Second Law of Thermodynamics - heat pumps and air conditioners

[elcaptain]

Homework Statement

A heat pump operates as an air conditioner, drawing 20000 Btu per hr from a building. It is operating between 21 degrees Celsius and 41 degrees Celsius.

a) If the COP (coefficient of performance) is 35% that of a Carnot air conditioner, what is the value of the effective COP (coefficient of performance)?

b) What is the value of the power (Watts) required for the compressor motor?

Homework Equations

Q_L = heat removed from the low temperature area (inside the building)
Q_H = heat removed from the high temperature area (outside the building)
W = work done to remove the heat
T_L = temperature of system at lower temperature (inside the building)
T_H = temperature of system at higher temperature (outside the building)

COP = Q_L/W

COP = Q_L/(Q_H - Q_L)

COP_ideal = T_L / (T_H - T_L)

The Attempt at a Solution

I first converted Btu (British thermal unit) to joules: 1Btu = 1055.056 J. Then I tried finding the ideal COP by using the above equation with the given temperatures and got 1.05, but I'm pretty sure the ideal COP is not calculated this way. I tried multiplying 1.05 by 0.35 but got the wrong answer for part a. Thanks for any help.

 

[elcaptain]

anything?

 

[nlightner]

I would approach this problem by first understanding the Second Law of Thermodynamics. This law states that heat will naturally flow from a hot object to a cold object, and that in order to reverse this flow, work must be done. This is the basis for how heat pumps and air conditioners work.

In this problem, we are given the COP (coefficient of performance) of the heat pump, which is defined as the ratio of the heat removed from the low temperature area (QL) to the work done to remove the heat (W). We are also given the temperatures of the system (TL = 21 degrees Celsius and TH = 41 degrees Celsius).

To solve for the value of the effective COP, we can use the equation COP = QL/W and plug in the given COP (0.35) and the temperature difference (TH-TL = 20 degrees Celsius). This gives us:

0.35 = QL/W
W = QL/0.35

To find the value of QL, we can use the equation COPideal = TL/(TH-TL) and plug in the given temperatures. This gives us:

COPideal = 21/(41-21)
COPideal = 21/20
COPideal = 1.05

Now we can substitute this value into the equation for W:

W = QL/0.35
W = (1.05)/0.35
W = 3 Joules

Therefore, the value of the effective COP is 3 and the power required for the compressor motor is 3 Watts. This means that for every 3 Watts of power used, the heat pump is able to remove 1 Joule of heat from the building.

In conclusion, the Second Law of Thermodynamics explains how heat pumps and air conditioners work by using work to reverse the natural flow of heat. By understanding this law and using the equations for COP and COPideal, we can solve for the value of the effective COP and the power required for the compressor motor.