Problem — Estimating the power of an air conditioner

Problem

Estimate the minimum power needed to condition a room into which $1000$ W of thermal power enters (from wall losses and sunlight). The desired indoor temperature is $25\,^\circ$C, the outside is at $35\,^\circ$C. (Use the Carnot limit for the COP.)

Solution

Absolute temperatures. Convert to kelvin:

$T_f = 25 + 273 = 298 \ \text{K}, \qquad T_c = 35 + 273 = 308 \ \text{K}$

Maximum COP (Carnot limit). An air conditioner is a refrigerator that extracts heat from the cold interior ($T_f$) and dumps it into the hot exterior ($T_c$). The best possible COP is:

$\beta_{\max} = \frac{T_f}{T_c - T_f} = \frac{298}{308 - 298} = \frac{298}{10} = 29{,}8$

Minimum power. In steady state the air conditioner must extract $Q_f$ at the same rate at which heat enters, i.e. $\dot{Q}_f = 1000$ W. Since $\beta = \dot{Q}_f / P$:

$P_{\min} = \frac{\dot{Q}_f}{\beta_{\max}} = \frac{1000}{29{,}8} \approx 33{,}6 \ \text{W}$

Order-of-magnitude estimate.

$\ev{P_{\min} \sim 30 \ \text{W} \ (\text{order } 10^1 \ \text{W})}$

The small thermal gap ($10$ K) makes the COP very high: only a few tens of watts of work are needed to move a kilowatt of heat. A real, non-ideal air conditioner consumes more (hundreds of watts), but the order of magnitude of the physical limit is this.

Collegamenti

Argomenti: Heat engines Concetti: Refrigerator and heat pump · Carnot cycle Competenze: Fermi estimates Oggetti: Refrigerator