modeling flow of fluid for pressure loss 4

[Spoonful]

Dear all,

Please see attached sketch, as shown fluid enters the chamber , then bend into an orifice, then bend again exit the chamber.
In order to work out the total pressure drop from inlet to out let, I have modeled it as following.


Using dH= K*V^2/ 2G K=pressure loss factor, V=speed
1. flow enter into the chamber, use pipe exit(sudden expansion of area), use K=1

2. flow enter into orifice, use orifice theory to find K

3. flow leave the chamber, as pipe entrance(sudden contraction of area), Use K=0.5.

My question is, the fluid have changed it flow direction twice, should this be taken into consideration of the pressure loss? If it is, how should it be modeled as? elbow? mitered bend?

Thanks for all your comments in advance.

Regards

Spoonful.

 

[katmar]

I agree with your 3 step procedure, and (if your sketch is more-or-less to scale) I would ignore the changes in direction. You could calculate K-values for the changes in direction using whatever method you choose, but when corrected back to the velocity in the entrance pipe the head loss will be negligible.

An area where you may have some difficulty is with the permanent loss from the orifice. You need a few pipe diameters downstream of the orifice to fully recover the pressure, and if your side offtake is as close to the orifice as it appears in the sketch you may not get as much recovery as the standard methods predict.

Katmar Software - Uconeer 3.0

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[Spoonful]

Hi Katmar,
Thanks for your reply.

What I try to model is a basket strainer, where fluid enters the body and flow down into the basket(orifice) and exit basket surface(treated as flat screen) and then exit the body. As the basket is in a cylinder shape. when the fluid pass through basket surface, it will be in the direction of perpendicular to basket surface, and then somehow all change direction at the exit.

Any suggestion on how to model this direction change?
Thanks in advance.

Regards
Spoonful

 

[katmar]

It would be much better to find data for actual basket strainers (Google will help you). But in most cases with a strainer you are less interested in the clean pressure drop than in the fouled condition. The clean strainer may have a pressure drop of only a psi or 2, but you need to design the system with sufficient head to allow it to operate with the strainer partially blocked. Otherwise you will be forever cleaning your strainer.

It is difficult to give any sort of advice regarding the pressure drop that should be allowed because it depends on your operating conditions and the nature and quantity of whatever it is that you want to remove from the fluid stream. It's the sort of knowledge that comes from hard-won experience in a particular situation, and can't really be predicted theoretically.

Katmar Software - Uconeer 3.0

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[Spoonful]

Hi Katmar,

The difference in dP of a clean and partially blocked strainer, would only be at the basket screen, as solid particle accumulate on the basket screen surface, its open area reduce, hence dP increase, where all other geometry of the strainer will remain the same. hence no change of dP. According to my research, dP at clean or even 50% blocked screen are pretty negligible compare to strainer entry/exit dP. At about basket 85% blocked, dP at basket becomes significant value but still less than dP at inlet/outlet.

As you have suggested to ignore dP for changing of direction. hence could we say, if allow 2 x of dP at inlet/outlet should cover the total dP across the strainer?

Regards
Spoonful.

 

[786392]

Spoonful though you seem to be basically correct as regards practical operation/flows at nominal or low rate(s)

but if it at all appears to be happening it is not due to no or little pressure drop at the strainer screen

but it is as per channelized flow via blocked strainer up-to 50% clogging or even more(at times)

whereas if the system in question is thoroughly sealed, no possibility of any channelization exist.

Then for any flow-rates at 50% or above the rated value will show "definite pressure drop increase(s) across the subject basket strainer at pump suction" and this will be visible if with proper tapping pressure gauges are in-place across subject strainer.

The same becomes verifiable for centrifugal pumps as on opening of discharge for increasing flow-rate results in actual load(motor amperage) decrease instead of increase.

I personally witnessed and diagnosed the similar suction strainer blockage issue; and trouble shoot at my previous organisation with a vertical submersible seven stages centrifugal pump in crude oil service!

Hopefully this clarifies the issue in detail. any further clarification will be provided with pleasure.


Best Regards
Qalander(Chem)

 

[katmar]

It all depends on your duty. If this strainer is on the suction of a pump and its function is to stop things like nuts and bolts or stones from damaging the pump then it will almost always run totally clean and you can design for a low pressure drop. But if it is in a pressure line and it is filtering fibrous or granular material the allowable pressure drop before it is cleaned could easily be 10 to 20 times the clean pressure drop.

If you allow a K-value of 1.5 for the entrance and exit and double it, it gives you a total K of 3. This puts you in the right ballpark for a strainer - I have seen values from 2 to 5 quoted, depending on the size. If the fluid flows with a velocity of 2 m/s in your pipe this gives you a velocity head of 2 kPa (assuming water is the fluid). So you can expect a clean pressure drop of 6 to 10 kPa. But it would not be unusual to see a dirty pressure drop of 100 kPa. Based on this I would not agree that you can disregard the pressure drop across the screen.

Katmar Software - Uconeer 3.0

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[786392]

Thanks katmar, being right on dot.


Best Regards
Qalander(Chem)

 

[ione]

Spoonful,

Are you talking about a commercial basket strainer? If so then the most reputable manufacturers have pressure drop charts for their products and must be aware of the maximum differential pressure a strainer can withstand before it fails.
Those charts are built assuming the strainer is clean and so one should handle them with care (it is assumed a strainer is fitted on a line for a purpose and for sure, earlier or later, a scheduled cleaning operation will be needed), but they could represent a starting point for your analysis.
Introduce then a multiplying safety factor which accounts for the increased pressure drop of the strainer when it becomes partially clogged (this is just for calculations purposes). As pointed out above by Katmar, the multiplying factor depends on the process and on the fluid involved and experience is the best teacher. Use a differential pressure transmitter to measure on the field the pressure difference trough the strainer. The differential pressure transmitter will reveal when a previously set pressure drop is reached and the strainer needs to be cleaned.

 

[Latexman]

ione has good advice. Some processes can handle going to the vendor's max. diff. pressure, and some processes cannot. "Soft" polymers that are oversized may extrude through the holes if dP is too large.

Good luck,
Latexman

 

[Spoonful]

Thanks for the helpful posts, understand a partially blocked strainer will have many times more dP than a clean one.
regardless the strainer is blocked or not. the total mass flow rate is not changed. hence dP at inlet/outlet and entry of basket(see as orifice), so where the extra dP come from, must be at the screen. But from my calculation, eg, dP for a 40% open area clean screen and 20% open area(when it is 50% blocked) screen, are still not signification compare to nozzle dP loss. please see below.

The method I use is Idelchik, (please see attached) Assume l/dh = 0.4, K factor for 40% and 20% open are 7.8 and 47. it might seems to be a big increase on K factor, but note the velocity to be used with the K factor are screen superficial velocity (not velocity through the hole). As normal strainer would have about 500-800% of open area ratio on the screen compare to the inlet area. let's say 500% as an example. At clean condition screen has 40% open area, and its total open area is 500% of the inlet area, hence, screen total area is 1250% of the inlet area. If 2m/s at inlet, flow rate fixed, hence velocity at screen total area becomes 0.16 m/s, with K factor given earlier, give us a dP of 0.1Kpa and 0.6Kpa, compare to dP at inlet/outlet with K=1.5 of 2m/s give us dP=3Kpa which both case are quite negligible.

Please correct me if I am wrong in any of above, if all correct, Can we conclude that upto 50% blocked screen, strainer will not have any signification change in dP ?

Further to this topic, what if gas is used rather liquid, as density of gas are much much lower than liquid, hence overall dP will be quite negligible?

Regards
Spoonful

 

[ione]

I cannot understand where your 1250% comes from (or maybe I know).
Let’s call Ai the inlet area and As the total open area of the screen. If As = 5 x Ai (that is 500%) than a 40% screen open area is equal to 2 x Ai.
With 2 m/s inlet velocity, at 40% screen open area the velocity will be 1 m/s and not 0.16 m/s.

 

[katmar]

As I said before - it all depends on what you are filtering. In my experience, you do not find a given percentage of the holes blocked. I suppose if you were filtering lead pellets they might sit at the bottom of the basket leaving the upper X% of holes fully open. But generally what happens is that the flow through the basket spreads the filtered material fairly evenly over the entire surface of the basket. This means that 100% of the holes are partially blocked and the pressure drop is no longer caused by the flow through the basket holes but by the flow through the layer of filtered material that has built up inside the basket. As this layer becomes thicker, and more compressed because of the increasing pressure drop across it, the pressure drop through the filter as a whole (not hole) increases.

Katmar Software - Uconeer 3.0

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[Spoonful]

Hi ione,

Let's call

Ai= inlet area
As= total open area on screen
Ass = total screen surface area
hence
Ai x 500% = As
Ass x 40% = As
hence
Ass = Ai x (500%/40%) = Ai x 1250%
yield,
total screen area/ inlet area = 1250%
hence, speed at total screen area is 2/12.5 = 0.16m/s

Please correct me if anywhere I made a mistake.

 

[Spoonful]

Hi Katmar,

I got your point, that the dirty/blocked pressure drop are mainly due to fouling on screen, as an approximation, I treat these as reducing of individual hole size, hence reduced screen total open area. Thats where I got the % of blocked come from. Any comment or advise on the approximation?