2500 Solved Problems In Thermodynamics Pdf
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
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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,
determine
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.
Result:
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.
Given Data:
V1 = 0.15 m3
p1 = 12 bar = 1200 KN/m2 p2 = 3.5 bar = 350 KN/m2
T1 = 250°C = 273+250 = 523 K
To find:
1)The final pressure,
2)Heat transfer
3)Change of entropy.
Solution:
Result:
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?
Given Data:
TL = - 15°C = 273 –15 = 258 K
TH = 30°C = 273 + 30 = 303 K
QS = 1.75 KW
To find:
Least power, (W)
Solution:
Result:
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.
Given Data:
W = 0.5 KW
Q2 = 1 KW
T2 = -15°C = 273 –15 = 258 K.
To find:
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)
Solution:
Result:
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.
1.A turbine operating under steady flow conditions receives steam at the following state: Pressure 13.8bar; Specific volume 0.143 Internal energy 2590 KJ/Kg; Velocity 30m/s. The state of the steam leaving the turbine is: Pressure 0.35bar; Specific Volume 4.37 Internal energy 2360KJ/Kg; Velocity 90m/s. Heat is lost to the surroundings at the rate of 0.25KJ/s. If the rate of steam flow is 0.38Kg/. What is the power developed by the turbine.
2.Derive the steady flow energy equation and reduce it for a turbine, pump, nozzle and a heat exchanger.
3.0.25kg of air at a pressure of 1 bar occupies a volume of 0.3 . If this air expands isothermally to a volume of 0.9 . Find
i.The initial temperature
ii.Final temperature
iii.External work done
iv.Heat absorbed by the air
Change in internal energy. Assume R=0.29 KJ/Kg K.
1.5kg of certain gas at a pressure of 8 bar and 20°C occupies the volume of 0.5 . It expands adiabatically to a pressure of 0.9 bar and volume 0.73 . Determine the work done during the process. Gas constant, ratio of specific heat, values of two specific heats, change in internal energy and change in enthalpy.
5.A cylinder contains 1 of gas at 100kpa and 100°C, the gas is polytropically compressed to a volume of 0.25 . The final pressure is 100kpa. Determine
a.Mass of the gas
b.The value of index ‘n’ for compression
c.Change in internal energy of the gas
d.Heat transferred by the gas during compression. Assume R = 0.287KJ/Kg K; γ= 1.4.