Question

An ideal Otto cycle has a compression ratio of 7. At the beginning of the compression process, P1 = 90 kPa, T1 = 27°C, and V1 = 0.004 m3. The maximum cycle temperature is 1127°C. For each repetition of the cycle, calculate the heat rejection and the net work production. Also calculate the thermal efficiency and mean effective pressure for this cycle. Use constant specific heats at room temperature.

Answer 1

Answer:

i) Heat rejection =  1.0288 KJ

  Network production = 1.212 kJ

ii) Thermal efficiency = 54.08%

   mean effective pressure = 353.5 kPa

Explanation:

Given data :

Compression ratio ( r )  = 7

P1 ( initial pressure ) = 90 kPa,

T1 ( initial temperature ) = 27°C + 273 = 300 K

V1 = 0.004 m^3

Max cycle Temperature = 1127°C for each repetition of the cycle

i) Determine the heat rejection and net work production

considering that process 1-2 is an Isentropic compression

Find ; T2 = T1 ( r )^1.4-1

      T2 = 300 ( 7 )^0.4  = 653.371 K

    P2 = P1( r )^k

          = 90 ( 7 )^1.4  = 1372.081 kPa

considering that process 3-4 is an Isentropic expansion

T4 = T3 / r^k-1

    = 1400 / 7^(1.4 -1 ) = 642.82 K

Next ; Calculate the value of m

m = P1V1 / RT1   = 90(0.004) / 0.287 ( 300 )

                      m = 4.18 * 10^-3 kg

Finally :

amount of heat rejected = mCv ( T4 - T1 )

                                          = 4.18 * 10^-3 ( 0.718 ) ( 642.82 - 300 )

                                    Qout  = 1.0288 KJ

amount of heat added ( Qin ) =  mCv ( T3 - T2 )

                                                = 4.18 * 10^-3 ( 0.718 ) ( 1400 - 653.371 )

                   hence           Qin  = 2.2408 kJ

Network production( Wnet) = Qin - Qout

                                  = 2.2408 - 1.0288 ) KJ

                                  = 1.212 kJ

ii) Determine the thermal efficiency and mean effective pressure of the cycle

Thermal efficiency = 1 - Qout / Qin

                               = 1 - ( 1.0288 / 2.2408 )  = 54.08%

mean effective pressure = Wnet / V1 ( 1 - 1/r )

                                         = 1.212 / 0.004 ( 1 - 1/7 )

                                         = 353.5 kPa