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Water Hammer 1
- Thread starter ECD40
- Start date
LittleInch
Petroleum
No attachment or formula.
Any equation is going to be pretty basic.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
Any equation is going to be pretty basic.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
- Thread starter
- #4
LittleInch
Petroleum
Did you click on the link in blue that says click here to..... and return to the forum?
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
- Thread starter
- #6
- Thread starter
- #7
The water hammer formula is:-
delta pressure (lb/inch sq.) = 0.070 x change in velocity (feet/sec.) x pipe length (feet) x 1/time (sec.).
The equation P = 0.07 (VL / t) shows the relationship of these three variables and their effect. “P” is the additional pressure generated by the shock wave, “V” is the flow velocity in ft/sec, “L” is the pipe length between the barriers in feet, and “t” is the valve closing time in seconds.
delta pressure (lb/inch sq.) = 0.070 x change in velocity (feet/sec.) x pipe length (feet) x 1/time (sec.).
The equation P = 0.07 (VL / t) shows the relationship of these three variables and their effect. “P” is the additional pressure generated by the shock wave, “V” is the flow velocity in ft/sec, “L” is the pipe length between the barriers in feet, and “t” is the valve closing time in seconds.
LittleInch
Petroleum
That looks like some sort of momentum force, but surge usually needs velocity squared at least. Don't understand the length number.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
LittleInch
Petroleum
Are you sure you didn't mean this?
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
Looks like an empirical formula; pipe length would be a surrogate for the moving water mass momentum, presumably.
Found same equation here
TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
Found same equation here
TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
That one doesn't make sense to me. Where did it come from.
IMO You should be using this one.
Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
IMO You should be using this one.
Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
Are you using a script blocker? You need to allow scripts to run, particularly from mathjax.org, and eng-tips.com end engineering.com . I think cloudinary.com is also required.
Also, if you see the link that is from the upload COPY AND PASTE THAT LINK YOURSELF INTO THE MESSAGE.
It looks like F* delta T = mass * delta V. The mass is on a per square inch basis so the pressure comes out that way.
Also, if you see the link that is from the upload COPY AND PASTE THAT LINK YOURSELF INTO THE MESSAGE.
It looks like F* delta T = mass * delta V. The mass is on a per square inch basis so the pressure comes out that way.
georgeverghese
Chemical
See page 6-44, Perry Chem Engg Handbook 7th edn for calcs on water hammer pressure with correction for valve closure time. That formula you've posted doesnt match eqn 6-206 in Perry.
HTURKAK
Structural
- Jul 22, 2017
- 3,350
Apparently the question in the post for slow valve closure case ,
F=m.a =P*A =ρ*L*A*( dv/dt) when you assume constant deceleration, P= ρ*L*V/t
The formula becomes for (FT-LB-SEC)
P= 0.0135*L*V/t
When F.S. =5 is applied for design calculations ,
P= 0.07*L*V/t + Pi
I hope this respond answers your question..
Tim was so learned that he could name a
horse in nine languages: so ignorant that he bought a cow to ride on.
(BENJAMIN FRANKLIN )
F=m.a =P*A =ρ*L*A*( dv/dt) when you assume constant deceleration, P= ρ*L*V/t
The formula becomes for (FT-LB-SEC)
P= 0.0135*L*V/t
When F.S. =5 is applied for design calculations ,
P= 0.07*L*V/t + Pi
I hope this respond answers your question..
Tim was so learned that he could name a
horse in nine languages: so ignorant that he bought a cow to ride on.
(BENJAMIN FRANKLIN )
- Thread starter
- #15
HTURKAK
Structural
- Jul 22, 2017
- 3,350
True !!
-If you are looking for actual water hammer pressure ; use P= 0.0135*L*V/t
- If you are looking for total internal pressure for design ; you may use P= 0.07*L*V/t + Pi
( notice that F.S. is applied only for wave pressure..)
EDIT= The O.P. was asking how the 0.070 factor in the Water Hammer pressure formula is derived.. and i tried to answer. I do not defense the subject formula and know that it is not the exact solution. The benchmark solution nowadays should be simulation of the piping system with enhanced softwares and i do not expect that the freshers should understand how the old calculations performed by hand , slide rule and Russich abakus...
Tim was so learned that he could name a
horse in nine languages: so ignorant that he bought a cow to ride on.
(BENJAMIN FRANKLIN )
LittleInch
Petroleum
Unless you know the original calculation this could be an empirical factor....
Any calculation is simply a guide/ estimate. Only a decent transient network analysis and valve simulation gives you a better number.
There's so much to this that a simple equation can't do it all
The calculation numbers do though tend to be conservative
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
Any calculation is simply a guide/ estimate. Only a decent transient network analysis and valve simulation gives you a better number.
There's so much to this that a simple equation can't do it all
The calculation numbers do though tend to be conservative
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
- Thread starter
- #19
This water hammer problem is at an old underground mine with many working levels and complex network of water supply pipes. A computer simulation will be meaningless or far too complex to be worth the bother. Is the noise level of the water hammer in the pipes in any way related to the magnitude of the pressure wave? Or is the noise generated only because of the physical inadequacies of the piping installation? In other words, what causes the noise of water hammer?
Who is the authority on the water hammer pressure formula? Is it the ASME or the Universal Plumbing Code or someone else?
Who is the authority on the water hammer pressure formula? Is it the ASME or the Universal Plumbing Code or someone else?
LittleInch
Petroleum
A network though normally reduces surge pressure as the pressure wave gets dissipated into other branches, especially if there is flow in those other branches.
Unless you have long single straight spurs you're going to get reflections from bends/elbows or some other element then those simplistic equations are, IMHO, not valid.
"Noise" is far too vague to use. Any change in flow could generate a small pressure force which translates to a force at a bend or tee and if the pipe is not well supported, then it can start to "clang" and bounce around a bit. The pressure rise might only a few psi or it could be dangerous levels of pressure rise. If you get hold of the pipe and try and shake it and it clangs then it needs more supports...
I don't believe anyone of those parties will provide an equation as there are simply far too many variables and the pressure surge is a complex thing. All anyone tends to do is point out that sudden changes or cessations in flow can cause pressure variations and you need to figure it out.
The easiest way is simply to replace any simple quarter turn ball valves with gate or globe valves which take several seconds to close.... similalry any simple water tank fill valve needs to be made soft close.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
Unless you have long single straight spurs you're going to get reflections from bends/elbows or some other element then those simplistic equations are, IMHO, not valid.
"Noise" is far too vague to use. Any change in flow could generate a small pressure force which translates to a force at a bend or tee and if the pipe is not well supported, then it can start to "clang" and bounce around a bit. The pressure rise might only a few psi or it could be dangerous levels of pressure rise. If you get hold of the pipe and try and shake it and it clangs then it needs more supports...
I don't believe anyone of those parties will provide an equation as there are simply far too many variables and the pressure surge is a complex thing. All anyone tends to do is point out that sudden changes or cessations in flow can cause pressure variations and you need to figure it out.
The easiest way is simply to replace any simple quarter turn ball valves with gate or globe valves which take several seconds to close.... similalry any simple water tank fill valve needs to be made soft close.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
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