## 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.

## 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.

### Review 3.5.2.1

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.

### Review 3.8.1.2.

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

At the picture for review 3.8.1.2.  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 4.1.1.3. Actually, in review 4.1.1.3. the format of cylinder’s parameters is correct.

### Review 3.8.2.1.

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:

$p_w=\frac{p_B}{SF}=\frac{5226.67}{4}=1307psi$

So, the correct answer should be c.

### Review 3.6.1.1. 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.

## April 2017 Fluidpower Journals

Hydraulics&Pneumatics UK,
April 2017

Hydraulics&Pneumatics,
April 2017

FluidPower World,
April 2017

FluidPower Journal,
2017 Directory

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