigid container with a capacity of 1000 litres and the temperature is raised from $27^{\circ}C \; to \; 37^{\circ}C$, the change in pressure of the contained gas (round offto two decimal places), assuming ideal gas behaviour, is ________ Pa. (R=8.314 J/mol.K)
 A 83.14 B 8.31 C 63.55 D 78.62
GATE ME 2019 SET-1   Thermodynamics
Question 7 Explanation:
Volume , $\mathrm{V}=1000 \mathrm{lit}=1 \mathrm{m}^{3}$
For ideal gas, $\mathrm{PV}=\mathrm{n} \overline{\mathrm{R}} \mathrm{T}$
Differntiating above expression we get,
$\mathrm{Vdp}=\mathrm{n} \overrightarrow{\mathrm{R}} \mathrm{d} \mathrm{T}$
$1 \times \mathrm{dp}=1 \times 8.314 \times 10$
$\therefore \mathrm{dP}=83.14 \mathrm{Pa}$
 Question 8
A calorically perfect gas (specific heat at constant pressure 1000 J/$kg\cdot K$ ) enters and leaves a gas turbine with the same velocity. The temperatures of the gas at turbine entry and exit are 1100 K and 400 K, respectively. The power produced is 4.6 MW and heat escapes at the rate of 300 kJ/s through the turbine casing. The mass flow rate of the gas (in kg/s) through the turbine is
 A 6.14 B 7 C 7.5 D 8
GATE ME 2017 SET-2   Thermodynamics
Question 8 Explanation:

\begin{aligned} c_{p}&1\mathrm{kJ}/\mathrm{kg-K}\\ V_{i}&=V_{e}(\text { Inlet and exit velocity same }) \\ \dot{W}_{C . V} &=4.6 \mathrm{MW}=4.6 \times 10^{3} \mathrm{kW} \\ \dot{Q} &=300 \mathrm{kW} \end{aligned}
Applying SFEE,
\begin{aligned} \dot{m}(h_{i}+\frac{1}{2} V_{i}^{2}+g z_{i})+ \dot{Q} &=\dot{m}(h_{e}+\frac{1}{2} V_{e}^{2}+g z_{e})+\dot{W}_{C . V} \\ \dot{m}[c_{p}(T_{i}-T_{e})]+\dot{Q}&=\dot{W}_{C . V} \quad \text { (Let } z_{i}=z_{e}) \\ \dot{m}[1 \times(1100-400)]+(-300)&=4.6 \times 10^{3} \\ \Rightarrow \quad \dot{m}&=7 \mathrm{kg} / \mathrm{s} \end{aligned}
 Question 9
A mass m of a perfect gas at pressure $P_{1}$ and volume $V_{1}$ undergoes an isothermal process. The final pressure is $P_{2}$ and volume is $V_{2}$. The work done on the system is considered positive. If R is the gas constant and T is the temperature, then the work done in the process is
 A $P_{1}V_{1}ln\frac{V_{2}}{V_{1}}$ B $-P_{1}V_{1}ln\frac{P_{1}}{P_{2}}$ C $RT\,ln\,\frac{V_{2}}{V_{1}}$ D $-mRT\,ln\,\frac{P_{2}}{P_{1}}$
GATE ME 2017 SET-2   Thermodynamics
Question 9 Explanation:
Work done on the system is considered as positive, so
$\delta W=-P d V$
(-ve sign is taken as dV is-ve during compression)
\begin{aligned} W &=-\int_{1}^{2} P d V=-\int_{1}^{2} m R T \frac{d V}{V} \\ &=-m R T \int_{1}^{2} \frac{d V}{V}=-m R T \ln \frac{V_{2}}{V_{1}} \\ &=-m R T \ln \frac{P_{1}}{P_{2}}=-P_{1} V_{1} \ln \frac{P_{1}}{P_{2}} \end{aligned}
 Question 10
Air contains 79% N2 and 21% $O_{2}$ on a molar basis. Methane ($CH_{4}$) is burned with 50% excess air than required stoichiometrically. Assuming complete combustion of methane, the molar percentage of $N_{2}$ in the products is _______________
 A 70 B 73.8 C 80 D 79.8
GATE ME 2017 SET-1   Thermodynamics
Question 10 Explanation:
Stoichiometrically balanced equation for combustion of ${\mathrm{CH}}_4$
${\mathrm{CH}}_4+2{\mathrm O}_2\rightarrow{\mathrm{CO}}_2+2{\mathrm H}_2\mathrm O$
Actual moles of ${\mathrm{O}}_2$ used $=2\times\left(\frac{150}{100}\right)=3$ moles of ${\mathrm{O}}_2$

Actual equation
${\mathrm{CH}}_4+3{\mathrm o}_2\rightarrow{\mathrm{Co}}_2+2{\mathrm H}_2\mathrm o+{\mathrm O}_2$

Consider Air instead of pure ${\mathrm{O}}_2$
${\mathrm{CH}}_4+\left[3{\mathrm o}_2+3\left(\frac{79}{21}\right){\mathrm N}_2\right]\rightarrow{\mathrm{Co}}_2+2{\mathrm H}_2\mathrm O+{\mathrm O}_2+\frac{79}7\mathrm N2$

Total moles in product $=1+2+1+\frac{79}7=15.285$
Molar percentage of ${\mathrm N}_2=\frac{11.285}{15.285}\times100=73.830\%$
There are 10 questions to complete.