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 |
Which one of the following is an intensive property
of a thermodynamic system?
Mass | |
Density | |
Energy | |
Volume |
Question 3 Explanation:
Intensive property -> Density
Mass, energy and volume are extensive properties.
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).
374.71 | |
352.24 | |
874.65 | |
458.32 |
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).
512 | |
248 | |
688 | |
560 |
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)
(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)
32^{\circ}C | |
37^{\circ}C | |
53^{\circ}C | |
76^{\circ}C |
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}
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
Isentropic | |
Polytropic | |
Isothermal | |
Adiabatic |
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.
\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
1.67 | |
3.33 | |
5.04 | |
6.66 |
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}
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
(i) They are boundary phenomena
(ii) They are exact differentials
(iii) They are path functions
both (i) and (ii) | |
both (i) and (iii) | |
both (ii) and (iii) | |
only (iii) |
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.
Answer should be (B).
\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.
Answer should be (B).
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 _______
334.36K | |
985.6K | |
452.6K | |
125.6K |
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.
great work,user friendly site, but one suggestion for some questions (like q1) ,please add some explanation like formula name or concept usedlike that 👍👏
Awesome👏
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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