In the hydraulic system, hydraulic oil will produce a lot of heat due to energy loss (such as pressure loss of pumps and valves, friction, etc.) in the working process. If the heat cannot be dissipated in time, the oil temperature will rise, which will affect the stability and efficiency of the hydraulic system and even cause equipment failure. Therefore, in the design of hydraulic system, it is very important to choose and configure the hydraulic cooler reasonably. This paper will introduce the calculation method of hydraulic cooler in detail.
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First, determine the system calorific value
To choose a suitable hydraulic cooler, it is necessary to accurately calculate the heat generated by the hydraulic system during operation. Usually, the heat in the hydraulic system comes from energy loss, which mainly includes:
1. Mechanical loss: such as the efficiency loss of pumps and motors;
2. Volume loss: such as valve leakage;
3. Throttle loss: such as pressure drop of directional valve and flow valve.
The total calorific value (q) of the system can be estimated by the following formula:
$$ Q = P_{in} – P_{out} $$
Among them:
-$ Q $: system calorific value (kW)
-$ P_{in} $: input power (kW)
-$ P_{out} $: output effective power (kW)
It can also be estimated by system efficiency:
$$ Q = P_{in} imes (1 – eta) $$
Among them:
-$ eta $: total system efficiency (usually between 0.7 and 0.85)
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Second, determine the cooling capacity required by the cooler
The key to the selection of cooler is that its heat dissipation capacity must be greater than or equal to the heat energy generated by the system. Therefore, the cooling power (Qc) of the cooler shall meet the following requirements:
$$ Q_c geq Q $$
In practical application, in order to ensure the reliability and redundancy of the system, it is generally suggested that the capacity of the cooler should be kept at 10%~20%.
In addition, the temperature of the cooling medium (such as air or water) should also be considered. For example, the higher the ambient temperature or cooling water temperature, the worse the cooling effect, so greater cooling capacity is needed.
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Third, calculate the cooling capacity according to the flow and temperature difference.
If the parameters such as flow rate and temperature difference of hydraulic oil are known, the cooling capacity can be calculated by thermodynamic formula:
$$ Q = m imes c imes Delta T $$
Among them:
-$ Q $: required cooling
capacity (kW)
-$ m $: mass flow of hydraulic oil (kg/s)
-$ c $: specific heat of hydraulic oil (about 1.9 kJ/(kg℃))
-$ Delta T $: temperature difference between inlet and
outlet of hydraulic oil (℃)
For example:
-the flow rate of hydraulic oil is 60 L/min (approximately equal to 60 kg/min, i.e. 1 kg/s).
-The temperature difference is 10℃
Then:
$$ Q = 1 imes 1.9 imes 10 = 19, ext{kW} $$
The cooling capacity of the cooler should be no less than 19 kW, and after adding the safety margin, the actual selection may be 22~23 kW.
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Four, the choice of cooler type
According to different cooling media, hydraulic coolers are mainly divided into:
-Air-cooled cooler: suitable for situations where the ambient temperature is low and space permits.
-Water-cooled cooler: high cooling efficiency, suitable for high-temperature or high-load systems.
The installation space, ambient temperature, maintenance cost and other factors should be considered comprehensively when selecting the cooler.
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V. Summary
Correctly calculating the capacity of hydraulic cooler is the key to ensure the long-term stable operation of hydraulic system. The calculation steps mainly include: determining the calorific value of the system, calculating the cooling demand according to the formula of flow rate and
temperature difference, considering environmental factors and selecting the appropriate type of cooler. In engineering practice, the final selection should also be combined with the selection manual provided by the manufacturer to ensure matching and reliability.
Through scientific calculation and reasonable selection, not only can the service life of hydraulic system be prolonged, but also the operation efficiency and safety of equipment can be significantly improved.