Calgary, Alberta, Canada mail@fluidpower.pro +1 (403) 471-7212

Hi,

Dzyanis Sukhanitski, P.Eng., CFPHS.

my name is Dzyanis Sukhanitski, I’m a Hydraulic/Mechanical Engineer (P.Eng., APEGA) and certified hydraulic specialist (CFPHS, IFPS) based in Calgary, Canada.

I have a passion for system design for different applications and open for any new opportunities:

• Design of the newest modern solutions for hydraulic and pneumatic systems in close collaboration with leading fluid power companies.
• Hydraulic system calculation and analysis; simulation of hydraulic process (MathCAD, Excell, Web-programming).
• Preparing of conceptual layouts/schematics of hydraulic and pneumatic systems in accordance with ISO 1219.
• Design and stress analysis of hydraulic manifolds (Creo, SolidWorks).
• Test and troubleshooting of hydraulic systems.
• Technical support for customers during the equipment installation, start-up problem troubleshoots.

This web site is my business card and in the same time a professional blog about hydraulic and pneumatic system design. Also, I’m creating online calculators to solve some routine hydraulic calculations. In addition, I try to catch all fluid power innovations in online/press media and review the new issues of fluid power magazines to reflect most interesting articles.

Welcome to fantastic world of fluid power system design!

Please feel free to leave any comments under the posts and contact me by E-mail if you have any questions.

My latest posts

Conduit calculations (pressure losses, velocity)

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:

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Pressure change due to temperature

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):

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

or:

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

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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.

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