B Thermal physics, COP upper bound for heat driven heat pumps

[sf1001]

TL;DR

I made a presentation where I derived, or at least attempted to derive, the formulae for the upper bounds of the efficiency/cop for a heat engine, work driven heat pump, and a heat driven heat pump with 3 thermal reservoirs, hot, warm, and cold. Could someone check my math?

I made a presentation where I derived, or at least attempted to derive, the formulae for the upper bounds of the efficiency/cop (cop = coefficient of performance) for a heat engine, work driven heat pump, and a heat driven heat pump with 3 thermal reservoirs, hot, warm, and cold. There are many online sources to compare my result for the efficiency/cop of a heat engine or work driven heat pump with 2 heat reservoirs, hot and cold, but I haven't found any for the last case, the heat driven heat pump. I am assuming all isothermal heat exchanges with the heat reservoirs, which I think yields a theoretical upper bound for the calculated efficiencies and cop. I intend to use this presentation for tutoring HS physics. Are there any mathematical errors in the linked presentation, particularly for my formulae for the cop of the heat driven heat pump? Also, is there a way I could make the presentation more succinct? Here is a link to the presentation:

https://docs.google.com/presentatio...ouid=113815052145113661511&rtpof=true&sd=true

 

[Andrew Mason]

Just a few comments.

1. It can get confusing if you use absolute values for heats when switching between the forward cycle (heat engines) and the reverse cycle (heat pumps and refrigerator). As as long as you make it clear that $\frac{Q_c}{Q_h}>\frac{T_c}{T_h}$ for the forward cycle and $\frac{Q_c}{Q_h}<\frac{T_c}{T_h}$ or $\frac{Q_h}{Q_c}>\frac{T_h}{T_c}$ for the reverse cycle, this is ok.*

*This follows from $\Delta S=Q_h/T_h+Q_c/T_c>0$ (where terms are actual, not absolute values). So when heat flow to the hot reservoir is negative (heat engine) the inequality is met only if $\frac{|Q_c|}{T_c}>\frac{|Q_h|}{T_h}$ and when heat flow to the cold reservoir is negative (heat pumps and refrigerators), the inquality is met only if $\frac{|Q_c|}{T_c}<\frac{|Q_h|}{T_h}$

2. In frame 4, an easier to follow derivation might be:
\eta = \frac{W}{Q_h}=\frac{Q_h-Q_c}{Q_h}=(1-\frac{Q_c}{Q_h})
and, since: $\frac{Q_c}{Q_h}>\frac{T_c}{T_h}$ for a heat engine (from 1. above) this means:
\eta = (1-\frac{Q_c}{Q_h})&lt;(1-\frac{T_c}{T_h})
3. Similarly in frame 6, it might be easier to show:
COP_h = \frac{Q_h}{W}=\frac{Q_h}{Q_h-Q_c}=\frac{1}{(1-\frac{Q_c}{Q_h})}
and, since: $\frac{Q_c}{Q_h}<\frac{T_c}{T_h}$ for a heat pump (from 1. above) this means:
COP_h=\frac{1}{(1-\frac{Q_c}{Q_h})}&lt;\frac{1}{(1-\frac{T_c}{T_h})}
4. And again for frame 7 the cooling coefficent of performance:
COP_c = \frac{Q_c}{W}=\frac{Q_c}{Q_h-Q_c}=\frac{1}{(\frac{Q_h}{Q_c}-1)}
and, since: $\frac{Q_h}{Q_c}>\frac{T_h}{T_c}$ for a refrigerator (from 1. above) this means:
COP_c=\frac{1}{(\frac{Q_h}{Q_c}-1)}&lt;\frac{1}{(\frac{T_h}{T_c}-1)}
5. You might also want to explain why a heat driven heat pump makes any sense. Why not just have heat flow directly from hot to warm?

AM