A two-stage compression refrigeration system operating between specified pressure limits is considered. The fraction of the refrigerant that evaporates in the flash chamber, the refrigeration and work input per unit mass, and the COP are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 The flash chamber is adiabatic.
Properties The enthalpies of the refrigerant at various states are determined from the refrigerant tables and are indicated on the T-s diagram.
Analysis The T-s diagram of the refrigeration cycle is shown in Fig. 11–15. We note that the refrigerant leaves the condenser as saturated liquid and enters the low-pressure compressor as saturated vapor.
(a) The fraction of the refrigerant that evaporates as it is throttled to the flash chamber is simply the quality at state 6, which is
x_{6}=\frac{h_{6}-h_{f}}{h_{f g}}=\frac{95.48-55.14}{196.78}=0.2050
(b) The amount of heat removed from the refrigerated space and the compressor work input per unit mass of refrigerant flowing through the condenser are
\begin{aligned}q_{L} &=\left(1-x_{6}\right)\left(h_{1}-h_{8}\right) \\&=(1-0.2050)[(239.19-55.14) \mathrm{kJ} / \mathrm{kg}]=146.3 \mathrm{~kJ} / \mathrm{kg}\end{aligned}
and
w_{\text {in }}=w_{\text {comp I,in }}+w_{\text {comp II,in }}=\left(1-x_{6}\right)\left(h_{2}-h_{1}\right)+(1)\left(h_{4}-h_{9}\right)
The enthalpy at state 9 is determined from an energy balance on the mixing chamber,
\begin{aligned}\dot{E}_{\text {out }} &=\dot{E}_{\text {in }} \\\text { (1) } h_{9} &=x_{6} h_{3}+\left(1-x_{6}\right) h_{2} \\h_{9} &=(0.2050)(251.93)+(1-0.2050)(255.95)=255.13 \mathrm{~kJ} / \mathrm{kg}\end{aligned}
Also, s_{9}=0.9417 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}. Thus the enthalpy at state 4\left(0.8 \mathrm{MPa}, s_{4}=s_{9}\right. ) is h_{4}=274.49 \mathrm{~kJ} / \mathrm{kg}. Substituting,
\begin{aligned}w_{\text {in }} &=(1-0.2050)[(255.95-239.19) \mathrm{kJ} / \mathrm{kg}]+(274.49-255.13) \mathrm{kJ} / \mathrm{kg} \\&=32.68 \mathrm{k} \mathrm{J} / \mathrm{kg}\end{aligned}
(c) The coefficient of performance is
\mathrm{COP}_{\mathrm{R}}=\frac{q_{L}}{w_{\mathrm{in}}}=\frac{146.3 \mathrm{~kJ} / \mathrm{kg}}{32.68 \mathrm{~kJ} / \mathrm{kg}}=4.48
Discussion This problem was worked out in Example 11-1 for a single-stage refrigeration system (COP =3.97) and in Example 11-4 for a two-stage cascade refrigeration system (COP =4.46 ). Notice that the COP of the refrigeration system increased considerably relative to the single-stage compression but did not change much relative to the twostage cascade compression.
