If a cooling system uses TR units, the total heat rejection is approximately TR multiplied by what factor?

Prepare for the ASME Code Standards Test. Study with flashcards and multiple-choice questions, each with hints and explanations. Get ready for your exam!

Multiple Choice

If a cooling system uses TR units, the total heat rejection is approximately TR multiplied by what factor?

Explanation:
Total heat rejection is the sum of the cooling effect removed from the space and the heat added by the compressor. One ton of refrigeration (TR) equals 12,000 BTU/hr of cooling. The compressor’s power input becomes heat that is also rejected to the surroundings, and W = Qc / COP. With a typical COP around 4, the compressor adds about 3,000 BTU/hr for each 12,000 BTU/hr of cooling. So the total heat rejected is 12,000 + 3,000 = 15,000 BTU/hr, i.e., 1.25 times the cooling capacity. Hence, the total heat rejection is approximately TR multiplied by 1.25.

Total heat rejection is the sum of the cooling effect removed from the space and the heat added by the compressor. One ton of refrigeration (TR) equals 12,000 BTU/hr of cooling. The compressor’s power input becomes heat that is also rejected to the surroundings, and W = Qc / COP. With a typical COP around 4, the compressor adds about 3,000 BTU/hr for each 12,000 BTU/hr of cooling. So the total heat rejected is 12,000 + 3,000 = 15,000 BTU/hr, i.e., 1.25 times the cooling capacity. Hence, the total heat rejection is approximately TR multiplied by 1.25.

More Multiple Choice Questions
Subscribe

Get the latest from Passetra

You can unsubscribe at any time. Read our privacy policy