# Thermodynamic System and Processes

 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).
 A 320 B 650 C 275 D 450
GATE ME 2022 SET-2   Thermodynamics
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}
 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).
 A 395 B 254 C 355 D 125
GATE ME 2022 SET-2   Thermodynamics
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
Which one of the following is an intensive property of a thermodynamic system?
 A Mass B Density C Energy D Volume
GATE ME 2022 SET-2   Thermodynamics
Question 3 Explanation:
Intensive property -> Density

Mass, energy and volume are extensive properties.
 Question 4
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   Thermodynamics
Question 4 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 5
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   Thermodynamics
Question 5 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 6
A gas is heated in a duct as it flows over a resistance heater. Consider a 101 kW electric heating system. The gas enters the heating section of the duct at 100 kPa and $27^{\circ}C$ with a volume flow rate of 15 $m^3/s$. If heat is lost from the gas in the duct to the surroundings at a rate of 51 kW, the exit temperature of the gas is

(Assume constant pressure, ideal gas, negligible change in kinetic and potential energies and constant specific heat; $C_p$=1 kJ/kg.K; R=0.5 kJ/kg.K)
 A $32^{\circ}C$ B $37^{\circ}C$ C $53^{\circ}C$ D $76^{\circ}C$
GATE ME 2019 SET-1   Thermodynamics
Question 6 Explanation:
$\dot{\mathrm{m}}\left[\mathrm{h}_{1}+\frac{\mathrm{V}_{1}^{2}}{2000}\right]+\frac{\mathrm{d} \mathrm{Q}}{\mathrm{dt}}=\dot{\mathrm{m}}\left[\mathrm{h}_{2}+\frac{\mathrm{V}_{2}^{2}}{2000}\right]+\frac{\mathrm{d} \mathrm{W}}{\mathrm{dt}}$
where, $\dot{\mathrm{m}}=\frac{\mathrm{P}_{1} \mathrm{V}_{1}}{\mathrm{RT}_{1}}=\frac{100 \times 15}{0.5 \times 300}=10$
Thus,
$10\left[\mathrm{c}_{\mathrm{p}} \mathrm{T}_{1}\right]+101-51=10 \times\left[\mathrm{c}_{\mathrm{p}} \times \mathrm{T}_{2}\right] \qquad\left[\because \mathrm{v}_{1}=0, \mathrm{v}_{2}=0 \text { and } \frac{\mathrm{d} \mathrm{W}}{\mathrm{dt}}=0\right]$
$10 \times 1 \times 300+50=10 \times 1 \times \mathrm{T}_{2}$
$\mathrm{T}_{2}=305 \mathrm{K}=32^{\circ} \mathrm{C}$
 Question 7
During a non-flow thermodynamic process(1-2) executed by a perfect gas, the heat interaction is equal to the work interaction ($Q_{1-2}=W_{1-2}$) when the process is
 A Isentropic B Polytropic C Isothermal D Adiabatic
GATE ME 2019 SET-1   Thermodynamics
Question 7 Explanation:
Given, $\mathrm{Q}_{1-2}=\mathrm{W}_{1-2}$
$\therefore \Delta \mathrm{U}_{1-2}=0$
$\Rightarrow \mathrm{c}_{\mathrm{v}}\left[\mathrm{T}_{2}-\mathrm{T}_{1}\right]=0$
$\Rightarrow \mathrm{T}_{1}=\mathrm{T}_{2}$
So, the process is isothermal.
 Question 8
The volume and temperature of air (assumed to be an ideal gas) in a closed vessel is 2.87 $m^{3}$ and 300K, respectively. The gauge pressure indicated by a manometer fitted to the wall of the vessel is 0.5bar. If the gas constant of air is $R=287\, J/kg\! \cdot \! K$ and the atmospheric pressure is 1 bar, the mass of air (in kg) in the vessel is
 A 1.67 B 3.33 C 5.04 D 6.66
GATE ME 2017 SET-2   Thermodynamics
Question 8 Explanation:
\begin{aligned} P_{g}&=5 \mathrm{bar}=50 \mathrm{kPa} \\ R&=287 \mathrm{J} / \mathrm{kgK}=0.287 \mathrm{kJ} / \mathrm{kg}^{\mathrm{K}} \end{aligned}
Ideas gas equation:
$P V=m R T$
Here P is absolute pressure:
\begin{aligned} P &=P_{a b s}=P_{\text {gauge }}+P_{\text {atm }} \\ &=50+101.325=151.325 \mathrm{kPa} \\ m &=\frac{P V}{R T}=\frac{151.325 \times 2.87}{0.287 \times 300}=5.04 \mathrm{kg} \end{aligned}
 Question 9
Which of the following statements are TRUE with respect to heat and work?
(i) They are boundary phenomena
(ii) They are exact differentials
(iii) They are path functions
 A both (i) and (ii) B both (i) and (iii) C both (ii) and (iii) D only (iii)
GATE ME 2016 SET-1   Thermodynamics
Question 9 Explanation:
Heat and work are boundary phenomenon and depend upon the path followed between initial and final states. Both heat and work are inexact differential since
$\begin{array}{l} \int_{1}^{2} d W \neq W_{2}-W_{1} \\ \int_{1}^{2} d Q \neq Q_{2}-Q_{1} \end{array}$
Hence statement (ii) alone is wrong.
 Question 10
Two identical metal blocks L and M (specific heat = 0.4 kJ/kg.K), each having a mass of 5 kg, are initially at 313 K. A reversible refrigerator extracts heat from block L and rejects heat to block M until the temperature of block L reaches 293 K. The final temperature (in K) of block M is _______
 A 334.36K B 985.6K C 452.6K D 125.6K
GATE ME 2014 SET-4   Thermodynamics
Question 10 Explanation:

Given data:
Specified heat,
\begin{aligned} c&=0.4 \mathrm{kJ} / \mathrm{kgK} \text{ each body}\\ \text{Mass:}m&=5 \mathrm{kg} \quad \text { each body } \end{aligned}
Initial temperature,
$T_{1}=313 \mathrm{K} \quad \text { each body }$
Final temprature of block L
$T_{2}=293 \mathrm{K}$
Let $\quad T_{f}=$ Final temperature of block M,
Entropy change of block L
$\Delta S_{L}=m c \log _{e} \frac{T_{2}}{T_{1}}$
Entropy change of block M
$\Delta S_{M}=m c \log _{e} \frac{T_{f}}{T_{1}}$
Entropy change of universe,
\begin{aligned} \Delta S_{\text {univ }} &=\Delta S_{L}+\Delta S_{M} \\ &=mc \log _{e} \frac{T_{2}}{T_{1}}+mc \log _{e} \frac{T_{f}}{T_{1}} \end{aligned}
For a reversible refrigeration,
\begin{aligned} \Delta S_{\text {univ }}&=0 \\ m c \log _{e} \frac{T_{2}}{T_{1}}&+m_{c} \log _{e} \frac{T_{f}}{T_{1}}=0\\ mc \log _{e} \frac{T_{2} T_{f}}{T_{1}^{2}}&=0\\ \text{or}\quad \log _{e} \frac{T_{2} T_{f}}{T_{1}^{2}}&=0\\ \text{or}\quad \frac{T_{2} T_{f}}{T_{1}^{2}} &=e^{0}=1 \\ T_{f} &=\frac{T_{1}^{2}}{T_{2}}=\frac{(313)^{2}}{293}=334.36 \mathrm{K} \end{aligned}
There are 10 questions to complete.

### 4 thoughts on “Thermodynamic System and Processes”

1. great work,user friendly site, but one suggestion for some questions (like q1) ,please add some explanation like formula name or concept usedlike that 👍👏

2. In qu-8:
Given, Patm = 1bar =100kpa and Pgauge = 0.5bar = 50kpa,V = 2.87m3,T = 300K
so, Pabs = Patm + Pgauge = 100+50 = 150kpa
Therefore, Mair = (150*1000*2.87)/(287*300) = 5kg