The prerequisite for a suitable optional
chiller is that you want a smaller chiller, which is simple and convenient to use; and where your location is relatively short of water resources; at the same time, there are restrictions on the area, and finally One point is that customers who should not put cooling water towers; if your actual situation meets the above four aspects, then choosing an air-cooled chiller will be a better choice. Now that you have decided to choose an air-cooled chiller, the next step is It is necessary to understand the estimation method as follows:
Selection is mainly based on the required cooling capacity. Attach the estimation method
1.Calculate the calorific value through the temperature difference between the inlet and outlet of the cooling water (oil) Q = SH * De * F * DT / 60 Q: Calorific value KW
SH: Specific heat of hot water is 4.2KJ / Kg * C (4.2 kilojoules / kg * degrees Celsius) Specific heat of oil is 1.97KJ / Kg * C (1.97 kilojoules / kg * degrees Celsius)
De: Specific gravity of water 1Kg / L (1kg / liter) Specific gravity of oil 0.88Kg / L (0.88kg / liter) F: Flow rate LPM (L / min liter / minute)
DT: Cooling water (oil) inlet and outlet temperature difference (outlet temperature-inlet temperature) Note: "/ 60" is used to change the flow rate / minute into liter / second; 1kW = 1kJ / s;
Example 1: Cooling water inlet water is 20 degrees, outlet water is 25 degrees, flow rate is 10 liters / min. Calorific value Q = 4.2 * 1 * 10 * (25-20) / 60 = 3.5KW
Example 2: The inlet of the cooling oil is 25 degrees, the water is 32 degrees, the flow rate is 8 liters / minute, and the heating value is Q = 1.97 * 0.88 * 8 * (32-25) / 60 = 1.62KW
2. Through the power and heat generation of the equipment, a. If it is used for spindle cooling, the cooling capacity of the refrigeration unit can be estimated according to 30% of the spindle motor power. Example: 7.5KW motor, optional 2.2kw or 2.8kw refrigeration unit; b. Injection molding machine can be estimated at 0.6KW per ounce
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 * degrees Celsius) Specific heat of oil is 1.97KJ / Kg * C (1.97 kilojoules / kg * degrees Celsius)
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: the 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 is 1000L. Maximum water temperature is 0.2 ° C / min. Calorific value Q = 4.2 * 1 * 1000 * 0.2 / 60 = 14KW
In fact, there is still a certain error between the exact data and the estimated value, because the exact data is also affected by five aspects, such as the cooling capacity of the chiller and the ambient temperature and the temperature of the water outlet. Changes in workpieces, molds, parameters, etc .; the temperature drops after using the
chiller , the temperature of the connecting pipes, water tanks, fuel tanks, molds, spindles, equipment surfaces will be lower than the ambient temperature, so it will absorb heat and increase the load; in industrial cooling Many situations in actual application cannot be accurately calculated using the above methods, and can only be estimated by empirical data, analogy of similar equipment, and so on. Any calculation method may have deviations, so that the actual selection of the refrigeration unit is too large or too small, so the above method is only for reference; friends who are not familiar with these estimation methods can call us to consult the professional Staff, recommend you the high quality chiller equipment for you!