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Þ |
ð18Þ |
QL ¼ γL τL τLC |
ð |
Þ |
ð19Þ |
Q0 ¼ 1 γH γL |
ð |
Þ τ0C 1 |
ð |
Þ |
ð20Þ |
QH ¼ B τst τH |
ð |
Þ |
ð21Þ |
QH þ |
QL |
Q0 ¼ 0 |
ð22Þ |
d |
Sin |
dθ ¼ |
Q0 |
τ0C |
QH |
τHC |
QL |
τLC |
ð23Þ |
d |
Sin |
dθ ¼ β1 τHC τ0C |
ð |
Þ þ β2 τ0C τLC |
ð |
Þ |
ð24Þ |
d |
Stot |
dθ ¼ |
Q0 |
QH |
τH |
QL |
τL |
ð25Þ |
dτL |
dθ ¼ γw 1 τL |
ð |
Þ þ |
Qload |
QL |
ð26Þ |
3 |
Thermodynamic Optimization |
The problem consists of integrating Eqs. (25) and (26) in time and solving the |
nonlinear system (18), (19), (20), (21), (22), (23), and (24) at each time step. The |
objective is to minimize the time θset to reach a specified refrigerated space |
temperature, τLset, in transient operation. To generate the results shown in Figs. 2, |
3, 4, 5, 6, and 7, some selected parameters were held constant and others were |
varied. The numerical method calculates the transient behavior of the system, |
starting from a set of initial conditions, then the solution is marched in time and |
checked for accuracy until a desired condition is achieved (temperature set points or |
steady state). The equations are integrated in time explicitly using an adaptive time |
step, 4th–5th order Runge–Kutta method (Yang et al. 2005). Newton–Raphson’s |
method with appropriate initial guesses was employed for solving the above set of |
nonlinear equations. |
During the integration of the ordinary differential equations, once the set of fixed |
parameters τH , τst, B, γH , γL , γW and |
Qload is defined Eqs. (18) and (18) give τHC. |
The system of Eqs. (18), (19), (20), (21), (22), (23), and (24), at each time step of |
integration of Eqs. (25) and (26), delivers |
Q0, |
QL, τ0C, and τLC. |
1184 |
B. Yasmina et al. |
To test the model and for conducting the analysis presented in this section, |
assuming a small absorption refrigeration unit with a low total thermal conductance |
(UA ¼ 400 W/K), we considered a total heat exchanger area A ¼ 4 m2 and an |
average global heat transfer coefficient U ¼ 0.1 kW/m2 K in the heat exchangers |
and Uw ¼ 1.472 kW/m2 K across the walls, which have a total surface area |
0 |
5 |
10 |
15 |
20 |
25 |
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