Page 74 - DJJ20063- Thermodynamics 1
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DJJ20063- Thermodynamics 1
The reversible constant volume process is shown on a P-V diagram in Fig. 2.5-1.
P
P2 2
P1 1
V
V1 = V2
Figure 2.5 -1 P-V diagram for reversible constant volume process
Example 2.14
3
o
3.4 kg of gas is heated at a constant volume of 0.92 m and temperature 17 C until
o
the temperature rose to 147 C. If the gas is assumed to be a perfect gas, determine:
a) the heat flow during the process
b) the beginning pressure of gas
c) the final pressure of gas
Given
Cv = 0.72 kJ/kg K
R = 0.287 kJ/kg K
Solution to Example 2.14
From the question
m = 3.4 kg
3
V1 = V2 = 0.92 m
T1 = 17 + 273 K = 290 K
T2 = 147 + 273 K = 420 K
Cv = 0.72 kJ/kg K
R = 0.287 kJ/kg K
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