FluidPower.Pro – Page 10 – Hydraulic and pneumatic professionals

Pump-Motor calculation

Dzyanis January 30, 2018 1 Calculators close loop, motor, pump 4028 Views

Pump-Motor calculations

Here you can find the online calculations for Pump-Motor in close loop.
These calculations are for variable displacement pump and fixed displacement motor.

Conduit calculations (pressure losses, velocity)

Dzyanis October 15, 2017 0 Calculators hose, losses, pipe, pressure drop 3599 Views

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:
- Darcy equation – for laminar flow.
- Churchill’s all-regime (1977) – the best approximation what I found for critical and transient flow.
- Haaland’s approximation (1983) to the Colebrook–White relation – for turbulent flow.

Pressure change due to temperature

Dzyanis June 10, 2017 0 Articles temperature 3173 Views

A change in temperature will cause hydraulic fluid to try to have a corresponding change in volume. If the fluid is trapped in a chamber and is unable to change volume, there will be a change in pressure.

The difference in pressure is based on the bulk modulus (stiffness) of the fluid. A mineral based oil may have a pressure difference of about 11 bar for each 1°C change in temperature (90 psi for each 1°F change in temperature):

## \triangle p=\triangle t\cdot k ##

where k= 90 (imperial units) or k= 11 (metrical units):

## \triangle p\;\lbrack PSI\rbrack\;=\triangle t\;{\lbrack^\circ F}\rbrack\cdot90 ##

what is equal to:

## \triangle p\;\lbrack bar\rbrack\;=\triangle t\;{\lbrack^\circ C}\rbrack\cdot90 ##

Hydrostatic Pressure

Dzyanis May 20, 2017 0 Articles density, pressure drop, Weight 1821 Views

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:

#\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#)

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.