The earth is absorbing energy all the time. At the surface this is mainly solar energy, but at greater depth factors such as rainfall, groundwater and heat transfer through the bedrock all play important roles. This energy is entirely renewable and as a result, the ground remains constant at approx. 10^oC when you go deeper than a few metres (as shown below). It is this constant, renewable source that geothermal heat pumps use to provide heating in your building.

 

 

Specific heat capacity

 The specific heat capacity of a substance is the amount of heat energy it requires to raise the temperature of 1kg of the substance by 1^oC. For water this is 4.2 kJ kg^-1 K^-1. Note that this does not change if you are heating the water from 10^oC to 11^oC or from 90^oC to 91^oC. Conversely, 4.2kJ of heat must be given out to cool 1kg of water down by 1^oC. If we look at the example below we see that all a heat pump is really doing is moving heat energy from one body of water to another.

1) Both bodies of water are at the same temperature (N.B. 1m³ of water weighs 1000kg)
 

 

2) The heat pump cools the larger body of water by approx. 4^oC. This gives out 168,000kJ of energy!!! (4.2kJ x 10000kg x 4^oC)
 

 

  3) The heat pump then uses this energy to heat the smaller body of water. As it is 10 times smaller, the temperature increase is 10 times greater and so the water is heated by 40^oC to 50^oC!!! (168000kJ ÷ 1000kg ÷ 4.2kJ)
 

 

 From this example you can see that by cooling a large body of water by a few degrees, it is possible to heat a small body of water to very high temperatures. This is the main principal behind geothermal heating. The large body of water is in the pipes underneath your garden while the small body of water at higher temperatures is the water in your underfloor heating pipes. However there is one obvious problem with the above example. We all know that heat travels from the hotter body to the cooler one until both temperatures are the same e.g. a hot cup of tea cools down to room temperature. Therefore how is the heat pump moving energy from two bodies that start at the same temperature? The answer lies in the common household fridge and the refrigeration cycle. 

 

Latent Heat

 To explain latent heat we use a simple experiment that many people will have carried out in the early years of secondary school. Ice is placed in a beaker with a thermometer and is heated using a constant heat source. Then the temperature is recorded a set time intervals and a graph plotted. An example of the apparatus is shown below.
 

 

 

 Obviously you will find that the ice melts at 0^oC and boils at 100^oC. If however you were to perform this experiment under strict conditions your results would look something like this. 

 

At point A, C and E the temperature is rising, as we would expect when we add heat. However at points B and D, the temperature is constant, even though we are heating the substance the whole time. However as we know point B and D are the melting and boiling point of water, the energy must be being used to change the state of the water (i.e. solid à liquid and liquid à gas). This is called the ‘latent heat’ and it is the energy required to overcome the molecular forces between the particles so the substance can change state. 

 If we now look at the change of state from liquid to gas (evaporation/vaporisation), we will get a better idea of how a heat pump works. The specific latent heat of vaporization is defined as ‘the amount of heat required to convert a unit mass of a liquid into the vapour without a change in temperature’. For water at its normal boiling point of 100 ºC, the latent specific latent heat of vaporization is 2260 kJ.kg^-1. This means that to convert 1 kg of water at 100ºC to 1 kg of steam at 100ºC, 2260 kJ of heat must be absorbed by the water. Conversely, when 1 kg of steam at 100ºC condenses to give 1 kg of water at 100 ºC, 2260 kJ of heat will be released to the surroundings. Very simply

        When something evaporates, heat is taken in e.g.

• You feel cold when you get out of a swimming pool. The water evaporating off your skin takes extra heat from your body and makes you feel cold even though the pool hall is hotter than a normal room.
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• You sweat when you’re warm. The sweat evaporates and takes heat away from your body, cooling you down.

        When something condenses, heat is given out e.g.

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• When you get steam from a kettle on your hand, it burns so badly. The steam condenses on your hand and gives out extra heat.

 These two processes are the foundation of refrigeration and heat pump technologies. 

 

Refrigeration Cycle

 

 

The above diagram shows the refrigeration cycle inside a Waterkotte heat pump. Instead of water however, the heat pump uses a refrigerant gas with a very low boiling point (-40^oC and lower, depending on the refrigerant used). This allows it to evaporate very easily at low temperatures.

  The refrigeration cycle is made up of 4 steps 
1)      The refrigerant is allowed to evaporate in the evaporator. This draws heat from the heat source water into the refrigerant
2)      The refrigerant is compressed using a mechanical motor. This increases the energy per unit volume and allows the heat pump to produce higher output temperatures. This is the largest electrical input into the heat pump.
3)      The compressed gas is allowed to condense in a condenser. This gives out the heat to the heating water. The more compressed the gas, the higher temperatures can be reached
4)      The gas is allowed to uncompress as it goes through an expansion valve and the cycle starts again To achieve maximum efficiency, the temperature of the heat source must be as high as possible and the flow temperature on the heating side must be as low as possible. In this way the compressor will not have to compress the refrigerant gas as much and less electricity will be used. This again highlights the importance of the design of the geothermal heat source and the heat distribution system.

With the correct design from heat source to heat distribution, Waterkotte heat pumps can achieve an efficiency of more than 450% (a co-efficient of performance of 4.5), i.e. for every 1kW of electrical power used in the compression stage, there is 4.5kW of useful heat output, 3.5kW of which comes free from the ground.