Thermodynamics: Carnot engines

So, in summary, the problem involves calculating the coefficient of performance for the refrigerator in Part 1 and using that to find the heat removed from the building. In Part 2, the efficiency of the generator and motor are considered, along with the energy lost during transmission, to determine the number of units of refrigeration obtained per unit of heat supplied.
  • #1
2DGamer
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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.

 
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  • #2
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].
 

1. What is a Carnot engine?

A Carnot engine is a theoretical heat engine that operates based on the principles of thermodynamics. It consists of two isothermal (constant temperature) and two adiabatic (no heat transfer) processes, and is used to demonstrate the maximum efficiency that any heat engine can achieve.

2. How does a Carnot engine work?

A Carnot engine works by taking in heat from a high temperature reservoir, performing work, and then releasing the remaining heat into a low temperature reservoir. This process is repeated in a cyclical manner, with the engine converting heat energy into mechanical work.

3. What is the efficiency of a Carnot engine?

The efficiency of a Carnot engine is given by the formula: efficiency = (T1 - T2) / T1, where T1 is the temperature of the high temperature reservoir and T2 is the temperature of the low temperature reservoir. This means that the efficiency of a Carnot engine is directly proportional to the temperature difference between the two reservoirs.

4. What is the significance of Carnot engines?

Carnot engines are significant because they provide a theoretical limit for the maximum efficiency of any heat engine. This means that no real-world heat engine can be more efficient than a Carnot engine. They also demonstrate the reversibility of certain processes in thermodynamics.

5. Can a Carnot engine be built in real life?

While Carnot engines are purely theoretical, certain real-world engines, such as Stirling engines, can closely approximate the efficiency of a Carnot engine. However, it is not possible to build a perfect Carnot engine due to the limitations of materials and processes in the real world.

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