Energy Unlimited by Victron

6.3. Refrigeration

6.3.1. Introduction

More often than not, refrigeration on board is a nightmare, or at least a headache.

On small yachts the refrigerator often takes more energy from the battery than all other equipment together. On medium sized yachts it is the refrigerator plus freezer that will drain the battery. And on larger yachts it is because of the air conditioning that a generator has to run day and night.

In order to understand why, and see whether anything can be done about it, some theoretical background is needed. This is the subject of the next section.

6.3.2. Theory of the heat pump

Nearly all refrigeration systems are of the compressor heatpump type.

Operation is as follows: The compressor, driven by a DC or AC electric motor compresses a gas (freon, until this was forbidden because it destroys the ozone layer in the upper atmosphere) which is cooled down in what is called the condenser. The condenser often is a small radiator with a fan in the cupboard under the sink, or it is a much larger naturally ventilated radiator at the back of the refrigerator (normal household type refrigerator), or it can be water-cooled. In the condenser the gas condenses to liquid and in that process a lot of heat is taken from it. The liquid then moves to the evaporator, which is the cold plate in the refrigerator or freezer. There the pressure is reduced and the liquid evaporates. To evaporate a lot of heat has to be absorbed; this heat is removed from the refrigerator or freezer. The gas then goes to the compressor, and so on.

The amount of energy needed for drawing a certain quantity of heat from the surroundings with a heat pump may be calculated with the formula;

CoP = n r

x n c

= n r

x Tlow / (Thigh – Tlow)

where CoP is the Coefficient of Performance, Tlow is the temperature of the evaporator expressed in degrees Kelvin (=°C + 273), Thigh is the temperature of the condenser, likewise expressed in degrees Kelvin, and n r is a factor (the efficiency, always less than 1) which gives the CoP in practice compared to the theoretical CoP n c . (Note: the CoP formula used here is a simplification of what happens in practice, but it is nevertheless an adequate tool to find out what measures can be taken to reduce electricity consumption)

An example for a refrigerator:

Temperature cold side: -5°C i.e. Tlow = 268°K (this is not the average temperature in the refrigerator but the temperature of the evaporator or cold plate in the refrigerator). Temperature hot side: 45°C i.e. Thigh = 318°K Efficiency: 25 %

Then the CoP is:

CoP = 0.25 x 268 / (318 – 268) = 1.34

This means that for every kWh of heat that leaks in through the refrigerator’s insulation, or is drawn away from food or drink put into the refrigerator while still warm, 1 / 1.34 = 0.75 kWh of electric energy is needed to “pump” this heat out again.

6.3.3. The refrigerator and freezer in practice

When running, the average compressor motor of a refrigerator or freezer takes about 50 W, or 4.2 A from a 12 V battery. The compressor motor is controlled by a thermostat that switches it on when the temperature increases to a pre-set value, and switches it of again after the temperature has been brought down to a few degrees below the pre-set value. The on / off ratio is called the duty cycle. A duty cycle of 100 % results in a daily capacity drain from a battery of 4.2 A x 24 h = 101 Ah. A nightmare! A duty cycle of 50% results in 50 Ah daily consumption and a duty cycle of 25 % translates to 25 Ah daily consumption.

What we want is low energy consumption. How can this be achieved?