Question 14.7: Predicting Part-Load Behavior Using the Degradation Factor W......
Predicting Part-Load Behavior Using the Degradation Factor
We will use the results of Example 14.4 to illustrate the use of the degradation factor approach to predicting part-load performance of a unitary chiller. The same chiller is operated under part load where the cooling load to be met is only 350 kW. What will be the corresponding electric draw of the compressor? The manufacturer has not specified a value for the degradation factor.
Given: \dot{Q}_{cool} = 350 kW, and from Example 14.4, \dot{W}_{comp,ss} = 111.55 kW and \dot{Q}_{cool,ss} = 500 kW.
Assume: C_{d} = 0.25
Find: \dot{W}_{comp} and COP
From Equation 14.27: PLR = (350/500) = 0.7
PLR = min \{ 1, \frac{\dot{Q}_{cool}}{\dot{Q}_{cool,ss}} \} (14.27)
From Equation 14.30: PLF = 1−0.25 × (1−0.7) = 0.925
PLF = 1 – C_{d} \times (1 – PLR) (14.30)
Finally, from Equation 14.29:
\dot{W}_{comp} = \dot{W}_{comp,ss} \times \left\lgroup \frac{PLR}{PLF} \right\rgroup (14.29)
\dot{W}_{comp} = 111.55 \times \left\lgroup \frac{0.7}{0.925} \right\rgroup = 84.42 kW
which results in a part-load COP = (350/84.42) = 4.15
The steady-state COP_{ss} = (500/111.55) = 4.48; thus, cycling has reduced the COP by 7.4%