Calculate heat generation Q=SH×De×F×DT/60 by the temperature difference between inlet and outlet of cooling oil

Q: heat generation KW (Note: the heat generation of 1P oil cooler is about 2.5KW)

SH:Specific heat of specific heat oil is 1.97KJ/Kg*C(1.97kJ/kg*C)

De:specific gravity of oil 0.88Kg/L(0.88kg/L)

F:Flow rate LPM(L/min liter/minute) DT:Temperature difference between import and export of cooling oil (export temperature - import temperature)

Example: cooling oil import is 18 degrees, the oil out 26 degrees, flow 15 liters / minute heat Q = 1.97 × 0.88 × 15 × (26-18)/60 = 2.6KW, the choice of chilled water (oil) machine cooling capacity can be increased by 20%-50% 2, through the power of the equipment, heat generation estimates a, such as for spindle cooling, can be estimated according to the spindle motor power 30% of the required refrigeration unit of the cooling capacity. Example: 15KW motor, optional 4.5kw or 5.8kw cooling capacity of the low-temperature chiller;

Calculate the heat generation Q=SH×De×V×DT/60 by the temperature rise of oil tank

Q:heat generation KWSH:specific heat of oil is 1.97KJ/Kg*C(1.97kJ/kg*C)

De:specific gravity of oil 0.88Kg/L(0.88kg/L)

V: oil capacity L (liters) including the total water capacity in the tank and pipeline DT:) comparative temperature rise in one minute Note: /60″ is used to change the temperature rise Celsius degree/minute to Celsius degree/s; 1kW=1kJ/s.