Table 1.

Equations for exergy and exergy analysis [45]

Component	Energy balance	Exergy balance
com.	${\dot{W}}_{\mathrm{com}.}={\dot{m}}_1\left(h_2,-,h_1\right)$ $={\dot{m}}_1\left(h_{2,\mathrm{is}},-,h_1\right)/{\eta }_{\mathrm{com}.}$	${\dot{\mathrm{\chi }}}_{\mathrm{D},\mathrm{com}.}={{\dot{m}}_{1}T}_0(s_2-s_1)$
con.	${\dot{Q}}_{\mathrm{con}.}={\dot{m}}_1(h_2-h_3)$	${\dot{\mathrm{\chi }}}_{\mathrm{D},\mathrm{con}.}={\dot{m}}_1{T}_0\left(s_3,-,s_2\right)+{\dot{Q}}_{\mathrm{con}}(1-T_0/T_3)$
EV1	$h_3=h_4$	${\dot{\mathrm{\chi }}}_{\mathrm{D},EV1}={{\dot{m}}_{5}T}_0(s_4-s_3)$
EV2	$h_5=h_6$	${\dot{\mathrm{\chi }}}_{\mathrm{D},EV2}={{\dot{m}}_{5}T}_0(s_6-s_5)$
EV3	$h_9=h_{10}$	${\dot{\mathrm{\chi }}}_{\mathrm{D},EV3}={{\dot{m}}_{9}T}_0(s_{10}-s_9)$
EV4	$h_{12}=h_{13}$	${\dot{\mathrm{\chi }}}_{\mathrm{D},EV4}={{\dot{m}}_{12}T}_0(s_{13}-s_{12})$
CE1	${\dot{Q}}_{CE1}={\dot{m}}_7\left(h_7,-,h_8\right)$ $={\dot{m}}_1\left(h_1,-,h_{17}\right)$	${\dot{\mathrm{\chi }}}_{\mathrm{D},CE1}={\dot{m}}_7\left[\left(h_7,-,h_8\right),-,T_0,\left(s_7,-,s_8\right)\right]+{\dot{m}}_1[\left(h_{17},-,h_1\right)-T_0\left(s_{17},-,s_1\right)]$
CE2	${\dot{Q}}_{CE2}={\dot{m}}_{11}\left(h_{11},-,h_{12}\right)$ $={\dot{m}}_{16}\left(h_{16},-,h_{15}\right)$	${\dot{\mathrm{\chi }}}_{\mathrm{D},CE2}={\dot{m}}_{11}\left[\left(h_{11},-,h_{12}\right),-,T_0,\left(s_{11},-,s_{12}\right)\right]+{\dot{m}}_{15}\left[\left(h_{15},-,h_{16}\right),-,T_0,\left(s_{15},-,s_{16}\right)\right]$
eva.	${\dot{Q}}_{\mathrm{eva}.}={\dot{m}}_{13}\left(h_{14},-,h_{13}\right)$	${\dot{\mathrm{\chi }}}_{\mathrm{D},\mathrm{eva}.}=T_0[\dot{m}\left(s_{14},-,s_{13}\right)-\frac{{\dot{Q}}_{\mathrm{eva}.}}{T_{\mathrm{ave}}+\mathrm{\Delta }\mathrm{T}}]$

where $T_{\mathrm{ave}}$ is the mathematic average temperature of the refrigerant at the inlet and outlet of the evaporator (K), and $\mathrm{\Delta }\mathrm{T}$ represents the temperature difference between the refrigerant and the cooled space, which is assumed to be 10 K