15 replies to this topic

[mixednuts]

Hi everyone, I hope someone can help me out with this problem I am facing currently.

 

Lets say I have a tank at an elevation of 7m from the ground. Water is flowing into the tank at a flow rate of 80m3/hr. At the bottom right and bottom left corners of the tank, there are 2 pipelines draining water out of the tank under gravity. These 2 pipes are inclined at an angle of 35 degrees from the ground, and after a short bend, they are both merged together. Subsequently, the single pipe undergoes a series of vertical followed by horizontal flow (something like a 'staircase-shape' flow) until it is discharged into an infinite-volume tank at ground level. Both tanks are open to atmosphere. The piping (drainage) system consists of 2 different materials and consists of pipes of varying diameter at the different intersections.

 

The question is: how do I know if the drainage system is effective such that the water is drained out fast enough. How do I calculate the velocity of water coming out into the ground tank. Will the water be flowing partially full in the pipe?

 

Hope to hear from some of you soon. Thank you

 

Mixednuts

[Dacs]

This is akin to a pipe network problem.

 

In this case, you have:

1. One single source (tank)

2. Two branches (let's call it B1 and B2)

3. It merged downstream to a single pipe (B3) that goes all the way to the destination (pond) at grade

 

If you're asking if the drainage is enough to accommodate 80m3/hr of flow, then you have to do a hydraulic calculation from the tank to the pond. Keep in mind however that you have two branches B1 and B2 that will compete with each other for the flow.

 

You have to brush up your knowledge on piping network analysis for you to ascertain the flowrates for B1 and B2. Then again, if both branches are perfectly symmetrical, you can do away with straight hydraulic calculation by assuming the flow that goes to either B1 or B2 at 50% of total flow going to B3.

 

Since it's not the case, then you basically have to solve simultaneous equations for flow and pressure loss on your network. Since we're just talking about 2 nodes and 2 branches, it's not that hard to solve.

 

That said, it boils down to the available static head (mainly the elevation of tank + tank liquid level) vs the pressure loss for the whole system. You just have to find out the flowrate that corresponds to the frictional head that then corresponds to your available static head (no pressure head exists since both are atmospheric).

 

It may be possible that you have sufficient head that you'll operate your tank level at tank bottom (which means a partially filled piping, or going to the other end that your tank elevation+maximum tank liquid level is not sufficient enough to force 80m3/hr of flow through your drain pipes (which means tank overflow)

 

Can't answer that without doing the number crunching though.

 

My 2 cents

 

 

[mixednuts]

okay, lets say the two branches are perfectly symmetrical, meaning that 40m3/hr goes down to each branch.

 

/QUOTE/

That said, it boils down to the available static head (mainly the elevation of tank + tank liquid level) vs the pressure loss for the whole system. You just have to find out the flowrate that corresponds to the frictional head that then corresponds to your available static head (no pressure head exists since both are atmospheric).

/QUOTE/

 

How do i find out the tank liquid level in this case? Also, i don't understand why there will be pressure loss for the whole system since both are open to atmosphere. Another thing is, how do i find out the flowrate that corresponds to the frictional head?

 

Another query is, is Manning's equation useful in this scenario?

 

Thanks for the reply!

[katmar]

Answering the question of whether the pipes will run full or not is extremely difficult when you have an inclined section, plus a "staircase" section. The way I would tackle this problem (assuming you do not have network analysis software) would be to assume that the pipes run full and use the Darcy-Weisbach formula. You know the total difference in static height (presumably 7 m) and you know the required minimum flow (80 m³/h). Calculate the pressure drop required to get 80 m³/h to flow through your pipe - if it is less than 7 m of head then you know the pipes will not run full and if it is more than 7 m plus the maximum level in the tank then you know the tank will overflow.

With these assumptions your problem becomes "how do I calculate a pressure drop through a pipe of varying diameter". The first part of this problem is how to calculate the pressure drop through the two parallel sections at the outlets from the tank. It sounds like they are symmetrical, so you can assume that 40 m³/h flows through each of them. You know the flow (40 m³/h), the length and pipe size so it is a trivial matter to calculate the pressure drop (using Darcy-Weisbach) from the tank outlet to the point where the lines merge.

Now to the staircase section of varying diameter. Since the flowrate is known (assumed to be 80 m³/h) and you know the diameter and roughness of each section you could calculate the individual pressure drops and add them all together. But here is another way that is more suitable if the flowrate is varying. It's a useful technique to have in your "toolbox".

 

In turbulent flow the pressure drop varies with approximately the 4.8th power of the diameter ratio. This knowledge allows you to substitute one section of pipe with another of a different diameter for calculation purposes. It is best (i.e. most conservative) to bring everything to the basis of the smallest diameter pipe. So the first step is to find the smallest diameter section and to measure the total length of pipe of that diameter. Then for each other section of pipe calculate the equivalent length of the base diameter pipe.

As an example, let us assume that the smallest diameter pipe you have is 100mm (4") nominal bore sched 40 - i.e. inside diameter is 102.3 mm. Now assume you also have a section of 20 m length of 150mm NB pipe with an inside diameter of 154.1 mm. This section of pipe can be calculated as length of 20 x (102.3/154.1)^4.8 = 2.8 m of 100 NB pipe. In this way you can convert all your pipe sections to equivalent lengths of 100 NB pipe. Add them all together to get the total equivalent length of 100 NB pipe.

Now you know the equivalent length of the pipe from the point of merging to the point of discharge and you know the flowrate (80 m3/h) so you can calculate the pressure drop for this section. Add this to the pressure drop through the parallel pipes and you have the total pressure drop. Now you can compare it to the 7 m available head to answer the question.

If you have pipe sections of differing surface roughness do the calculation twice - once assuming it is all at the roughest and again at the smoothest. Use a bit of engineering judgement to interpolate between the two.

What often happens in a situation like this is that the pipes do not run full, but when the tank level runs below the top of the outlet nozzle air is drawn into the pipe and this decreases the driving force head and the flow decreases. The level in the tank then rises until there is enough head to drive out the air in the pipe and the level drops until it starts drawing air into the outlet piping again and the cycle repeats. I have often seen this situation when the tank level is not controlled.

 

It is better to use Darcy-Weisbach than Manning for this calculation, but if you then determine that there are horizontal or sloped section that are not running full you can use Manning to determine the level in that pipe section.

[Dacs]

okay, lets say the two branches are perfectly symmetrical, meaning that 40m3/hr goes down to each branch.

 

/QUOTE/

That said, it boils down to the available static head (mainly the elevation of tank + tank liquid level) vs the pressure loss for the whole system. You just have to find out the flowrate that corresponds to the frictional head that then corresponds to your available static head (no pressure head exists since both are atmospheric).

/QUOTE/

 

How do i find out the tank liquid level in this case? Also, i don't understand why there will be pressure loss for the whole system since both are open to atmosphere. Another thing is, how do i find out the flowrate that corresponds to the frictional head?

 

Another query is, is Manning's equation useful in this scenario?

 

Thanks for the reply!

 

That's the whole point of the exercise, to find the tank level that's required to push 80m3/hr of fluid through your piping

 

You have to do a hydraulic balance to realize that while you don't have any pressure delta (both are atmospheric), you'd have static head and this will be your driving force to push the fluid through your piping.

 

The whole idea is when you have flow in a pipe, it will generate frictional losses. Basically you'd want to balance this frictional loss on your available driving force (which in this case is your static head).

 

So you can set your flow at 80m3/hr and calculate the overall frictional loss, and from that value, you can back calculate the required height, which will correspond to the required tank height.

 

For the calculation of overall pressure loss considering the split from the tank, I'll give you some pointers that will help you nudge in the right direction (B1 and B2 are branch 1 and 2 respectively and B3 is the common downstream line going to pond):

1. You can do 2 separate calculations for line B1->B3 and line B2-B3

2. Fix the flow for B3 at 80m3/hr

3. Assume a certain flow for B1 (which of course is lower than 80m3/hr)

4. Flow for B2 would be B3-B1 (obviously)

5. Calculate the pressure drop for both lines using the flows above

6. The goal is to match exactly the pressure loss for both pipes. If one line is higher (say B1->B3), reduce the flow for B1 (and increase the flow for B2) and do step 5 until both pressure losses match

 

I have to give it to katmar though, he has tackled the scenarios that might happen out in the field. Although I tend to think that this is more of an exercise to do hydraulic calculation than dealing with an actual issue

[katmar]

I have to give it to katmar though, he has tackled the scenarios that might happen out in the field. Although I tend to think that this is more of an exercise to do hydraulic calculation than dealing with an actual issue

 

Dacs, I am sure that you are correct and that Mixednuts' professor is just expecting him to calculate the hydraulics and this is not a real world situation. But the whole point of studying is to prepare oneself to be able to work out in the real world one day, and if we can drop in a comment or two that will go beyond just helping the student through the asignment and help them develop as an engineer then that is a good thing. 

[mixednuts]

Thanks Katmar and Dacs for the valuable input.

 

 

According to what you guys said, i have to assume that pipe is flowing full with a flow rate of 80m3/hr and then go on to find the frictional losses associated with this flow rate using DarcyWeisbach. And if the head loss due to friction is less than the elevation (in this case 7m), it means that the pipe will not be running full and if the head loss due to friction is more than the elevation plus the tank height, the tank wil overflow.

 

Can I verify if what I said in the above paragraph is correct?

 

Another thing is, assuming that I had calculated that the head loss due to friction is lesser than the elevation (very likely scenario), how do i calculate the outlet velocity of the water to the pond. If you ask me, I would say use Bernoullis equation between the tank level and the pipe outlet. I know the pressure terms would both be cancelled out. Velocity term for tank level would be cancelled out too. So it will be gz1 = gz2 + u2(sq)/2 + headloss due to friction?

 

Can I confirm if what i said in the above is correct?

 

Another thing is, how do i size the source tank to ensure that the water does not overflow? This is because according to what Katmar said, the tank level will go up and down.

 

By the way, this is not an assignment question. I am doing internship and this is a problem that i am faced with right now. So, much help would be greatly appreciated. Once again, cheers to the both of you for the valuable inputs.

 

Mixednuts

[Dacs]

According to what you guys said, i have to assume that pipe is flowing full with a flow rate of 80m3/hr and then go on to find the frictional losses associated with this flow rate using DarcyWeisbach. And if the head loss due to friction is less than the elevation (in this case 7m), it means that the pipe will not be running full and if the head loss due to friction is more than the elevation plus the tank height, the tank wil overflow.

Yep you got it right. If the frictional loss is lower than 7m (head equivalent), then you'll run into having partially filled liquid in your pipe and if more, then you'll have tank overflow.

 

For the partially filled pipe, this is dynamic in nature (as katmar has pointed out) so at every point in time, you'd have varying conditions inside your pipe.

 

I suppose you can size your manifold for self-venting so that you'd minimize the carryover of air in your piping, but is level control out of the question? A simple on-off LC should suffice for this kind of system.

Another thing is, assuming that I had calculated that the head loss due to friction is lesser than the elevation (very likely scenario), how do i calculate the outlet velocity of the water to the pond. If you ask me, I would say use Bernoullis equation between the tank level and the pipe outlet. I know the pressure terms would both be cancelled out. Velocity term for tank level would be cancelled out too. So it will be gz1 = gz2 + u2(sq)/2 + headloss due to friction?

I think velocity head is negligible for this system, at any rate: delZ + delV = F (delZ is static head and delV is velocity head) and F is frictional losses
 

Another thing is, how do i size the source tank to ensure that the water does not overflow? This is because according to what Katmar said, the tank level will go up and down.

This is more difficult to answer because we're dealing with pressure losses from mixed phase (air and water), which is quite a PITA to solve.

 

Thing is however, since you pretty much have a free hand on your design, you have come up with a design that will bring the least trouble for the operators. A steady state system is preferred (which means less operator intervention).

 

If I may suggest, size your lines that can accommodate the maximum expected drainage flowrate (which I think is 80m3/hr). Then size your tank to hold a certain amount of holdup (5 to 10 mins is reasonable). Then place an on-off level controller with an on-off valve (at your pipe discharge) that opens up whenever you have high level in your tank and closes when you have low level.

 

 

 

[mixednuts]

One quick question: i had calculated the head loss due to friction and it only comes up to a value of 12-13mm. the total length of pipes are approximately 22m. By the way, the pipes are a mixture of Stainless steel pipes and PVC pipes. Am i on the right track? the values of the frictional head loss seems too small from what i feel.

[katmar]

Lots of good advice and comments from Dacs, and I will also give you 10/10 for your first paragraph summary. It's good to know this is a real world problem - it's much more satisfying fixing real problems than solving academic exercises.

 

Calculating the exit velocity can either be very easy, or very difficult.  If the piping is running full then all you need is the diameter of the pipe to get the cross sectional area and you know the volumetric flowrate so the velocity is simple to calculate.  If the outlet pipe is not running full then it is quite tricky to calculate.  Probably the easiest estimate would be to take Manning's Equation and use it to calculate the required slope.  The problem is that in the usual formulation of Manning it is assumed that the pipe is sloped and the level of liquid inside the pipe is constant with length.  The more likely scenario is that your pipe is horizontal and there will be a liquid level gradient in it.  You have to use a bit of judgement here and once you have an estimate of the level in the pipe you can work out the cross sectional area of the liquid (not the whole pipe) and again from the volumetric flowrate you can calculate the velocity.

 

There is some good practical stuff on part full flow at http://www.cispi.org/CISPI/handbook/100.htm  It used to be in a nice downloadable format, but you can browse through it from this address.  Go to the tables of liquid level vs slope for the different sized pipes and you can make an estimate from there.

 

With your short pipeline length (22 m) I would not worry too much about the fluctuating level.  The important thing is to not have any high points where air can be trapped and cause a lock.  From my experience the level is likely to fluctuate by a small amount (less than 0.5m) and sizing the tank as advised by Dacs should be fine.  You can work out the water flowrate required to flush out the air (provided no high points) by using a minimum Froude number of 0.5 for the horizontal sections and 0.65 for the vertical (downwards) sections. Once you know this minimum flowrate you can estimate the pressure drop and the tank level change, but it is not an exact science.

 

It isn't really possible for us to comment on how accurate your calculated pressure drop is without having details of the piping.  But back-calculating a diameter from the flowrate of 80 m³/h and a pressure drop of 13 mmH₂O gives a pipe ID of 265 mm.  Is this the sort of size you have?  If the piping is this size you certainly do not need to worry about fluctuating levels - the pipe will never see more than a trickle along the bottom.

[mixednuts]

_{Yes Katmar, you are quite right in saying that my system has a pipe ID of ~265mm. I must say both of you have been of great help.}

_{A few things im still unsure of:}

_{How do I go about estimating the level of liquid in the last section of my horizontal pipe?}

_{Is there any way to know what is the slope of water if the last section of my pipe is a horizontal pipe flowing into a tank and the section preceding is a vertical pipe?}

_{I am asking these 2 questions, as you may have guess, in order to solve the mannings equation to obtain my outlet pipe velocity.} 

 

_{Thanks to the both of you once again.}

 

_Mixednuts

[breizh]

Katmar ,

Can you confirm the link ? 

Thanks

Breizh 

[Shivshankar]

breizh,

 

http://www.cispi.org...andbook/100.htm it will divert towards following link

 

http://www.cispi.org...ndbook/100.htm    make sure to delete %C2%A0

 

 

Regards

Shivshankar

[breizh]

Thanks Mate

Breizh

[Ajay S. Satpute]

Dear Mixednuts,

 

Can you post isometric of the piping arrangement? It will help solve the problem quanitatively.

 

Regards.

 

Ajay

[mixednuts]

i cant post the isometrics of the arrangement and i actually dont see the need to because i just need a method to estimate the outlet pipe velocity.

 

The scenario is like this: there is water flowing under gravity down a vertical pipe followed by a horizontal pipe. Like what Katmar had said, it is different from the usual formulation for Mannings because there will be a water level gradient inside the horizontal pipe. What I am asking and hope that some of you guys can shed light on is, is there any ways to determine the liquid gradient and liquid depth inside the pipe? If there is, I will just do an estimation of the slope, the cross-sectional area and the wetted perimeter in order to get my outlet pipe velocity through Mannings.

 

Hope my case is clear and hope to hear from you guys too.

 

Regards,

Mixednuts

Edited by mixednuts, 25 April 2013 - 09:33 PM.