Specifications

Method of selection of Oil Cooling Unit
T2 ℃
T
1℃
V
sm
3
/h
M
Q＝2.778×10
-7
Cp・γ・Vs・△T
0
t2
t1
0.5 1.0 1.5 2.0 2.5 3.0
η
Q＝H・
100
9
AKZ
SERIES
28
Supporting
information
Supporting information
Method of selection of Oil Cooling Unit
Unit conversion formula
●
1kW＝860kcal/h
1.Select Oil Cooling Unit having a cooling capacity 20 to 30% larger than the heat release value from the
machine tool.
2.Since the cooling capacity of Oil Cooling Unit varies with the change of liquid temperature （inlet liquid
temperature） and room temperature, it is necessary to clarify the liquid temperature and room temperature
conditions to select appropriate Oil Cooling Unit.
3.Three methods are shown below as a guide for estimating the heat release value from the machine tool.
For determining the heat release value eventually, it is necessary to conduct tests and determine the
exact heat release value for selecting appropriate Oil Cooling Unit.
●Calculation method of heat release value from main machine
for the selection of appropriate Oil Cooling Unit （as a general guide）
In the case of cooling of main shaft of machining center
1.Select Oil Cooling Unit having a cooling capacity 20 to 30% larger than the heat release value from the
machine tool.
2.Since the cooling capacity of Oil Cooling Unit varies with the change of liquid temperature （inlet liquid
temperature） and room temperature, it is necessary to clarify the liquid temperature and room temperature
conditions to select appropriate Oil Cooling Unit.
3.Three methods are shown below as a guide for estimating the heat release value from the machine tool.
For determining the heat release value eventually, it is necessary to conduct tests and determine the
exact heat release value for selecting appropriate Oil Cooling Unit.
●Method 1: To estimate the heat release value from the temperature
difference between the supply oil and return oil
Heat release value（kW）
Constant pressure specific heat（Ｊ/kg℃）・・・1967.4Ｊ/kg℃
Weight volume ratio（kg/m
3
）・・・876kg/m
3
Oil flow rate（m
3
/h）
Temperature difference（℃）・・・T
2ーT1
Q
Cp
γ
Vs
△T
：
：
：
：
：
Main machine
Oil tank
Oil Cooling
Unit
E.g.） When “Vs” is 18m
3
/h（30L/min） and “△T” is 5℃
Q＝2.778×10
-7
×1967.4×876×1.8×5
＝0.479×1.8×5≒4.3kW
●Method 2: To estimate the heat release value from the increase rate of
oil temperature in the tank
Find the maximum gradient of oil temperature increase
To find the maximum gradient of the oil
temperature, it is necessary to measure △t every
one minute during the first 10 minutes.
E.g.） When the total oil quantity is 300L （0.3m
3
） and “△t”is 10℃.
Q＝2.778×10
-7
×1967.4×876×0.3×10
＝0.479×0.3×10≒1.4kW
Heat release value（ｋW ）
Constant pressure specific heat（Ｊ/kg℃）・・・1967.4Ｊ/kg℃
Weight volume ratio（kg/m
3
）・・・876kg/m
3
Total oil quantity（m
3
）
Temperature difference（℃）・・・t
1ーt 2
Time（h）
Q
Cp
γ
V
△t
H
：
：
：
：
：
：
Time（h）
Oil temperature（℃）
（ ）
Tank oil
temperature
Ambient
temperature
Q＝2.778×10
-7
Cp・γ・V・△t/H
●Method 3:
When motor output loss is considered to be the heat release value
When the output loss is 30% for the motor output 7.5 kW → The output
loss is 30% or so in general （Cooling of main shaft head）
Ｑ＝7.5×0.3＝2.3ｋＷ
E.g.）
Heat release value（kW）
Motor output（kW）…For driving the main shaft
Motor output loss（%）
Q
H
η
：
：
：