Question

Determine the minimum work per unit of heat transfer from the source reservoir that is required to drive a heat pump with thermal energy reservoirs at 460 K and 540 K.

Answer 1

Answer:

The minimum work per unit heat transfer will be 0.15.

Explanation:

We know the for a heat pump the coefficient of performance ($C_{HP}$) is given by

$C_{HP} = \dfrac{Q_{H}}{W_{in}}$

where, $Q_{H}$ is the magnitude of heat transfer between cyclic device and    high-temperature medium at temperature $T_{H}$ and $W_{in}$ is the required input and is given by $W_{in} = Q_{H} - Q_{L}$, $Q_{L}$ being magnitude of heat transfer between cyclic device and low-temperature $T_{L}$. Therefore, from above equation we can write,

$&& \dfrac{Q_{H}}{W_{in}} = \dfrac{Q_{H}}{Q_{H} - Q_{L}} = \dfrac{1}{1 - \dfrac{Q_{L}}{Q_{H}}} = \dfrac{1}{1 - \dfrac{T_{L}}{T_{H}}}$

Given, $T_{L} = 460 K$ and $T_{H} = 540 K$. So,  the minimum work per unit heat transfer is given by

$\dfrac{W_{in}}{Q_{H}} = \dfrac{T_{H} - T_{L}}{T_{H}} = \dfrac{540 - 460}{540} = 0.15$