Calculating the Efficiency of a Heat engine

In summary, the efficiency of the heat engine with an adiabatic compression followed by a linear expansion back to the original point is being calculated. The attempt at a solution involves finding the temperature and pressure during expansion, as well as the heat flow in terms of the heat capacity. The net work and heat absorbed are found to be equal, leading to an efficiency of 1. However, this cannot be correct and further calculations are needed to find the correct efficiency.
  • #1
grandpa2390
474
14

Homework Statement


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...

Homework Equations


##e = \frac{W}{Q_H}##

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.
 
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  • #2
grandpa2390 said:

Homework Statement


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...

Homework Equations


##e = \frac{W}{Q_H}##

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.
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
 
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Likes grandpa2390
  • #3
hold on
 
  • #4
Andrew Mason said:
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

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
 
  • #5
grandpa2390 said:
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
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?

(hint: find the temperatures at the beginning and end of the expansion. got this. 571.6 K -> 300 K
Ok. What is the T at the beginning of compression?
What is the pressure during expansion? it changes linearly with volume.
You should be able to read it off the PV graph.
What is the heat flow in terms of the heat capacity for that kind of change? I'm not familiar with this approach.
Hint: what is Cp?

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
?? 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
 
  • #6
Andrew Mason said:
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.
I did all that. I found the work done during compression and subtracted it from the work done during the expansion. I found T.
Is there any work done during the constant volume part?
There is no constant volume part. there is an adiabatic compression followed by linear expansion. that is the entire cycle.
Ok. What is the T at the beginning of compression?
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
?? 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?
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...
 
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  • #7
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
 
  • #8
Andrew Mason said:
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

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 :(
 
  • #9
grandpa2390 said:
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.
Ok, I see that. See my previous post. (It would help if you could post the PV graph).

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...
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
 
  • #10
Andrew Mason said:
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.
I know that. It was just a typo. I know how to calculate the net work.

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
ok. I will have to write an equation for that line, and find where T is at a max.
 
  • #11
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
 
  • #12
Andrew Mason said:
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

I don't know what you are talking about. What isotherm?
 
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  • #13
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
 
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  • #14
vela said:
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.
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
 
  • #15
Andrew Mason said:
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
vela said:
You want to find the point where 0 = dQ/dV = dU/dV ± dW/dV (not sure what sign convention you're using).
View attachment 195546
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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  • #16
I can scan my work and upload it if necessary?
 
  • #17
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
 
  • #18
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?
 
  • #19
grandpa2390 said:
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
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
 
  • #20
Andrew Mason said:
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

how do I do that?
 
  • #21
grandpa2390 said:
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...
Some of that heat is converted to work, so there's a net decrease in internal energy and therefore temperature.

What do I do. I know this is a lot to ask. I don't think I learned what you are talking about.
Express U as a function of V and then differentiate to find dU/dV, and then follow what Andrew explained in post 19.
 
  • #22
vela said:
Express U as a function of V and then differentiate to find dU/dV, and then follow what Andrew explained in post 19.

How do you do that. I am searching all over the internet. And the youtube videos and nobody seems to know how to do it. Can you give the expression.
I don't know how to write U as a function of V for a process such as this. If it were an isobaric process, etc. I could break down U = W+Q, but nobody seems to know the formulas Q with respect to V for this linear process.
 
  • #23
Hint: It's an ideal gas.
 
  • #24
vela said:
Hint: It's an ideal gas.

well W = integral(-p(V)) so perhaps Q = integral(p(V))
 
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  • #25
vela said:
Hint: It's an ideal gas.

I need to have this figured out. you all are assuming that I have been taught this information already and just need a reminder, but I am having to figure it out as I go. If you don't want to break it down for me, can you point me to a video or something that will? The textbook is no help. Maybe there is a youtube video or something you can point me to? I need someone to walk me through this. I don't have any idea what the ideal gas law is telling me about the heat transfer.
 
  • #26
grandpa2390 said:
how do I do that?
You have to solve dU/dV = -P

U = nCvT = nCv(PV/nR) = Cv(PV/R)

It will get messy because you will end up with a quadratic equation to solve.

AM
 
  • #27
Andrew Mason said:
You have to solve dU/dV = -P

U = nCvT = nCv(PV/nR) = Cv(PV/R)

It will get messy because you will end up with a quadratic equation to solve.

AM

That's the formula for an isochoric process. Shouldn't it be U = nCvdT, and Why do we use that one rather than the formula for an isobaric?
 
  • #28
grandpa2390 said:
That's the formula for an isochoric process. Shouldn't it be U = nCvdT, and Why do we use that one rather than the formula for an isobaric?
For a given quantity of an ideal monatomic gas, internal energy is a function of temperature only. (1) U = (3/2)NkT. At T = 0, U = 0. So ΔU = (3/2)NkΔT. This is always true and does not depend on the process. This is derived from molecular theory. k is the Boltzmann constant and N is the number of particles. Nk = nR where n is the number of moles and R is the molar gas constant.

The reason the proportionality constant, (3/2)Nk (= (3/2)nR) is also the specific heat at constant volume is apparent from the first law. By definition the mole specific heat at constant volume is Cv = ΔQ/nΔT. From the first law, ΔQ = ΔU + PΔV. When ΔV = 0, ΔQ = ΔU and so applying (1): ΔQ/ΔT = (3/2)Nk = (3/2)nR = nCv.

AM
 
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  • #29
Andrew Mason said:
For a given quantity of an ideal monatomic gas, internal energy is a function of temperature only. (1) U = (3/2)NkT. At T = 0, U = 0. So ΔU = (3/2)NkΔT. This is always true and does not depend on the process. This is derived from molecular theory. k is the Boltzmann constant and N is the number of particles. Nk = nR where n is the number of moles and R is the molar gas constant.

The reason the proportionality constant, (3/2)Nk (= (3/2)nR) is also the specific heat at constant volume is apparent from the first law. By definition the mole specific heat at constant volume is Cv = ΔQ/nΔT. From the first law, ΔQ = ΔU + PΔV. When ΔV = 0, ΔQ = ΔU and so applying (1): ΔQ/ΔT = (3/2)Nk = (3/2)nR = nCv.

AM

Someone else told me that I have to draw a t-s diagram? is that what we are doing?

edit: I keep looking at your hint, and I'm not getting what I am supposed to do.
what does
solve dU/dV = -P mean?

i am thinking that U = nCvdT . and U = W + Q . So I can find the U and W during the temperature rise, subtract, and I will be left with Qh.

but v isn't constant. so I take the integrate with respect to V...
 
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  • #30
grandpa2390 said:
Someone else told me that I have to draw a t-s diagram? is that what we are doing?
Essentially, yes. We are trying to determine the relationship between P, V (or T) and heat flow (Q). A T-S diagram will show the relationship between temperature and heat flow. A T-S diagram will be a vertical line for the adiabatic portion and a curve on the right side (looking like a D shape overall) for the expansion portion. So if you can draw a T-S diagram for the expansion, you look for the point where S is maximum (i.e. the point farthest to the right). The area below the lower part of the curve from at that point is Qc and the area below the upper part of the curve is Qh. The area inside the curve is Qh-Qc. We are trying to find the point at which S is maximum, ie. where dS/dT = 0. This should also be where dQ/dV = 0 (in which case dU/dV = -P).
edit: I keep looking at your hint, and I'm not getting what I am supposed to do.
what does
solve dU/dV = -P mean?
U = Cv(PV/R)
Work out dU/dV and set it equal to -P. The solution will give you the point on the line where dQ/dV is zero. Before that point, heat flow is in. After, heat flow is out.

By the product rule:
dU/dV = (Cv/R)(P + VdP/dV)
Work out dP/dV from the relationship between P and V that you derived.

AM
 
  • #31
Andrew Mason said:
Essentially, yes. We are trying to determine the relationship between P, V (or T) and heat flow (Q). A T-S diagram will show the relationship between temperature and heat flow. A T-S diagram will be a vertical line for the adiabatic portion and a curve on the right side (looking like a D shape overall) for the expansion portion. So if you can draw a T-S diagram for the expansion, you look for the point where S is maximum (i.e. the point farthest to the right). The area below the lower part of the curve from at that point is Qc and the area below the upper part of the curve is Qh. The area inside the curve is Qh-Qc. We are trying to find the point at which S is maximum, ie. where dS/dT = 0. This should also be where dQ/dV = 0 (in which case dU/dV = -P).

U = Cv(PV/R)
Work out dU/dV and set it equal to -P. The solution will give you the point on the line where dQ/dV is zero. Before that point, heat flow is in. After, heat flow is out.

By the product rule:
dU/dV = (Cv/R)(P + VdP/dV)
Work out dP/dV from the relationship between P and V that you derived.

AM
##\frac{dU/dV}{C_V/R} - VdP/dV = P##

I don't know what to do with this. thanks for trying, but my assignment is due. Time to embrace my F. I wasted an entire week trying to figure out this one problem, my time would have been better spent studying for other classes :(
it's why i pleaded to be directed towards a video lecture or something that could walk me through to understand the concepts I need to solve this problem.
I do not have the knowledge I need to follow hints. I need instruction.
 
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  • #32
grandpa2390 said:
##\frac{dU/dV}{C_V/R} - VdP/dV = P##
Set dU/dV = -P to determine where heat flow direction changes (dQ/dV = 0 or dS/dT = 0)
ie. ##dU/dV = \frac{Cv}{R}(P + V\frac{dP}{dV}) = -P##. Substitute dP/dV from the equation for your expansion line (hint: it is a line of constant, negative slope and dP/dV is the slope).

##\frac{Cv}{R}(P + mV) = -P##

3) ##V = \frac{P}{m}\left(\frac{Cv}{R} + 1\right)##

Since the expansion line is of the form: P = mV + b substitute mV + b for P in 3). That will give you the value of V where dQ/dV = 0. I didn't say it was easy.
I don't know what to do with this. thanks for trying, but my assignment is due. Time to embrace my F. I wasted an entire week trying to figure out this one problem, my time would have been better spent studying for other classes :(
This is a very difficult problem. I doubt that you were expected to solve it. Give it your best shot. Look at it as a good learning experience.

AM
 
  • #33
Andrew Mason said:
Set dU/dV = -P to determine where heat flow direction changes (dQ/dV = 0 or dS/dT = 0)
ie. ##dU/dV = \frac{Cv}{R}(P + V\frac{dP}{dV}) = -P##. Substitute dP/dV from the equation for your expansion line (hint: it is a line of constant, negative slope and dP/dV is the slope).

##\frac{Cv}{R}(P + mV) = -P##

3) ##V = \frac{P}{m}\left(\frac{Cv}{R} + 1\right)##

Since the expansion line is of the form: P = mV + b substitute mV + b for P in 3). That will give you the value of V where dQ/dV = 0. I didn't say it was easy.
This is a very difficult problem. I doubt that you were expected to solve it. Give it your best shot. Look at it as a good learning experience.

AM
I don't know. We took a test and I didn't do so well. Apparently nobody else must have either because he sent this problem out for us to do for an extra three points if correctly done.

And he said it was OK to ask for help from people and so forth otherwise I wouldn't of asked. But it was due today but now it's too late. I wrote down all the information I found it figured out. And explained the problem I ran into. I probably failed. But maybe I'll get enough partial credit to pass

I'm already having to do my best to teach myself on the book and these problems are not like anything in the homework or otherwise

Take home questions ought to be difficult. But to ask questions that require concepts not taught... :(
 

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A heat engine is a device that converts thermal energy into mechanical energy. It typically uses a working fluid, such as steam or gas, to transfer heat from a high temperature source to a low temperature sink in order to produce work.

2. How is efficiency calculated for a heat engine?

The efficiency of a heat engine is calculated by dividing the amount of work output by the amount of heat input. This can be expressed as a percentage or decimal value.

3. What factors affect the efficiency of a heat engine?

The efficiency of a heat engine is affected by several factors, including the temperature difference between the hot and cold reservoirs, the type of working fluid used, and the design and construction of the engine.

4. Can the efficiency of a heat engine be greater than 100%?

No, the efficiency of a heat engine cannot be greater than 100%. This would violate the first and second laws of thermodynamics, which state that energy cannot be created or destroyed, and that heat always flows from hot to cold.

5. How can the efficiency of a heat engine be improved?

The efficiency of a heat engine can be improved by increasing the temperature difference between the hot and cold reservoirs, using a more efficient working fluid, and optimizing the design and construction of the engine to minimize energy losses.

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