Refrigeration Numerical Problem 2

TL;DR

Determining the coefficient of performance of a refrigerator based on heat removal and power input.

Transcript

hello students today let us discuss a problem on the refrigerator so here they have given that the refrigerator maintains the food compartment at 3 degree celsius by removing heat at a rate of 450 kilojoules per minute so if the required power input to the refrigerator is 2.2 kilowatt they are asking us to determine the coefficient of performance o... Read More

Key Insights

• 🥵 The COP of a refrigerator indicates its efficiency in removing heat and providing cooling.
• 🥵 Heat removal rate and power input are crucial parameters in calculating the COP.
• ✊ Converting power units ensures consistency in calculations for energy efficiency assessments.
• 💨 The COP formula provides a straightforward way to evaluate refrigeration system performance.

Questions & Answers

Q: What is the significance of determining the coefficient of performance of a refrigerator?

Calculating the COP helps analyze how efficiently a refrigerator removes heat and provides cooling, crucial for energy efficiency and cost-saving evaluations.

Q: How is the COP formula derived and applied in refrigeration problems?

The COP formula compares the output heat removed to the input power consumed, enabling engineers to assess the effectiveness of the refrigeration system.

Q: What factors can influence the coefficient of performance of a refrigerator?

The COP can be impacted by factors like the design of the refrigerator, temperature differentials, insulation quality, and overall efficiency of the cooling system.

Q: Why is it essential to convert power units from kilowatts to kilojoules per second in the given problem?

Converting power units to a common measurement (kilojoules per second) allows for accurate calculations and comparison of heat removal rates to power input.

Summary & Key Takeaways

• The problem involves calculating the coefficient of performance of a refrigerator providing a food compartment at 3 degrees celsius.

• Heat removal rate is given as 450 kilojoules per minute, and the power input is 2.2 kilowatts.

• The coefficient of performance formula is utilized to determine the efficiency of the refrigerator.