Fake POU performances
If you go though the catalogues of different makers you will see that very seldom the performances of the machines on the market are comparable.
As a matter of fact different measures with little meaning, some indeed very amusing, have appereared on the market and this is due to the fact that unlike other products there is no unique test protocol.
We can quite safely say that the claimed performaces are not often in line with the technical specs of the machines and this can cause afterward commercial issues for example if the machine is not suitable for a certain “claimed” output.
The basics of the concept is that to define the quantity of heat subtracted to water we need to know two variables:
- The quantity of dispensed water
- The ∆t defined as the difference between inlet and outlet water temperature
If the compressor cooling capacity is 130 W and the COP (Coefficient of Performance) is say 1,6 the cooling capacity is 208 W or 748800 J in an hour. Since we need 4186 J to increase the temperature of water by 1 °C the maximum cooling capacity is 180 °C or with a ∆t =15°C around 12 liters.
Generally speaking the machines for Horeca have an ice-bank technology so they have a reservoir of ice while the machines for household and offices have a small reservoir. Therefore to calculate the total cooling capacity of a machine you have to take into account the reservoir of ice or the reservoir of water.
Suppose you have 2 kg of ice in the ice bank. Since a kg of ice @ 0°C needs 334400 J to melt we have 5,34 liters of water with a ∆t =15°C – given by 334400/(4186*15). So an ice bank of 2 Kg of has a reservoir of cool of 10,6 liters with a ∆t =15°C (inlet temperature 25°C outlet temperature 10°C).
Then with a ∆t =15°C every kg of ice produces 5,34 liters of water and with a ∆t =10°C 8 liters of water.
To have an easy way to assess the cooling capacity of an ice bank you can use the following table:
| COP | ∆t = 10°C | ∆t = 15°C |
|---|---|---|
| 1,00 | 0,086 | 0,057 |
| 1,10 | 0,095 | 0,063 |
| 1,20 | 0,103 | 0,069 |
| 1,30 | 0,112 | 0,075 |
| 1,40 | 0,120 | 0,080 |
| 1,50 | 0,129 | 0,086 |
| 1,60 | 0,138 | 0,092 |
| 1,70 | 0,146 | 0,097 |
| 1,80 | 0,155 | 0,103 |
| 1,90 | 0,163 | 0,109 |
| 2,00 | 0,172 | 0,115 |
Multiply the parameter between the COP and the ∆t for the power of the compressor in Watts and then add the reservoir of ice.
Example:
A machine with a compressor power of 160 W Cop 1,6 and 6 kg of ice in the ice-bank
160 * 0,092 + 6*5,34 = 46,76 @ ∆t =15°C
160 * 0,138 + 6*8,00 = 70,08 @ ∆t =10°C
Of course this is a theoretical calculation. Big variations might exist if the whole refrigeration system is not properly designed.
On the other hand the machines by GWS (Global Water Service) keep the promises and the stated capacity is a real output like our best seller the MF50.