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Calculating the Efficiency of a Heat engine

  1. Apr 17, 2017 #1

    grandpa2390

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    1. The problem statement, all variables and given/known data
    I am trying to calculate the efficiency of this heat engine that has two step. an adiabatic compression, followed by a linear expansion back to the original point. I keep getting an efficiency of 1, which I know can't be right...

    2. Relevant equations
    ##e = \frac{W}{Q_H}##

    3. The attempt at a solution

    Is the W the net Work or the Work done by the gas. (I am guessing the Work done by the gas)
    and Q_H is the heat absorbed. which is 0 for the adiabatic process. and since Q=0 for an adiabatic process, then Q_H is equal to the negative of the net work.

    this is giving me a efficiency greater than 1 though.
     
    Last edited: Apr 18, 2017
  2. jcsd
  3. Apr 18, 2017 #2

    Andrew Mason

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    On the PV diagram, the useful mechanical work output of the engine, W, is the area above the adiabatic curve and the below the expansion line (i.e. calculate the entire area in the rectangle below the expansion line and subtract the area below the adiabatic compression curve). The engine does useful work during expansion. The work done during compression is done on the system. During what part of the process is heat absorbed ? That is the Qh. You should be able to find Qh easily enough (hint: find the temperatures at the beginning and end of the expansion. What is the pressure during expansion? What is the heat flow in terms of the heat capacity for that kind of change?)

    AM
     
    Last edited: Apr 18, 2017
  4. Apr 18, 2017 #3

    grandpa2390

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    hold on
     
  5. Apr 18, 2017 #4

    grandpa2390

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    I found Qh for the system. and the net work. My problem is that they are equal.
    They are equal because Q = -W
    and since Q=0 during the adiabatic step,
    Qh = -W

    so when I say W/Qh, I get 1. I know this can't be right. but the numbers are correct. Q (heat absorbed) = -W for this problem. is that the wrong Q or something.


    (hint: find the temperatures at the beginning and end of the expansion. got this. 571.6 K -> 300 K
    What is the pressure during expansion? it changes linearly with volume.
    What is the heat flow in terms of the heat capacity for that kind of change? I'm not familiar with this approach.

    The way I did it was by calculating the work done by each process. I know that Q=0 for adiabatic expansion/compression. so Q = -W which makes Qh = -W since Qc = 0
     
  6. Apr 18, 2017 #5

    Andrew Mason

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    Study my previous post. There are 3 parts to the cycle. There is work done on the system during the adiabatic compression. Since there is no heat flow, what is that work equal to ( hint: apply the first law)? If you know the temperature at the beginning and end of the compression you can determine the work done (hint: how is ΔU related to ΔT?). You will need to determine that work during the adiabatic compression because you have to subtract that from the work done in the forward expansion part of the cycle. Is there any work done during the constant volume part?

    Ok. What is the T at the beginning of compression?
    You should be able to read it off the PV graph.
    Hint: what is Cp?

    ?? This is not correct. You must apply the first law: Q = ΔU + W where W is the work done BY the system. If Q = 0 what is W?

    AM
     
  7. Apr 18, 2017 #6

    grandpa2390

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    I did all that. I found the work done during compression and subtracted it from the work done during the expansion. I found T.
    There is no constant volume part. there is an adiabatic compression followed by linear expansion. that is the entire cycle.
    at the beginning of compression it is 300 K. it goes through adiabatic compression and ends up at 571.6 K. then it goes through linear expansion back to 300 K
    that's what I did. Or I am not seeing how I didn't do that.

    dU = Q + W and since dU = 0 {for the entire system}, Q {for the entire system} = -W {of the entire system}
    and since Q=0 for 1 of the processes, that means Q of the other process must be equal to -W...
     
    Last edited: Apr 18, 2017
  8. Apr 18, 2017 #7

    Andrew Mason

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    When you said expansion was a straight line, I thought you meant at straight horizontal line (i.e at constant pressure to the original volume) and then a constant volume cooling (vertical line) to get to the original point.

    Finding the heat flow in, Qh, is a bit tricky. For the first part of the expansion, heat flows into the gas (if no heat flowed, the pressure would be lower ie. it would follow the adiabatic curve). For the last part of the expansion, heat flows out. The trick is to find where that point is.

    Let me think about this a bit more.

    AM
     
  9. Apr 18, 2017 #8

    grandpa2390

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    ok... I get what you are saying. rather than looking at the expansion with a net heat in or out, you are looking for just the heat that comes in. because while Q may equal 60 J for the expansion. it could be 90 J in and 30 J out? I know from similar problems that there is a Max T in the middle somewhere. But you need to think about whether that's what we want to find.

    I sure hope not because I did a similar problem on the test with isothermal instead of adiabatic, and it worked out... unless the Qh involves this extra step :(
     
  10. Apr 18, 2017 #9

    Andrew Mason

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    Ok, I see that. See my previous post. (It would help if you could post the PV graph).

    You are confusing work done on the system with work done by the system. There is no net work done on the system during a whole cycle in any heat engine since the system always returns to its original state. This does not mean that there is no work done by the system. The net work output is the area between the compression and expansion lines on the PV graph. This area is necessarily equal to Qh-Qc since ΔU = 0 for the system over an complete cycle.

    The trick here is to find the point at which heat stops flowing into the system on that linear portion and starts flowing out.

    AM
     
  11. Apr 18, 2017 #10

    grandpa2390

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    I know that. It was just a typo. I know how to calculate the net work.

    ok. I will have to write an equation for that line, and find where T is at a max.
     
  12. Apr 18, 2017 #11

    Andrew Mason

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    The isotherm curve depicts an expansion at constant U (assuming the gas is ideal). The curve is defined by the ideal gas law: P = nRT/V

    If you were to plot the isotherm curve for the system at the maximum temperature 571.6 K over the same volume expansion, that curve would descend below your expansion line for the first part and then it would cross the line (where T = 571.6) and stay above the line until it reached the final volume.
    Prior to crossing that isotherm, the rate at which work is done by the system (ie. P: dW/dV = d(PdV)/dV = P) is greater than if the system expanded isothermally (ΔU = 0). This could only occur if net heat flow into the system has occurred up to that point. After crossing the isotherm, the rate at which work is done is less than the rate that work would have been done if the system expanded isothermally. This could only occur if net heat flow out of the system occurrs. So if you can determine where your line crosses the isotherm (ie.,the point on the line where P = nRT/V) you should be able to find Qh.

    AM
     
  13. Apr 18, 2017 #12

    grandpa2390

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    I don't know what you are talking about. What isotherm?
     
    Last edited: Apr 18, 2017
  14. Apr 18, 2017 #13

    vela

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    Here's a pV diagram that might help. The blue curve is an adiabat, and the green one is an isotherm. I believe Andrew is saying the isotherm divides the orange line between the two endpoints into two segments, and on the top segment, the engine absorbs heat while on the bottom segment, it rejects heat.

    You want to find the point where 0 = dQ/dV = dU/dV ± dW/dV (not sure what sign convention you're using). I'm not convinced that occurs where the isotherm crosses the line, though.

    Note: The units on the axes of the diagram aren't relevant to this problem. I just made up numbers to get a reasonable looking diagram. Also, my calculations show that the crossover between heat absorption and heat rejection wasn't where the isotherm intersected the line.

    pv.png
     
    Last edited: Apr 18, 2017
  15. Apr 18, 2017 #14

    Andrew Mason

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    I was not saying necessarily that the point where dQ/dV = 0 is the intersection with the isotherm. I was just saying that up to that point there has been net heat flow into the system and after that point net heat flow is negative (out of the system).

    I am not sure where dQ/dV = 0 but I agree that this is what you want to find. Up to that point, dQ/dV>0 (heat flow is in) and after, dQ/dV<0 (heat flow is out). From the first law, dQ/dV = dU/dV + dW/dV = dU/dV + P (W is the work done by the gas). So you want to find the point on the line where dQ/dV = 0 i.e. where: dU/dV = -P.

    AM
     
  16. Apr 19, 2017 #15

    grandpa2390

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    I may be mistaken. but I am pretty sure that positive is work done on the gas, and negative is work done by the gas.
    Positive is heat absorbed and negative is heat removed.

    A thermodynamic cycle consists of only 2 processes and uses a monatomic ambient gas, First, the monatominc gas is compressed adiabatically from the initial pressure Pi=1 bar to Pf=5 bar starting at temperature Ti=300 K The initial volume of the gas before the compression is Vi=10 liters. In the second process gas expands along straight line on the PV diagram. Find the Efficiency.
    ?temp_hash=91fb6a9a2e12ccb3eff64a03c876b164.png

    Here's what I have got so far:
    Pi=10^5 Pa
    Vi = .01 m^3
    ni = .4 mol
    Ti = 300 K

    Pf = 5*10^5
    Vf = .0038 m^3
    nf = .4 mol
    Tf = 571.6 K

    Wtotal = 1354.2 J (adiabat) + -1860J (linear) = -505.8 J
    Qi->f = 0

    Qf->i = 505.8 J = Qh + Qc
    Part of this is the sum of heat absorbed and heat given off. this doesn't make a whole lot of sense to me because the process absorbs more heat than it exhausts, but the temperature drops...

    I found the eq for the linear expansion to be ##-6.45*10^7V + 745161##

    knowing this and I found the temperature during expansion to be ##T = \frac{-6.45*10^7V^2 + 745161V}{.4*8.31}##

    and the max temperature would be 647.308 K at V = .00577644 m^3

    Seems strange that the temperature would rise during expansion first... but it isn't for long. .0038 -> .0058 is pretty short between .0038 and .01

    What do I do. I know this is a lot to ask. I don't think I learned what you are talking about.
     

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    Last edited: Apr 19, 2017
  17. Apr 19, 2017 #16

    grandpa2390

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    I can scan my work and upload it if necessary?
     
  18. Apr 19, 2017 #17

    grandpa2390

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    could i divide the linear process into two. separate it where the T reaches max, and then calculate the work done on either side of that line. the work would be equal to Qin and Qout
     
  19. Apr 19, 2017 #18

    grandpa2390

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    could i divide the linear process into two. separate it where the T reaches max, and then calculate the work done on either side of that line. the work would be equal to Qin and Qout

    between .003... and .005... x amount of work is done which is Qin (heat being absorbed)
    and between .005... and .01 y amount of work is don which is Qout(heat being exhausted)

    I'm still stuck on how the net Q can be heat absorbed if the temperature is dropping. or is my sign wrong?
     
  20. Apr 19, 2017 #19

    Andrew Mason

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    The maximum T (maximum U) just tells you where dU/dV = 0. When dU/dV = 0, dQ/dV = dU/dV + dW/dV = dU/dV + P = 0 + P. So dQ/dV is still positive (heat flow is into the system). What you want to find is where dQ/dV = 0. After that, heat flow is out of the system.

    AM
     
  21. Apr 19, 2017 #20

    grandpa2390

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    how do I do that?
     
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