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Homework Help: Thermodynamics: Carnot engines

  1. Feb 18, 2010 #1
    1. There are two parts to this problem. I'm having trouble with the second part, but I'll include the first part as well since it's probably needed for the second part.

    Part 1:
    "A building is cooled by a Carnot engine operated in reverse (a Carnot refrigerator). The outside temperature is 35°C and the temperature inside the building is 20°C. If the engine is driven by a 12x10^3 Watt electric motor, how much heat is removed from the building per hour?"

    Part 2:
    "The motor is supplied with electricity generated in a power plant which consists of a Carnot engine operating between reservoirs at 500° C and 35° C. Electricity (transmitted over a 5 ohm line), is received at 220 volts. The motors operating the refrigerator and generator at the power plant each have an efficiency of 90%. Find the number of units of refrigeration obtained per unit of heat supplied."

    2. Coefficient of performance: c = TC/(TH-TC)
    also c = QC/Win
    Efficiency: n = 1 - TC/TH = Wout/QH
    TH is the hot reservoir
    TC is the cold reservoir
    QH is the heat being added to or removed from the hot reservoir
    QC is the heat being added to or removed from the cold reservoir

    3. Here's my solution for Part 1:
    Ok, I treated the two temperatures as the hot and cold reservoir and converted them to Kelvin, and found the coefficient of performance to be 19.53. Then I used the 12x10^3 Watt as the "work in", so I multiplied that by 19.53 to find the heat removed from the building which is 2.34x10^5 Watts. However, Watts is Joules per sec, and the question asked per hour. So the final answer I got is 8.42x10^8 J/hr.

    Here's what I have for Part 2:
    Ok, there are a few things that confuse me about this part of the problem. It says that the efficiencies of the fridge and generator are 90%. But it also says that the generator operates between reservoirs of 773K and 308K. With those temperatures, the efficiency is less than 90%, more like 60%. How can it be both? Also refrigerators don't have efficiencies, but they have coefficients of performance. Anyways, I'm just really confused as to how to put the pieces together in the second part. Also isn't watts = (volts*volts)/ohms? Any info that can help clarify what I'm supposed to be doing for Part 2 would be greatly appreciated.

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 18, 2010 #2


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    The Carnot engine produces a certain amount of work, and that work is converted to electrical energy with a 90% efficiency. Similarly, the motor driving the refrigerator converts electricity to work with 90% efficiency.

    There's also some energy lost during transmission of the electricity from the generator to the refrigerator. As you noted, the power dissipated by the resistance of the line is given by [itex]V^2/R[/itex].
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