A simple selection formula for chiller factory tutors

Release time：

2023-05-11 13:03

Customers often ask a question when calling to consult industrial chillers, that is, what type of refrigerator should be used in the production process of their industry, they do not understand what kind of machine can achieve their refrigeration effect, or what kind of equipment they need in their workshop. Below, chiller manufacturers teach you a simple selection formula to help you understand this problem.

(4) How to choose an industrial chiller and a small chiller suitable for yourself, in fact, it is very simple to have a selection formula: cooling capacity = chilled water flow * 187.<> * temperature difference * coefficient

1. Chilled water flow refers to the cold water flow required during the work of the machine, and the unit needs to be converted to liters / second;

2. Temperature difference refers to the temperature difference between the inlet and outlet water of the machine, the unit is °C;

3. 4.187 is quantitative (specific heat capacity of water), the unit is KJ/kg.°C;

4. When choosing an air-cooled chiller, you need to multiply the safety factor of 1.3, and when choosing a water-cooled chiller, you need to multiply the safety factor of 1.1.

5. Select the corresponding machine model according to the calculated cooling capacity.

6. The unit of calculating the cooling capacity is kj.h, 1kw=3600kj.h It is generally customary to use P to calculate how large the chiller should be, but the main thing is to know the rated cooling capacity, and if the general air-cooled 2.5KW looks like, choose a machine with 1P. Therefore, the selection of industrial chillers is important to find the rated cooling capacity.

(2) The calculation method of cold water mechanism

1. Volume (liters) × liters of temperature÷ heating time (min) × 60÷0.86 (coefficient) = (w)

2. Volume (tons or cubic meters) × liters of temperature÷ when heating up (time) ÷ 0.86 (coefficient) = (kw)

3. Example of cooling capacity of chilled water machine: 4 hours (heating time) 10000L× (15-7) ÷4h÷0.86=23255w=23.255kw5 hours(heating time) 10 tons×(15-7)÷5h÷0.86=18.604kw

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