MODULE 4 Cooling Production Equipment and Systems
The calculation examples are shown in an accordion-style display. Some questions are accompanied by equations and/or tables for quick reference. Solve the problem on your own and then click on the question to view the solution. This will help you make sense of the work as you read through the problem-solving process.
1.
A vapor compression refrigeration machine uses 30 kW of electric power to produce 50 tons of cooling. What is its COP?
COP is the ratio of energy removed to energy used.
COP =
(50 × 12,000)⁄
(30 × 3413) = 5.9.
2.
An absorption refrigeration machine uses 30 kW and 1200 lbs. per hour of steam to produce 100 tons of cooling. What is its COP?
COP is the ratio of energy removed to energy used.
COP =
(100 x 12,000)⁄
((30 x 3,413) + (1200 x 1000)) = 0.92
3.
Approximately how much chilled water storage would be required to reduce chiller load by 250 tons for a period of 6 hours? How much ice storage would be required?
Six hours at 250 tons equates to 1500 ton-hours of cooling. Depending on water temperature rise, the required volume of chilled water will be 70 to 100 gallons per ton-hour. Thus, 1500 ton-hours would need 105,000 to 150,000 gallons. Ice requires about one-tenth the volume of water, or 10,500 to 15,000 gallons.