I. Connection of oil chiller to spindle and hydraulic system   
Notes:

1. Oil chiller shall be used in a well-ventilated place without dust and corrosive gas. Do not put any sundries around the chiller.

2. When connecting the oil chiller with the machine tool, you should fix a filter on the oil system. The mounting position is shown above. 

II. Model Selection

1. Model selection of precision oil chiller for spindle cooling

P _(Heat)= P _(Motor)·η

P _(Heat)――Spindle heating value  (kW)
P _(Motor)――Spindle motor power   (kW)
 η ――Heat loss efficiency of the spindle
For general mechanical spindles, heat loss efficiency can be calculated as η=5～8%.
For high-speed motorized spindle, heat loss efficiency is can be calculated asη=20～30%
e.g.: The power of a spindle motor (general mechanical spindle) is 22kW, the spindle heat value can be calculated as:

P (Heat): 22x（5%～8%）=1.1～1.76kW
When oil temperature is 30℃ and ambient temp is 30℃, the cooling capacity of MCO-20C is 1.9kW, which is greater than spindle heating power 1.76kW. When oil supply pressure is 0.3MPa, oil flow is 16L/min. 

2. Model selection method for hydraulic pressure station   

(1) Oil temperature: the oil temperature of hydraulic pressure station shall be controlled about 30～50℃ (ambient temperature 30℃)

(2) Oil pressure: the needed oil supply pressure is 0.3～0.5MPa 

(3) Heating power can be calculated based on oil pump power  

 P _(Heat)= P_(Motor)·η1·η2·η3
 P_(Heat) ――Heating value  (kW)
 P_(Motor)――Total power of motors in the hydraulic pressure station (kW)
 η1 ――Spindle efficiency (70%～80%)
 η2 ――Load ratio of the motor, with value is determined by different conditions 
 η3 ――Heat loss efficiency (70%～85%)

If hydraulic pump keeps working under high pressure, η should be taken the upper limit value in calculation.
e.g.: In a hydraulic pressure station, oil pump power is 9 kW. When the pump works for 20 minutes and stops (or goes into standby mode) for 10 minutes, the heating value can be calculated as:
 P(Heat)= P(Motor)·η1·η2·η3 =9×80%×（20/30）×80%=3.84  kW

(4) Heating power is deduced from the rising rate of oil temp in oil tank
 P_(Heat)= CP ·ρ·Vs·ΔT/t 
 P_(Heat)    Heating power  (kW)
 1kW=860kcal/h, 1 kcal/h =1.16×10-3 kW
 CP: specific heat capacity at constant pressure  (kJ/kg·℃)

For hydraulic oil, CP=1.9674  kJ/kg·℃ 
 ρ: specific gravity (density) kg/L For cooling oil, ρ=0.876 kg/L
 Vs   total oil volume  (L)
 ΔT: temperature rise  (℃) 
 t: time   (S)
 e.g. A hydraulic pressure station has a total oil volume of 200L. If the hydraulic pressure station works for 10 minutes, oil temperature will rise from 25℃ to 35℃. Then, the heating value can be represented as:

 P (Heat)= CP ·ρ·Vs·ΔT/t  = 1.9674  ×0.876×200×(35-25)/(60×10)=5.745 kW
 According to the model selection principle, cooling capacity of oil chiller should be 20% greater than that of heating value. Therefore, oil chiller with cooling capacity being 7kW will be a good choice. 

III. Pipe Resistance Calculation  

When oil pipe size is determined, please figure out the pipe resistance according to the following formula:

 △P=0.595ν·Q·L/D4 (For ordinary hydraulic or lubricating oil)
P: pipe resistance (MPa)

ν: kinematic viscosity (mm2/s)             Q: quantity of flow (L/min) 
 L: pipeline length (m)                     D: inner diameter of the pipeline (mm)