A Two-Stage Cascade Refrigeration Cycle Consider a two-stage cascade refrigeration system operating between the pressure limits of 0.8 and 0.14 MPa. Each stage operates on an ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid. Heat rejection from the lower cycle

Question

A Two-Stage Cascade Refrigeration Cycle

Consider a two-stage cascade refrigeration system operating between the pressure limits of 0.8 and 0.14 MPa. Each stage operates on an ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid. Heat rejection from the lower cycle to the upper cycle takes place in an adiabatic counterflow heat exchanger where both streams enter at about 0.32 MPa. (In practice, the working fluid of the lower cycle is at a higher pressure and temperature in the heat exchanger for effective heat transfer.) If the mass flow rate of the refrigerant through the upper cycle is 0.05 kg/s, determine (a) the mass flow rate of the refrigerant through the lower cycle, (b) the rate of heat removal from the refrigerated space and the power input to the compressor, and (c) the coefficient of performance of this cascade refrigerator.

Accepted Answer

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,

[latex]\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}[/latex]

(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:

[latex]\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}[/latex]

[latex]\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}[/latex]

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

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

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.

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

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