# Calculating Minimum Power for Air Conditioner

• S_fabris
In summary: in summary, if the temperature difference goes to zero, it should ideally take no work to move the heat from one resevoir to another.
S_fabris
A home is kept cool by an air conditioner. The outside temperature is 311.75K and the interior of the home is 288.55K. If 127kJ/h of heat is removed from the house, what is the minimum power that must be provided to the air-conditioner? answer in kJ/h
my work

COP = Th / (Th - Tc)
= 311.75 / (311.75 - 288.55)
= 13.44

W = 127kJ/h / 13.44
= 9.45 kJ/h this is incorrect according to the homework server

Anybody know where i went wrong on this one?
any help is appreciated!

Sergio

There's no way the electrical input can be less than the power removed! You need to multiply, not divide.

Th, Tc and Qc are given in the problem...isolate Qh using Qh/Th = Qc/Tc
to find work
W = Qh-Qc

thank you for your help marcusl i realized i was going about it the wrong way :)

marcusl said:
There's no way the electrical input can be less than the power removed! You need to multiply, not divide.

I think that is not true. If the temperature difference goes to zero, it should ideally take no work to move the heat from one resevoir to another. I think the only problem with the original calculation is that the OP was using the COP for heating instead of the COP for cooling.

http://en.wikipedia.org/wiki/Coefficient_of_performance

S_fabris said:

Th, Tc and Qc are given in the problem...isolate Qh using Qh/Th = Qc/Tc
to find work
W = Qh-Qc

thank you for your help marcusl i realized i was going about it the wrong way :)

Qh/Th = Qc/Tc
Qh = Qc(Th/Tc)
W = Qh - Qc = Qc(Th/Tc) - Qc = Qc[(Th/Tc) - 1] = Qc(Th- Tc)/Tc = Qc/COP_cooling

COP_cooling = Tc/(Th- Tc) = 288.55/(311.75 - 288.55) = 12.44

W = 127kJ/h/12.44 = 10.21kJ/h

ahhh ok i see what i did wrong!

Perfect thank you very much for your time!

Sergio :D

## 1. What is the purpose of an air conditioner?

An air conditioner is a device that cools and dehumidifies indoor air to create a more comfortable and livable environment. It works by circulating refrigerant through a closed system, absorbing heat from the indoor air and releasing it outside.

## 2. How does an air conditioner work?

An air conditioner works by pulling in warm air from the room and passing it through a refrigerant-filled coil. The refrigerant absorbs the heat from the air, which is then pumped outside through another coil and released. The cooled air is then blown back into the room, creating a comfortable temperature.

## 3. What factors affect the efficiency of an air conditioner?

The efficiency of an air conditioner can be affected by several factors, including the size and type of the unit, the temperature and humidity levels in the room, and the quality of the insulation in the space. Regular maintenance, such as cleaning or replacing air filters, can also impact its efficiency.

## 4. How often should I clean or replace the air filters in my air conditioner?

It is recommended to clean or replace air filters in an air conditioner every 1-3 months, depending on the usage and the quality of the air in the room. Dirty or clogged filters can reduce the efficiency of the unit and also negatively impact air quality.

## 5. Can an air conditioner help improve indoor air quality?

Yes, an air conditioner can help improve indoor air quality by filtering out dust, pollen, and other airborne particles. However, it is important to regularly clean or replace the air filters to maintain the effectiveness of this function. Additionally, using a high-quality air filter and keeping the unit well-maintained can also contribute to better air quality.

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