# Hydraulic Oil Cooler Selection

Data required
Pv: Heat load generated in the system(Kw)
P1: Specific cooling capacity(Kw/℃)
V: Oil volume(L)
p: Oil Density(normally):0.85Kg/L
C:Specific heat of oil(normally):2.15KJ/Kg℃
△T:Tempreature increase(℃)
T1:Expected temperature(℃)
T2:Ambient temperature(℃)
H:Operating time(second)
n1:Safety margin(power of heat exchanging),generally=1.1
Q:Flow rate(L/min)

Hydraulic Oil Cooler Selection
The choosen type shall meet the requirements of system flow and heating power at the same time.The selection of the type can be made by the following calculating methods.

The heat load can be easily calculated by measuring the increase in temperature in the tank over a specific period of time.Usually it is calculated by the following formula:
Pv=p*V*C*△T/H

i.e:Measure a hydraulic system with the temperature  rising from 30℃ to 48℃ within 10minutes,and the capacity of the oil tank is 80L.The generated heat load is:
Pv=0.85*80*2.15*(48-30)/(10*60)=4.39Kw

Then,specific cooling capacity is calculated according to the best expected oil temperature under the normal operation of the system:
P1=Pv/(T1-T2)*n1

If:T1=55℃,T2=35℃
P1=4.39/(55-35)*1.1=0.242Kw/℃=242w/℃
Power loss 242w/℃ must be dissipated by a cooler.
CHANCE suggestion:Model HD1018T at 50L/min. 