Calculation of heat generation by temperature rise of the tank

      Q=SH x De x V x DT/6

      Q: Heat generation kW

      SH: Specific heat of oil 1.97kJ/kg*°C (1.97 kJ/dry g*C)

      De: the specific gravity of oil is 0.88kg/L (0.88 dry grams/liter)

      De: specific heat of water is 4.2x10³ J/kg°C

      V: oil / water capacity L (liter) including the tank and the total water capacity in the pipeline

      DT: the maximum temperature rise in one minute

Note: "/60" is used to change the temperature rise in degrees Celsius/minute to degrees Celsius/second; 1kW = 1kJ/s;

Note: When measuring, the temperature of the tank should be slightly lower than the ambient temperature; and the equipment is working under the maximum load.

Example: 1 tank volume of 3000L maximum water temperature or oil temperature 0.6 degrees / minute

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

Supplementary note: Selection of oil cooler cooling capacity can be appropriately increased by 20% -50% can be selected.

Estimate the heat generation power according to the motor power of the hydraulic station.

Hydraulic oil as a transmission medium, the system energy loss will be mostly in the form of its heat.

Our company according to specific practical experience, summed up the hydraulic system at different working pressures, the corresponding energy loss coefficient is as follows:

P heat = 1.2 x (P motor * n)

P motor All motor power of hydraulic station: n Energy loss coefficient

Note: 1kcal/h = 1.163W 1kW = 860kcal/h

Estimation of spindle heat generation power:

P heat generation power = P motor * λ 

λ - spindle heat loss efficiency

For general mechanical spindle can be calculated according to the heat loss λ = 5-8%, for high-speed spindle heat loss λ = 20-30% calculation.

Example: a machine tool general mechanical spindle, motor power 45kW, estimated heat:

P heat = P motor * λ = 45 * 8% = 3.6kW