Pressure Drop - how do you "guess" the value? 3

[ddkm]

We are planning to construct a pipe for overflow (at a given flowrate). As such, we need to determine the minimum diameter for this pipe to cater for this overflow.

Looking at the calculation:

Diameter D = Unknown (to be determined)
Velocity u = Unknown

and Pressure Drop is required input to determine the above using Reynolds number and Friction Factor charts.

My question is: what kind of assumptions to be made to "guess" the value of Pressure Drop?

---engineering your life---

 

[ddkm]

Sorry, the picture turns out a bit small. And note that the drawing is NOT to scale and exaggerated sizing at the oveflow piping!

But basically, the numbers are:

Height 10000mm (or 10m) of the tank
Height 300mm of the tank roof above the overflow line
Both tank and collection pot are at atmospheric
There is an INPUT material from the plant flowing at 60m3/hr

My Concept
The size of the overflow pipe should be determined based on the fact that we DON'T WANT the contents in the Tank to reach the Roof. This means the overflow pipe should be large enough to cater for the INCOMING 60m3/hr while up to an ELEVATION of Z1-Z2.
As such, in tandem with RB's earlier post, it seems that Z1-Z2 should be equal to the height or elevation of the Tank Roof ABOVE the Overflow Line. i.e. about 300mm.



Any comments?


(Please note that I have used arbitrary figures for the above problem, instead of my plant's actual flow data. Hopefully, there's no unrealistic numbers used)


---engineering your life---

 

[Latexman]

Read the last line of the quote from rbcoulter's post you cited. Your sketch suggests dH = 10 meters, but your words suggest dH = 300 mm. Which is it?

Don't forget to include a term in the Bernouilli's equation for frictional losses. What you have so far is for frictionless flow.

Good luck,
Latexman

 

[ddkm]

Latex, thanks for the quick response.

Your sketch suggests dH = 10 meters, but your words suggest dH = 300 mm. Which is it?

This is precisely what I'm trying to clarify, as it was not so clear from RB's post. Like you, he seems to suggest to use the overall elevation as the dH - in my example, this would be 10m. HOWEVER, I feel that the dH should actually be the small height ABOVE the overflow line, because this is the only height that is PUSHING the fluid across the Overflow line. Everything else BELOW the overflow line can't contribute to the oveflow, right??

Appreciate your comments.


Note: Your second point on catering for frictional losses would be my next step in this problem. I'm trying to solve the uncertainties one by one. Thanks.



---engineering your life---

 

[katmar]

ddkm, the sketch makes the whole question so much clearer. It is now obvious what your details are, and even more importantly, what information is still missing.

Try this mental experiment. Imagine a vertical piece of 2 inch pipe, say 5m long, where you can somehow control the amount of water being fed into the upper end and the water simple falls out the bottom end onto the floor. Now start feeding it at 1 litre per minute. I'm sure we would all imagine that this small amount of water would simply fall through the pipe without filling it. Now gradually increase the feed rate. There will come a time where the water perfectly fills the pipe and it runs out the bottom at the same rate that it comes in. If you increase the feed rate any more the top end will simply overflow because no more can go down the pipe.

This equilibrium point is when the driving force (which is the head of water in the pipe) exactly matches the resistance (which would be the friction in the pipe, plus any entrance or exit effects you want to take into account). Note that although I said a 5m length of pipe, the length is actually irrelevant. If you double the length you double both the driving force and the resistance (neglecting end effects). For a 2" sched 40 pipe with water this equilibrium flowrate is about 55 m³/h.

But you have a complicating factor that has only come to light in your sketch and was not mentioned before. The horizontal section of pipe between the tank and the collection pot will also have a resistance, and the driving force of the head in the vertical pipe has to overcome this as well. What is this length?

In your case the driving force would be due to the height of 9700 mm of vertical pipe. I would not include the 300 mm ullage here - that is your safety margin. Let's guess that the horizontal distance is 20 m and there are just the two bends at the top and bottom of the vertical section. Your driving force is 9700 mm of water (=95 kPa) and the resistance is the two bends plus entrance and exit effects plus friction in 29.7 m of pipe with a flow of 60 m³/h. To get the driving force and resistance to match you would need a 70 mm ID pipe. (This is the trial and error calculation we discussed before.)

It is unlikely that you will find a pipe of exactly the diameter you calculate, so you have to use your engineering judgement of what size you will actually use. How likely are upsets where you get more than 60 m³/h flowing? What is the impact of the tank overflowing? Etc, etc. You may want to go one size bigger and use a 3" pipe, or maybe a 4" pipe - perhaps even bigger if the risks are great. If your pipe ID is larger than the calculated value it simply means that the level in the vertical section will settle out somewhere less than 9700 mm.

In practice I would make one other adaptation. Whatever size pipe I decided on, I would make the actual nozzle in the tank one size bigger, with the first bend that same size. At the top of the vertical section I would put a reducer to get to the pipe size I had decided on. Why would I do this? I just like it that way.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[ddkm]

Katmar, thanks so much for your lengthy description. It's a big big help, especially to understand how the 9700mm becomes the driving force instead of the 300mm which I have been suggesting.

Imagine a vertical piece of 2 inch pipe, say 5m long, where you can somehow control the amount of water being fed into the upper end and the water simple falls out the bottom end onto the floor. Now start feeding it at 1 litre per minute. I'm sure we would all imagine that this small amount of water would simply fall through the pipe without filling it. Now gradually increase the feed rate. There will come a time where the water perfectly fills the pipe and it runs out the bottom at the same rate that it comes in. If you increase the feed rate any more the top end will simply overflow because no more can go down the pipe.

I'm getting close to see what you mean, but still need some time to digest all the info. Thanks.

By the way, do you think that the linear velocity could be determined using the Bernoulli equation? Like what I've described earlier?


---engineering your life---

 

[rbcoulter]

As the others have pointed out, you will need the frictional term in the Bernoulli's equation. That is the Darcy head loss expression with friction factor from Colebrook or Moody. The delta H in your case is 10.3 meters assuming your pipe is completely internally flooded with no air leaks.

I bet the Hazen-Williams formula will give you the answer within 5 %.

 

[katmar]

Yes, Bernoulli is the way to tie it all together.

Theoretically you can calculate the velocity from the Bernoulli acceleration term. The kinetic energy due to the velocity is usually termed the "exit" loss.

In practice (with liquids) the energy lost to friction is usually much more than the acceleration term, so you would have to factor this into your calculation of the velocity.

I suspect that rbcoulter is correct in saying that Hazen-Williams will come very close to the right answer, but I find that when I am trying to get my mind around the concepts of all the elements that go into a problem applying Bernoulli is a good way to ensure that you have covered all the bases and interpreted the problem correctly.

Once you have understood it - then you just wack it with the software!






Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[ddkm]

Katmar/RB, I understand the need to include the frictional losses. Both of you have described it clearly and confirmed my own understanding of it.

My problem now is how to include this term inside the calculations. In the Bernoulli equation itself?? Or in the Darcy-Weisbach equation for pressure loss determination (=f*rho*L*u^2/(2d)) ? Note: to answer Katmar's earlier question, the length of the horizontal pipe is L = 8m.

Let me illustrate what I've done with the Bernoulli equation so far. For simplicity, I've taken the 9700mm as the figure to be used (we could argue that 10000mm or even the 10,300mm could be the better number).

Calculation done as follows:

http://i76.photobucket.com/albums/j24/dd_km/EngTipsPic1.jpg

My question is:
a) Have I used the numbers correctly, example: Z1=9.7m?
b) Where does the frictional losses come into the above calculation/equation?


[small]
My own answer (Correct me if I'm wrong):
- Use Darcy-Weisbach equation and somehow, by making some arbitrary assumptions, obtain the Pressure Loss
- Put this estimated Pressure Loss value into the Bernoulli equation in the P2 term; this means that P1-P2 becomes negative, since P1=atmospheric.
- Recalculate iterative until all the values can match each other

Looks really far-fetched!
[/small]


---engineering your life---

 

[quark]

The procedure is ok but not the logic. You should never consider the elevation of tank (particularly for overflow pipe) as the total head that causes the primary flow.

Further, the velocity you got seems to be ridiculously high and it would take a 250 mm pipe to cancel the frictional loss by taking credit of the 10m pipe elevation.

First, you have to calculate the flowrate with whatever small head that is available above the overflow pipe bottom (the maximum is 300 mm). On a safe side, I would take 300mm as the head. This is about 1.94 m/s. If you go with a 3" pipe, the pressure drop is approximately 0.8 meters and you have no problem.

Then using the calculated flowrate, size the piping based on 10m (or 9.7m elevation)elevation. You can reduce the pipe size as long as the frictional losses equal the elevation of 9.7 meters.

But, apart from the available static head of liquid in tank, you have another potential(that causes flow) acting here i.e pumping. So, if you are flowing in the tank at a cosntant rate of 60cu.m/hr, the 3" pipe can take care of it by just consuming 2.2 meters of elevation, leaving out 7.8 meters of available head untouched. Go with a 2.5" piping and you will have 4 meters of available head untouched. 2" is too tight for this application.

If subscript 2 is the end state then you have to add the frictional loss to the RHS of the equation.

 

[itdepends]

The problem appears to be very similar to calculations for standard tank overflows- which are normally two calculations.

For a standard overflow- where the tank has an internal weir or a standard nozzle followed by a vertical downleg the calculation falls into two parts.

The first one determines the head required to push the flow over a weir (if it is present)- in your case this would be compared to the 300mm clearance you have).

The second calculation (and the only one if you don't have a weir) is to determine the head loss in the exit nozzle and first bend from the tank- again this is compared to the 300mm of clearance you have available to drive it.

Thats it- assuming that the outlet piping (which is vertical) is the same size as your outlet nozzle- if you can get it out of the tank it'll fall vertically no problems.

The full 10000mm is NOT available to drive the flow unless you've formed a siphon as noted above- i.e. fully flooded pipe. Designing overflow piping using the full 10000mm will give you grief- in particular you'll find that air locks, air entrainment and surging will reduce the capacity of the line.

In your case I'd check the capacity of the outlet nozzle and bend with respect to your 300mm. Then do another calc based on the same sized piping with the full 9700mm as the driving force to the overflow collection pot. In the abscence of a software package that can perform calculations for launder flow assume the pipe is half full (i.e. double the velocity for your calculation). Unless your overflow collection pot is a looong way away you should have plenty of head available (the outlet nozzle on the tank should be the point that dictates the line size). If the pipe/nozzle size you calculate is a bit close- go up one size.

Lastly- I'm assuming the purpose of the excercise is to ensure the tank can overflow safely in an emergency and be contained. If it is designed to be a continuous process be aware that you are going to get significant air entrainment in the overflow line.

Cheers.

 

[katmar]

ddkm, you have ignored the friction term in your Bernoulli set-up. The classic Bernoulli equation is for a frictionless system and you have to modify it. Check your Fluids Textbook. The terms of the Bernoulli equation must cover :

1. Pressure change - zero in your case as you have atmospheric both sides
2. Velocity head - to be calculated
3. Static head - 9.7 m in your case
4. Friction head - to be calculated
5. Internal energy change - negligible
6. Shaft work - not applicable to your system
7. PxV work - not applicable to liquids

Ignoring the zeroes and the N/A's you have
Static head + velocity head + friction head = 0

The math to solve this is ugly (for an old man like me) so that is why I said that once you have understood it, you just wack it with the software.

As the tank fills up, the level should never get above the top of the overflow pipe. That is why I said earlier that the actual nozzle on the side of the tank should be one size larger than the rest of the overflow piping. Itdepends has highlighted a very important aspect here and the outlet nozzle probably needs to be designed with more attention than just to say "one size bigger".

I agree completely with Itdepends that there are two separate calculations - one for the outlet nozzle and one to drive the liquid to the collection pot. This is probably what has caused the confusion over whether we should use the 300 mm or the 9700 mm as the driving force. When you separate it into the two components then it makes sense. 300 mm is the maximum driving for to get the water through the outlet nozzle, and 9700 mm is the maximum driving force to get the water to the collection pot. The Bernoulli analysis above applies to the second part of the problem.

However, I disagree that you can use all of the 300 mm as driving force to get the liquid through the exit nozzle. If the level in the tank gets above the top of the outlet nozzle then you will start to get air locks and surging.

In my previous posts I have deliberately ignored the aspect of making the vertical section of pipe self-venting because we have enough confusion as it is. But once you have sorted out what the situation is in terms of pressures, flows and velocities you will realize that the vertical leg is not running full if you select a standard size pipe. The section of pipe above the liquid level in the vertical leg must be wide enough to allow any entrained gas to escape back up the pipe, or it will reduce the density in the vertical leg and lower the driving force and therefore also lower the flow. If the exit nozzle runs flooded it will not be possible for this air to escape easily - hence my argument that the liquid level should be below the top of the nozzle. If the air cannot get out it would lead to oscillating flowrates as the vertical leg gradually filled up and then syphoned out.

By my calculations an 8" outlet and vertical leg down to the liquid level would be about the smallest you could use, but a 10" would be safer. The lower part of the vertical leg and the horizontal section could be 3". It is the requirement for self-venting that makes the pipe so big. If you don't require steady flow and you could put up with the surging you could make the whole line 3", but it would be hard to predict how much of the 300 mm ullage you would use during the air lock phase of the surge cycle.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[Latexman]

This thread is beyond this old one, thread164-129548 , but for completeness I'll pass it along.

Good luck,
Latexman

 

[itdepends]

Regarding operating overflows fully flooded- I have seen this in action in continuous circuits- basically you get lots of air entrainment and slugging above around 50% flooded until the overflow nozzle is fully submerged- by the time the overflow nozzle is submerged you've used up most of the capacity of the nozzle. i.e. a partially submerged outlet nozzle does not guarentee the line is self venting- the velocity of the fluid will drag air with it.

However- you may find it very difficult to get the required flow through an outlet nozzle of a reasonable size without using the available ullage in the tank (in your case 300mm) as the driving force- in fact you may need to increase the ullage. This is one of the common reasons for installing a weir- it allows you to install the outlet pipe at a lower level in the tank- thus giving you more driving force through the nozzle. The top of the weir is relatively high- giving you more live volume in the tank. The crest height over a weir will be small relative to the height required to drive the flow through the overflow nozzle.

 

[Latexman]

u₁²/2g_c + P₁/[ρ] + gz₁ = (fL/D)u₂²/2 + ([∑]K)u₂²/2 + u₂²/2g_c + P₂/[ρ] + gz₂

simplifies to

gz₁ = (fL/D)u₂²/2 + ([∑]K)u₂²/2 + u₂²/2g_c


Good luck,
Latexman

 

[ddkm]

Guys, thanks for the help so far. I'm still digesting the vast info here, but have a few "side" questions:

1) Latex, in your equation above, you have a "sum of K". What is that referring to?

2) In my sample calculation

http://i76.photobucket.com/albums/j24/dd_km/EngTipsPic1.jpg

I acknowledge that it's incomplete because frictional losses have not been accounted for. But my question is this: is my assumption of u1=0 correct?? (the basis for that is the liquid is sitting at the top of the overflow line and therefore at "rest")



---engineering your life---

 

[katmar]

1) "Sum of K" is the total resistance of the fittings in the friction calculation.

2) Yes, u1 is zero for this calculation.

3) If you are asking questions this basic then you need to go back to your books and do a bit of revision of what you studied at college.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[Latexman]

ddkm,

I agree with katmar on all points. I also recommend you buy Crane's Technical Paper 410 and study that. It can be purchased here:

http://www.cranevalve.com/tech.htm

Good luck,
Latexman

 

[UmeshMathur]

katmar:

In your post of 15 Apr 06 6:51, you said:
"Quark, I simply substitute the first 5 terms of the Colebrook expansion in place of Churchill's term "A", raised to the correct power of course. This general procedure (i.e. using different expressions for Churchill's "A", "B" and "C") is described by Chandra Verma, Hydrocarbon Processing, Aug 1979, p122-124."

Would you be so kind as to write completely your modified expression for A in Churchill's equation? This may hold the key to removing some discrepancies between the Colebrook and Churchill formulas. (You might recall we had exchanged some remarks in another forum about the desirability of using Churchill's formula to avoid discontinuities in iterative calculations).

Thanks very much.

 

[katmar]

Umesh,

For the turbulent regime Churchill (Chem Eng. Nov 7, 1977) estimated the friction factor with his equation (14)

f ^-1/2 = 2.457 * Ln(1/((7/Re)^0.9 + 0.27e/D))

Note that in his original paper Churchill was using the Stanton form of the fiction factor, which is 1/8 of the more frequently used Darcy-Weisbach form.

This expression, raised to the value of 16, is what Churchill terms "A" in his overall equation.

Colebrooks estimate for the friction factor, also in the turbulent regime, is

f ^-1/2 = -5.65685 * log10( (e/D)/3.7 + (0.88742 * f ^-1/2 / Re ));

Note that I have converted this to the Stanton form as well, so the constants will be a bit different from the commonly seen ones.

The basic forms of these two equations are very similar, except that Churchill's form is explicit while the Colebrook formulation has f ^-1/2 on both sides of the equation.

In pseudo-code my calculation for "A" is simply

A := "initial guess" {I use 14.14 as initial guess}
For I := 1 to 5 do
A := -5.65685 * log10( (e/D)/3.7 + (0.88742 * f ^-1/2 / Re ))
A := A^16

And then I use this "A" in Churchill's equation (18). By experimenting, I found that 3 or 4 iterations was sufficient to get this trial and error equation to converge to an acceptable degree. I simplified my code by leaving out the convergence test and always doing 5 iterations. I'm sure the purists will find it a bit crude, but it works very well.

I have made plots of the friction factor against Re and e/D and it is remarkable just how smoothly Churchill's method works. A true work of genius in my opinion. All I have done is take advantage of the speed of modern PC's to bring Churchill's method a bit closer to the Colebrook numbers. As various people have stated, Colebrook may not be perfectly accurate but it is the standard that others are judged against. If you would like a copy of this plot, contact me via my signature below and I will email it to you with pleasure.

regards
Harvey




Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[UmeshMathur]

Many thanks, Harvey. I appreciate the clarification, and it does seem to be a very creative way to eliminate the discrepancy against the Colebrook formula.