(1) How to choose the most suitable industrial chillers and small chillers? In fact, there is a simple selection formula:
Refrigeration capacity = chilled water flow *4. 187* temperature difference * coefficient
1. The chilled water flow refers to the cold water flow required when the machine works, and the unit needs to be converted into liters/second;
2. Temperature difference refers to the temperature difference between the inlet and outlet water of the machine;
3, 4. 187 is quantitative (specific heat capacity of water);
4. When air-cooled chiller is selected, the multiplication coefficient is 1.3, and when water-cooled chiller is selected, the multiplication coefficient is 1. 1.
5. Select the corresponding machine model according to the calculated cooling capacity.
Generally speaking, P is used to calculate the size of water chillers, but the most important thing is to know the rated cooling capacity. If the air-cooled 9.07KW looks like it, choose a threesome, and so on. Therefore, when choosing industrial water chillers, the most important thing is to find out the rated cooling capacity.
(2) Calculation method of refrigerating capacity of water chillers
Calculation method of refrigerating capacity of water chillers, refrigeration principle of water chillers, 20kw can save calculation method:
1: volume (liter) × heating degree ÷ heating time (minute) ×60÷0.86 (coefficient) =(w)
2: volume (ton or cubic meter) × heating degree ÷ heating time (hours) ÷0.86 (coefficient) =(kw)
Your data includes the calculation method of cooling capacity of water chillers. If the principle of chiller refrigeration comes out, it can save 4 hours 1000L× (15-7) ÷ 4 hours ÷ 0.86 = 23255W = 23.255KW5 hours 10 ton× (15-7).
(c) Selection method of cooler
1, temperature difference flow method q = cp.r.vs. △ t.
Q: heat load (KW) Cp: specific heat at constant pressure (kj/kg. ℃) ...4. 1868kj/kg。 ℃
R: specific gravity (kg/m3) ... 1000kg/m3 vs: water flow (m3/h) for example: 5.1m3/h.
△T: water temperature difference (℃)...△ T = T2 (inlet and outlet temperature) -T 1 (inlet temperature) Example: =5℃
For example: q = CP.r.vs. △ t = 4.1868 *1000 * 5.1* 5/3600 = 29.6565 (kw).
When the cooling capacity is 29.8kw, the temperature difference is 5 degrees, and the specific heat is 4. 1868, the effluent flow can also be calculated as 29.8 * 3.6/(4.1868 * 5) = 5.1247.
2. Time temperature rise method Q = Cp.r.V.△T/H
Q: heat load (KW) Cp: specific heat at constant pressure (kj/kg. ℃) ...4. 1868kj/kg。 ℃
R: specific gravity (kg/m3) ... 1000kg/m3 V: total water volume (m3) Example: 0.5m3
△T: water temperature difference (℃)...△ t = T2-t 1 example: =5℃ H: time (h) example:1h.
Example: q = CP.r.v. △ t/h = 4.1868 *1000 * 0.5 * 5/3600 = 2.908 (kw).
3. energy conservation method Q=W in -W out
Q: heat load (KW) W in: input power (KW) Example: 8KW out: output power (KW) Example: 3KW.
For example: Q=W input -W output =8-3=5 (kW)
4. Common methods of rubber and plastic: q = w * c * △ t * s.
Q= required chilled water energy kcal/h W= plastic raw material weight KG/H Example: w = 31.3kg/h.
C= specific heat kcal/KG℃ of plastic raw materials Example: polyethylene PE C=0.55 kcal/KG℃.
△T= the temperature difference between the plastic melting temperature and the plastic molding temperature of the product℃ is generally (200℃), and S= safety factor (1.35-2.0), generally 2.0.
For example: q = w * c * △ t * s = 31.3 * 0.55 * 200 * 2.0 = 6886 (kcal/hour)
For the calculation formula of air and steam enthalpy, please refer to Trane Central Air Conditioning Technology Blog.