A Carnot engine uses air as the working substance, receives heat at a temperatur
ID: 1997612 • Letter: A
Question
A Carnot engine uses air as the working substance, receives heat at a temperature of 315oC, and rejects it at 65oC. The maximum possible cycle pressure is 6.0MPa and the minimum volume is 0.95 liters. When heat is added, the volume increases by 250%. Determine the pressure and volume at each state in the cycle. A Carnot engine uses air as the working substance, receives heat at a temperature of 315oC, and rejects it at 65oC. The maximum possible cycle pressure is 6.0MPa and the minimum volume is 0.95 liters. When heat is added, the volume increases by 250%. Determine the pressure and volume at each state in the cycle.Explanation / Answer
Initial temperature T = 315 o C = 315 +273 = 588 K
Initial pressure P = The maximum possible cycle pressure = 6.0MPa = 6 x10 6 Pa
Initial volume V = minimum volume = 0.95 liters =0.95 x10 -3 m 3
State 2 : from 1 to 2 is isothermal process
Temperature T' = T = 588 K
Volume V ' = 250 % of V = 2.5 V = 2.5 x 0.95x10 -3 = 2.375 x10 -3 m 3
Pressure P ' = ?
In isothermal process , P' V ' = PV
P ' = PV / V '
= (6MPa)(V) / 2.5 V
= 2.4 MPa
State 3 : from 2 to 3 is adiabatic expansion
Temperature T " = 65 o C = 65 + 273 = 338 K
Volume V " = ?
In adiabatic process , TV r-1 = constant
T " V " r-1 = T ' V ' r-1
T " / T ' = ( V ' / V " ) r-1
Where r = ratio of specific heats = 1.4
So, 338 / 588 = (2.375 / V " ) 0.4
0.5748 = (2.375 / V " ) 0.4
(2.375 / V " )= 0.5748 1/0.4
= 0.2505
V " = 9.48 lit
Pressure P " = ?
In adiabatic process , PV r = constant
P "V" r = P'V' r
P " = P ' ( V ' / V " ) r
= (2.4 MPa)(2.375 /9.48) 1.4
= 0.3456 MPa
State 4 : 3 to 4 is isothermal
So, Temperature T"' = T" = 338 K
1 to 4 is adiabatic
Similarly,
T"' V "' r-1 = TV r-1
From this you find V"'
Using this V"' you find P"' by using P"' V"' r = PV r
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