Heat pumps and coeffiecent of performance

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SUMMARY

This discussion focuses on calculating the work required to operate a heat pump with a coefficient of performance (COP) of 3.8, delivering 7.6 MJ of heat per hour to maintain an indoor temperature of 22°C while the outdoor temperature is -5°C. The relevant equation for COP is defined as COP = |Q|/W, where Q is the heat delivered and W is the work input. To find the work needed, the heat delivered must be converted to SI units, and the relationship between the temperatures of the hot and cold reservoirs can be applied using Carnot principles.

PREREQUISITES
  • Understanding of thermodynamics and heat pump operation
  • Familiarity with the coefficient of performance (COP) concept
  • Knowledge of SI unit conversions, particularly for energy
  • Basic grasp of Carnot cycle principles in thermodynamics
NEXT STEPS
  • Calculate the work required using the formula W = |Q|/COP
  • Explore the Carnot efficiency for heat pumps and its implications
  • Research the impact of temperature differences on heat pump efficiency
  • Learn about real-world applications and limitations of heat pumps
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Students studying thermodynamics, engineers working with HVAC systems, and anyone interested in the efficiency of heat pumps and their operational mechanics.

rmorelan
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Hi everyone, I was hoping to ask a thermodynamics problem, specificially one about heat pumps and coeffiecent of performance. I am wondering if I am missing something.

I am given that heat pump is heat a building. Inside temperature is to be maintained at 22 degree celsius, outside is -5.0 celsius. The coefficient of performance is 3.8, and the the pump delivers 7.6MJ of heat each hour to the building. I need to find out at what rate work needs to be applied to run the pump.

Homework Equations



Now I believe the equation of the the coeffiecient of performance (CP) is equal to:

cp = |Q|/W

Now this is where I am confused, as I don't understand what (if anything) I am supposed to do with the temperatures given. I know cp, and I know Q, so I can solve for W needed to run the pump can't I? I would need to convert the heat produced to be in SI in order to get Watts as a result of course.

thanks for your help!
rob
 
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If we treat the heat pump as a Carnot heat pump we can say that;

\frac{Q_2}{Q_1} = \frac{T_2}{T_2}

Hence;

\eta^{HP}_{C} = \frac{Q_1}{W} = \frac{Q_1}{Q_1-Q_2} = \frac{T_1}{T_1-T_2}

Where T1 and T2 are the temperatures of the hot and cold reservoirs respectively.
 

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