The type selection must meet the requirements both of the system flow and the heating power.
1.Calculation of powerloss:
Estimate the power loss of the existing equipment: measure the temperature rise of the oil in a time period, then calculate the power loss according to the oil temp rise, usually with the following formula:
Pv=ΔT XCoil XroilXV/t/60[KW]

Where:
Pv=Power loss[kW]
V=oil tank capacity[L]
Coil= Equivalent heat capacity [KJ/kgk]，and for mineral oil:1.88KJ/kgk
roil=oil density[kg/L]，and for mineral oil:0.915kg/L
△T=system temperature rise[℃]
 t=operating time[min] 

For example: When the oil temperature of a hydraulic system rises from 20℃ to 45℃ within 20 minutes, and the oil tank capacity is 100L, then the heat power is:
p=(45-20)X1.88X0.915 x100/20/60=3.58[kw]
And, calculate the equivalent cooling power as per the optimal expected oil temperature under the normal operation:
P01=Pv/(T1-T3)Xh[kW/℃]

Where:
P01=Equivalent cooling power 
T1=Expected temperature[℃]  
T3=ambient temperature[℃]
h=safety coefficient, typically 1.1

When the optimal expected oil temperature of the system is55℃ and the ambient temperature is 35℃.
P01=3.58/(55-35)X1.1=1.97[kW/℃]
Finally, select the matching cooler as pert the equivalent cooling power.

2.Estimate Calculation of heating power: 1/3 of the total system power is generally taken as the cooling power of the cooler.

3.Calculation of flow: 

1 For return oil system cooling 
Q=L X S X h 
S=A1/A2

2 For cooling oil leakage circuit or independent cooling circuit 
Q=L X h
Where:
Q=flow through the cooler[L/min]   L=flow out of the oil pump[L/min]
S=effective area ratio
A1=effective area of cylinder without lever: 
A2=Feffective area of cylinder with lever 
h=safety coefficient(1.5~2):
Note: Generally accept 1.8 ,the higher the hydraulic oil viscosity, the larger the safety coefficient.