18 replies to this topic

[Afshin4451]

Dear members,

 

I have been asked about a hypothetical question as per following but I can't find any logical answer:

 

Question #

 

There is a vertical pipe with 1,000 m length and 12 inch dia. The pipe is full of water, two valves have been installed in elevations of 1 m and 1,000 m.  When the top valve is in the closed condition, if the bottom valve has been opened, answer the following questions:,

 

 

1-How much water will drain from the pipe?

 

2- Will all of the water drain from the pipe?

 

3- What is pressure in elevation of 1,000 m just after opening of valve?

 

4- What is the water velocity during draining? 

 

 

Appreciated if you could share your knowledge and experience about above question.

[cea]

You appeared to be fresh graduate. Give a logical thought to these questions. Your attempts will demonstrate how much chemical engineering knowledge you have grasp till date, or else those attempts will drive you to think logically. Either way, you will be winner. All the best.

[samayaraj]

Hi,

 

I am giving you an example that will be a key for your problem.

 

Take a closed vessel partly filled with water. Connect a vacuum pump to the vessel and start pulling the vacuum. Will you get full vacuum? You will not get full vacuum. You can pull the vacuum up to the vapor pressure of water at that temperature. In your case, if water starts draining, there will be reduction in pressure at the top elevation and the above said hint is your answer for question 1 to 3. Now you say whether water will drain or not, what will be the pressure at the top elevation of pipe.

 

For question no. 4, flow rate depends on the head available. In simple head h = v²/2g. You can calculate the velocity by this formula. But actual velocity will be different and it depends on other factors.

 

Hope you will find the answer.

Edited by samayaraj, 02 August 2016 - 04:39 AM.

[Afshin4451]

cea:

 

Yes you right I am fresh graduate, but my own answer seems not logical for others.

I share this question to know other opinions.

[Afshin4451]

samayaraj:

 

I fully understand your explanation, but in my case vacuum will produce by gravity not pump. As I understand after opening bottom valve pressure in pipe will be atmospheric, in this case about 10m water column above bottom pressure should be around water vapor pressure, then what should be pressure above that point till top of pipe? What is pressure profile in vertical pipe?

 

For using h = v²/2g formula I think both top and bottom of pipe should be in same pressure (by simplification of Bernoulli equation), please correct me if i am wrong.

[samayaraj]

I mention about vacuum pump here for an example and it not related to your problem. You did not understand my example.

 

The water filled in vertical pipe starts to flow out after opening bottom valve by gravity. So the volume reduction due to water draining in pipe creates vacuum in the top elevation and not by any pump. The minimum pressure can be seen on the top elevation is the vapor pressure of water at corresponding temperature. Hence all the water will drain completely.

 

I did not understand your point "As I understand after opening bottom valve pressure in pipe will be atmospheric, in this case about 10m water column above bottom pressure should be around water vapor pressure." Is your water is near boiling in this case?

 

Pressure profile you can do with respect to flow rate (reduction head due to flow)

 

You statement "For using h=v^2/2g formula I think both top and bottom of pipe should be in same pressure (by simplification of Bernoulli equation)" is correct. I have given head h in terms of simplification after accounting top & bottom pressure (h = head of water + pressure head at top - pressure head at bottom neglecting pressure drop)

 

[Afshin4451]

samayaraj:

 

I understand vacuum pump concept and water vaporization in vacuum condition in ambient temperature. Let me explain my concern by attaching a sketch.

 

1- Before opening of bottom valve pressure in end of pipe is hydraulic pressure due to 1 km water level which is approximately 100 bar ( assume top of pipe is Atm.)

 

2- After bottom valve will open pressure in end of pipe will decrease to Atm.

 

3- If we assume water flow out in form of full flow in 12" pipe, then there is reduction in pressure along of vertical pipe and finally pressure can reach to vapor pressure of water in amb. condition (let say 0.03 bara).

 

4- Then approximately in elevation of 10 m above end of pipe pressure reach to water vapor pressure which is minimum accessible pressure.     

   

5- With above assumption it should be a two-phase regime in elevation between 10 m and 1000 m in 0.03 bara.

 

 All of this assumption is correct if we assume there is no chance for air entertainment in water from bottom of pipe.

 

What I expect is know other experts idea about my opinion.

Attached Files

•  sketch.pdf   95.57KB   29 downloads

[samayaraj]

My reply is highlighted in red color.

 

1- Before opening of bottom valve pressure in end of pipe is hydraulic pressure due to 1 km water level which is approximately 100 bar ( assume top of pipe is Atm.)

Yes

 

2- After bottom valve will open pressure in end of pipe will decrease to Atm.

Yes. The difference in pressure is converted into velocity.

 

3- If we assume water flow out in form of full flow in 12" pipe, then there is reduction in pressure along of vertical pipe and finally pressure can reach to vapor pressure of water in amb. condition (let say 0.03 bara).

Yes. At 25 Deg. C, final pressure will be vapor pressure of water at 25 Deg. C. (i.e. 0.032 bar a)

 

4- Then approximately in elevation of 10 m above end of pipe pressure reach to water vapor pressure which is minimum accessible pressure.     

Yes. Draining will stop once the Patm - Vap pressure = head of water column. This is nearly 10m of water column.   

 

5- With above assumption it should be a two-phase regime in elevation between 10 m and 1000 m in 0.03 bara.

Yes.

 

All of this assumption is correct if we assume there is no chance for air entertainment in water from bottom of pipe.

Yes. If there is no air ingress into water, water level will not fall below 10m. But in general air will enter and increases the pressure above water level to atm. pressure and hence all the water will drain completely.

[Afshin4451]

samayaraj:

 

It seems we both agree in explaining of what will happen in pipe, what we assumed is same concept will happen in straw when is full of liquid and somebody stop liquid flowing by putting finger in top of straw. The difference is straw diameter is only few millimeters and there is no chance for air ingress but in our case pipe diameter is big enough for air ingress.

 

As far as I know to answer this question froude number should be calculated and if this number will be low (let say less than 1) air can ingress from bottom of pipe. For calculating froude number first velocity should be calculated which was my last question.

[samayaraj]

For velocity, you can calculate using v = sqrt(2*g*h)

 

where h = (diff in water column + (pr. at top elev - atm. pr)/(rho x g) - head losses)

 

From the calculated velocity, find the flow rate, Q = v x CSA of pipe

 

Find the water level drop due to flow by taking time interval, level drop = Q x time / CSA

 

Again calculate the h, velocity, flow rate, drop in level and it will go in cycle. Do it in excel sheet for fixed interval of time.

 

With each interval calculate froude number and find at which point the air starts to ingress.

Edited by samayaraj, 03 August 2016 - 02:50 AM.

[Afshin4451]

samayaraj:

 

I agree with your equation, which is derived from Bernoulli equation.  Just could you explain how diff in water column can be calculated in each interval?  We know pressure at top elev. = 0.03 bara but we don't know in which elevation the pressure can reach value.  In this regard, I suggest that the Froude number can be used to calculate directly by inserting the velocity formula in the Froude number which can result as per below:

 

Fr = sqrt (2*h/D)       which D is pipe diameter in meter

[samayaraj]

Drop in level = vol of water drained / CSA of pipe

 

Vol of water drained = Flow rate x time interval 

(assume time interval as 0.5 sec. Lesser the time interval you get more accurate solution. But it leads to lengthy calculation)

 

In simple, drop = velocity x time interval

 

Due to draining, head of water will fall and thus the velocity too. Hence after a time interval, the velocity will change.

 

Since the pipe is filled full of water, the moment water starts draining, the top end pressure equals vap. pressure of water.

Edited by samayaraj, 04 August 2016 - 03:26 AM.

[Afshin4451]

samayaraj:

 

It seems all of your proposed equations are correct  - except the last one which should change to "drop = velocity * time interval".  Let's say that the time interval is 1 second.  Could you solve it for one iteration?

[samayaraj]

Yes it a typo error I made. I have corrected it.

 

First calculate the initial velocity

 

Assume the initial velocity (after opening drain valve) V1 = 10 m/sec

 

After one second, the drop in water level is 10 m

 

Subtract 10m from the initial height, then again calculate velocity after 1 seconds, find the level drop.

 

repeat the same for each interval.

 

Note: Smaller the time interval, more accurate is your answer.

Edited by samayaraj, 04 August 2016 - 03:25 AM.

[Afshin4451]

samayaraj

 

Thank you for your advice.  I solved the above equation using time interval method, but initial velocity directly can calculate from V = sqrt (2*g*(L-10)) which L is water level in pipe which decrease by time.  Refer to the attached for my calculation results for the time interval of 0.5 second.

Attached Files

•  Vertical pipe drain.xlsx   19.76KB   35 downloads

[samayaraj]

Yes, it seems your calculation is correct.

 

head after opening valve h = L-10, where 10 is the difference in head between vap space and atm pressure right? Then it correct.

[Afshin4451]

samayaraj:

 

Yes.  Its correct.  The result also shows after the water level in the pipe reaches nearly 10 m, the Froude Number will be less than 1 - which means air ingress will happen from the bottom of the pipe, which drain all of last water amount in pipe.  

[samayaraj]

In your calculation, when time = 0, velocity, level drop will be zero. After the time interval, you have to calculate the velocity. That is the change required.

 

Look at the excel sheet I have prepared. I have shown the calc in base and I made it for both with and without pressure drop.

 

I assumed friction factor be constant (thought it will vary as velocity varies) to make the calculation simple. You can also include friction factor calculation for each interval to get accurate solution.

Attached Files

•  Pipe drain time calcualtion.xlsx   43.87KB   27 downloads

[samayaraj]

I have made a mistake in calculating drain time with pressure drop. As the level of level of water decreases, friction coeff. will decrease (K is proportional to length of pipe. Here the length of pipe is level of water present. So L = h_W). This I have indicated in yellow.

 

Sorry for the inconvenience caused.

Attached Files

•  Pipe drain time calcualtion R1.xlsx   47.19KB   38 downloads