Water Hammer Formula 3

[jeffvalve]

Hi All

Does anyone know of a formula for calculating the minimum closing time of a butterfly valve to prevent water hammer?
I have a vague recollection of there being such a beast but cannot find any references

Regards

 

[hydrae]

Jeff

Marks handbook has a short bit on waterhammer

Some rules of thumb on water hammer
For every 1 ft/sec of velocity there is a 50 psi surge on a sudden stoppage, so that give you the magentude of the surge. As you close the valve a pressure wave develops as a higher pressure upstream and a lower pressure down stream, the time it takes this wave to reach a point where the water does not have velocity is based upon the speed of sound in water (about 2000 ft/sec) For example, two tanks with water flowing from the upper tank to lower tank through a 3000 ft pipe. (diameter, pressure and flow do not matter for amount of surge pressure) A valve is 1000 ft from the upper tank, the water is flowing at 10 ft/sec.
Lets look at upstream for now, assume the system can withstand an additional 50 psi upstream of the valve (static pressure 100 psi and system rated for 150). The fastest you can slow the water down is 1 ft/sec in a sudden change in flow, it will take 1/2 second for the pressure wave to reach the upper tank and another 1/2 second later the wave will return while that wave is propigating the pressure will read 150 just upstream of the valve, since the valve is still mostly open the wave will then travel through the valve and a small amount will be reflected back, at this point you can close the valve to decrease the flow another ft/sec. but as the valve continues to close more of each pressure wave will be reflected instead of passing through, this is when the closing rate needs to be slowed down so the waves to do not stack upon each other. When the valve is fully closed the wave will bounce back and forth between the valve and the upper tank until it is disapated in 10 to 100 cycles.
Downstream has the same thing but the pressure wave is negative, for the example the cycle is 1 second down and 1 second back and the pressure will read 50 psi for those 2 seconds. The down stream case does not seam as damaging but if the static pressure was 35 psi, the pressure will fall below the vapor pressure of the fluid at minus 10 psi for cold water, then the water will undergo volumetric boiling also known as cavitation, and cavitation can destroy anything with enough time, other considerations: thin wall steel pipes that are not designed to take a vaccumm can colapse under even mild negative pressures, and leaks in the system can suck back in contaimination.
So for this example the operator should not close the valve in less than one minute, and to be safe take five minutes...


Hydrae

 

[jeffvalve]

Thanks Hydrae.... Most helpful

 

[metalonis]

This site might be informative:

http://www.setra.com/tra/app/app_har.htm

 

[budvolz]

This site may be helpful in eliminating hammer conditions. Check out this site to solve your problem.

http://www.dft-valves.com

 

[bulkhandling]

Check this link for a formula:

http://www.plastomatic.com/water-hammer.html

I got the following from somewhere else:
The maximum closure time can be estimated from
tmax=2*L/a
where:-
L=Developed length of the piping system
a=Speed of sound of the fluid
If the pump or valve closure time is less than tmax, then a hammer analysis should be performed.
But I think this judgment is not conservative enough, just give you some idea.

 

[BRIS]

As explained by Hydrae "2 * L/a" is the time for the pressure wave to travel from the valve to the tank and back again to the tank.

If the valve is closed faster than this time it is termed a rapid closure and the pressure wave generated is equal to that generated by an instantaneous closure.

For water the pressure rise due to instantaneous closure is V.a/g (in SI units for water in a rigid pipe - v = initial velocity m/sec, a is the wave speed 1100 m/sec and g is 9.81 m/sec/sec).

If the valve is closed slower than "2 * l/a" then it is termed a slow closure and the pressure rise will be less than v.a/g. For a slow closure the reflected pressure wave will return to the valve before it is closed. The sign of the reflected pressure wave is opposite to the propagated wave. On valve closure a positive pressure wave is propagated upstream. When it reaches the tank it is reflected back at an equal magnitude but opposite sign (negative). It is this fact that results in the pressure rise being less for slow valve closure.

If the pump or valve closure time is rapid (less than 2 * L/a) then you really don't need to do a complex waterhammer analyses because the pressure rise can be calculated from v * a/g. If the closure time is slow then you need a complex analyses which examines the interaction of the propagating and reflected pressure waves is needed.

Brian

 

[BRIS]

Whoops
Last sentence should read

If the closure time is slow then the pressure rise will be less than that for instantaneous closure but, in order to determine the pressure rise envelope a complex analysis which examines the interaction of the propagating and reflected pressure waves will be needed.


Back to your original question - the formula is 2 x l/a but it will not prevent waterhammer - it will reduce it. The amount of reduction also depends on the valve characteristics. For example with a butterfly valve 80 % of the flow will be controlled by the last 20% of opening and to give any significant reduction in the waterhammer pressure the last 20% of valve opening must be closed in a period longer than 2 * L/a.

 

[xhpipe]

Water Hammer is a comlex time based phenomena, which one simple formula or another do not fully adress the range of adverse possibilities. This is why it is wise to have an outside expert firm perform water hammer analysis of a system.

The formula above adesses a valve which has linear charecteristics, most butterdly valves are non-linear! Depending upon the buterfly valves charactersitics you may need to mutiply the time by 4 or more.

Also this adresses only hammer due to its opening or closure, what about pump trip, or flow increase from another pump?

If the systems reliabilty is an important factor added study is warranted.

Regards,
XHPIPE

 

[stanier]

The problem with closure of a butterfly valve is that there can be substansive flow even when the valve is 80% closed. the time to fully close from this position is what is of concern.

I agree with XHPipe in that you cannot simplify the study of waterhammer. Other factors not mentioned to date are:-

The characteristics of any pump non return valve ( refer Prof Thorley Fluid Transients in Pipeline Systems)

Column separation that can cause even higher pressures than predicted by Joulowsky

Performance of air valves


Suggest that a formal risk assessment be carried out as to whether you need to invest in a waterhammer analysis or not. What is the likelihood? You dont know because you havent doen the analysis. What are the consequences of 1) line break? 2) Long term fatigue damage? 3) Legal Implications/Embarassment of of your design being wrong?

I specialise in waterhammer analysis so push my own barrow. I do however belive that there is a lot of ignorance out there in respect of this topic.

 

[BRIS]

I agree with Stanier and xhpipe - Waterhammer is complex and the descriptions given by myself and Hdrae are very much simplifications focused at your original question which was "Does anyone know of a formula for calculating the minimum closing time of a butterfly valve to prevent water hammer?"

The answer is no - there is no formula for a minimum closing time to prevent water hammer. For a simple system a waterhammer will start to reduce if the closing time is greater than 2l/a but the reduction depends on many factors - mainly the valve characteristics. For the majority of valves the closure time will need to be several times 2l/a to achieve a significant reduction.

And you may have to consider stepped closure - eg rapid closure to say 80% opening then slow closure for the remaining 20%.

True if column separation occurs wathammmer pressures can exceed va/g (Joukowsky equation). The celerity of the pressure wave (a) is also a significant variable and for water can vary between 400 m/sec to over 1200 m/sec depending on pipe material and the conclusion is waterhammer calculations can often be simplified but there are many factors and you need an expert to assess the extent of the investigation needed. -

(I could say that some 50% of waterhammer analyses undertaken by specialist agencies are unnecessary but it needs a specialist to know this so I won’t say it).

Regards Brian

 

[jeffvalve]

Thanks everyone for your efforts on this. I have contracted the problem out to a hydraulic analysis company. Let them have the headaches and the responsibility. Thanks again all!

Jeff

 

[LSThill]

jeffvalve (Mechanical)

Water Hammer Calculation. Hydraulic Transient Analysis.

http://www.lmnoeng.com/WaterHammer/WaterHammer.htm

INTRODUCTION
Rapidly closing or opening a valve causes pressure transients in pipelines, known as water hammer. Valve closure can result in pressures well over the steady state values, while valve opening can cause seriously low pressures, possibly so low that the flowing liquid vaporizes inside the pipe. Our calculation computes the maximum and minimum piezometric pressures (relative to atmospheric) in each pipe in a pipeline as well as the time and location at which they occur. The calculation is helpful in design to determine the maximum (or minimum) expected pressures due to valve closure or opening. If you are investigating the cause of pipe rupture, the calculation can provide insight as to what the pressure may have been in the pipeline during the rupture.

Our calculation simulates water hammer in a pipeline flowing full, bounded upstream by a large reservoir and bounded downstream by a valve which discharges to the atmosphere. The reservoir is assumed to be large enough to absorb changes in pressure and remain at the same elevation during the transient. There can be up to three pipes in series having different lengths and diameters but the same pipe material. The time to close (or open) the valve is entered. Also, the calculation allows you to enter an intermediate time and % open so that the valve curve can be represented by two piece-wise linear equations, rather than a step function or single linear function. If the valve is being closed, you must also enter the initial (time=0) minor loss coefficient (K) for the valve. Additional information can be found in Equations, Variables, and Discussion.

EQUATIONS Top of Page
Mass conservation and momentum conservation are the fundamental equations used to analyze hydraulic transients (water hammer). The boundary conditions consist of a large reservoir at the upstream end of the pipeline and a valve at the downstream end discharging to the atmosphere. The equations, subject to the boundary conditions, are not readily solved analytically - a numerical solution is required. We used the method of characteristics to obtain a solution. The method of characteristics is a finite difference technique where pressures are computed along the pipe for each time step. Our calculation automatically sub-divides the pipe into sections (i.e. reaches or intervals) and selects a time interval for computations. Computational accuracy is enhanced by having lots of pipe sections and time steps; however, 2 or 3 digit accuracy is generally obtained in a pipe having as few as 5 sections while 6 digit accuracy is typically obtained with 50 sections. The number of pipe sections used is shown in the calculation next to the pipe number after the calculation runs. You can "force" the calculation to have more pipe sections by decreasing Tmax, the total calculation time. Our calculation uses 100,000 pipe sections times time steps, with a maximum of 1000 sections per pipe. The product 100,000 was selected to provide maximum accuracy and fast computations. The Courant stability criterion is used to determine the number of sub-sections versus time steps.

The following equations can be found in Chaudhry (1987), Fox (1989), Hwang and Houghtalen (1996), and Wylie and Streeter (1978).

Mass Conservation and Momentum Conservation (|V| is absolute value since velocity changes direction during the transient). Courant stability condition for method of characteristics solution.



For each pipe:



Boundary and Initial Conditions:



If valve initially closed: Q(x,t=0)=0. If valve initially open: Q(x,t=0)=Qmax

Valve closing or opening curves:



where Y(t) = 100 [K/K(t)]1/2. Our calculation allows you to enter the valve curve as two piecewise linear curves. See Discussion for further explanation.


VARIABLES: Top of Page
Dimensions: F=Force, L=Length, M=Mass, T=Time

A = Pipe cross-sectional area [L2]. Up to three pipes can be used, each with its own length and area. An = Area of the pipe furthest from the reservoir. Computed.
c = speed of pressure wave (celerity) [L/T]. Different for each pipe. Computed.
D = Diameter of each pipe [L]. Dn = Diameter of pipe furthest from reservoir. Entered.
E = Composite elastic modulus [F/L2]. Computed.
Ef = Elastic modulus of fluid [F/L2]. Computed.
Ep = Elastic modulus of pipe material [F/L2]. Computed.
f = Darcy-Weisbach friction factor (obtained from Moody diagram) for each pipe. Computed automatically using same method as in Design of Circular Liquid or Gas Pipes.
g = Acceleration due to gravity = 9.8066 m/s2.
h = Piezometric head in pipe (elevation + static head) [L] as a function of time and distance relative to the valve elevation and relative to atmospheric pressure. Computed.
H = Head in reservoir relative to elevation of valve [L]. Vertical distance between valve and reservoir surface. Entered.
K = Minor loss coefficient for valve when flowrate is Qmax. Entered if valve being opened.
K(t) = Minor loss coefficient as a function of time while valve is being opened or closed. Computed.
L = Length of each pipe [L]. Ln = Length of last pipe (i.e. pipe furthest from reservoir). Entered.
P = Piezometric pressure in pipe (elevation pressure + static pressure) [F/L2] relative to the valve elevation and relative to atmospheric pressure. Function of time and distance. Computed.
Pmin, Pmax = Minimum and maximum piezometric pressures in each pipe [F/L2] relative to the valve elevation and relative to atmospheric pressure. Computed.
Q, Q(t) = Flowrate in pipe as a function of time [L3/T]. Computed.
Qmax = Steady state maximum flowrate in pipe [L3/T], when Y=100%. Entered.
t = Time [T].
Tmid = Intermediate time for user to enter Ymid. Tmid must be between 0 and Tvlv. Entered.
Tvlv = Valve opening or closing time. Time to achieve Y=100% for opening or Y=0 for closure. Entered.
Tmax = Maximum time for the calculation. Can be less than Tvlv. Pmin and Pmax are found for each pipe in the time period up to Tmax.
V = Velocity in pipe as a function of time and distance [L/T]. Computed.
w = Wall thickness of each pipe [L]. Entered.
x = Distance along pipes measured from the reservoir [L].
Y, Y(t) = % that valve is open. 100% does not have to be a completely open valve, but it is the most that the valve is open for the transient. See Discussion for further information. Computed.
Ymid = % that the valve is open at Tmid. Entered.



DISCUSSION: Top of Page
Reservoir Elevation, H
Physically, the reservoir should have a large enough area such that H does not change significantly during the transient. H is considered to be constant.

Composite Elastic Modulus, E
The equation shown above for E assumes that the pipe undergoes no longitudinal stress (Hwang and Houghtalen, 1996, p. 121). Though this may not seem realistic due to the liquid pulling on the pipe walls, very little difference in the value for E occurs in our equation compared to more complicated equations (see References) which account for expansion joints, fixed pipe ends, etc.

Pipes - Number and Orientation
You may enter up to 3 pipes in series. They do not have to be telescoping (e.g. pipe 2 can have a larger diameter than pipe 1), and they don't have to be horizontal. The calculation is valid for horizontal and non-horizontal pipes. All pipes must be made of the same material.

User can enter location to compute maximum and minimum pressure
Our calculation is set up to compute the overall Pmax and Pmin for each pipe. However, say in the case of an actual pipe rupture, you may need to know Pmax or Pmin at a specific location which may not necessarily be where the overall Pmax or Pmin occurs. You can do this by entering two pipes with the same diameter, with one of them being very short. The two pipe lengths must add up to the actual pipe's length. Have the short pipe located in the vicinity of the rupture.

Piezometric Head and Piezometric Pressure
Our calculation uses and computes piezometric heads and piezometric pressures. Even though the calculation says "Min. Pressure in Pipe" and "Max. Pressure in Pipe", these are piezometric pressures - not static pressures. (Due to space considerations, the word "piezometric" was omitted from the headers.) Head (h) has units of elevation (meters, feet). Pressure (P) has units such as Pascal (N/m2), psi (lb/in2), or bars. We also allow pressure to be output as meters of fluid or feet of fluid - which is actually a head rather than a pressure. Pressure and head are related to each other by: P=pgh (p is the greek letter "rho" representing fluid density). Piezometric head = elevation head + static pressure head. Piezometric pressure = elevation pressure + static pressure. Static pressure is what people are usually referring to when they ask, "What's the pressure in the pipe?" In our water hammer calculation, the valve discharges to atmospheric pressure at an elevation of 0.0 m. Since the valve discharges to atmospheric pressure, the static pressure at the discharge is 0.0 Pa (we use pressures relative to atmospheric - also known as "gage pressure&quot.
If your pipe is horizontal, then the elevation pressure and elevation head are 0.0 everywhere in all pipes, and the pressures computed by the calculation are the static pressures in the pipes. However, if the pipes are sloping, you need to subtract off the elevation pressure from the computed piezometric pressure in order to obtain static pressure. For example, let's say you run the calculation and the maximum piezometric pressure in Pipe 1 is computed to be 166.3247 m at x=500 m. From your pipe drawings, you know that this location is at an elevation of 50 m above the valve. The static pressure head at x=500 m is thus: 166.3247 - 50 = 116.3247 m. You can convert this to static pressure by (998.2 kg/m3)(9.8066 m/s2)(116.3247 m)(1 N-s2/kg-m)=1,138,696.5 N/m2 (assuming the fluid is water at 20oC with a density of 998.2 kg/m3). Alternatively, instead of selecting "Compute pressures in m of fluid", you could select "Compute pressures in N/m2 or Pa". Then, the output maximum piezometric pressure in Pipe 1 is 1,628,144.2 Pa at x=500 m. Then the static pressure is computed as 1,628,144.2 N/m2 - (998.2 kg/m3)(9.8066 m/s2)(50 m) = 1,138,696.8 N/m2, which is the same (barring round-off error).

Pmax and Pmin at multiple locations along the same pipe
Pmax or Pmin may occur at multiple locations along a pipe even though the calculation only indicates one location. Usually this happens only when the computed Pmax or Pmin has the same value as H (can be detected if reservoir elevation and pressures have the same units). Splitting one pipe into two pipes can provide additional values of Pmax and Pmin and their locations for verification.

Discharge and Friction Factor
The valve is located at the end of the last pipe and discharges to the atmosphere. The "steady Q at Y=100%" field that you enter is used to compute the initial loss across the valve if the valve is initially open (i.e. if the valve is being closed) and to compute the friction factor (f) for both the valve opening and closing scenarios. Chaudhry (1987, p. 37) indicates that accuracy is sufficient if the maximum steady state flowrate is used to compute f, rather than re-computing f every distance and time increment as flowrate changes.

Valve
The valve minor loss coefficient (K) only needs to be entered if the valve is being opened. If the valve is being closed, then the reservoir elevation (H), atmospheric pressure boundary beyond the valve, and Qmax provide sufficient information to compute pressures.
If the valve is only going to be opened half-way, enter K for a half-open valve. If the valve will be opened completely, enter K for a fully open valve.
To determine the valve curve: mathematically the valve curve is Y(t) = 100 [K/K(t)]1/2. However, physically it may be very difficult to determine how K changes with time as the valve is opened or closed. A method for computing K(t) is to run several steady state flow tests through the piping system with different valve openings: measure pressure (P) just upstream of the valve and flowrate (Q) for each valve setting. Then,

where A is the pipe area (not the flow area inside the valve).
If it is more convenient, using Y(t)=Av(t)/Av is an adequate approximation for the valve curve - where Av is the valve area through which water flows (not the pipe area). If finding Av(t) or K(t) is too difficult, you can enter various valve curves and see if there is much difference in the computed pressures.

Courant Stability Criterion
The Courant stability criterion shown in equations is a necessary criterion for numerical solutions that vary in space and time, like the water hammer calculation. Our program automatically satisfies this criterion. However, satisfying the criterion can result in reducing the number of distance intervals (sections) into which each each pipe is split. This can decrease accuracy. There is a trade-off between time intervals and distance intervals with the product being 100,000, which we set to maximize accuracy while keeping the calculation fast. For 2 or 3 digit accuracy, try to have at least five pipe sections in each pipe (the number of pipe sections is output next to the pipe number when the calculation is run). 50 sections gives about 6 digit accuracy. The calculation will run if there are at least 2 sections in each pipe. Reducing Tmax will increase the number of pipe sections.


FLUID PROPERTIES, PIPE PROPERTIES, MINOR LOSS COEFFICIENTS, and PIPE WALL THICKNESS Top of Page

Fluid Properties
Fluid density, viscosity, and elastic modulus provided by the drop-down menus in the calculation, have been compiled from the closed conduit pipe flow references shown on our literature web page.

Pipe Properties
Pipe material roughness, provided by the drop-down menus in the calculation, have been compiled from the closed conduit pipe flow references shown on our literature web page. Pipe elastic moduli have been compiled from the references shown below.

Minor Loss Coefficients References for K values

Valve Minor Loss Coefficient, K Valve Minor Loss Coefficient, K
Globe, 100% open 10 Angle, 100% open 5
Gate, 100% open 0.2 Ball, 100% open 0.05
Gate, 75% open 1.2 Ball, 67% open 5.5
Gate, 50% open 5.6 Ball, 33% open 210
Gate, 25% open 25


Pipe Wall Thickness (adapted from Glover, 1996)
Glover (1996) provides a fairly complete list of pipes and their wall thickness. The tables below are a sampling adapted from Glover. Additional listings of wall thickness can be found on the web at sites such as J.B. Smith, Ductile Iron Pipe, Ohio Department of Transportation: Steel, aluminum, and plastic pipe; Concrete and clay pipe.

Plastic Pipe: PVC & CPVC Schedule 40
Plastic Pipe: PVC & CPVC Schedule 80

Nominal size (inch) Actual inside diameter (inch) Wall thickness (inch) Nominal size (inch) Actual inside diameter (inch) Wall thickness (inch)
0.5 0.622 0.109 0.5 0.546 0.147
1 1.049 0.133 1 0.957 0.179
2 2.067 0.154 2 1.939 0.218
4 4.026 0.237 4 3.826 0.337
8 7.981 0.322 8 7.625 0.500
12 11.938 0.406 12 11.376 0.687

Steel Pipe: Schedule 40

Nominal size (inch) Actual inside diameter (inch) Wall thickness (inch) Nominal size (inch) Actual inside diameter (inch) Wall thickness (inch)
0.5 0.622 0.109 1 1.049 0.133
2 2.067 0.154 4 4.026 0.237
8 7.981 0.322 12 11.938 0.406
16 15.000 0.5 20 19.250 0.375
24 22.626 0.687 30 28.500 0.750
36 34.500 0.750 42 40.5 0.750

Copper Pipe: Type K
Copper Pipe: Type L

Nominal size (inch) Actual inside diameter (inch) Wall thickness (inch) Nominal size (inch) Actual inside diameter (inch) Wall thickness (inch)
0.5 0.527 0.049 0.5 0.545 0.040
1 0.995 0.065 1 1.025 0.050
2 1.959 0.083 2 1.985 0.070
4 3.857 0.134 4 3.905 0.110
8 7.583 0.271 8 7.725 0.200
12 11.315 0.405 12 11.565 0.280


ERROR MESSAGES given by calculation Top of Page
"Q, T's, Res, Moduli, Dens, Visc must be >0". Flowrate, times, reservoir elevation, elastic moduli of pipe and fluid, fluid density, and viscosity must be positive.
"K must be >0". Valve minor loss coefficient must be positive.
&quot;Ymid must be: 0<Ymid<100%&quot;. The intermediate point on the valve curve must be in this range.
&quot;Tmid must be: 0<Tmid<Tvlv&quot;. The intermediate time on the valve curve must be greater than 0 but less than the valve opening (or closure) time.
&quot;Pipe 1: L, D, Wall Thick must be >0&quot;. Pipe length, diameter, and wall thickness must be positive.
&quot;Pipe 2: L, D, Wall Thick must be >0&quot;. Same as above.
&quot;Pipe 3: L, D, Wall Thick must be >0&quot;. Same as above.
&quot;Tmax must be >1e-10 sec&quot;. Pressures must be computed for greater than 10-10 sec.
&quot;Re or e/D out of range&quot;. Friction factor (f) can only be computed if Reynolds number and roughness/diameter ratio are in the range of the Moody diagram.
&quot;Reduce Tmax&quot;. Tmax (maximum time for computation of pressures) must be reduced in order to have sufficient distance sections (i.e. intervals, reaches). There is a trade-off between time intervals and distance intervals with the product being 100,000. In all cases, the program will be sure that the Courant stability condition is satisfied. For decent accuracy, try to have at least five pipe sections in each pipe (the number of pipe sections is output next to the pipe number when the calculation is run). The calculation will run if there are at least 2 sections in each pipe. If there are less than 2 sections per pipe, the message will be displayed.
&quot;Tmax too low&quot;. This message rarely occurs but has to do with the precision of the calculation. We use double precision (the most precise) but idiosyncrasies can still occur. Increase Tmax to eliminate this message.
&quot;Reservoir elev too low&quot;. The reservoir elevation is too low for the flow and valve settings to be physically achievable.
&quot;Warning: Pmin<Patm/2&quot;. At least one of the computed minimum pressures is less than half of atmospheric pressure (-101325/2 N/m2 gage). The equations used in the calculation don't care if pressure drops below 0.0 absolute, even though in reality such pressures can never occur. The warning message indicates that the results are not accurate below about half of atmosheric pressure. Note that Pmin and Pmax are gage pressures (relative to atmospheric pressure).

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