Application Center - Maplesoft

App Preview:

Analysis of a Vapor Compression Refrigeration Cycle

You can switch back to the summary page by clicking here.

Learn about Maple
Download Application




Analysis of a Vapor-Compression Refrigeration Cycle

Introduction

This application analyzes the following refrigeration cycle, and calculates the coefficient of performance.

 

 

Additionally, the thermodynamic cycle will be plotted on a pressure-enthalpy-temperature chart.

 

The compressor, condenser, throttle and evaporator are analyzed in sequence with this equation, a statement of the conservation of energy,

 

q - w= Δh + ΔKE + ΔPE

where

 

• 

w is the work done by the component

• 

ΔKE and ΔPE are the changes in kinetic and potential energy

• 

Δh is the change in specific enthalpy

• 

q is the heat transferred to the system

 

restart

with(ThermophysicalData); with(Units[Standard]); with(plots)

Compressor

Consider the energy flows in the compressor.  For an adiabatic process, q = 0. Also ΔKE = 0 and ΔPE = 0. Hence -w =  Δh

 

 

Enthalpies at points 1 and 2

P2 := 1000*10^3*Unit('Pa')

h1 := Property(H, temperature = 253*Unit('K'), Q = 1, R134a)

386461.5358*Units:-Unit(J/kg)

(2.1)

h2 := Property(H, temperature = 343*Unit('K'), pressure = P2, R134a)

451844.2441*Units:-Unit(J/kg)

(2.2)

The work done by the compressor (w)

workCompressor := h1-h2

-65382.7083*Units:-Unit(J/kg)

(2.3)

Pressure at point 1

P1 := Property(P, temperature = 253*Unit('K'), Q = 1, R134a)

131876.9284*Units:-Unit('Pa')

(2.4)

Condenser

For the condenser, w = 0, ΔKE = 0 and ΔPE = 0. Hence q = Δh

 

 

Enthalpy at point 3

h3 := Property(H, temperature = 303*Unit('K'), pressure = P2, R134a)

241499.5190*Units:-Unit(J/kg)

(3.1)

h3-h2

-210344.7251*Units:-Unit(J/kg)

(3.2)

Throttle

For the throttle, q = 0, w = 0, ΔKE = 0 and ΔPE = 0. Hence Δh = 0

 

 

Enthalpy at point 4

h4 := h3

241499.5190*Units:-Unit(J/kg)

(4.1)

Saturation pressure of R134a at 253 K

P4 := P1

131876.9284*Units:-Unit('Pa')

(4.2)

Quality at  P = press4 and H = h4

Property(Q, pressure = P4, H = h4, R134a)

.3194910312

(4.3)

Evaporator

For the evaporator, w = 0, ΔKE = 0 and ΔPE = 0. Hence q = Δh

 

 

Heat extracted by evaporator

heatEvaporator := h4-h1

-144962.0168*Units:-Unit(J/kg)

(5.1)

Coefficient of Performance

heatEvaporator/workCompressor

2.217130807

(6.1)

Plot the Refrigeration Cycle on a P-h-T Chart

phtChart := PHTChart(R134a, 100*Unit('kPa') .. 4100*Unit('kPa'))

pts := `~`[`~`[convert]]([[h1, P1], [h2, P2], [h3, P2], [h3, P4], [h1, P1]], unit_free)

cycle := pointplot(`~`[`*`](0.1e-2, pts), connect = true, color = "DarkRed", thickness = 5)

display(phtChart, cycle)