Refrigeration and heat pumps: running it backwards
Introduzido| English | Significado |
|---|---|
| heat pump | heat pump |
| vapor-compression cycle | A refrigeration-cycle model using compression, condensation, throttling and evaporation. |
| compressor | The only work input. |
| evaporator | evaporator |
| reversed Carnot cycle | reversed Carnot cycle |
| coefficient of performance/ˌkəʊɪˈfɪʃənt ɒv pəˈfɔːməns/ | coefficient of performance |
Running it backwards
- Reverse the power cycle and it moves heat uphill: a refrigerator keeps the cold side cold, a heat pump heats a house, and one skeleton - the vapor-compression cycle 蒸气压缩循环 - serves both: compressor 压缩机 to condenser to throttle to evaporator 蒸发器.
- Performance is benefit over cost: $\mathrm{COP}_R = q_L/w_{in}$ for cooling, $\mathrm{COP}_{HP} = q_H/w_{in}$ for heating. The reversed Carnot cycle 逆卡诺循环 bound $\mathrm{COP} \le T_{cold}/(T_{hot} - T_{cold})$ caps both - in absolute kelvin, always.
Which port answers the question?
Route each request to the component that owns it.
The coefficient of performance is defined as:
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COP is benefit over cost; the reversed-Carnot bound caps it.
An R-134a refrigerator
- Evaporator at $-10^\circ\mathrm{C}$: $h_1 = 394.6\ \mathrm{kJ/kg}$; compressor to $800\ \mathrm{kPa}$ isentropically: $h_2 = 432.9$ (superheated - read it there, not from the saturated row); condenser exit saturated liquid $h_3 = 241.8$; the throttle keeps $h_4 = 241.8$.
- The ledger: $q_L = h_1 - h_4 = \mathbf{152.8}$, $w_{in} = h_2 - h_1 = \mathbf{38.3}$, so $\mathrm{COP}_R = 152.8/38.3 = \mathbf{3.99}$. The reversed-Carnot bound between $-10$ and $31.3^\circ\mathrm{C}$ is $263.2/41.3 = \mathbf{6.37}$ - the real machine sits below it, as it must.
Refrigeration effect q_L (one decimal, kJ/kg)?
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394.6 - 241.8.
Compressor work per kilogram (one decimal, kJ/kg)?
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432.9 - 394.6.
COP of refrigeration (two decimals)?
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152.8/38.3.
Reversed-Carnot bound between -10 and 31.3 C (two decimals)?
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263.2/(304.5 - 263.2).
The compressor exit state of this cycle is superheated vapour.
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h2 = 432.9 sits outside the dome at 800 kPa - read it from the superheated table.
The same machine heating
- Read the same ports as a heat pump 热泵: $q_H = h_2 - h_3 = 432.9 - 241.8 = \mathbf{191.1}$, and $\mathrm{COP}_{HP} = 191.1/38.3 = \mathbf{4.99}$.
- The identity holds by definition: $\mathrm{COP}_{HP} = \mathrm{COP}_R + 1 = 3.99 + 1 = 4.99$ - the coefficient of performance 性能系数 changes with the question, not with the machine.
Heating effect q_H (one decimal, kJ/kg)?
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432.9 - 241.8.
COP as a heat pump (two decimals)?
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191.1/38.3 = 3.99 + 1.
Three refrigeration slips
- Inverting COP. It is benefit over cost, never $w/q$.
- Using temperature differences in the reversed-Carnot bound. Absolute temperatures: $263.2/41.3$, not $10/41.3$.
- Reading $h_2$ from the saturated-vapour row. The compressor exit is superheated.
Recap
- Compressor, condenser, throttle, evaporator - the compressor is the only work input.
- $q_L = 152.8$, $w_{in} = 38.3$, $\mathrm{COP}_R = 3.99$ against the 6.37 ceiling.
- $\mathrm{COP}_{HP} = \mathrm{COP}_R + 1 = 4.99$: same machine, different port.