## Imports¶

:

from thermostate import State, Q_, units, EnglishEngineering as EE
from thermostate.plotting import IdealGas, VaporDome


## 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.

The cascade refrigeration cycle is consistes of two smaller cycles both with 4 processes:

R22 cycle:
1. Isentropic compression
2. Isobaric heat exchange
3. Isoenthalpic expansion
4. Isobaric heat exchange

R134A cycle:
1. Isentropic compression
2. Isobaric heat exchange
3. Isoenthalpic expansion
4. Isobaric heat exchange


The following properties are used to fix the four states:

State

Property 1

Property 2

1

$x_1$
$T_1$

2

$p_2$
$s_2=s_1$

3

$p_3=p_2$
$x_3$

4

$p_4=p_1$
$h_4=h_3$

5

$T_5=T_3 + \Delta T_5$
$x_5$

6

$p_6$
$s_6=s_5$

7

$p_7=p_6$
$x_7$

8

$p_8=p_5$
$h_8=h_7$
:

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


Plotting the T-s diagrams of both cycles,

:

R22_vapordome = VaporDome(sub_hiT, ("s", "T"))
R134A_vapordome = VaporDome(sub_loT, ("s", "T"))  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.