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laser1
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- Efficiency Of Stirling Cycle - Is it the same as the Carnot Engine? Why is my described engine not the same as Stirling engine?
Ok, the problem I was given is this:
One mole of a perfect gas goes through a quasistatic cycle consisting of the following four stages:
1. isothermal expansion from V1 to V2 at temperature TH,
2. cooling at constant volume V2 from TH to TC,
3. isothermal compression from V2 to V1 at TC,
4. heating at constant volume until the system is back in its initial state
Wikipedia states this for a Stirling cycle:
Ok, so it seems to me they are the same. (I am just including this part because my flaw might be in the assumption that they are the same cycle).
This guy (https://www.youtube.com/watch?v=sklO_qZTBeY&t=371s) explains how to calculate the efficiency of the Stirling engine. His logic all looks fine. His final answer is given by η=1−TLTH Now I have a problem because for the original question, I worked out $$\eta = \frac{W}{Q_H} = \frac{RT_C \ln\left(\frac{V_f}{V_i}\right) (T_H - T_C)}{RT_H \ln\left(\frac{V_f}{V_i}\right) + C_V (T_H - T_C)}$$ which is definitely not the same as the efficiency of the Stirling cycle. Explaining in a bit more detail how I achieved the above expression: η is simply work you get over work you pay for.
One mole of a perfect gas goes through a quasistatic cycle consisting of the following four stages:
1. isothermal expansion from V1 to V2 at temperature TH,
2. cooling at constant volume V2 from TH to TC,
3. isothermal compression from V2 to V1 at TC,
4. heating at constant volume until the system is back in its initial state
Wikipedia states this for a Stirling cycle:
Ok, so it seems to me they are the same. (I am just including this part because my flaw might be in the assumption that they are the same cycle).
This guy (https://www.youtube.com/watch?v=sklO_qZTBeY&t=371s) explains how to calculate the efficiency of the Stirling engine. His logic all looks fine. His final answer is given by η=1−TLTH Now I have a problem because for the original question, I worked out $$\eta = \frac{W}{Q_H} = \frac{RT_C \ln\left(\frac{V_f}{V_i}\right) (T_H - T_C)}{RT_H \ln\left(\frac{V_f}{V_i}\right) + C_V (T_H - T_C)}$$ which is definitely not the same as the efficiency of the Stirling cycle. Explaining in a bit more detail how I achieved the above expression: η is simply work you get over work you pay for.