Technical News | Oil Cooler Selection Basis

Type selection method 1

        Calculate the calorific value from the temperature rise of the oil tank

        Q = SHxDexVxDT/60

        Q: Calorific value KW

        SH: The specific heat of the oil is 1.97KJ/Kg°C (1.97kJ/kg Celsius)

        De: Specific gravity of oil 0.88Kg/L

        De: The specific heat of water is 4.2x103J/kg°C

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

        DT: Maximum temperature rise in one minute

        Note: “/60″ is used to convert the temperature rise in degrees Celsius/minute into degrees Celsius/second; 1kW = 1kJ/s;

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

        Example: 1 tank volume 3000L maximum water temperature or oil temperature 0.6 degrees Celsius/min

        Calorific value Q= ( 1.97 x 0.88 x 3000 x 0.6) /60 = 52KW


Supplementary instructions: when choosing the cooling capacity of the oil cooler, it can be appropriately increased by 20%-50%.

Oil cooler selection basis1


Type selection method 2

        The heating power is estimated according to the motor power of the hydraulic station.

        Hydraulic oil is used as a transmission medium, and the energy loss of the system will mostly exist in the form of heat. Based on the actual experience of the substrate, our company has summed up the corresponding energy loss coefficients of the hydraulic system under different working pressures as follows:

Oil cooler selection basis2

P heat = 1.2x (P motor n)

P motor hydraulic station all motor power: n energy loss coefficient

Remarks: 1 Kcal/h=1.163W 1 KW=860Kcal/h

Post time: Nov-04-2022