Question 8.3: Sketch the air-standard Diesel cycle on a PV diagram, and de...
Sketch the air-standard Diesel cycle on a PV diagram, and derive an equation giving the thermal efficiency of this cycle in relation to the compression ratio r (ratio of volumes at the beginning and end of the compression step) and the expansion ratio r_e (ratio of volumes at the end and beginning of the adiabatic expansion step).
The air-standard Diesel cycle is the same as the air-standard Otto cycle, except that the heat-absorption step (corresponding to the combustion process in the actual engine) is at constant pressure, as indicated by line DA in Fig. 8.10.
On the basis of 1 mol of air in its ideal-gas state with constant heat capacities, the heat quantities absorbed in step DA and rejected in step BC are:
Q_{DA} = C_{P}^{ig} ( T_A − T_D ) and Q_{BC} = C_{V}^{ig} ( T_C − T_B )The thermal efficiency, Eq. (8.3), is:
\eta=\frac{-W(\text { net })}{Q_{D A}}=\frac{Q_{D A}+Q_{B C}}{Q_{D A}} (8.3)
\eta=1+\frac{Q_{B C}}{Q_{D A}}=1+\frac{C_V^{i g}\left(T_C-T_B\right)}{C_P^{i g}\left(T_A-T_D\right)}=1-\frac{1}{\gamma}\left(\frac{T_B-T_C}{T_A-T_D}\right) (A)
For reversible, adiabatic expansion (step AB) and reversible, adiabatic compression (step CD), Eq. (3.23a) applies:
T ( V^{ ig} )^{γ − 1} = const (3.23a)
T_A V_A^{\gamma-1}=T_B V_B^{\gamma-1} \quad \text { and } \quad T_D V_D^{\gamma-1}=T_C V_C^{\gamma-1}
By definition, the compression ratio is r \equiv V_C^{i g} / V_D^{i g} , and the expansion ratio is r_e \equiv V_B^{i g} / V_A^{i g} . Thus,
T_B=T_A\left(\frac{1}{r_e}\right)^{\gamma-1} \quad T_C=T_D\left(\frac{1}{r}\right)^{\gamma-1}
Substituting these equations into Eq. (A) gives:
\eta=1-\frac{1}{\gamma}\left[\frac{T_A\left(1 / r_e\right)^{\gamma-1}-T_D(1 / r)^{\gamma-1}}{T_A-T_D}\right] (B)
Also P_A = P_D , and for the ideal-gas state,
P_D V_D^{i g}=R T_D \quad \text { and } \quad P_A V_A^{i g}=R T_A
Moreover, V_C^{i g}=V_B^{i_g} , and therefore:
\frac{T_D}{T_A}=\frac{V_D^{i g}}{V_A^{i g}}=\frac{V_D^{i g} / V_C^{i g}}{V_A^{i g} / V_B^{i g}}=\frac{r_e}{r}
This relation combines with Eq. (B):
\eta=1-\frac{1}{\gamma}\left[\frac{\left(1 / r_e\right)^{\gamma-1}-\left(r_e / r\right)(1 / r)^{\gamma-1}}{1-r_e / r}\right]
or \eta=1-\frac{1}{\gamma}\left[\frac{\left(1 / r_e\right)^\gamma-(1 / r)^\gamma}{1 / r_e-1 / r}\right] (8.7)
