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