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System Design

Articles around Fluid Power

SANFAB Injector

What is a wonderful gadget I met today! Will try to use it in next applications. It used for oil recirculation, but it seems like you can make a close loop system with open loop pump, and get almost all benefits of the close loop:

  • reduced tank size
  • lower weight
  • higher pump work rpm

With benefits above this device is really welcome at the mobile applications where open loop pumps need to be involved. For example, at the project I’m working now, I got a situation where the open loop pump has to work at high 2500 rpm (2100 diesel engine + PTO ratio) and by some reasons open loop pump I need to use. And it isn’t easy to find not expensive open loop pump for so hight cont. work RPM.

If it is interesting here the link for the gadget:

Free Fluid Power Books

Fluid Power Basics

By this link you anybody can free download (or just read online) first couple chapters of “Fluid Power Basics” by Alan Hitchcox. This handbook is really just a basics and probably will be good enough for beginners. Unfortunately, full book is available for $59. But you always can read previous edition (from 2007) online by this link.

Pneumatics. Practical Guide

By clicking this link you can download very well illustrated eBook about pneumatic system design from Automation Direct. And it is really good book for beginners.

Electrohydraulic Control Systems

Here is a free eBook what I tried to use for proportional valve calculations. In the reality this book was not useful enough, but you can read it at least because it free for download. The book contains a lot of ads at the pages, so, probably, this is the reason why it free.


Very basics book for beginners from Yuken.

Hydraulic tank volume

The main rule for hydraulic tank sizing is: “bigger is better”, but because there is not always exist a possibility to find a lot of space for the tank (mostly in mobile applications) we need to know and follow min requirements for system calculation.

Just wanted to summarize all info I have for estimation of hydraulic tank volume:

Min. value Recommendation
Industrial application – Mineral Oil
2.5 times of all pumps flow + 10% for air cushion 3..5 times of all pumps flow + 10% for air cushion
Industrial application – HFC/HFD
5 times of all pumps flow + 10% for air cushion 8 times of all pumps flow + 10% for air cushion
Mobile application – Open loop pumps
1.5..2 times of all pumps flow + 10% for air cushion 2.5 times of all pumps flow + 10% for air cushion
Mobile application – Close loop pumps
1..2 times of all CHARGE pumps flow + 10% for air cushion 1.5..2 times of all pumps flow + 10% for air cushion

Please correct me if I’m wrong.

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


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


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)



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:


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.

HS Certification – Mistakes in Study Manual

IFPS issued very good Study Manual for preparing to HS Certifications. I have read the manual, tried to solve all reviews and found couple mistakes in formulas and review answers. I mean the edition 03/29/17 of the Study Manual.

I just want to share all mistakes I found and ask somebody who is preparing to this certification exam, keep in mind info below and check, am I right or not. Of course, I have already notified IFPS about found, but didn’t receive any confirmation am I right or not.



The answer b is correct, but there is a mistake in the answer solution. First of all, the wet area is calculated wrong. The correct calculation is:

4 ft x 2 ft + 3/4 * ( 4 ft * 2 ft * 2 pcs. +  2 ft * 2 ft * 2 pcs.) = 26 ft^2

Because the wet area of the bottom has to be calculated fully not 3/4 of the bottom.

Next, we have to determine the power: P = 0.001 * 100 * 26 = 2.6 hp

Next, when in solution they convert hp  to Btu, they multiply to 2454. They have to multiply 2545:

2.6 hp * 2545 Btu/hr = 6617 Btu/hr.

So, only with this way you can get correct answer 6617 Btu/hr.



The normal practice at the schematic for parameters of cylinder is format: [Bore Diam.] x [Rod Diam.] x [Stroke]

At the picture for review  these parameters are mixed and is can be confused. As result  – wrong answer for review!

The task has to be more clear, like it done, for example, in the review Actually, in review the format of cylinder’s parameters is correct.



The answer in study manual is d. 5227 psi

This is a wrong answer, because using Eq. 3.28 you can calculate bursting pressure. The review asks to determine the working pressure. For that, in addition, you have to apply Wq. 3.27 and using safety factor 4:1 you can find working pressure:


So, the correct answer should be c.


Review and formula °C to K

The algorithm of solving and the answer is correct, but the formula for converting from Celsius to Kelvin (at the page 3-45 of the manual) is wrong:

Instead: °C to K: K = °C + 273.7

Should be: °C to K: K = °C + 273.15

The source example.

So, the correct solving way has to be:

V_2=\frac{(6.9+0.1) \cdot 4 \cdot (65 + 273.15)}{(12+0.1) \cdot (27+273.15)}=2.61 liters

Only in this case we can get the answer 2.61 liters for volume V2


Formula Eq. 3.35

There are mistakes in the formula 3.35:

Q=\frac{V\cdot A}{K}

  1.  At the page 147 of the HS Certification Study Manual:
    – Here the convert coefficient K For metrical units should be: K = 16.667 (instead wrong one 0.06).
  2.  At the page 26 of the Fluid Power Math for Certification handbook the convert coefficient K should be:
    – For metrical units: K = 16.667 (instead wrong one 0.06)
    – For imperial units: K = 3.85 for in./sec. (instead wrong one 0.3208) or K = 0.3208 for ft./sec. (instead wrong one 3.85)

The same issue metrical units you can find in formula Eq.# N.9 of Fluid Power Math for Certification handbook (at the page 25).

Moreover, at the end of the page 167 of the HS Certification Study Manual you also can find Eq. 3.35. But the correct number of this Eq. is 3.25 and the convert coefficient K for imperial units should be:

K = 0.204 for in./sec. (instead wrong one 3.85) or K = 2.45 for ft./sec. (instead wrong one 0.3208).

So, be careful.


Please check and let me know if I’m wrong.