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Tag Archive: pressure drop

How to select a hydraulic cooler

At this article, I show my vision on how to:

  • determine the value of the heat needs to be rejected from the system;
  • calculate and select the right cooler size.

…and will provide an example of cooler calculation and selection.

What is a cooler and what is a heat exchanger?

The hydraulic cooler is one of the heat exchangers type. But, what is a heat exchanger? The best definition of the heat exchanger is:

Heat exchanger is a device that transfers heat between two fluids.

The simple sentence but a very good description, because fluids can be either oil, air, water, etc., and because transferred heat can be for either cooling or heating target.

There are two most popular types of heat exchangers in hydraulic systems:

Plate heat exchangers

Plate heat exchanger

Plate heat exchanger

This type has the best value of efficiency/reliability and designed for both cooling and heating applications and a very good for low-viscosity fluids. Pairs of plates can be removed individually for maintenance, cleaning, or replacement. Another advantage of the plate is exchangers is their low initial cost, as well as easy and inexpensive operation.

Fan radiators (air-cooled)

Fan radiators (air-cooled)

Fan radiators (air-cooled)

This type is the only option where water is unavailable or expensive for a delivery. From benefits: low maintenance and operating costs and the only option for oil cooling in mobile applications.

Conduit calculations (pressure losses, velocity)

Pressure losses in the conduit

Pressure losses in the conduit

These calculations can help you to estimate the pressure losses, and flow velocity in the conduit (hose, pipe or tube) and check/correct conduit ID.

Calculations Notes:

  • The recommended flow velocity in conduits you can find at the article Recommended flow velocity.
  • The assumption: pressure losses on elbows, fittings at calculated conduit is zero.
  • Height difference between IN and OUT points needs to calculate hydrostatic pressure what will be added to pressure losses, if IN point is below OUT (use positive value), or subtracted from pressure losses, if IN point is upper than OUT (use negative value). Use value “0” if height difference can be neglected.
  • To find Darcy friction factor there are different formulas used:

Hydrostatic Pressure

In one application I’m working on I need to supply hydraulic pressure to the unit what placed at the height approx 100 ft. I decided to calculate the pressure losses created by the height only for the pump at the land HPU.

The height difference creates hydrostatic pressure (what is a pressure drop for the pump) what can be calculated from Pascal’s law:

\triangle p=\rho\cdot g\cdot(h_2-h_1)=\gamma\cdot(h_2-h_1)

here:

\begin{array}{l}\rho\;-\;oil\;density\;(slugs/ft^3)\\g\;-\;acceleration\;due\;to\;gravity\;(32.174\;ft/s^2)\\\gamma\;-\;specific\;weight\;of\;the\;oil\;(lb_s/ft^3)\end{array}

The specific weight (the weight per unit volume of a material) of the mineral oil on Earth is approx.: 53.57\;lb_s/ft^3 = 0.031\;lb_s/in^3 (at environment temperature 40°C for oil with ISO Grade 46)

So, for 1 ft (for 12 in) of height we get pressure drop in PSI:

\triangle p=0.031\;\frac{lb_s}{in^3}\cdot12\;in\;=\;0.372\;psi

Therefore, for 100 ft the pressure drop is 37.2 psi = 2.57 bar only.

The calculations look very easy, but one problem I found in internet: many sources mix density and specific weight. For example, at The Engineering Toolbox you can find value of density of the water 62.4 \;lb_s/in^3. But it is not right, because this value is a specific weigh of the water, not a density! The density of the water is:

\rho_{water}=\frac{\gamma_{water}}g=\frac{62.4\;lb/ft^3}{32.174\;ft/s^2}\;\;=\;1.94\;slugs/ft^3

It has to be clear: the density is mass per unit volume, this is constant at any places (on Earth, on Moon, on any height., etc.) and it is different from specific weight.