Question

A Carnot engine uses air as the working substance, receives heat at a temperature of 315^oC, and rejects it at 65^oC. 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.

Answer 1

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