Solution to spindle and hydraulic pressure station cooling

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): 22x5%8%=1.11.76kW
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 3050℃ (ambient temperature 30℃)

(2) Oil pressure: the needed oil supply pressure is 0.30.5MPa

(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)

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