 
Dongguan Guansheng Cold Water Machinery Co., Ltd. finds that many friends in the industry do not know where to start when choosing a chiller, and they do not know what kind of chiller is more suitable for themselves? It is essential to understand your specific situation, and it is understandable to have the necessary common knowledge of the chiller. Let ’s take a look at the selection method and specific calculation method of the chiller!
How to choose a chiller
First, the standard chiller temperature difference is 5 degrees (20 degrees down 15 degrees), if an NWS10AC chiller has a standard chiller flow rate of 3.5 cubic meters per hour. Secondly, then when you try this chiller , you can't just look at the flow of the chiller, but also the temperature you need. If the temperature you tried is lowered from 28 degrees to 7 degrees, If the machine temperature difference is large, then the hourly chilled water flow rate will be reduced, from 3.5 cubic meters to 1 cubic meter. Then if you want 3.5 cubic meters per hour you have to choose a bigger model. Finally, when you try a chiller, it depends on whether you are cooling directly or indirectly. Direct cooling is to directly enter water into the chiller, and chilled water comes out to cool the product you need directly. This effect will be better. For indirect cooling, for example, put water in a basin, place a small basin on the basin, and put the product to be cooled in the small basin. The effect is definitely worse. When you choose the machine, you must consider this issue It all affects the size.
And if you want to choose a chiller through calculation, you can refer to the following formula and calculation guide:
1.Calculate the calorific value by the temperature difference between the inlet and outlet of the cooling water (oil)
Q = SH × De × F × DT / 60
Q: Calorific value KW (Note: The calorific value of Guansheng 1P chiller is about 3KW)
SH: Specific heat of hot water is 4.2KJ / Kg * C (4.2 kilojoules / kg * ° C) Specific heat of oil is 1.97KJ / Kg * C (1.97 kilojoules / kg * ° C)
De: Specific gravity of water 1Kg / L (1kg / L) Specific gravity of oil 0.88Kg / L (0.88kg / L)
F: Flow LPM (L / min liters / minute)
DT: Cooling water (oil) inlet and outlet temperature difference (outlet temperatureinlet temperature)
Note: "/ 60" is used to change the flow rate / minute into liter / second; 1kW = 1kJ / s;
Example 1: The cooling water inlet is 15 degrees, the outlet is 20 degrees, and the flow rate is 20 liters / minute.
Calorific value Q = 4.2 × 1 × 20 × (2015) / 60 = 7KW
When choosing the chiller cooling capacity, it can be appropriately increased by 20% 50%
Example 2: Cooling oil inlet is 18 degrees, oil output is 26 degrees, flow rate is 15 liters / minute
Calorific value Q = 1.97 × 0.88 × 15 × (2618) / 60 = 2.6KW
20% 50% can be appropriately increased when selecting cold water (oil) engine cooling capacity
2.Estimated by the power and calorific value of the equipment
a. If used for spindle cooling, the cooling capacity of the refrigeration unit can be estimated according to 30% of the spindle motor power.
Example: 15KW motor, optional 4.5kw or 5.8kw refrigeration unit;
b. The injection molding machine can be calculated according to the cooling capacity of 0.6KW per ounce or the cooling capacity of 3KW per 80 tons;
3. Calculate the calorific value by the temperature rise of the water (oil) tank
Q = SH × De × V × DT / 60
Q: Calorific value KW
SH: Specific heat of hot water is 4.2KJ / Kg * C (4.2 kilojoules / kg * ° C) Specific heat of oil is 1.97KJ / Kg * C (1.97 kilojoules / kg * ° C)
De: Specific gravity of water 1Kg / L (1kg / L) Specific gravity of oil 0.88Kg / L (0.88kg / L)
V: Water capacity L (liter) includes the total water capacity in the water tank and pipeline
DT: Maximum temperature rise of water (oil) in one minute
Note: "/ 60" is used to change the temperature rise from Celsius / minute to Celsius / second; 1kW = 1kJ / s;
Note: When measuring, the temperature of the water (oil) tank needs to be slightly lower than the ambient temperature; and the equipment is working under the maximum load.
Example: Water tank volume 3000L, maximum water temperature 0.6 degrees / minute
Calorific value Q = 4.2 × 1 × 3000 × 0.6 / 60 = 126KW
When choosing the cooling capacity of the chiller, it can be appropriately increased by 20% 50%, then CBE50W can be used;
Supplementary note:
1. The cooling capacity of the chiller varies with the ambient temperature and the outlet temperature;
2. The actual heating value of the equipment will also change due to different workpieces, molds, parameters, etc .;
3. After using the chiller, the temperature will drop, and the surface temperature of the connecting pipes, water tanks, fuel tanks, molds, spindles, and equipment will be lower than the ambient temperature, so it will absorb heat and increase the load;
4. In the practical application of industrial cooling, many situations cannot be accurately calculated using the above methods. At this time, it can only be estimated by empirical data, analogy of similar equipment and other methods.
5. Any calculation method may have deviations, so that the actual selected refrigeration unit is too large or too small, so the above method is for reference only; aftersales service
In summary, you only need to master the above calculation methods, and then understand the supplementary instructions in this chapter in detail to better handle the chiller selection problem. At the same time, friends who do n’t know much about this can also directly contact Contact us and we will recommend the most suitable model for you! Contact: 076983867468 Mr. Dai

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