Hydraulic Fluids with minus temperature TD2K response

[axial]

In response to TD2K (chemical). The calculation that I am using is as follows.

First is finding the Reynolds number. Nr = 3162xQ / ud
Nr= Reynolds number
u = Viscosity in Cst (33.1) 3162 = constant
Q = flow rate (GPM)
d = ID (inches) of hose or tube

Second is finding the delta P. delta P = .000215fpLQ**2)
delta P = psig per ft loss --------------
f=friction factor 64/Nr when Nr <2000 d**5
or f = 0.316/Nr**0.25 for values Nr
>2000
P = Density(62.4 LBS/FT
L = 1ft.

I am working with Mobil Oil 424 Tractor fluid Cst range from 4.0 Cst @150 degrres C to 10,000 Cst @ -32 degrees C.

I have recorded minus pressures at 7000 Cst @ -28 degrees C
and pull a vacum of 13.5 in. Hg.

If I run the above calculation and plug in my hose length that I am using @ the 7000 Cst for the -28 degrees C.
Hose length= 9.7 ft.
ID = 1
I would come up with 135.5 in. Hg.
This is obviously why off from what I recorded
This is why I would like to know what may happen that throws the numbers off so far.

TD2K or anyone can you help.

TD2K thanks for the response.

 

[axial]

I forgot to add that my GPM is 4 GPM.

 

[butelja]

From the way you worded your question, I'm assuming that this is a pump suction line. If this is a pump suction line, you have problems. What is going to happen is cavitation, since you have inadequate net positive suction head (NPSH) available. It looks like it would be impossible to flow much more than ~0.88 GPM if your resivoir is at atmospheric pressure. Your options are to either use a different oil, heat the resevoir, enlarge the suction line, or slow down the pump.

 

[axial]

butelja(mechanical)
Thanks for your reply.
You are correct and it is a pump suction line. We can operate short term to 14 in. Hg. and that is how we are appling pump in this aplication. The in. Hg never stays there due to the system heating up. What problem I run into is trying to calculate for below freezing or minus temperatures. The calculation that I posted is for pressure drop through hosing and piping. As you can see I am having problems geting the numbers correct for minus temperatures. I do not know of any other calculation to use that will give me the correct numbers. Thanks in advace to all who responds to this.

 

[ChasBean1]

You're off by a factor of 10. Check units & conversions! Look at the following.

• Check the math for that Reynolds number constant.
• Did you skip a decimal in the pressure drop calc?
• Should you use absolute temps? 150°C = 423K, -32°C = 241K
• Did you plug in 1 for the ID representing 1 foot, or is it 1 inch (e.g., should it have been .0833 ft?)

 

[axial]

In response to ChasBean1(Mechanical).
You are correct in being off by a Factor of 10. See example below. The calculation is as stated above and I have converted into absolute pressure and than to in. Hg.

The below are

For the Reynolds (Nr) number
3162 constant x 4 gpm / 7000 Cst x 1 d = Nr 1.80686

Delta p from below calculation = 6.7

(constant)
0.000215 ((f=35.42xp=62.4 x.88 for Sg x L=1 x Q**2= 16))
________________________________________________________
d = 1

Conversion to absolute and than to in. hg from below calculation.

psig ger ft loss from above is 6.7 and than multiply that number by .06804 to convert to absolute pressure x 29.92 to convert into in Hg = 13.63 in Hg.

Now here is where the problem comes in. I have 9.7 ft of 1 inch id hose and as you can see the calculation is for psig per ft loss. If I multiply 13.63 x 9.7 = 132.3 in.hg.

The weird part about this is that the 13.63 in Hg. is about what I recorded when i tested at minus temperatures.

Can someone shed some light on this because I am out of ideas.

 

[butelja]

Once you have inadequate NPSH for your pump, the liquid flow rate slows down to match the available pressure drop. The pump is moving a constant volume of 2-PHASE FLUID. The liquid phase flow rate decreases, and vapor/dissolved gases flash to make up the remainder of the volumetric flow rate. These vapor bubbles re-collapse inside the pump when they reach a region of higher pressure. This WILL damage the pump if done long term.

Cavitation in a gear pump sounds different from cavitation in a centrifugal pump. In a centrifugal pump, it sounds like &quot;pumping gravel.&quot; In a gear pump with slight cavitation, the pump will become noisier, emitting a typical whining gear pump sound, only louder than under favorable conditions. If the suction is further starved to send the pump into deep cavitation, the pump will actually quiet down. However, the volumetric flow rate of liquid, as well as the discharge pressure, will be decreased. Is this what you are seeing?

 

[axial]

In response to betelja.(mechanical)

It very well could be since the viscosity of the oil is high the slower the movement of cylinders etc..

As far as pump damage is concerned short term is ok for the axial piston unit I am using and I agree during initial start-up of the unit cavitation is heard for about 3 to 5
seconds.

As far as seeing the actual flow being slowed down we have not instrumented it. We know things will move slow.

WE ARE TRYING TO CACULATE UP FRONT BEFORE WE INSTUMENT THE PRESSURE DROP / in. Hg. to size the hose for velocity but also what may happen at minus temperatures. Any and all my resources that I have asked the question to cannot find a reason why the calculation is not workig for higher viscosity oil. This is why I am looking on this site.

Thanks to all who may and have responded to this problem

 

[TD2K]

Sorry axial to have taken so long.

As others have suggested, I don't think you are moving 4 gpm of liquid through that hose when it is cold, I just don't match the dPs you are seeing. What's likely happening is, as Butelja suggests, is that you are getting flashing of the fluid which is going to affect the hydraulics significantly so you are 'really' moving 4 gpm total of vapor and liquid. As more ‘warm’ liquid enters the line, the viscosity starts to drop and the pump begins to move all liquid as intended.

Laminar flow (and oh boy, do we have laminar flow here at the start), is pretty well studied, I just can't believe the formulas are out by that much. Granted, some of the error could be due to the viscosity (at -28C, I estimate the viscosity to be about 6300 cSt using a correlation I have versus your 7000 cSt). Is the hose really 1&quot; ID? If it was 7/8&quot; ID, the dP for a given flow would be 2x what you calculated for a 1&quot; but none of these account for ANYTHING close to the differences you are seeing versus the calculated values.

First, your Re formula is correct (my reference has the constant as 3160 but that's a trival difference). Your dP formula is also correct though if you want to say the dP is psi/ft, the L term on the other side shouldn’t be there.

Anyway, to move 4 gpm of 7000 cst fluid through 9.7 feet of 1” ID hose using your formula takes almost 74 psi. My spreadsheet comes up with 74.0 psi. You are measuring about 7 psi vacuum so I would conclude the flowrate of liquid is much less than 4 gpm initially.

I also did some checking on the Web. The specific gravity of Mobil 424 at 60F is about 0.882. At -28C, that would likely be about 0.911.

 

[TD2K]

Axial, something about your March 14th post isn't clear to me.

You calculate the dP as 6.7 which per your formula (which matches my formulas) should be in units of psi. You then multiply by 0.06804 and then by 29.62 inHg (the 29.62 I understand, I don't understand the 0.06804 to get to absolute pressure as you put it).

Basically, your equation says the dP is 6.7 psi per foot. Over 9.7 feet, the dP would be 75 psig. As this is impossible (unless you have a pressurized suction), you must be flowing much less than 4 gpm.

 

[axial]

In response to TD2K
The 0.06804 converts the PSIG(which I belive is correct in the calculation I posted, please correct me if i am wrong) to PSIA. What are your thoughts on this.

 

[axial]

Also in addition to your response TD2K, the psig that you see is the loss per ft of hose length. At the point recorded it showed 13.5 In Hg. so the loss I am trying to calculate is related to psig per ft. loss in pressure. I am trying to convert that to a In. Hg. calculation. Thanks for your help.

 

[TD2K]

Thanks Axial, the 0.06804 (1/14.695) converts psi to atmospheres but doesn't change gauge to absolute. To go from psig to psia, simply add 14.7. Then you can multiply by 0.06804 to get the number of absolute atmospheres and then if you multiply again by 29.92, you get the number of absolute inches of mercury.

If you simply do the same exercise but don't add 14.7 in the first place, it's still right but you remain either with units of psi (for differential) or psig.

The formula you are using gives you the pressure drop in units of psi, not psig or psiA. It's a subtle difference perhaps but to take the pressure drop and get a pressure reading in psig or psia means you need to know the pressure at one end of the hose.

 

[axial]

Thanks TD2K. I just ran some rough numbers with your answer to my question. If I add the 14.7 to the 6.7 psi I do get a more accurate calculation. This is maybe what I need to do. I may be seeing absolute in negative units (In.Hg i guess?). The 6.7 + 14.7 x 0.06804=1.46 psia ? loss per ft. x 9.7 ft of hose I come up with 14.1 psia ?. This is extremly close to what I recorded. I did not as you can see add in the 29.92 for absolute inches of mercury. Do you think this could be right. If so you have solved the mystery that I have been trying to get the answer on. Thanks

 

[TD2K]

Don't add 14.7 to the 6.7 psi, if you do it's a meaningless number.

The 6.7 psi is the calculated pressure drop PER FOOT of hose. You have 9.7 feet of hose so the calculated pressure drop over the entire hose should be 65 psi. The 6.7 psi pressure drop can not be converted to an absolute pressure, it's just the difference between the pressure at the inlet to the hose and at the outlet.

Let's say instead the pressure at the inlet to the hose was 100 psig. The pressure at the outlet of the hose would then be 100 - 65 = 35 psig. You could say the inlet pressure was 100 psig or 114.7 psia (by adding 14.7 to the gauge pressure). The outlet pressure could be said to be 35 psig or 49.7 psia. But the differential pressure would always be 65 psi (114.7 psia - 49.7 psia) or (100 psig - 35 psig). Note both the inlet and outlet pressures has to be in the same units, gauge or absolute. The 65 psi is just the difference between the two points. The 65 psi by itself does't tell you at all if the pressures at the two points are high or low.

You measured a pressure of 13.5 inches mercury vacuum which I assume was at the pump suction at the end of the 9.7 feet of hose. Inches of vacuum is typically measured on a pressure gauge relative to atmospheric pressure (eg. 29.92 inches vacuum is a perfect vacuum). 13.5 inches mercury is therefore about -6.6 psig. I assume the pressure at the inlet to the hose is about atmospheric pressure or 0 psig so the measured pressue drop is about 6.6 psi (the fact it is almost the same as you calculated per foot of hose is just a coincidence). If the pressure drop through the hose is calculated to be 65 psi (6.7 psi/ft * 9.7 ft) and you have atmospheric pressure at the inlet to the hose (0 psig or 14.7 psia), you would have a negative absolute pressure at the outlet which is impossible.

You can convert static pressures to absolute pressures but differential pressures are just a difference.

 

[axial]

TD2K, let me ask you this. I did record right at the pump inlet. The recorded vacum was 13.5 In Hg. Can it be possible with the calculation that instead of adding the 9.7 ft. of hose to the final number at the end of the calculation that the psi per ft loss is all I need to plug in. What I am getting at, maybe is another coincidence, but if i take my 6.7 x 0.06804 x 29.92 x 1 = 13.64 In.Hg. This number is extremly close to what I recorded. What I mean it is within a few tenths.

 

[TD2K]

No, it's just a coincidence that they are that close.