Finding efficiency of an engine (with attempt)

  • Thread starter Thread starter RandiSS
  • Start date Start date
  • Tags Tags
    Efficiency Engine
Click For Summary
The discussion revolves around attempts to calculate engine efficiency, with the first attempt yielding a minimal result and the second leading to undefined efficiency due to canceled equations. The user references a process involving heat absorbed and work done across multiple steps but questions the accuracy of the numbers used. An update indicates a potential breakthrough in understanding the calculations, specifically regarding isothermal expansion at 283 K. The conversation also touches on unit conversions, noting that while one participant converted to kiloliters, it ultimately does not affect the ratio. The focus remains on clarifying the correct method for determining engine efficiency.
RandiSS
Messages
5
Reaction score
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?
 
Physics news on Phys.org
Update:
A7538912-430F-4FE2-8E1B-40BA7E591C49.jpeg

I think I might have figured it out but I’m not sure.
 
RandiSS said:
Update:View attachment 324922
I think I might have figured it out but I’m not sure.
I got it!
 
where you get 0.0315 and 0.0178 from?
 
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.
 
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;
 
If have close pipe system with water inside pressurized at P1= 200 000Pa absolute, density 1000kg/m3, wider pipe diameter=2cm, contraction pipe diameter=1.49cm, that is contraction area ratio A1/A2=1.8 a) If water is stationary(pump OFF) and if I drill a hole anywhere at pipe, water will leak out, because pressure(200kPa) inside is higher than atmospheric pressure (101 325Pa). b)If I turn on pump and water start flowing with with v1=10m/s in A1 wider section, from Bernoulli equation I...

Similar threads

Replies
14
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
4K
Replies
3
Views
3K
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K