B Heat pump as a heat engine that operates in reverse

[FranzDiCoccio]

TL;DR Summary

I'm trying to clarify the relation between heat pumps and engines, and to understand the inequalities involved in their "Carnot cycle" versions.

The efficiency of a heat engine is calculated as $\eta = |W|/|Q_h| = 1- |Q_c|/|Q_h|$. If this engine operates between the temperatures $T_c$ and $T_h$, then Carnot's theorem states that $\eta<\eta_C = 1-T_c/T_h$. This means $T_c/T_h < |Q_c|/|Q_h|$.

Now assume that the heat engine is reversed to obtain e.g. a heat pump. Its efficiency is described by the coefficient of performance $COP = Q_h/|W|$. This efficiency should be less than that of a "Carnot" heat pump,
$$ \frac{|Q_h|}{|Q_h|-|Q_c|} < \frac{T_h}{T_h-Tc} $$
After a little algebra this gives $T_c/T_h > |Q_c|/|Q_h|$, i.e. the opposite inequality as before. The same result is obtained if the COP for a refrigerator is considered.

I'm not entirely clear what this means, or even whether this line of reasoning makes sense. I naively assumed that after reversing the cycle the quantities remained the same, except for their signs (I used absolute values also when not strictly necessary, to be on the safe side).

Does this simply mean that $|Q_h|$ and $|Q_c|$ in a heat pump are not quantitatively the same as in the heat engine that has been reversed to obtain it? That is, we use the same symbols but their values are not the same?
In other words, the "reversing" of the engine is not as symmetric as one might naively think?

(sorry I do not seem to be able to compile LaTeX formulas)

 

[256bits]

Isn't that what you wrote?
So how does COP become the efficiency of heat engine when it is already stated as above?
@FranzDiCoccio

 

[FranzDiCoccio]

Hi, thanks for your time.
That is indeed the inequality I wrote.
I'll try to clarify my question. Correct me if I'm wrong

• if the cycle of the heat pump is reversed a refrigerator cycle is obtained.
• Now suppose that this cycle is used in a heat pump (a refrigerator would have a different COP, but the final inequality would be the same). The COP measures how good that heat pump is. So it is a measure of the efficiency of the pump.
• Such COP would be smaller than that of a heat pump working between the same temperatures on a (reverse) Carnot cycle
• the inequality ${\rm COP}<{\rm COP_C}$ results in the equality $$ \frac{T_c}{T_h}> \frac{|Q_c|}{|Q_h|} $$ that appears to contradict the equality obtained for the thermal engine (i.e. for the same cycle running clockwise) $$ \frac{T_c}{T_h}<\frac{|Q_c|}{|Q_h|}$$

I was trying to understand this apparent contradiction.

 

[FranzDiCoccio]

I have just realized that more than five years ago I asked a highly related question I completely forgot about.

The conclusion of that discussion was that the work $L_e$ obtained from a heat engine that absorbs $Q_h$ at $T_h$ is smaller than the work $L_p$ required by a heat pump for transferring $Q_h$ back to the source at $T_h$. The inequality $L_e<L_p$ derives from the two inequalities in my original post. Since we are assuming that $Q_h$ is the same, $Q_c=Q_h-L$ must be different in the two cases.

More in detail, we can assume $Q_{he} = Q_{hp} = Q_h$ and, recalling that $L_e<L_p$, $Q_{ce} = Q_{he}-L_e = Q_{hp}-L_e > Q_{hp}-L_p = Q_{cp}$ (where all quantities are assumed positive).
This means that
$$ \frac{Q_{ce}}{Q_{he}}>\frac{Q_{cp}}{Q_{hp}}$$
In other words, the above apparent contradiction disappears if we do not use the same symbols for the two cycles (because the corresponding value are not the same)

$$ \frac{Q_{ce}}{Q_{he}}>\frac{T_c}{T_f}>\frac{Q_{cp}}{Q_{hp}}$$

I have to assimilate this, otherwise I'm going to ask the same question in five years :)

 

[demaha123]

I have just realized that more than five years ago I asked a highly related question I completely forgot about.

The conclusion of that discussion was that the work $L_e$ obtained from a heat engine that absorbs $Q_h$ at $T_h$ is smaller than the work $L_p$ required by a heat pump for transferring $Q_h$ back to the source at $T_h$. The inequality $L_e<L_p$ derives from the two inequalities in my original post. Since we are assuming that $Q_h$ is the same, $Q_c=Q_h-L$ must be different in the two cases.

More in detail, we can assume $Q_{he} = Q_{hp} = Q_h$ and, recalling that $L_e<L_p$, $Q_{ce} = Q_{he}-L_e = Q_{hp}-L_e > Q_{hp}-L_p = Q_{cp}$ (where all quantities are assumed positive).
This means that
$$ \frac{Q_{ce}}{Q_{he}}>\frac{Q_{cp}}{Q_{hp}}$$
In other words, the above apparent contradiction disappears if we do not use the same symbols for the two cycles (because the corresponding value are not the same)

$$ \frac{Q_{ce}}{Q_{he}}>\frac{T_c}{T_f}>\frac{Q_{cp}}{Q_{hp}}$$

I have to assimilate this, otherwise I'm going to ask the same question in five years :)

Heyy. I have encountered the same problem as you guys. My only catch with this explanation is that from the T-s diagram we can show the magnitude of both Qh, QL and W to be same. Would be helpful if you guys can help me figure out on what gives?

 

[Andrew Mason]

In the reverse cycle of a real engine it takes more work than was produced in the forward cycle to reverse the heat flow to the hot register. So if you ran the engine one complete forward cycle and saved the output (e.g. by lifting a weight) that stored energy would not be enough to return the hot register to its original state.

It is always the case that $W=|Q_h|-|Q_C|$. So in the reverse cycle $|Q_c|$ will be less than $|Q_c|$ in the forward cycle if $|Q_h|$ is the same.

AM

 

[Pisica]

I am also looking for the answer.
If you need it for school, what you wrote is good.

If you need it to apply in practice, like me, I don't think it's a good formula.
Where does the heat pump get its heat and what does it do with it?
It takes the heat from outside and puts it in the house, the same heat. It transfers heat.
Q1=Q2
or
Qhot=Qcold
effCarnot=1-abs(Qcold/Qhot)=0 !!!!!

 

[Pisica]

But a Stirling engine can work in reverse as a heat pump or as a refrigerator.
Let's draw the p-V diagrams! How do you see the situation?

 

[Andrew Mason]

For any heat engine or heat pump $W=|Q_h|-|Q_c|$. Since W is not 0 in either a heat engine or heat pump, $|Q_h| \ne |Q_c|$

 

[Pisica]

"It takes the heat from outside and puts it in the house, the same heat. It transfers heat.
Q1=Q2
or
Qhot=Qcold"

"For any heat engine or heat pump W=Qh-Qc"

My mistake, I apologize.
And yet I feel that something is missing.....

It is true that a Stirling engine, put in reverse, behaves like a refrigerator.

 

[russ_watters]

You can use a sterling cycle for refrigeration/heat pump, yes.

 

[Andrew Mason]

“For any heat engine or heat pump W=Qh-Qc"

My mistake, I apologize.
And yet I feel that something is missing.....

In a reversible Carnot cycle beginning with the forward heat engine direction, if one stores the output work and then applies that work to the reverse cycle, one would return to the initial states of system+surroundings - in which case |Qh| is the same in both as is |Qc|, with the heat flow Qh<0 and Qc>0 in the forward direction and Qh>0 and Qc<0 in the reverse direction.

But in a real cycle this can never be achieved. The stored energy from the work produced in the forward cycle is never enough to get the system + surroundings back to its original state in the reverse direction: ie it is not reversible. So one needs to supply more work in the reverse direction.

This means that if $|Q_h|$ is the same in both directions and $|W_{fwd}|<|W{rev}|$, since $|W|=|Q_h|-|Q_c|$, then $|Q_c|$ in the reverse direction will be less than in the forward direction (for complete forward and reverse cycles). So the magnitude of the heat flow to the cold reservoir in the forward cycle is greater than that from the cold reservoir in the reverse cycle. Entropy of the cold reservoir increases over a complete forward and reverse cycle.

 

[Herman Trivilino]

TL;DR Summary: I'm trying to clarify the relation between heat pumps and engines, and to understand the inequalities involved in their "Carnot cycle" versions.

Its efficiency is described by the coefficient of performance

No, it's not. Coefficient of performance does not describe the efficiency.

 

[Pisica]

No, it's not. Coefficient of performance does not describe the efficiency.

Coefficient of performance=1/efficiency ?

 

[Herman Trivilino]

Coefficient of performance=1/efficiency ?

No. Both can be described as the ratio of what you get to what you pay for. In the case of efficiency that would be $\frac{|W|}{|Q_h|}$, in the case of the coefficient of performance that would be $\frac{|Q_c|}{|W|}$.

The coefficient of performance is not an efficiency.

 

[Pisica]

https://en.wikipedia.org/wiki/Coefficient_of_performance

 

[russ_watters]

I'll have to check a thermo textbook when I get home to see if they take a hard line on the definition of "efficiency", but there's a lot of sources that don't and neither do I, as long as one understands the difference:

By the strictest definition efficiency is a like-for-like comparison of how well you convert an input to an output, and by definition can't be above 100%. A heat pump isnt doing an energy conversion and therefore doesn't have an "efficiency" per se. But colloquially COP is often called "efficiency", including by regulators:

https://www.energystar.gov/products/central_air_conditioners

 

[Herman Trivilino]

View attachment 359388
https://en.wikipedia.org/wiki/Coefficient_of_performance

Okay, my mistake. I was thinking of the cop of a refrigerator when you had specifically referred to heat pumps, and even stated in your OP the cop of a heat pump. So the cop of a heat pump is indeed the multiplicative inverse of the efficiency of a heat engine.

TL;DR Summary: I'm trying to clarify the relation between heat pumps and engines, and to understand the inequalities involved in their "Carnot cycle" versions.

Now assume that the heat engine is reversed to obtain e.g. a heat pump.

Okay, but reversing a heat engine can also refer to a refrigerator.

Its efficiency is described by the coefficient of performance COP=Qh/|W|.

I still don't agree with this claim. COP is not an efficiency. If it weren't there would be no need to create a separate name for it. For example, as @russ_watters points out, efficiency can never be more than $1$ and the COP of my refrgerator is around $5$.

This efficiency should be less than that of a "Carnot" heat pump,
|Qh||Qh|−|Qc|<ThTh−Tc

I don't see how you got this relation.

 

[Herman Trivilino]

The same result is obtained if the COP for a refrigerator is considered.

No, because the COP of a refrigerator has a different definition.

 

[russ_watters]

Regarding the terminology, I did check my thermo book. Cengel & Boles, "Thermodynamics, an Engineering Approach" has this to say:

"The efficiency of a refrigerator is expressed in terms of coefficient of performance (COP)...In fact, one reason for expressing the efficiency of a refrigerator by another term - the coefficient of performance - is the desire to avoid the oddity of having efficiencies greater than unity." [their emphasis]

So yeah, I would say that "COP" is a measure/type of efficiency, it's just a different type of efficiency vs a heat engine. But I don't want to get bogged down in terminology; the meaning/how they are calculated is what matters.

So, regarding other terms; heat pump vs refrigerator/air conditioner/chiller. They are the same device, but the term is generally used to denote what you are trying to achieve; heating or cooling. Heat pump = heating, refrigerator/chiller/air conditioner = cooling. And if you want both, you might buy a "heat pump chiller" or "heat recovery chiller".

Then, regarding the question of the efficiencies of heat pumps and heat engines being the inverse of each other, the answer depends on which cycle you are looking at. An air conditioner's efficiency/COP uses Q_L (heat absorbed) whereas a heat pump's efficiency is based on Q_H (heat rejected). A heat pump has the benefit of converting the input mechanical work into heat and using that, so Q_H = Q_L + W. Carnot heat engine efficiency is the inverse of the heat pump efficiency(COP), not refrigeration efficiency(COP).

Connecting the above, you can also define the efficiency of a heat pump chiller based on the useful output of both ends: 2Q_L + W or Q_H + Q_L.