Save money with using a refrigerator correctly - blowing cold air to the back of it?

Question

We are heating up our houses in winter to ex.: 21C to then have an insulated box inside the house, which we cold down to 5 and -15C, and we call it a refrigerator.. :)

The situation:

• we would buy a new refrigerator and put it in a house having 21C constantly
• we would also have a "heat-pump boiler" for heating water.
• the "heat-pump boiler" outputs colder air than what is there in the house normally, that is how "heat-pump boilers" work, ok.

Like if I would put out the whole refrigerator to outside, where there is ex.: day: 5C night: -5C, I would be sure that the electricity consumption (operation cost) would go down, because it wouldn't have to battle with the indoor 21C.

The question: would blowing cold air from the "heat-pump boiler" to the back (to the bottom or top of the back?) of a refrigerator help reduce the operational cost for the refrigerator?

Answer 1

Foundations

The CoP of a refrigerator uses the cold heat flow and the work input, expressed as $\dot{q}_c / \dot{w}$. The Carnot expression based on temperature is $T_c / (T_h - T_c)$, where $T_c$ is the internal cold temperature and $T_h$ is the external hot temperature. Decreasing $T_h$ will increase the CoP, meaning less work flow $\dot{w}$ will be needed to remove the same amount of heat $\dot{q}_c$.

Application to System at Hand

In a picture of a control volume for the system at hand, $T_h$ can be taken as the temperature of the coils used to remove the heat. The heat flow to the air that surrounds the coils happens outside of the control volume.

The value of $T_h$ is controlled by two factors. The first is the amount of heat that is being pumped out of the refrigerator in to the cooling fluid. The second is the amount of heat that is being dumped from the coil to the air.

When the first factor is constant, anything that can be done to increase the second (heat flow from coil to air) will decrease $T_h$. Two approaches are to decrease the temperature of the air $T_{air}$ and to increase the flow rate of air. The first approach increases the temperature difference in the convection coefficient $h$ in the equation $\dot{q} = hA(T_h - T_{air})$, thereby increasing the heat flow from the coils, thereby decreasing $T_h$. For the same $T_{air}$, the latter approach increases $h$ of the air, thereby increasing the heat that is removed from the coil, thereby decreasing $T_h$.

Proposed Approach

Based on your overview of the system, you might try a passive convection approach to move the cool air from the heat-pump boiler to the back of the refrigerator. Basically, the passive convection systsem is a channel where the hot air rising from the back of the refrigerator "sucks in" the cooler air from the heat-pump boiler. The better that you seal the channels, the better will be the operation. No forced convection fan needed, therefore no cost is lost by balancing the power drain to run the fan versus improving the cooling efficiency at the back of the refrigerator. The only investment is the cost in materials and labor.

The aesthetics are a different story.

Answer 2

I would design the fridge's heat exchanger in such a way that the exothermic side is kept wetted and exposed to a natural air stream. The resulting evaporation will have a greater cooling effect than just cool air.