Stirling Refrigeration
ME 498 - Senior LaboratoryNovember 16, 2004
Nicholas Taylor
Overview
Background and theory of the Stirling Cooler
Experimental Set-up and Procedure
Results
Objectives
Operate Global Cooling Model 100B Stirling Cooler
Use thermodynamic principles to analyze the performance
Calculate COP, Qrejected, Qlifted
Background
Model 100B free piston Stirling Cooler
High conversion between mechanical and thermal energy
AC linear motor drives the piston
Working gas – helium
Background
Theory
Coefficient of Performance
Stirling
LiftedR
W
QCOP
dt
dTmcQ plifted
liftedrejectedstirling QQW
Theory
Finding Qrejected
)( inletoutletprejected TTcmQ
Vm water
Experimental Apparatus
Water ToSink
Water FromSink
DigitalPowerMeter
Wire Box
Variac Auto-Transformer
Variac Auto-Transformer
Fluke 77DMM
Fluke 87DMM
OmegaTemperature
Controller Omega TemperatureMeasurement Device
Model 100BStirlingCooler
Flowmeter
Thermocouple
ThermofoilHeater
Procedure
Measure flow rate Prepare data acquisition system Increase voltage to the stirling cooler, to
begin cooling Record data until the temperature
reaches -30°C, activate the heater and reach equilibrium at -20°C
Turn off equipment
Results
Temperature vs. Time
COPR for 5°C, 0°C, -10°C, and -23°C
Compare COP found to COP from Global Cooling
COPR, Wstirling, and Qrejected at equilibrium
Compare COP at equilibrium with COPcarnot
Uncertainty Analysis
Results
Temperature vs. Time
-40.00
-30.00
-20.00
-10.00
0.00
10.00
20.00
30.00
0 200 400 600 800 1000 1200 1400 1600 1800
Time (sec)
Tem
per
atu
re (
C)
Results
Temperature vs Time(during cooling)
y = 281.45e-0.0003x
0.00
100.00
200.00
300.00
400.00
0 100 200 300 400 500 600 700 800
Time (sec)
Tem
per
atu
re (
K)
Results
Global Cooling
Temp Time Cp m dT/dt Qlifted Wstirling COP Input COP % diff.
(deg C) (sec) (J/kJ*C) kg (W) (W) (W)
5 79 380.23 0.27 -0.087 8.88 9.22 0.963 6 2.33 58.66
0 118 379 0.27 -0.088 8.954 9.2 0.973 5 2.06 52.75
-10 219 376.42 0.27 -0.090 9.167 9.11 1.006 3 1.06 5.07
-23 419 372.8 0.27 -0.096 9.633 9.68 0.995 4 1.18 15.67
Results
Equilibrium Test
Qreject 32.318
Qlifted 12.204
Wstirling 20.114
COP 0.607
COP carn. 2.809
% difference 78.398
Uncertainty
Uncertainty of COP
Uncertainty of Qlifted
Uncertainty of Qrejected
Uncertainty of Vdot
Uncertainty
% ωCOP = 1.2%
ωQrejected = 4.61 (32.318)
ωVdot = 2.85E-8
ωQlifted = 0.11 (12.204)
ωCOP = 6.18E-3 (.607)
Uncertainty
Qlifted
Component Value Uncertainty
V 10.8 0.01
I 1.13 0.01
Component Qlifted n (n+dn) Qlifted(n+dn) dQlifted/dn
V 12.204 10.8 10.8 12.204 1.13
I 12.204 1.13 1.13 12.204 10.80
ωQlifted 0.11
Conclusions
Learned a practical use of a Stirling Cooler
Found COP, Qrejected, and Qlifted
Found the uncertainty in measurements
Recommend comparing to a traditional refrigeration cycle
References
http://www.globalcooling.com
http://132.235.18.152/seniorlab
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