Operating Conditions of an Absorption Refrigeration Plant
The following diagrams give a rough idea of the conditions under which an ammonia absorption refrigeration plant (ARP) can operate and the coefficient of performance (COP) that can be be achieved. For a general explanation of the absorption cycle look on our page: “Working Principle“. The operating conditions of an absorption refrigeration plant are determined by a constellations of three temperatures.
Three Temperatures Determine the Process
There are three external temperatures which have influence on the the absorption cycle :
- The temperature of the Chilled Fluid
- The temperature of the Cooling Water
- The temperature of the Heat Supply in the desorber
If two of these temperatures are fixed, the third one can be calculated, or roughly estimated with this diagram. The following examples explain how to use this diagram:
- To get an evaporation temperature of -10°C with cooling water of 20°C the heat supply must have a minimum temperature of 102°C.
- If the heat supply temperature is 130°C and the cooling water is 25°C an evaporation temperature of -18°C can be achieved.
- To get an evaporation temperature of -30°C with a heat supply of 150°C, the cooling water temperature must not be higher than 25°C.
Coefficient of Performance COP
The thermal COP is the refrigeration capacity divided by the heat input. With a given COP the amount of heat supply can be calculated, which is required to produce a certain refrigeration capacity. The COP depends mainly on the operating conditions of an absorption refrigeration plant. This diagram gives a rough indication of the relationship. For example:
- For an evaporation temperature of -10°C and cooling water between 20°C and 30°C, the COP will be roughly 0,53, which means that 530 kW of refrigeration capacity can be produced when a heat input of 1000 kW is available.
- For an evaporation temperature of -50°C and cooling water between 20°C and 30°C, the COP will be roughly 0,28.
