Selection Method 1

Calculate the heat generated based on the temperature rise of the fuel tank.     

Q = SH × De × V × DT/60

Q: Heat output (kW)

SH: Specific heat of oil is 1.97 kJ/kg·℃ (1.97 kilojoules per kilogram per degree Celsius)

De: Specific gravity of oil is 0.88 kg/L (0.88 kilograms per litre)

De: Specific heat of water is 4.2 × 10³ J/kg·℃

V: Oil/water capacity in litres (L), including the total water capacity in the tank and pipes

DT: Maximum temperature rise per minute

Note: ‘/60’ is used to convert temperature rise in degrees Celsius per minute to degrees Celsius per second; 1 kW = 1 kJ/s;

Note: When measuring, the tank temperature should be slightly lower than the ambient temperature; and the equipment should be operating at maximum load.

Example: For a tank volume of 3000 L, the maximum water temperature or oil temperature is 0.6°C per minute

Heat output Q = (1.97 × 0.88 × 3000 × 0.6) / 60 = 52 kW

Additional note: When selecting the cooling capacity of a cold oil machine, it is appropriate to increase the capacity by 20%–50%.

Main shaft heat generation power estimation:

P heat generation = P motor * λ

λ - Main shaft heat loss efficiency

For general mechanical main shafts, heat loss λ = 5-8% can be used for calculation. For high-speed main shafts, heat loss λ = 20-30% can be used for calculation.

Example: For a standard mechanical spindle on a machine tool with a motor power of 45 kW, the estimated heat generation is:

P heat generation = P motor × λ = 45 × 8% = 3.6 kW