1 Determination of input data 1.1 Heat Rejection
N = Installed power in the hydraulic system in kW
Q = Heat Rejection (heat to be dissipated) in kW
Hydraulic pump driven by an electric motor, then
Q = ¾N or 0.75 x N
If there is hydraulic motor in the system, then
Q = N
1.2 V = Oil flow in ℓ/min
1.3 T oil = Oil inlet temperature or maximum allowed Oil temperature in °C
1 4 T amb = Cooling air flow temperature or ambient temperature in °C
1.5 ETD = Entering temperature difference
ETD = T oil ‐ T amb
2 Specific Heat Rejection
Q sp = Specific heat rejection, the required performance of the heat exchanger
in kW/°C or kcal/h°C
Q sp = Q / ETD
3 Select according to diagram
Determination of Oil Temperature Difference from selected cooler
Q act = Actual heat rejection in kW
Q sp act = Actual specific heat rejection in kW/°C or kcal/h°C (from diagram)
ΔT = Oil temperature difference in °C
Q act = Q sp act x ETD
Thus,
ΔT = 33 x (Q a c t / V)
5 Pressure Drop
Obtain ΔP = Pressure drop @30 cSt oil viscosity from diagram
4.1 Obtain pressure drop for other oil viscosities
The pressure drop curves are based on viscosity of 30 cSt. Thus, use conversion factor f to calculate pressure drop at other viscosities.
|
cSt
|
10
|
20
|
30
|
40
|
50
|
60
|
80
|
|
f
|
0.5
|
0.75
|
1
|
1.2
|
1.4
|
1.6
|
2.1
|
Pressure drop of type AH0608T is 0.55 bar @30 ℓ/min & 30 cSt.
Assume an oil type ISO VG46 is used @60°C having a viscosity of 20 cST
Then,
New ΔP = 0.55 x 0.75
= 0.41 bar