Finding efficiency of an engine (with attempt)

In summary, the conversation discusses a failed attempt at finding efficiency through a specific process, followed by a potential solution that involves finding work and heat for each step and dividing to get the total efficiency. There is also a mention of converting units and potential sources for the numbers used in the calculation.
  • #1
RandiSS
5
1
Homework Statement
An engine using 1 mol of an ideal gas ini- tially at 17.8 L and 283 K performs a cycle consisting of four steps:
1) an isothermal expansion at 283 K from 17.8 L to 35.1 L;
2) cooling at constant volume to 125 K ;
3) an isothermal compression to its original volume of 17.8 L; and
4) heating at constant volume to its original temperature of 283 K .
Find its efficiency. Assume that the heat capacity is 21 J/K and the univer- sal gas constant is 0.08206 L · atm/mol/K = 8.314 J/mol/K.
Relevant Equations
ɛ=W/Q
W=nRTln(Vf/Vi)
Q=nCΔT
4D5E1E81-4AEF-4944-9A34-02D1ABF814BB.jpeg

This was my first attempt which was not right. But it was at least a percent unlike my 2nd attempt.
64735F59-2206-42A3-AE8D-BB003A89B58E.jpeg

This was my second which I found on another forum website. It said it found the work of step 1, heat absorbed of step 2, work of step 3, and absorbed heat of step 4. However, the answers to the equations they provided just canceled out when finding the total heat absorbed and work done, which gave me an undefined efficiency. I know I need to find the work and heat for each step, combine, then divide total W by total Q. Am I using the wrong numbers or is this just not the right process to find efficiency?
 
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  • #2
Update:
A7538912-430F-4FE2-8E1B-40BA7E591C49.jpeg

I think I might have figured it out but I’m not sure.
 
  • #3
RandiSS said:
Update:View attachment 324922
I think I might have figured it out but I’m not sure.
I got it!
 
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  • #4
where you get 0.0315 and 0.0178 from?
 
  • #5
soulpassion said:
where you get 0.0315 and 0.0178 from?
"an isothermal expansion at 283 K from 17.8 L to 35.1 L"
For some reason, @RandiSS converted to kilolitres, but since only the ratio matters it makes no difference.
 
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  • #6
soulpassion said:
where you get 0.0315 and 0.0178 from?
Maybe from here?
RandiSS said:
1) an isothermal expansion at 283 K from 17.8 L to 35.1 L;
 

FAQ: Finding efficiency of an engine (with attempt)

1. What is the formula for calculating the efficiency of an engine?

The efficiency of an engine, denoted as η (eta), is calculated using the formula: η = (Work Output / Heat Input) * 100%. This formula gives the efficiency as a percentage.

2. How do you measure the work output of an engine?

The work output of an engine can be measured using a dynamometer, which measures the torque and rotational speed of the engine. The work output is then calculated as Work Output = Torque * Angular Velocity.

3. What is the significance of the Carnot efficiency in engine performance?

The Carnot efficiency represents the maximum possible efficiency that an engine operating between two temperatures can achieve. It is given by the formula: η_Carnot = 1 - (T_cold / T_hot), where T_cold and T_hot are the absolute temperatures of the cold and hot reservoirs, respectively. This sets an upper limit on the efficiency of real engines.

4. How do you account for heat losses in calculating engine efficiency?

To account for heat losses, you need to measure the total heat input to the engine and subtract the heat losses to the surroundings. The effective heat input is then used in the efficiency formula: η = (Work Output / Effective Heat Input) * 100%. Heat losses can be measured using calorimetry or inferred from temperature and insulation data.

5. What are common sources of inefficiency in an engine?

Common sources of inefficiency in an engine include frictional losses, heat losses to the environment, incomplete combustion of fuel, and mechanical losses in components such as bearings and gears. Improving these aspects can help enhance the overall efficiency of the engine.

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