Question 14.10: DOE-2 Model for Part-Load Performance Find the electric powe......
DOE-2 Model for Part-Load Performance
Find the electric power drawn by a water-cooled chiller operating at 50% capacity under a condenser water inlet temperature of 80°F and chilled water setpoint temperature of 46°F. Use the data in Table 14.12. The rated kW/ton is 0.6 kW/ton and the full-load rated cooling capacity is 1.2 MBtu/h (100 tons).
Given: rated condition: kW/ton^{*} = 0.6 kW/ton and \dot{Q}_{cool}^{*} =1.2 MBtu/h, T_{ch,out} = 46°F and T_{cd,in} = 80°F.
Find: \dot{W}_{comp}
| TABLE 14.12 | |||||
| Polynomial Model Regression Coefficients for the DOE-2 Model for Water-Cooled Generic Centrifugal Chillers | |||||
| CAP_FT | EIR_FT | EIR_FPLR | |||
| Terms | Coefficients | Terms | Coefficients | Terms | Coefficients |
| Intercept | −0.29861976 | Intercept | 0.51777196 | Intercept | 0.171493 |
| T_{ch,out} | 0.029961 | T_{ch,out} | −0.00400363 | PLR_{part-load} | 0.588202 |
| T^{2}_{ch,out} | −0.00080125 | T^{2}_{ch,out} | 0.00002028 | PLR^{2}_{part-load} | 0.237373 |
| T_{cd,in} | 0.017363 | T_{cd,in} | 0.00698793 | ||
| T^{2}_{cd,in} | −0.00032606 | T^{2}_{cd,in} | 0.0000829 | ||
| T_{ch,out} T_{cd,in} | 0.000631 | T_{ch,out} T_{cd,in} | −0.00015467 | ||
Source: From LBL, DOE-2 Engineers Manual, Lawrence Berkeley Laboratory Report, Berkeley, CA, 1982.
All temperatures are in °F.
First, we calculate the power at rated conditions:
\dot{W}^{*}_{comp} = 100 tons × 0.6 kW/ton = 60 kW
As described in step (b) earlier, we use the model coefficients of Table 14.12, and find the following model:
CAP_FT = -0.29861976 + 0.02996076 \times T_{ch,out}
-0.00080125 \times T^{2}_{ch,out} + 0.01736268 \times T_{cd,in}
– 0.00032606 \times T^{2}_{cd,in} + 0.00063139 \times T_{ch,out} \times T_{cd,in} (14.38)
Replacing numerical values for T_{ch,out} and T_{cd,in} in Equation 14.38 yields CAP_FT = 1.00987.
Similarly, we find EIR_FT = 0.89693.
Next, using PLR =0.5, we can compute EIR_FPLR = 0.52494.
One can directly conclude that though the cooling load has reduced by half from the off-design conditions, the power draw due to part-load cycling operation did not reduce as much since its value is now 0.52494, i.e., about a 5% penalty compared to the value of PLR = 0.5.
Finally, we use Equation 14.37 to find the electric power draw:
\dot{W}_{comp} = \dot{W}_{comp}^{*} \times (CAP\_FT × EIR\_FT × EIR\_FPLR) (14.37)
\dot{W}_{comp} = 60 kW \times 1.00987 \times 0.89693 \times 0.52494= 60 kW × 0.4755 = 0.0951 MBtu/h = 28.5 kW
Finally, it would be instructive to illustrate the regression model building process itself in case an analyst has performance data of a particular chiller.