A refrigerator in a hospital removes 40 kW of heat from a sp

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SUMMARY

A refrigerator in a hospital removes 40 kW of heat from a blood storage space, operating with a coefficient of performance (COP) of 8. The work required to operate the refrigerator continuously for one day is calculated using the formula W in = Qin/COP. The correct calculation involves converting kilowatts to kilojoules by applying the conversion factor, resulting in W in = (40 kW * 24 hr) / 8, which equals 120 kJ. The discussion highlights the importance of unit conversion in thermodynamic calculations.

PREREQUISITES
  • Understanding of thermodynamics principles
  • Familiarity with the concept of Coefficient of Performance (COP)
  • Knowledge of unit conversions between kilowatts and kilojoules
  • Basic mathematical skills for performing calculations
NEXT STEPS
  • Research thermodynamic cycles and their applications in refrigeration
  • Learn about the principles of heat transfer in medical refrigeration systems
  • Study advanced unit conversion techniques in engineering
  • Explore the design and efficiency optimization of hospital refrigeration units
USEFUL FOR

Engineers, medical facility managers, and students studying thermodynamics or refrigeration technology will benefit from this discussion.

hc23881
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A refrigerator in a hospital removes 40 kW of heat from a space used for blood storage. The coefficient of performance of the refrigerator is 8. What is the work (kJ) required to operate the refrigerator continuously for one day?

What I attempted to do:

W in = Qin/COP therefore, Win = 40kW(24hr)/8 = 120 kJ

but that is wrong..
 
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You have a unit error:

(kW)*(Hr) ≠ kJ

A conversion factor on the LHS is required..
 
billy_joule said:
You have a unit error:

(kW)*(Hr) ≠ kJ

A conversion factor on the LHS is required..
ahh i got it! thanks!
 

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