Calculate the ideal (Carnot) efficiency of a heat engine

In summary: The heat rejected (heat flow out to cold reservoir) plus the work done has to be equal to the heat flow into the engine.So, the efficiency is equal to Q_H/Q_C, which is equal to T_H/T_C.
  • #1
lilkrazyrae
64
0
Calculate the ideal (Carnot) efficiency of a heat engine operating between 23.0 degrees and 515 degrees. How much heat would be rejected by the engine if 1.00 x 10^6 calories were taken from the high temperature resivoir.

Would you just use e=1-(T(c)/T(H))?
And for the second part i don't have any clue how to get started.
 
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  • #2
You have to use Kelvin ... .

efficiency is not "1" if your Energy input is split between 2 (or more) outputs. You have a million cal input, and only "e %" of them will be used for what you had intended. Once you get Tc/Th , you know what fraction are wasted.
 
  • #3
lilkrazyrae said:
Calculate the ideal (Carnot) efficiency of a heat engine operating between 23.0 degrees and 515 degrees. How much heat would be rejected by the engine if 1.00 x 10^6 calories were taken from the high temperature resivoir.
Would you just use e=1-(T(c)/T(H))?
And for the second part i don't have any clue how to get started.
The efficiency is defined as the ratio of work output to heat input:

Efficiency[tex] = W/Q_H = (Q_H-Q_C)/Q_H = 1 - Q_C/Q_H[/tex]

It can be shown that for a Carnot engine (no entropy change), [itex]Q_C/Q_H[/itex] is equal to [itex]T_C/T_H[/itex].

That should enable you to answer the question.

AM
 
Last edited:
  • #4
So for efficiency e =1-T(c)/T(h)=1-(298.18/788.15)=62.171%
And for the heat rejected .62171=x/1.00*10^6=6.22*10^5 cal.

Is this right
 
  • #5
Or would you have to include the 1 in the efficiency equation again?
 
  • #6
No Body knows?!?
 
  • #7
lilkrazyrae said:
So for efficiency e =1-T(c)/T(h)=1-(298.18/788.15)=62.171%
And for the heat rejected .62171=x/1.00*10^6=6.22*10^5 cal.
Is this right
You are trying to find the heat rejected. You have found the work done: W = 6.22 x 10^5 cal. The heat rejected (heat flow out to cold reservoir) plus the work done has to be equal to the heat flow into the engine.

[tex]Q_C = Q_H - W = 1 \times 10^6 - 6.22 \times 10^5 = 3.78 \times 10^5 cal.[/tex]

AM
 

What is a heat engine?

A heat engine is a device that converts heat energy into mechanical work, typically by using a fuel source to generate heat and powering a piston or turbine.

What is the ideal efficiency of a heat engine?

The ideal efficiency of a heat engine is known as the Carnot efficiency, named after French physicist Nicolas Léonard Sadi Carnot. It is the maximum possible efficiency for a heat engine operating between two temperatures, and is given by the ratio of the difference between the temperatures to the higher temperature, expressed as a percentage.

How is the ideal efficiency of a heat engine calculated?

The ideal efficiency of a heat engine can be calculated using the formula: Efficiency = (Th - Tl) / Th * 100%, where Th is the absolute temperature of the hot reservoir and Tl is the absolute temperature of the cold reservoir.

What factors affect the ideal efficiency of a heat engine?

The ideal efficiency of a heat engine is affected by the temperatures of the hot and cold reservoirs, as well as the type of working fluid and the design of the engine. In general, higher temperature differences and more efficient working fluids result in higher efficiencies.

How does the ideal efficiency of a heat engine compare to the actual efficiency?

The ideal efficiency of a heat engine is a theoretical limit and is not achievable in real-world systems. Actual efficiencies are always lower due to factors such as friction, heat loss, and imperfect design. However, engineers strive to increase the efficiency of heat engines as close to the ideal as possible through advancements in technology and design.

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