Question 11.3: A Two-Stage Cascade Refrigeration Cycle Consider a two-stage...

A cascade refrigeration system operating between the specified pressure limits is considered. The mass flow rate of the refrigerant through the lower cycle, the rate of refrigeration, the power input, and the COP are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 The heat exchanger is adiabatic. Properties The enthalpies of the refrigerant at all eight 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. 11–11. The topping cycle is labeled cycle A and the bottoming one, cycle B. For both cycles, the refrigerant leaves the condenser as a saturated liquid and enters the compressor as saturated vapor.
(a) The mass flow rate of the refrigerant through the lower cycle is determined from the steady-flow energy balance on the adiabatic heat exchanger,

\begin{aligned}\dot{E}_{\text {out }}=\dot{E}_{\text {in }} \longrightarrow \dot{m}_{A} h_{5}+\dot{m}_{B} h_{3} &=\dot{m}_{A} h_{8}+\dot{m}_{B} h_{2} \\\dot{m}_{A}\left(h_{5}-h_{8}\right) &=\dot{m}_{B}\left(h_{2}-h_{3}\right) \\(0.05 kg / s )[(251.88-95.47) kJ / kg ] &=\dot{m}_{B}[(255.93-55.16) kJ / kg ] \\\dot{m}_{B} &= 0 . 0 3 9 0 kg / s\end{aligned}

(b) The rate of heat removal by a cascade cycle is the rate of heat absorption in the evaporator of the lowest stage. The power input to a cascade cycle is the sum of the power inputs to all of the compressors:

\begin{aligned}&\dot{Q}_{L}=\dot{m}_{B}\left(h_{1}-h_{4}\right)=(0.0390 kg / s )[(239.16-55.16) kJ / kg ]=7.18 kW \\&\dot{W}_{\text {in }}=\dot{W}_{\text {comp I,in }}+\dot{W}_{\text {comp II,in }}=\dot{m}_{A}\left(h_{6}-h_{5}\right)+\dot{m}_{B}\left(h_{2}-h_{1}\right)\end{aligned}

\begin{aligned}=&(0.05 kg / s )[(270.92-251.88) kJ / kg ] \\&+(0.039 kg / s )[(255.93-239.16) kJ / kg ] \\= & 1 . 6 1 kW\end{aligned}

(c) The COP of a refrigeration system is the ratio of the refrigeration rate to the net power input:

COP _{ R }=\frac{\dot{Q}_{L}}{\dot{W}_{\text {net, in }}}=\frac{7.18 kW }{1.61 kW }=4.46

Discussion This problem was worked out in Example 11–1 for a single-stage refrigeration system. Notice that the COP of the refrigeration system increases from 3.97 to 4.46 as a result of cascading. The COP of the system can be increased even more by increasing the number of cascade stages.