制冷与热泵：反向运行

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.

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.

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.

Three refrigeration slips

Recap