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%.
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:
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