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1. Two kg of air at 500kPa, 80°C expands adiabatically in a closed system until its volume is doubled and its temperature becomes equal to that of the surroundings which is at 100kPa and 5°C.
For this process, determine
1) the maximum work
2) the change in a availability and
3) the irreversibility.
Take, Cv = 0.718 KJ/kg K, R = 0.287 KJ/kg K.
2. A reversible heat engine receives 3000 KJ of heat from a constant temperature source at 650 K . If the surroundings is at 295 K,
i) the availability of heat energy
ii) Unavailable heat.
A = Q1 –T0 (ΔS)
= 3000 –295 (3.17)
= 2064.85 KJ.
Unavailable heat (U.A) = T0 (ΔS)
= 295 (3.17)
= 935.15 KJ.
1) The availability of heat energy (A) = 2064.85 KJ.
2) Unavailable heat (U.A) = 935.15 KJ.
3. Air in a closed vessel of fixed volume 0.15 m3 exerts pressure of 12 bar at 250 °C. If the vessel is cooled so that the pressure falls to 3.5 bar , determine the final pressure, heat transfer and change of entropy.
V1 = 0.15 m3
p1 = 12 bar = 1200 KN/m2 p2 = 3.5 bar = 350 KN/m2
T1 = 250°C = 273+250 = 523 K
1) The final pressure,
2) Heat transfer
3) Change of entropy.
1) The final pressure, T2 = 152.54 K.
2) Heat transfer, Q = - 446.78 KJ.
3) Change of entropy,-1.06KJ/K. ΔS =
4. A domestic food freezer maintains a temperature of - 15°C. The ambient air is at 30°C. If the heat leaks into the freezer at a continuous rate of 1.75 KJ/s, what is the least power necessary to pump the heat out continuously?
TL = - 15°C = 273 –15 = 258 K
TH = 30°C = 273 + 30 = 303 K
QS = 1.75 KW
Least power, (W)
Least power necessary to pump heat, W = 0.305 KW.
5. A refrigerator working on reversed Carnot cycle requires 0.5 KW per KW of cooling to maintain a temperature of -15°C. Determine the following:
a) COP of the refrigerator
b) Temperature at which heat is rejected and
Amount of heat rejected to the surroundings per KW of cooling.
W = 0.5 KW
Q2 = 1 KW
T2 = -15°C = 273 –15 = 258 K.
1) COP of the refrigerator (COP)
2) Temperature at which heat is rejected (T1)
3) Amount of heat rejected to the surroundings per KW of cooling (Q1)
1) COP of the refrigerator (COP) = 2
2) Temperature at which heat is rejected (T1) = 387 K
Amount of heat rejected to the surroundings per KW of cooling (Q1) = 1.5 KW.
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