Dual Cycle Net Work

You are given the dual cycle from the figure with the following data: compression ratio 1:15, $P_1=14.4 PSI$, $T_1=60 F$. The volume ratio $V_4/V_3=2:1$ and the pressure ratio $P_3/P_2=1.5:1$. You are given $C_p=0.24 [Btu/(lbm*R)]$, $C_v=0.171 [Btu/(lbm*R)]$ and $k=1.4$. please find:
1. Maximum T 2. Maximum P 3 Total specific heat 4. Total heat out of cycle 5. Specific net work of the cycle.

Dual Cycle

Given data $P_1=14.4 [psi] =10^5 [Pa]$, $T_1=60 F=288.7[K]$  ,$r=V_1/V_2 =15$,   $β=V_4/V_3 =2$,   $α=P_3/P_2 =1.5$ $c_p=0.24[Btu/(lbm*R)]=1004.8[J/(kg*K)]$    and $c_v=0.171[Btu/(lbm*R)]=715.94[J/(Kg*K)]$, $c_p/c_v =k=1.4$
$c_p$ and $c_v$ are constants

Analysis:
1-2: adiabatic (isentropic)
$T_2/T_1 =(V_1/V_2 )^{k-1}=r^{k-1}$   and
$P_2/P_1 =(V_1/V_2 )^k=r^k$   that is
$T_2=T_1*r^{k-1}$  and
$P_2=P_1*r^k$

2-3: constant volume
$P_2/T_2 =P_3/T_3$    that is
$T_3=P_3/P_2 *T_2=αr^{k-1}*T_1$    and
$P_3=αP_2=αP_1*r^k$

3-4: constant pressure
$V_3/T_3 =V_4/T_4$      that is
$T_4=V_4/V_3 *T_3=βαr^{k-1}*T_1$

4-5: adiabatic (isentropic)
$T_5/T_4 =(V_4/V_5 )^{k-1}=(V_4/V_3 *V_3/V_5 )^{k-1}=(V_4/V_3 *V_2/V_1 )^{k-1}=(β/r)^{k-1}$
$T_5=T_4*(β/r)^{k-1}=αβr^{k-1}*T_1*(β/r)^{k-1}=αβ^k*T_1$

Answers:
Maximum temperature is
$T_4=αβr^{k-1}*T_1=1.5*2*15^{0.4}*288.7=2558.6 K=4145.8 [F]$
Maximum pressure is
$P_3=αr^k*P_1=1.5*15^1.4*10^5=6.647*10^6 [Pa]=964 [Psi]$
Specific heat into cycle is
$Q_1=c_v (T_3-T_2 )+c_p (T_4-T_3 )=c_v T_1 [(αr^{k-1}-r^{k-1} )+k(αβr^{k-1}-αr^{k-1} )]$
$Q_1=c_v T_1 [r^{k-1}*(α-1)+kαr^{k-1}*(β-1) ]=$
$=715.94*288.7*15^{0.4}*[0.5+1.4*1.5*1]=1.588*10^6  J/kg=1504.7 [Btu/kg]$

Specific heat out the cycle is
$Q_2=c_v (T_5-T_1 )=c_v T_1 (αβ^k-1)=715.94*288.7*(1.5*2^{1.4}-1)=$
$=6.12*10^5  J/kg=579.60 [Btu/kg]$
Specific net work is
$W=Q_1-Q_2=976000 [J]$


valentin68

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