Problem — Heat pump for heating (295-273 K)

Problem

A heat pump keeps a house at $T_c = 295\;\text{K}$ by absorbing heat from a garden at $T_f = 273\;\text{K}$. For every $5\;\text{kWh}$ of electrical work at the compressor, how much heat does it deliver to the house?

Solution

Coefficient of performance (Carnot model). For an ideal heat pump the COP is

$\beta_\text{pdc} = \frac{Q_c}{L} = \frac{T_c}{T_c - T_f} = \frac{295}{295 - 273} = \frac{295}{22} = \ev{13{,}4}$

Heat delivered. From $\beta_\text{pdc} = Q_c/L$ we get $Q_c = \beta_\text{pdc}\, L$, with $L = 5\;\text{kWh}$:

$Q_c = 13{,}4 \times 5\;\text{kWh} \approx \ev{67\;\text{kWh}}$

With just $5\;\text{kWh}$ of electricity the pump supplies $\approx 67\;\text{kWh}$ of heat: the remaining $\approx 62\;\text{kWh}$ are drawn for free from the garden. The small temperature gap ($22\;\text{K}$) makes the COP very high — heat pumps perform best when indoors and outdoors are close in temperature.

Collegamenti

Argomenti: Macchine termiche Concetti: Frigorifero e pompa di calore · Ciclo di Carnot Competenze: Impostazione simbolica Oggetti: Frigorifero