# Cascade Refrigeration Cycle Example¶

## Imports¶

:

from thermostate import State, Q_, units, EnglishEngineering as EE


## Definitions¶

:

sub_hiT = 'R22'
sub_loT = 'R134A'

T_1 = Q_(-30.0, 'degF')
x_1 = Q_(1.0, 'dimensionless')

p_2 = Q_(50.0, 'psi')

p_3 = p_2
x_3 = Q_(0.0, 'dimensionless')

delta_T_5 = Q_(-5.0, 'delta_degF')
x_5 = Q_(1.0, 'dimensionless')

p_6 = Q_(250.0, 'psi')

p_7 = p_6
x_7 = Q_(0.0, 'dimensionless')

Qdot_in = Q_(20.0, 'refrigeration_tons')


## Problem Statement¶

A two-stage cascade vapor-compression refrigeration system operates with R22 as the working fluid in the high-temperature cycle and R134A in the low-temperature cycle. For the R134A cycle, the working fluid enters the compressor as a saturated vapor at -30.0 fahrenheit and is compressed isentropically to 50 lbf/in.2 Saturated liquid leaves the intermediate heat exchanger at 50 lbf/in.2 and enters the expansion valve. For the R22 cycle, the working fluid enters the compressor as saturated vapor at a temperature 5 fahrenheit below that of the condensing temperature of the R134A in the intermediate heat exchanger. The R22 is compressed isentropically to 250 lbf/in.2 Saturated liquid then enters the expansion valve at 250 lbf/in.2. The refrigerating capacity of the cascade system is 20 ton_of_refrigeration. Determine

1. the power input to each compressor, in BTU/min
2. the overall coefficient of performance of the cascade cycle

## Solution¶

### 1. the power input to each compressor¶

The power input to each compressor can be found by

$\dot{W}_{cv} = \dot{m}(h_i - h_e)$

To find the enthalpies, we need to fix all of the states.

:

st_1 = State(sub_loT, x=x_1, T=T_1)
h_1 = st_1.h.to(EE.h)
s_1 = st_1.s.to(EE.s)
p_1 = st_1.p.to(EE.p)

s_2 = s_1
st_2 = State(sub_loT, p=p_2, s=s_1)
h_2 = st_2.h.to(EE.h)
T_2 = st_2.T.to(EE.T)

st_3 = State(sub_loT, p=p_3, x=x_3)
T_3 = st_3.T.to(EE.T)
h_3 = st_3.h.to(EE.h)
s_3 = st_3.s.to(EE.s)

h_4 = h_3
p_4 = p_1
st_4 = State(sub_loT, p=p_4, h=h_4)
T_4 = st_4.T.to(EE.T)
s_4 = st_4.s.to(EE.s)
x_4 = st_4.x

T_5 = T_3 + delta_T_5
st_5 = State(sub_hiT, T=T_5, x=x_5)
h_5 = st_5.h.to(EE.h)
p_5 = st_5.p.to(EE.p)
s_5 = st_5.s.to(EE.s)

s_6 = s_5
st_6 = State(sub_hiT, s=s_6, p=p_6)
h_6 = st_6.h.to(EE.h)
T_6 = st_6.T.to(EE.T)

st_7 = State(sub_hiT, p=p_7, x=x_7)
T_7 = st_7.T.to(EE.T)
h_7 = st_7.h.to(EE.h)
s_7 = st_7.s.to(EE.s)

h_8 = h_7
p_8 = p_5
st_8 = State(sub_hiT, p=p_8, h=h_8)
T_8 = st_8.T.to(EE.T)
s_8 = st_8.s.to(EE.s)
x_8 = st_8.x


Summarizing the states:

State T p h s x phase
1 -30.00 degF 9.86 pound_force_per_square_inch 162.30 btu / pound 0.4196 btu / degR / pound 100.00% twophase
2 58.23 degF 50.00 pound_force_per_square_inch 176.42 btu / pound 0.4196 btu / degR / pound gas
3 40.27 degF 50.00 pound_force_per_square_inch 88.65 btu / pound 0.2442 btu / degR / pound 0.00% twophase
4 -30.00 degF 9.86 pound_force_per_square_inch 88.65 btu / pound 0.2482 btu / degR / pound 22.96% twophase
5 35.27 degF 76.59 pound_force_per_square_inch 174.42 btu / pound 0.4175 btu / degR / pound 100.00% twophase
6 146.75 degF 250.00 pound_force_per_square_inch 187.12 btu / pound 0.4175 btu / degR / pound gas
7 112.76 degF 250.00 pound_force_per_square_inch 110.14 btu / pound 0.2834 btu / degR / pound 0.00% twophase
8 35.27 degF 76.59 pound_force_per_square_inch 110.14 btu / pound 0.2876 btu / degR / pound 26.55% twophase

The mass flow rate of the R134a is found from the refrigeration capacity

$\dot{m}_{\text{R134A}} = \frac{\dot{Q}_{in}}{h_1 - h_4}$

while the mass flow rate of the R22 is found from the intermediate heat exchanger

$\dot{m}_{\text{R22}} = \frac{\dot{m}_{\text{R134A}}\left(h_2 - h_3\right)}{h_5 - h_8}$
:

mdot_r134a = (Qdot_in/(h_1 - h_4)).to('lb/min')
Wdot_loT = (mdot_r134a*(h_1 - h_2)).to('BTU/min')

mdot_r22 = mdot_r134a*(h_2 - h_3)/(h_5 - h_8)
Wdot_hiT = (mdot_r22*(h_5 - h_6)).to('BTU/min')


Answer: The compressor work for the low-temperature cycle is $$\dot{W}_{\text{R134A}} =$$ -766.52 btu / minute and the work for the high temperature cycle is $$\dot{W}_{\text{R22}} =$$ -941.38 btu / minute

### 2. The coefficient of performance¶

The coefficient of performance of this refrigeration cycle is defined as

$\beta = \frac{\dot{Q}_{in}}{\lvert\dot{W}_{\text{R134A}} + \dot{W}_{\text{R22}}\rvert}$
:

beta = (Qdot_in/abs(Wdot_loT + Wdot_hiT)).to('dimensionless')
beta_max = (T_4/(T_7 - T_4)).to('dimensionless')


Answer: The coefficient of performance is $$\beta =$$ 2.34, while the maximum coefficient of performance for a refrigeration cycle operating between $$T_C =$$ -30.00 degF and $$T_H =$$ 112.76 degF is $$\beta_{\text{max}} =$$ 3.01 dimensionless. The actual cycle has a lower COP, so it is possible.