#### What is in this article?:

- Extra fluid and cooling for compact hydraulic systems
- Perhaps a cooler?

Strategic reservoir design can help maximize thermal performance in scaled-down power units.

Higher power densities and shrinking dimensions dominate today’s hydraulic system designs, making thermal performance all the more critical. In many systems, the reservoir is the only heat sink that provides cooling. It’s thus become impractical to sustain the rule-of-thumb volume equal to 2½ to 3 times the pump flow, which allows for this heat exchange. Now designers are investigating new limits to optimize systems while maintaining thermal performance.

It’s difficult to understate the importance of operating at the proper temperature. For example, the sensitivity of oil viscosity to temperature influences lubrication and leakages. A 20°C rise in temperature can reduce the viscosity by one-half, which compromises the immediate performance and longevity of the system’s components.

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Temperature vs pressure model

A model based on the conservation of mass and energy better illustrates how hydraulic systems respond to changes in temperature and pressure, Figure 1. The primary heat sink is the reservoir or cooler and the main source is the loading valve, which are described in Equations 1 and 2, respectively.

(1) d*T*/d*t* = ( *m* × *C _{p}* × ∆

*T*–

*q*) / (

*M*×

*C*)

_{p}(2) d

*T*/d

*t*=[

*Q*× ∆

_{th}*P*+(

*m*×

*C*× ∆

_{p}*T*)] / (

*M*×

*C*)

_{p}where

*T*= temperature

*t*= time

*m*= mass flow rate

*C*= specific heat capacity

_{p}*q*= rate of heat transfer

*M*= control mass

*Q*= Flow (adjusted for thermal expansion)

_{th}*P*= pressure.

Equation 1 defines the pressure drop across the reservoir as negligible, and simplifies to the heat stored minus the heat lost to the environment. Equation 2 considers the pressure drop across the valve, the heat stored, and the heat lost to the environment (for simplicity, the second and third terms are neglected). Other potential sources include the pump and piping, which can be described similarly to the pump and reservoir, but they also are neglected for this purpose.

Using the model in Figure 1, rising oil temperature leads to decreases in pressure and oil viscosity during the warm-up phase (Figure 2, top). In another case with the model, a discrete shift in the flow-control valve causes the characteristic lagging response of temperature (Figure 2, middle). Repeated shifting of the valve presents another scenario of the temperature’s effect on pressure (Figure 2, bottom). The two key points illustrated with these graphs are that system temperature can’t ever reach equilibrium and peak load isn’t maintained.

When evaluating the “2½ to 3 times the pump flow for volume” rule of thumb, the primary concern becomes the rate of heat transfer, q, which is a function of the environmental condition and very difficult to precisely predict. By viewing the heat and calculating the overall heat transfer coefficient, *U*, as shown in Figure 2, you can assume that the internal wall temperature is the same as the that of the oil. This means *R _{C}* = 0 in Figure 3.

(3) *U* = 1 ÷ [Σ(*L _{i}*/

*k*) + Σ(1/

_{i}*h*)]

_{j}= 1 ÷ (

*A*•

*R*)

_{th}*L*= length

*k*= thermal conductivity

*h*= film coefficient

*A*= area

*R*= thermal resistance

_{th}(4) Q = U •A •∆T

*where:**U* = overall coefficient of heat transfer.

A further assumption of 12-gauge steel results in *R _{B}* being negligible. This simplifies Equation 3 so that only the external film coefficient, h, is considered. The result is that

*U*ends up in the 30- to 40-

*W*/

*m*

^{2}

*k*range.

Overall heat transfer coefficient has a large range. Consequently, machine designers often push the higher values by designing a reservoir that maximizes surface area, is in the path of air flow, and is away from heat sources. Interestingly, while the change of reservoir material from steel to aluminum drastically reduces *R _{B}*, the change isn’t significant enough to achieve the desired thermal performance.