12-02-2009, 11:48 AM
King12 wrote:-
Hello guys,
Kindly have a look on the rough sketch of this municipal water supply pipe. The sump tank is 6 feet below ground level. I need to uplift the water to ground level ( so + 6 feet) by attaching another pipe to this municipal pipe. Of course without any pump. Is there any natural way? The pipe is on a natural slope of 2-3 deg which cannot be altered.
Kindly if anybody has any idea please share.
Regards,
Contribution:-
If the municipal system is meant to supply your area of interest, then there should be enough head to provide for your supply. The height difference of 6’ between your (private) connection line to the supply system is very small as such no problem will arise. If the system was meant to supply an area in which your area of interest does not form part of, then you could have a problem. It seems this is the case and if so, it will not be possible for you to lift water directly from the sump to a height of 6’ (about 2m) above the sump level without impacting energy on water that is stored in the sump. This could be achieved by the use of pump or any other source that could in whatever way possible impact energy on the water as to overcome the forces of friction, gravity etc. Like you said that the option of using a pump is not possible or that you have not considered that option(if I understood you right), your only possible option (on the condition that the delivery head at your point of connection is enough to lift water to your point(s) of need(s)) will be to connect your feed pipe directly to the source (s) without letting it to drop into the sump or unto any other storage but channeled to first servicing your needs, then returned to the sump. This is the only foreseeable option which is subject to the availability of head as to drive the water from the point at which you are connect to the system to your point(s) of need(s). This option is most likely to be unacceptable as the suppliers (the municipal) will not allow you to connect directly on to the supply line in such a way to by-pass the sump as to supply your needs first before re-directing the supply to the sump. It seems to me an impracticable option as this practice will definitely violet the local regulations. For you to decide on the capacity of the pump which you will eventually use to lift water from the sump to your point(s) of need(s), you needed to estimate your water requirement (demand), estimate the overall distance of your point(s) of need(s) from the sump, the overall height difference between the sump and you delivery point(s), the available head from the supply; as to aid you in estimating the losses, the head that you are to overcome, coupled with the expected flow velocity of your delivery. The difference between the available head and the sum of all these will make up your hydraulic head which is to be provided by the pump. Use Bernoulli’s equation of continuity, Hazen and/or Darcy’s equations that related the kinetic energy of fluid (in this case water) to the potential energy (due to its relative height in this case the height of 6’), putting into consideration the losses that are to be encountered. Basically:-
0.5MV^2 =ℓgH + losses
M = mass of water to be delivered/unit time
V = velocity of water
ℓ = specific mass of water
H =height difference
Losses = sum of all losses due to height difference, frictional losses in the hydraulic system etc.
You had already said that water rises to the height of 1.5’ in a ½” diameter pipe. This means that the system cannot force water up to the level at which it will serve you (6’) without resorting to pumping. Considering that ½” pipe should be at the lowest end of piping to serve any useful purpose (in terms of water supply), you should consider a situation in which you have to increase the pipe diameter say to 1” and in such a situation, what will happen?. It is obvious that the height to which water will rise to in this (new) situation will be related to the height to which it rose to for the ½” diameter pipe. Consider again the equation:
0.5MV^2 =ℓgH + losses. This implies that H = (0.5MV^2 – losses)/ℓg. Discharge Q is a product of cross section area of the pipe with the discharge velocity i.e. Q = Av. If A1v1 is the cross sectional area and velocity of flow for the ½” pipe and A2v2 is the cross sectional area and velocity of flow for the 1” pipe, then if the discharge is to remain constant Q = A1v1 = A2v2, thus v2 = A1v1/A2 = v1 x Pi x 0.5^2/ ( Pi x 1^2) = 0.25v1. This show that water will rise to a lower height in a pipe with larger diameter as in this case from the diameter of ½” to 1”. If we consider this case further, H1/H2 = 0.5Mv1^2/0.5mv2^2 = v1^2/v2^2 = v1^2/(0.25v1)^2 = 1/0.0625 = 16. It means that water will fall from a height of 1.5’ (in the ½” pipe) to a height of 1.5’/16 = 0.09’ or 1.5’ x 0.25v1 (in the 1” pipe) if we did not consider other factors including the losses (that are assumed constant). The head under this situation is practically equal to zero. The situation is such that you must have to pump the water as to get it to the destination. Whether the hydraulic head at the point that you took your measurement is actually equal, greater or less than that observed is really an academic question that will not resolve your case. The case could be resolved by running an analysis which in your case will not be difficult, considering that this is a very little project.
If you cannot calculate this manually, use the software called ‘EPANET’. It is a nice and a well tested software for simulation of fluid flow and it is free (use Google search to locate it and download). It will tell you what you needed to do as to solve the problem after you have simulated the problem.
Remember to keep it in your mind that ‘water does not climb hills or does not flow in reversed direction’ (a reason for that is best explained by you as an engineer). You can only have the water delivered to you without any energy input only and only if you are locate below the level of the sump in which case instead of the water climbing hills will in that case descend hill, which ‘water will be too happy to do’.
If you still have any problem after this, please do post again.
Good luck
Regards
Teddy
Hello guys,
Kindly have a look on the rough sketch of this municipal water supply pipe. The sump tank is 6 feet below ground level. I need to uplift the water to ground level ( so + 6 feet) by attaching another pipe to this municipal pipe. Of course without any pump. Is there any natural way? The pipe is on a natural slope of 2-3 deg which cannot be altered.
Kindly if anybody has any idea please share.
Regards,
Contribution:-
If the municipal system is meant to supply your area of interest, then there should be enough head to provide for your supply. The height difference of 6’ between your (private) connection line to the supply system is very small as such no problem will arise. If the system was meant to supply an area in which your area of interest does not form part of, then you could have a problem. It seems this is the case and if so, it will not be possible for you to lift water directly from the sump to a height of 6’ (about 2m) above the sump level without impacting energy on water that is stored in the sump. This could be achieved by the use of pump or any other source that could in whatever way possible impact energy on the water as to overcome the forces of friction, gravity etc. Like you said that the option of using a pump is not possible or that you have not considered that option(if I understood you right), your only possible option (on the condition that the delivery head at your point of connection is enough to lift water to your point(s) of need(s)) will be to connect your feed pipe directly to the source (s) without letting it to drop into the sump or unto any other storage but channeled to first servicing your needs, then returned to the sump. This is the only foreseeable option which is subject to the availability of head as to drive the water from the point at which you are connect to the system to your point(s) of need(s). This option is most likely to be unacceptable as the suppliers (the municipal) will not allow you to connect directly on to the supply line in such a way to by-pass the sump as to supply your needs first before re-directing the supply to the sump. It seems to me an impracticable option as this practice will definitely violet the local regulations. For you to decide on the capacity of the pump which you will eventually use to lift water from the sump to your point(s) of need(s), you needed to estimate your water requirement (demand), estimate the overall distance of your point(s) of need(s) from the sump, the overall height difference between the sump and you delivery point(s), the available head from the supply; as to aid you in estimating the losses, the head that you are to overcome, coupled with the expected flow velocity of your delivery. The difference between the available head and the sum of all these will make up your hydraulic head which is to be provided by the pump. Use Bernoulli’s equation of continuity, Hazen and/or Darcy’s equations that related the kinetic energy of fluid (in this case water) to the potential energy (due to its relative height in this case the height of 6’), putting into consideration the losses that are to be encountered. Basically:-
0.5MV^2 =ℓgH + losses
M = mass of water to be delivered/unit time
V = velocity of water
ℓ = specific mass of water
H =height difference
Losses = sum of all losses due to height difference, frictional losses in the hydraulic system etc.
You had already said that water rises to the height of 1.5’ in a ½” diameter pipe. This means that the system cannot force water up to the level at which it will serve you (6’) without resorting to pumping. Considering that ½” pipe should be at the lowest end of piping to serve any useful purpose (in terms of water supply), you should consider a situation in which you have to increase the pipe diameter say to 1” and in such a situation, what will happen?. It is obvious that the height to which water will rise to in this (new) situation will be related to the height to which it rose to for the ½” diameter pipe. Consider again the equation:
0.5MV^2 =ℓgH + losses. This implies that H = (0.5MV^2 – losses)/ℓg. Discharge Q is a product of cross section area of the pipe with the discharge velocity i.e. Q = Av. If A1v1 is the cross sectional area and velocity of flow for the ½” pipe and A2v2 is the cross sectional area and velocity of flow for the 1” pipe, then if the discharge is to remain constant Q = A1v1 = A2v2, thus v2 = A1v1/A2 = v1 x Pi x 0.5^2/ ( Pi x 1^2) = 0.25v1. This show that water will rise to a lower height in a pipe with larger diameter as in this case from the diameter of ½” to 1”. If we consider this case further, H1/H2 = 0.5Mv1^2/0.5mv2^2 = v1^2/v2^2 = v1^2/(0.25v1)^2 = 1/0.0625 = 16. It means that water will fall from a height of 1.5’ (in the ½” pipe) to a height of 1.5’/16 = 0.09’ or 1.5’ x 0.25v1 (in the 1” pipe) if we did not consider other factors including the losses (that are assumed constant). The head under this situation is practically equal to zero. The situation is such that you must have to pump the water as to get it to the destination. Whether the hydraulic head at the point that you took your measurement is actually equal, greater or less than that observed is really an academic question that will not resolve your case. The case could be resolved by running an analysis which in your case will not be difficult, considering that this is a very little project.
If you cannot calculate this manually, use the software called ‘EPANET’. It is a nice and a well tested software for simulation of fluid flow and it is free (use Google search to locate it and download). It will tell you what you needed to do as to solve the problem after you have simulated the problem.
Remember to keep it in your mind that ‘water does not climb hills or does not flow in reversed direction’ (a reason for that is best explained by you as an engineer). You can only have the water delivered to you without any energy input only and only if you are locate below the level of the sump in which case instead of the water climbing hills will in that case descend hill, which ‘water will be too happy to do’.
If you still have any problem after this, please do post again.
Good luck
Regards
Teddy