System Design

Articles around Fluid Power

Parker 50P filter will be discontinued soon

Just got a feedback from one of my Parker suppliers about 50P series filters: this series will be discontinued soon. I used them in a lot of my projects in the past and still continue to use them now. The supplier informed me about the next:

• These filters are a pretty old design, from middle 1960’s and this is why this is a time to stop using these filters.
• The suggested replacement for new projects (by pressure and flow rate) is WPF series
• The price for 50P filters will be increased soon to stimulate customers to start to use WPF.
• No more expedite option even with the price increasing for 50P filters.

The main problem with WPF filters is a design: Base-ported (50P) vs head-up (WPF):

In some applications where I use 50P filters now, will not be so easy to replace filters with new WPF series. There are a lot of factors like access to replace filter element in the current place, different mounting brackets will be required, extra hours for our mechanical engineers to design these brackets and etc.

Moreover, in the applications, we have already done for the years this replacement will not be possible at all.

This is why the best way if 100% replacement can be found.

Take a look at filter Schroeder KC50:

• Mounting dimensions are 100% the same (5″ x 4.5″):
• Schroeder KC50 is approx. 2.5″ higher
• The same rated pressure 3500 psi (240 bar)
• The same flow rating: up to 100 GPM
• BONUS: Filter elements are interchangeable!
You can check Parker’s filter elements for 50P series here and you will be surprised: they matched Schroeder’s KC50 series!

So, if you take a look closer on Parker 50P and Schroeder KC50 you can find these products are 99% identical and we can use Parker’s filter elements in Schroeder filters and opposite.

Really, these filters are suspiciously similar! Therefore I made a request to Schroeder about KC50 filters and got a confirmation they were designed to be interchangeable. The only thing they noticed to be aware is the 50P had an option where they have a reverse flow check option for hydrostats in the housing and the Schroeder version of that filter assy is a KFH50. Also, for 3000 psi pressure (210 bar) you can save a little bit money and use identical filter KF30 series (with the same flow rate).

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.

Free Fluid Power Books

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.

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.

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):

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

or:

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

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:

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

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.