MCQOPTIONS
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MCQOPTIONS
| 1. |
The COP of simple air cooling system is given by?T6 = Inside temperature of cabinT5′ = Exit temperature of cooling turbineT3′ = Temperature at the exit of compressorT2′ = Stagnation temperature |
| A. | COP = \(\frac{(T_6 – T_{5′})}{(T_3′ – T_{2′})}\) |
| B. | COP = \(\frac{(T_6 + T_{5′})}{(T_3′ – T_{2′})}\) |
| C. | COP = \(\frac{(T_6 + T_{5′})}{(T_3′ + T_{2′})}\) |
| D. | COP = \(\frac{(T_6 – T_{5′})}{(T_3′ + T_{2′})}\) |
| Answer» B. COP = \(\frac{(T_6 + T_{5′})}{(T_3′ – T_{2′})}\) | |