{"id":2806,"date":"2020-05-05T10:32:40","date_gmt":"2020-05-05T15:32:40","guid":{"rendered":"http:\/\/www.jmcampbell.com\/tip-of-the-month\/?p=2806"},"modified":"2020-05-05T10:32:40","modified_gmt":"2020-05-05T15:32:40","slug":"piping-vibration-causes-limits-remedies","status":"publish","type":"post","link":"https:\/\/www.jmcampbell.com\/tip-of-the-month\/2020\/05\/piping-vibration-causes-limits-remedies\/","title":{"rendered":"Piping Vibration: Causes, Limits & Remedies"},"content":{"rendered":"

Piping vibration is a major cause of concern in process plants, particularly in the oil and gas industry where the loss of containment could be catastrophic. This Tip of the Month explains the root causes of piping vibration, natural frequencies and how they may be changed using appropriate structural supports and layouts.<\/em><\/p>\n

 <\/p>\n

Imagine sitting in your office having a well-earned cup of coffee. The phone rings and on the other end is an irate production supervisor screaming that the plant is about to explode because the new section of pipe you installed last week is galloping up and down. Vast amounts of natural gas will be released into the plant when the pipe breaks.\u00a0 You calm him down and arrange to quickly take the section of pipe out of service. Now that everything is safe, you must figure out what went wrong. We are embarking on a voyage of discovery, that entails us calling at the following way stations:<\/p>\n

PART\u00a01<\/strong>\u00a0\u2013 Calculate the natural frequency of a pipe<\/p>\n

PART 2<\/strong>\u00a0\u2013 Calculate VIV (Vortex Induced Vibration) affecting the pipe<\/p>\n

PART 3<\/strong>\u00a0\u2013 Calculate the effect of flow induced vibration as flow rates change<\/p>\n

PART 4<\/strong>\u00a0\u2013 Determine the severity of the vibration: Is it acceptable or does it need modifications?<\/p>\n

 <\/p>\n

<\/h3>\n

PART\u00a01: NATURAL FREQUENCY OF A PIPE<\/strong><\/h3>\n

<\/h3>\n

Why do pipes gallop?<\/em><\/h3>\n

Maybe galloping isn\u2019t the right word; I would prefer to say the pipe is vibrating.\u00a0 But why is it vibrating so much?\u00a0 One word \u2013\u00a0resonance<\/strong>.<\/p>\n

 <\/p>\n

If you seat a child on a swing and you give a gentle nudge to the swing every time the child swings back towards you, you will quickly be able to get the swing higher and higher.\u00a0 You only need a very small force because you waited until the swing had reached its limit of movement (the limit of the swing) and only then did you give that gentle push.\u00a0 But you did it EVERY time the swing moved back towards you.\u00a0A couple of interesting things are happening here.<\/p>\n

1.\u00a0<\/strong>You are pushing at\u00a0exactly<\/u>\u00a0the same frequency as the frequency at which the child is swinging.\u00a0 Don\u2019t forget that if you are troubleshooting a pipe vibration problem the natural frequency will depend not only on the steel of the pipe but also the mass of the fluid inside.<\/p>\n

2.\u00a0<\/strong>You are pushing at the same position in the swing cycle \u2013 in other words you have \u201clocked phase\u201d with the swing.<\/p>\n

 <\/p>\n

The occurrence of these two \u201cinteresting\u201d points means you are in RESONANCE with the swing.\u00a0 When a pipe is vibrating heavily it is almost always because there is a resonance issue.\u00a0 The swing has a natural frequency because it has a period of oscillation that depends on the mass of the swing and the length of the pendulum.\u00a0 This is the reason why a pendulum is used on a clock.\u00a0 If you can work out the period of the cycle back and forth, then you know the frequency.<\/p>\n

 <\/p>\n

If the time period of oscillation is one second, then the frequency is one per second or 1Hz.\u00a0 If the time period is half a second, then the frequency is 2Hz and so on.\u00a0 By the same reasoning the pipe has a natural period of oscillation and so it has a natural frequency.\u00a0 The natural frequency of the pipe depends on its stiffness and its mass; the stiffer the pipe the higher the frequency, the more mass the pipe (including contents) has, the lower the natural frequency.<\/p>\n

 <\/p>\n

To calculate the natural frequency of a pipe with rigid supports use the following formula:<\/p>\n

<\/p>\n

Where:<\/strong><\/p>\n

fn<\/sub><\/strong>\u00a0= natural frequency of the pipe (Hz)
\nE<\/strong>\u00a0= Young\u2019s modulus of elasticity (200 GPa or 30E6 psi for steel \u2013 approximately but close enough)
\nI<\/strong>\u00a0= 4th<\/sup>\u00a0polar moment of inertia for the pipe (0.049*[OD4<\/sup>-ID4<\/sup>]) in inches or metres
\n\u00b5<\/strong>\u00a0= mass per unit length of the pipe (remember to include the mass of the fluid) lbs\/inch or kg\/m
\nL<\/strong>\u00a0= distance between pipe supports (inches or metres)<\/p>\n

 <\/p>\n

Let\u2019s find the natural frequency of a 12\u201d (300 mm) pipe made of A-106 GrB schedule 80 that is first empty and then filled with water.\u00a0 I\u2019ll use SI units to make the math easier.\u00a0 Pipe supports are 5 m apart.<\/p>\n

 <\/p>\n

Given:<\/strong><\/p>\n

Pipe OD<\/strong>\u00a0= 323.8 mm
\nPipe ID<\/strong>\u00a0= 288.84 mm
\nMass\/length empty<\/strong>\u00a0= 132.05 kg\/m
\nMass\/length full<\/strong>\u00a0= 132.05 + \u03c0 (0.28884)2<\/sup>\/4 * 1000 = 197.57 kg\/m
\nE<\/strong>\u00a0= 200 x 109<\/sup>\u00a0Pa<\/p>\n

 <\/p>\n

Calculate I:<\/strong><\/p>\n

<\/p>\n

 <\/p>\n

 <\/p>\n

Natural Frequency:<\/strong><\/p>\n

<\/p>\n

 <\/p>\n

Now you know the natural frequency of the pipe you ask your vibration techs to take a vibration measurement on the pipe when it\u2019s in operation.\u00a0 If the frequency is the same as your calculated frequency, then the pipe has a resonance problem and the next step is to identify what is the force that is exciting the natural frequency.<\/p>\n

 <\/p>\n

 <\/p>\n

PART\u00a02 \u2013 CALCULATE VIV (VORTEX INDUCED VIBRATION)<\/strong><\/h3>\n

In part 1 we determined the natural frequency of the pipe so now we know the pipe natural frequency, but you are not standing there pushing it \u2013 so what is?\u00a0It could be any of a number of things:<\/p>\n

\u25baVibration at the same frequency coming from a pump or compressor (usually speed related).\u00a0This could be caused by unbalance, misalignment, something may have come loose or just about any fault on a machine that causes a vibration at the same frequency as the natural frequency.<\/p>\n

\u25baFlow induced vibration.\u00a0Now, this could be from the internal flow of the fluid through the pipe or even from wind flowing across the outside of the pipe.\u00a0We have all seen the effect of wind induced vibration on street lamps or poles so when the wind hits a particular speed the pole starts to sway.\u00a0That\u2019s because the oscillation force associated with the wind is a function of the pipe outside diameter and the wind speed so the wind vortexes or swirls on the downwind side of the pipe and the vortexing induces KARMAN vibration.<\/p>\n

Check this out<\/a> for more information.<\/p>\n

 <\/p>\n

I usually start with the easiest option.\u00a0Have a look around the pipe and see if there is any rotating equipment that has a run speed (or a harmonic of run speed) that is very close to the pipe natural frequency.\u00a0If there is can either change the speed of the machine (if possible, because that is the easiest option) or change the natural frequency of the pipe.\u00a0An easy way to change the natural frequency of the pipe is change the value of L \u2013 in other words change the location of the pipe supports or maybe just add another support.\u00a0If you add another support be careful that you put it at a location of high amplitude \u2013 in other words an antinode of vibration.<\/p>\n

 <\/p>\n

You can see from the following image that the modifier we used (a=22.4) is only applicable to the first mode of vibration of a \u201cclamped-clamped\u201d beam or pipe.\u00a0If you install another pipe support halfway between existing supports but the pipe is vibrating at the second mode (a=61.7) the amplitude of vibration will be unaffected.\u00a0I usually hammer in a stout piece of wood as a (very) temporary measure to see if that indeed reduces the vibration.<\/p>\n

 <\/p>\n

If you find that there is no rotating equipment nearby that could affect your pipe we could see if there are any other possible causes and an easy one to check is vortex induced vibration.<\/p>\n

 <\/p>\n

For the same reason that a flag flutters, a pipe (or any object) will experience an oscillatory force when placed in a fluid flow.\u00a0As the wind flows across the pipe there are tiny differences in air pressure from one external side of the pipe to the other, so the wind finds slightly less resistance on one side and more wind flows towards the lower pressure.\u00a0As more wind flows towards the lower pressure side that side experiences an increase in air pressure so the flow of wind flips over to the other side. The other side experiences the increase in air pressure so the flow flops back again.\u00a0 The flow of wind is now flip flopping back and forth causing a transverse oscillating force on the pipe.<\/p>\n

 <\/p>\n

<\/p>\n

\"\"
By Cesareo de La Rosa Siqueira – http:\/\/www.mcef.ep.usp.br\/staff\/jmeneg\/cesareo\/vort2.gif, Copyrighted free use<\/figcaption><\/figure>\n

 <\/p>\n

 <\/p>\n

Going back to our flag analogy, the pipe flutters as the wind passes the flagpole and vortexes on the downwind side.\u00a0The vortex is traveling along the flag and the flag \u201cflutters.\u201d<\/p>\n

 <\/p>\n

We, though, are interested in what is happening to our pipe.\u00a0According to Strouhal and Karman there is a distinct relationship between the speed of the wind, the diameter of the pipe and the frequency of the oscillating force.<\/p>\n

 <\/p>\n

St = fD \/ V<\/strong><\/p>\n

Where:<\/strong><\/p>\n

f<\/strong>\u00a0<\/strong>is the frequency (Hz)
\nD<\/strong>\u00a0is the diameter of the pipe
\nV<\/strong>\u00a0is the wind velocity.<\/p>\n

St<\/strong>\u00a0is the Strouhal number.\u00a0This does vary somewhat with Reynolds number, but we can assume it to be 0.22.<\/p>\n

 <\/p>\n

So, if we have a wind velocity of 10 m\/s and we know that our pipe has an OD of 0.3238 m (see Part 1 where we calculated natural frequency) the VIV frequency is 0.22*10\/0.3238 = 6.79 Hz.\u00a0 Easy isn\u2019t it?<\/p>\n

 <\/p>\n

However, the vortexing frequency effect is not limited to external wind.\u00a0 You will get the same effect from fluid flowing inside the pipe as it flows across an obstruction.\u00a0 So if we have a gate valve with a non-rising stem with a stem diameter of 3.5 cm and a fluid flow of 10 m\/s we would have a vortexing frequency of 62.8 Hz.\u00a0 Remember that our pipe has a natural frequency of 63.77 Hz which is close enough to ensure resonance.<\/p>\n

 <\/p>\n

In our next article, we will examine the effect of changing the fluid velocity and see how that affects resonance in our pipe.<\/p>\n

 <\/p>\n

PART\u00a03 \u2013 VIBRATION FROM FLOW VELOCITY<\/strong><\/h3>\n

In Part 1 we figured out the natural frequency of a pipe and in Part 2 we looked to see if the resonance excitation was from a nearby rotating equipment or perhaps vortex induced vibration.\u00a0 If neither of these options came close to identifying the forcing frequency, we need to look at slightly more exotic causes.<\/p>\n

 <\/p>\n

Most people who work in process, power, oil & gas or refining will have come across a problem in which a perfectly normal section of piping with no significant vibration \u201csuddenly\u201d starts to vibrate for no apparent cause apart from a slight change in flow rate.\u00a0But that doesn\u2019t seem to make any sense.\u00a0 We have made no change to the mass or stiffness of the pipe, so the natural frequency hasn\u2019t changed.\u00a0We have not changed any run speed of nearby equipment and even if we check VIV vibration it doesn\u2019t even come close to the problem frequency.\u00a0 What the heck could it be?<\/p>\n

 <\/p>\n

Let\u2019s take a trip down to the train station.\u00a0The express through train is coming down the track and as it passes us, we hear a definite change in pitch.\u00a0A high pitch as the train travels towards us and a lower pitch as the train moves further away.\u00a0We are talking about the DOPPLER effect.<\/p>\n

 <\/p>\n

To understand Doppler, we need a good understanding of noise and sound.\u00a0When we speak the sound, we make travels through the air at about 330 m\/s.\u00a0That doesn\u2019t mean we are expelling air from our mouth at that speed \u2013 that would be rather unpleasant.\u00a0As our voicebox vibrates it creates an area of high air density as it pushes onto the air molecules. That high-density rams into the air next to it and bounces back transmitting the energy to the adjacent air molecules.\u00a0The rate at which the energy is transmitted to adjacent air molecules is the speed of the sound.\u00a0Remember from high school physics class that the speed of sound is function of frequency and wavelength?<\/p>\n

 <\/p>\n

<\/p>\n

Where:<\/strong><\/p>\n

C<\/strong>\u00a0<\/strong>is the speed of sound in the fluid
\nf\u00a0<\/strong>\u00a0is the frequency of the sound
\n\u03bb\u00a0<\/em><\/strong>is the wavelength between the high-pressure pulsations of the sound<\/p>\n

 <\/p>\n

As a stationary observer of a stationary object that is making a noise the speed of sound is set by the air density and the air pressure.\u00a0The pitch or frequency is determined by the distance between the high-pressure pulsations.\u00a0So, let\u2019s see what happens if we start moving the object making the noise.<\/p>\n

 <\/p>\n

<\/p>\n

When our train is stationary the sound moves away from the train at 330 m\/s in all directions.\u00a0 The wavelength is the same as the same travels in all directions. When the train starts to move, we have the sound AND the train traveling in the same direction to the front of the train but moving in OPPOSITE directions when viewed from behind the train.\u00a0The speed of the sound hasn\u2019t changed but because the train and the sound are traveling in the same direction at the front of the train the wavelength is compressed.\u00a0 Using C = f\u03bb as the wavelength is compressed and the speed of sound stays the same then frequency must increase.\u00a0The opposite effect happens as the train moves away from us, so we hear a lower frequency or deeper tone.<\/p>\n

 <\/p>\n

Factor C is a physical value depending on the properties of the fluid.\u00a0 However, the wavelength is affected by the speed of the fluid flow through the pipe –\u00a0the Doppler effect<\/a>.<\/p>\n

 <\/p>\n

But we have a pipe with fluid traveling along the inside of the pipe.\u00a0What does this have to do with moving trains?\u00a0Only that in both cases we have to think about Doppler.<\/p>\n

 <\/p>\n

<\/p>\n

 <\/p>\n

Let\u2019s move to wind musical instruments.\u00a0When air is blown into a trumpet or trombone you only get a distinct tone if you blow into the mouthpiece at a particular rate \u2013 that is why trumpeters \u201cpurse\u201d their lips to get the right airspeed.\u00a0When you get the correct airspeed the wavelength becomes the same as the length of tubing, so you get a\u00a0standing wave<\/u><\/strong>\u00a0and the instrument sounds the desired note.\u00a0The length of the tubing can be adjusted on a trumpet using valves or on a trombone using the slide.\u00a0In effect, you are changing the wavelength of the air and that is changing the frequency.\u00a0All of us have\u00a0tried blowing across a part empty bottle and we get a tone if the speed of the blow is \u201cjust right.\u201d<\/p>\n

 <\/p>\n

The nearest our pipe is to the analogy of a trumpet is a section of pipe between two bends.\u00a0The half wavelength is actually longer than the distance between the bends (add about 15%).\u00a0The part open bottle analogy equates to a dead leg at a tee and in that case, we have a quarter wavelength.<\/p>\n

 <\/p>\n

So now let\u2019s combine the standing wave and the Doppler effect in our pipe.\u00a0Instead of a train moving we now have liquid or gas moving along the pipe.\u00a0In effect, we are causing the wavelength to change by changing the flow velocity relative to the speed of sound.\u00a0This can get us into trouble in one of two ways:<\/p>\n

1.<\/strong>\u00a0If you have a forcing vibration at the same frequency you get resonance of the fluid inside the pipe.\u00a0This happened to me once on a pump running at 2970 rpm discharging into a line that had a length between the discharge flange and the next tee that equated to the standing wave that had a frequency of 50 Hz.\u00a0Vibration at about 20 mm\/s rms and a bearing life of 3 months.\u00a0We changed the configuration of the discharge piping and vibration came down to less than 2 mm\/s and no more bearing failures.<\/p>\n

2.<\/strong>\u00a0If the frequency of the standing wave is close to the natural frequency of the mechanical section of pipe you have resonance.\u00a0This happened on the discharge of a large blower in China.\u00a0The distance between the discharge flange of the blower and the NRV gave a standing wave with a frequency that was very close to the natural frequency of the piping.\u00a0We moved the NRV and the vibration problem disappeared.<\/p>\n

 <\/p>\n

<\/p>\n

 <\/p>\n

Remember that standing waves can occur not only at fundamental frequency but also at \u201covertones\u201d.\u00a0 Don\u2019t ignore the overtones.<\/p>\n

 <\/p>\n

We still use C = f\u03bb to calculate the fundamental frequency so with a speed of sound of 330 m\/s and an end-corrected wavelength of 10 m we would have a fundamental tone of 33 Hz.\u00a0 But as soon as fluid flows through the pipe we have to modify that wavelength.\u00a0 So, if we flow at 33 m\/s that means a 10% change in apparent speed of sound so we would get a new frequency of (330+33)\/10 = 36.3 Hz.<\/p>\n

 <\/p>\n

Our pipe is carrying 22 MW natural gas at 80 \u05afC with a speed of sound of 365 m\/s.\u00a0 Our section of piping has a length of 8.5 m between bends which equates to an open\/open pipe of 8.5 m length.\u00a0\u00a0 which gives a wavelength of 17 m. Add 15% to that length for \u201cend effect\u201d correction and we get a fundamental tone or frequency of 365\/ (10*1.15) = 18.54 Hz.<\/p>\n

 <\/p>\n

But we now flow gas through the pipe at 20 m\/s which, when considering the Doppler effect, changes that frequency to 19.72 Hz.\u00a0 But what makes it really worrying is that the first overtone is 39.45 Hz, the second overtone is 49.31 Hz but the 3rd<\/sup>\u00a0overtone is 78.9 Hz which is rather close to our mechanical natural frequency of 78 Hz.<\/p>\n

 <\/p>\n

On very large diameter piping there is a possibility of shell wall resonance but that is quite rare and tends to happen on trunking rather than pipework (large diameter and thin wall) so we won\u2019t get into it here.<\/p>\n

 <\/p>\n

PART\u00a04 \u2013 LIMITS FOR PIPING VIBRATION<\/strong><\/h3>\n

Okay\u00a0\u2013 let\u2019s recap.\u00a0We have a pipe with a natural frequency and a force with the same frequency. This means high amplitude vibration.\u00a0So what?<\/p>\n

 <\/p>\n

If we leave a vibrating pipe in place long enough and the vibration is severe enough the pipe will develop a crack and we get a leak.\u00a0 We are talking about fatigue failure.\u00a0 To make things easy for us there are several versions of fatigue limits we can apply to piping and the one I will mention is API STD 618.\u00a0 Now before you start jumping up and down complaining that is a standard for reciprocating compressors let me say that yes, you are right.\u00a0 But this section of the standard works for all steel piping because it is VERY conservative.<\/p>\n

 <\/p>\n

Let\u2019s look at some of the detail.<\/p>\n

Section\u00a07.9.4.2.5.2.4 Piping Design Vibration Criteria<\/strong><\/p>\n

The predicted piping vibration magnitude shall be limited to the following:<\/p>\n

\u25ba\u00a0A constant allowable vibration\u00a0amplitude<\/u><\/strong>\u00a0of 0.5 mm peak-to-peak (20 mils peak-to-peak) for frequencies below 10 Hz (the frequency of 10 Hz is also according to ISO 10816).<\/p>\n

\u25ba\u00a0A constant allowable vibration\u00a0velocity<\/u><\/strong>\u00a0of approximately 32 mm\/s peak-to-peak (1.25 in.\/s peak-to-peak) for frequencies between 10 and 200 Hz.<\/p>\n

\u2022\u00a0We need to be aware that 32 mm\/s pk-pk is the same as 16 mm\/s peak and 11.3 mm\/s rms.\u00a0 To convert displacement to velocity:<\/p>\n

\u2022\u00a0V = 2\u03c0. Displacement. Frequency<\/p>\n

\u2022 So\u00a00.5 mm at 10 Hz gives a velocity of 2\u03c0*0.5*10 = 31.42 mm\/s<\/p>\n

\u2022 Vibration measurements are almost always displayed in terms of peak-peak for displacement (total movement) but velocity readings are only ever shown as zero to peak or even rms so don\u2019t be confused by the API 618 limits. To see the relationship between pk-pk, peak and rms look at the image below.<\/p>\n

\n

<\/p>\n

 <\/p>\n

The rms. or\u00a0root mean square<\/em>\u00a0velocity is a quantitative measure of the effective velocity and reflects the power or energy being used to vibrate the machine mass.\u00a0 Peak value is the maximum amplitude seen during the measurement referenced to zero velocity and peak to peak is a measure of the total movement so is usually only used for displacement.<\/p>\n

<\/p>\n

 <\/p>\n

This image illustrates the movement of a pendulum.\u00a0 The figure above shows that at position B and C, the velocity is zero, and at position A the velocity is maximum, first to the right, then to the left.\u00a0 The negative peak velocity differs only in direction, not magnitude.\u00a0 The rate of change of displacement is the velocity, therefore if D is expressed in terms of mm, instead of the usual micron, then the product 2pfD will be the velocity in mm\/s which are the units used for velocity in vibration work.<\/p>\n

 <\/p>\n

This relationship between velocity and displacement is an important factor when considering severity of piping vibration \u2013 if you want to be accurate.\u00a0 I really don\u2019t care about vibration amplitudes or movement in pipes, what I care about is the stresses that have been imparted to the pipe.\u00a0 If we consider a resonant pipe, then the actual forces are quite low but the physical movement (or displacement) could be quite high.<\/p>\n

 <\/p>\n

<\/p>\n

 <\/p>\n

If we consider the stress-strain diagram for carbon steel,\u00a0A106GrB has a tensile stress specified at 415 MPa and yield stress of 240 MPa (see below).<\/p>\n

<\/p>\n

 <\/p>\n

We need to make sure the stress value due to the bending effect of the piping vibration is well below the yield value and we can use our high school physics knowledge\u00a0to do this.<\/p>\n

 <\/p>\n

<\/p>\n

Where:<\/strong><\/p>\n

E<\/strong>\u00a0= Youngs modulus<\/p>\n

D<\/strong>\u00a0= pipe outside diameter<\/p>\n

\u0394<\/strong>\u00a0= peak to peak displacement<\/p>\n

L<\/strong>\u00a0= length of pipe between supports<\/p>\n

So, let\u2019s say our 5 m pipe has a maximum vibration amplitude of 0.5 mm pk-pk.\u00a0 The bending stress is<\/p>\n

S = 8 * 200E9 Pa * 0.3238 m * 0.5 \/ 52<\/sup>\u00a0= 10.36 MPa<\/strong><\/p>\n

 <\/p>\n

That sounds quite reasonable but what happens if the vibration is 5 mm pk-pk?\u00a0 The stress is now 103 MPa.\u00a0Compare that to our yield stress of 240 MPa and you would think our pipe should be fine, but API 618 warns us that if there is a cyclic stress level in the piping that stress level must not exceed the endurance limits of the piping materials. So, what is the endurance level?\u00a0 It is the value of the stress below which a material can presumably endure an infinite number of stress cycles, that is, the stress at which the S-N diagram becomes and appears to remain horizontal. The existence of a fatigue limit is typical for carbon and low alloy steels.<\/p>\n

 <\/p>\n

Looking at the image below we have a plot of applied stress against the number of cycles the metal endured before rupturing of a carbon steel of a known UTS.\u00a0It is possible to predict (very approximately) the life to failure of the equipment if we have this plot.\u00a0Our A-106 pipe steel follows the dashed line fairly closely so we can see that if we have a stress of 103 MPa the steel would be expected to fail due to cyclic loading at about 200,000 cycles.<\/p>\n

 <\/p>\n

If our pipe is vibrating at 67 Hz that means 67 cycles per second, so we have a life before failure of 200,000 \/ 67 = 2958 seconds \u2013 less than an hour.<\/p>\n

<\/p>\n

To operate our pipe safely for an extended period then the vibration MUST induce a stress that is less than the endurance limit.\u00a0 And that is why the API recommendation is so conservative.\u00a0 But if you really need to operate the equipment you can carry out a fatigue analysis as we have done or you could use API 579-1\/ASME FFS-1 which can guide you through a step by step procedure to determine if you can use the pipe for an extended time.<\/p>\n

In conclusion, if your pipe is vibrating, you should do the following:<\/p>\n

\u25baFind out if the vibration is resonant \u2013 that means find the natural frequency<\/p>\n

\u25baIf the vibration is resonant find out what is the forcing frequency \u2013 that could be a nearby piece of equipment, external effects such as wind or fluid flow induced vibration<\/p>\n

\u25baDetermine if the vibration levels are acceptable \u2013 you could use API 618 for that or carry out a full fatigue analysis.<\/p>\n

 <\/p>\n

To remedy the problem, you simply have to separate the natural frequency from the forcing frequency.<\/p>\n

 <\/p>\n

To change the natural frequency, you must change either the stiffness or the mass.\u00a0\u00a0 It is usually much easier to change the effective mass by changing the length between supports either by moving one of the existing supports or by adding a new one.<\/p>\n

 <\/p>\n

Otherwise, you have to attack the forcing frequency.\u00a0 If it\u2019s a rotating machine you can change the speed, isolate the machine from the pipework or fix the vibration issue such as unbalance or misalignment.\u00a0 If the problem is flow induced or VIV you will have to do the analysis to positively identify the culprit and make changes in flow rate, piping arrangement or even production procedures.<\/p>\n

 <\/p>\n

 <\/p>\n

By: Ron Frend,\u00a0Head of Facilities Training<\/em><\/p>\n

<\/p>\n

\n

To receive Tips of the Month directly to your inbox, simply sign up below!<\/em><\/p>\n

I hope you find this series of posts useful. To learn more about similar cases and how to minimize operational troubles, we suggest attending our\u00a0ME41<\/strong>\u00a0(Piping Systems \u2013 Mechanical Design and Specification)<\/strong><\/u><\/a>\u00a0<\/strong>and\u00a0PF49<\/strong>\u00a0(Troubleshooting Oil and Gas Processing Facilities)<\/a>\u00a0courses.<\/p>\n