One mole of a diatomic ideal gas `(gamma=1.4)` is taken through a cyclic process starting from point A. The process `AtoB` is an adiabatic compression, `BtoC` is isobaric expansion, `CtoD` is an adiabatic expansion, and `DtoA` is isochoric. The volume ratios are `V_A//V_B=16 and V_C//V_B=2` and the temperature at A is `T_A=300K`. Calculate the temperature of the gas at the points B and D and find the efficiency of the cycle.

Explanation

The respective cyclic process is shown in Figure-3.52. The expansion and compression ratio are given as
`V_A/V_B=16` and `V_C/V_B`=2
In adiabatic process AB, we have
`T_AV_A^(gamma-1)=T_BV_B^(gamma-1)`
or `T_B=T_A(V_A/V_B)^(gamma-1)`
`=300xx(16)^(0.4)`
=909 K
Similarly for isobaric process BC, we have
`T_C/V_C=T_B/V_B`
or `T_C=T_B(V_C/V_B)` ....(3.119)
= 909 x 2 = 1818 K
Similarly for adiabatic process CD we have
`T_CV_C^(gamma-1)=T_DV_D^(gamma-1)`
or `T_D=T_C(V_C/V_D)^(gamma-1)`
As we have `V_D=V_A`, thus
`V_C/V_D=V_C/V_A=V_C/V_BxxV_B/V_A`
`=2xx1/16=1/8`
Thus from equation-(3.119)
`T_D=1818xx(1/8)^(0.4)`
=791.3 K
The efficiency of cycle can be given as
`eta=1-|Q_"out"/Q_"in"|`
Here we know in process AB and CD, as being adiabatic processes no heat exchange takes place and in isobaric process BC as temperature ofgas increases, heat is taken in (say `Q_1` ) and it is given as
`Q_1=nC_P(T_C-T_B)`...(3.120)
Similarly in isochoric process DA as temperature of gas decreases heat is rejected (say `Q_2`) and it can be given as
`Q_2=nC_P(T_D-T_A)` [ A magnitude of rejected heat]
Now efficiency of this cycle can be given as
`eta=1-|Q_"out"/Q_"in"|`
`=1-Q_2/Q_1`
`=1-(C_V(T_D-T_A))/(C_P(T_C-T_B))`
`=(T_D-T_A)/(gamma(T_C-T_B))`
`=((791.3-300))/((1.4)(1818-909))`
`=1-491.3/1272.6=0.614`=61.4%