Piping Presentation

PIPING DESIGN: ROUTING, FLOW AND

STRESS ANALYSIS

A COMPLETE

INTRODUCTION

TO PIPING

COURSE DISCUSSIONS: • Process Calculations:

• Bernoulli. • Flow/Head. • Friction Losses.

• Pump Selection: • Positive Displacement vs Centrifugal. • Pump Curves. • Combination of Pumps.

• Pipe Stress Calculations: • Static. • Dynamic. • Caesar II.

• General Routing and Safety: • Guidelines for safe operation, stress

reduction and supporting of pipes. • Guidelines for flow control, safety

devices and maintenance.

BERNOULLI: In general Bernoulli’s law is given as follows: 𝑃 + ½ρ𝑣² + ρ𝑔𝑕 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

However, when we consider piping systems, we like to refer to each item as an equivalent height of static liquid or “head”. If we alter the equation for two points on a system,

and express it as a function of “head”, we arrive at the following: 𝑃1

ρ𝑔+

𝑣1²

2𝑔+ 𝑕1 =

𝑃2

ρ𝑔+

𝑣2²

2𝑔+ 𝑕2

If we take this one step further, and consider the effects of friction losses and pumps, we can obtain this equation:

𝑃1

ρ𝑔+

𝑣1²

2𝑔+ 𝑕1 + 𝑃𝑕 =

𝑃2

ρ𝑔+

𝑣2²

2𝑔+ 𝑕2 + 𝑘𝑣²

Where Ph represents the pressure “head” supplied by any pumps between the two points and kv² represents the total “head” loss between the two points under consideration.

EXAMPLE 1:

A1 = 0.2 m² V1 = 1 m/s P1 = 70 kPa H1 = 5 m ρ = 1000 kg/m³

A2 = 0.5 m² V2 = ? m/s P2 = ? kPa H2 = 7 m

Ignoring friction and pump heads for now:

𝑉2 = 𝑉1.𝐴1

𝐴2

= 0,4 𝑚/𝑠

Then using Bernoulli’s equation:

70 000

1000 ∗ 9,81+

1²

2 ∗ 9.81+ 5 =

𝑃2

1000 ∗ 9,81+

0,4²

2 ∗ 9,81+ 7

P2 = 50,8 kPa

EXAMPLE 2: V1 = 0 m/s P1 = 0 kPa H1 = 50 m ρ = 1760 kg/m³ k = 0,2

V2 = ? m/s P2 = 0 kPa H2 = 10 m D2 = 100 mm Q2 = ? m³/h

Using Bernoulli’s equation: 0

1760 ∗ 9,81+

0²

2 ∗ 9.81+ 50 + 0 =

0

1760 ∗ 9,81+

𝑉2²

2 ∗ 9,81+ 10 + 0,2 ∗ 𝑉2²

V2 = 14,015 m/s

𝑄2 = 3600 ∗ (14,015 ∗0,12

4∗ 3,14) = 396,064 𝑚3/𝑕

EXAMPLE 3:

V1 = 0 m/s P1 = 0 kPa H1 = 12 m ρ = 1200 kg/m³ k = 0,32

P2 = 0 kPa V2 = 5 m/s H2 = 60 m D2 = 75 mm Ph = ? m

Using Bernoulli’s equation: 0

1200 ∗ 9,81+

0²

2 ∗ 9.81+ 12 + 𝑃𝑕 =

0

1200 ∗ 9,81+

5²

2 ∗ 9,81+ 60 + 0,32 ∗ 5²

Ph = 57,274 m/s

FRICTION FACTOR “k”: In the previous examples, we used set values for k to determine the friction losses in the systems analysed. It is however not usually quite as simple to determine this value as it is dependant on quite a number of things: • Velocity of the fluid in the pipe. • Properties of the fluid in the pipe (such as the density and viscosity). • Diameter of pipe. • Shape, material and relative roughness of pipe. • Length of pipe. • Number of bends and turns in the piping system. • Equipment such as control valves and flow meters in the pipe. • Fabrication tolerances. • Local effects in the piping system (such as possible damage, low pressure zones

and hot spots). • Age of piping system. • Possible build up or obstructions in the flow of the pipe.

We are going to run through a few of these scenario’s and create some guidelines to estimate as accurately as possible what the friction losses in a piping system may be.

STRAIGHT SECTIONS OF PIPE AND THE DARCY-WEISBACH EQUATION:

There have been many estimates over the centuries of liquid transportation to determine the exact amount of flow in a section of pipe or trench. Some of these methods date back to the Roman period. Most of these were empirical estimates, even though some provided near accurate results. It was only in 19th century that Julies Weisbach and Henry Darcy proposed the derived formula we know today:

𝑕𝑓 = 𝑓𝐷 ∗𝐿

𝐷∗

𝑣²

2𝑔

In this equation the section we can now substitute:

𝑘 = 𝑓𝐷 ∗𝐿

𝐷∗

1

2𝑔

Many documents however refer to the more common form:

𝐾 = 𝑓𝐷 ∗𝐿

𝐷

Which allows easier calculations in the Bernoulli equation where:

𝑕𝑓 = 𝐾 ∗𝑣2

2𝑔

This allows for simplification with the kinetic energy term: 𝑣2

2𝑔

FRICTION COEFFICIENT fD:

It is important to note that, even though the Darcy-Weisbach equation is derived from first principles, the friction factor (fD) is usually empirical in nature. The best estimation currently available today is obtained using the Moody diagram where the Reynolds number and relative roughness values are calculated and related to a graphical chart to determine the friction factor. The Reynolds number can be calculated as:

𝑅𝑒 =ρ𝑣𝐷

𝑢

Where 𝑢 is the dynamic viscosity of the fluid. The relative roughness is given as 𝑒/D where 𝑒 is the absolute roughness of the material.

SOME TYPICAL ABSOLUTE ROUGHNESS VALUES:

Surface

Absolute Roughness Coefficient

- 𝑒 - (m) 10-3 (feet)

Copper, Lead, Brass, Aluminium (new) 0.001 - 0.002 3.33 - 6.7 10-6

PVC and Plastic Pipes 0.0015 - 0.007 0.5 - 2.33 10-5

Stainless steel 0.015 5 10-5

Steel commercial pipe 0.045 - 0.09 1.5 - 3 10-4

Stretched steel 0.015 5 10-5

Weld steel 0.045 1.5 10-4

Galvanized steel 0.15 5 10-4

Rusted steel (corrosion) 0.15 - 4 5 - 133 10-4

New cast iron 0.25 - 0.8 8 - 27 10-4

Worn cast iron 0.8 - 1.5 2.7 - 5 10-3

Rusty cast iron 1.5 - 2.5 5 - 8.3 10-3

Sheet or asphalted cast iron 0.01 - 0.015 3.33 - 5 10-5

Smoothed cement 0.3 1 10-3

Ordinary concrete 0.3 - 1 1 - 3.33 10-3

Coarse concrete 0.3 - 5 1 - 16.7 10-3

Well planed wood 0.18 - 0,9 6 - 30 10-4

Ordinary wood 5 16.7 10-3

EXAMPLE 4:

We can now calculate the friction loss in a pipe section of uniform shape, size and velocity:

D = 114 mm L = 700 m V = 2.5 m/s P = 998 kg/m³ 𝑢 = 0,001 Ns/m² 𝑒 = 0,001 m

𝑅𝑒 =ρ𝑣𝐷

𝑢= 998 ∗ 2,5 ∗

0,114

0,001= 284430

Relative Roughness =𝑒

𝐷=

0,001

0,114= 0,008772

𝑓𝐷~0,039

𝑕𝑓 = 𝑓𝐷 ∗𝐿

𝐷∗

𝑣2

2𝑔= 0,039 ∗

700

0,114∗

2,52

2 ∗ 9,81= 76,28 m = 746,8 kPa

ELBOWS, REDUCERS & TEE’S: Due to the change in flow direction, velocity and/or obstructions found inside pipe fittings, there is usually an additional head loss above and beyond the standard friction loss. The simplest approach to solving the additional loss is to use documented “equivalent length” sections of pipe. For example the head loss in a 90° bend in a 4” pipe with screwed fittings can be equated to the same amount of loss experienced in a section of straight pipe of length of 4m. Consider the following example of two pipes of 3m. One line has an expansion loop of 1m while other does not:

If we equate that the 4 elbows each have an equivalent length of 4m and there are 5 straight sections of (just less than) 1m each, then the total effective length of the section with the expansion loop is roughly 19,6m, while the straight section is only 3m effective length. The losses in the top pipe would thus be roughly 6,5 times higher than in the bottom pipe.

ELBOWS, REDUCERS & TEE’S:

Typical values for standard pipe fittings and their equivalent lengths are commonly available and can be easily obtained using resources like the internet. Below is just one example for flanged fittings. Source: http://www.engineeringtoolbox.com

VALVES, FLOWMETERS AND OTHER EQUIPMENT: The actual measured losses in a valve can usually be obtained from the supplier either directly or through the use of their online information, brochures and catalogues. See below excerpt from a valve supplier: Source: http://www.velan.com

VALVES, FLOWMETERS AND OTHER EQUIPMENT:

The flow factor 𝐾𝑣 (or 𝐶𝑣) expresses the amount (in m³/h) of water that can flow through a valve with a pressure drop of 1 bar. It can be equated as follows:

𝐾𝑣 = 𝐹𝑆𝐺

∆𝑃

𝐾𝑣 = 0,865𝐶𝑣

Where: 𝑆𝐺 = Specific Gravity ∆𝑃 = Difference in Pressure 𝐹 = Rate of Flow

Note that this value of 𝐾𝑣 cannot be directly substituted into the Bernoulli equation. It can be used in the Bernoulli equation by the following transformation:

𝐾 =162𝑒6𝜋²𝐷4

𝐾𝑣²

Easier to use (but less accurate) equivalent lengths can also be found through online sources as per standard pipe fittings.

FULL SYSTEM EVALUATION – FRICTION LOSS: Consider a system of 200NB pipe of length 150m, followed by three sections of 150NB of length 50m:

Mas Flow 350 m³/h

Viscosity 0,001 Ns/m²

Temperature 19 °C

Density 998 kg/m³

Section Description Qty Diameter

(m) Velocity

(m/s) Effective

Length (m) Re Relative

Roughness Friction Factor K

Head Loss (m)

1 200NB Pipe 1 0,202742 3,011525 150 609342,7 0,00044391 0,0218 16,12884 7,455507

2 200x150 Reducer 1 0,202742 3,011525 1,3 609342,7 0,00044391 0,0218 0,139783 0,064614

3 150NB Pipe 3 0,154076 5,214416 150 801809,5 0,00058413 0,0212 20,63917 28,60253

4 150NB Elbow 2 0,154076 5,214416 5,4 801809,5 0,00058413 0,0212 0,74301 1,029691

37,1524

FULL SYSTEM EVALUATION – BERNOULLI: If we now consider the very same system from the previous slide with the following additional information: Point A (Reservoir): Point B (Fire Hose): H1 = 150 m P2 = Atmosphere P1 = 200 kPa D2 = 0,08 m What must be the height of the fire hose exit to produce the 350m³/h required if there is no pump? Using Bernoulli’s equation from the information we have:

𝑃1

ρ𝑔+

𝑣1²

2𝑔+ 𝑕1 + 𝑃𝑕 =

𝑃2

ρ𝑔+

𝑣2²

2𝑔+ 𝑕2 + 𝑘𝑣²

200 000

9,81 ∗ 998+

0

2 ∗ 9,81+ 150 + 0 =

0

9,81 ∗ 998+

19,342²

2 ∗ 9,81+ 𝑕2 + 37,15

𝑕2 = 114,21 m

FULL SYSTEM EVALUATION – SYSTEM CURVE:

In most cases, the system information is well known such as the static heads, static pressures and the velocities at various points. We have also shown that using this information, we can easily calculate the friction losses. What needs to be calculated still is the pumping requirements. If we rewrite Bernoulli’s equation to isolate the pumping requirements, we can obtain the following:

∆𝑃

ρ𝑔+ ∆𝑕 +

∆𝑣²

2𝑔+

𝐾𝑣²

2𝑔= 𝑃𝑕

If we evaluate the equation above, we realise that there are two components to the required head of a pump:

∆𝑃

ρ𝑔+ ∆𝑕 = Static Head

∆𝑣²

2𝑔+

𝐾𝑣²

2𝑔= Dynamic Head

SYSTEM CURVE CONTINUED: If we would calculate the static and dynamic heads for a given system at various flow rates, we would obtain a result similar to the following:

Where the 500m head at zero flow represents the static head, and the dynamic head ranges from 0 to 1300 going up exponentially.

COMBINED SYSTEM CURVES: Consider the following problem:

A B

C

D

Two end points, C and D, are commonly fed through a single line AB. Drawing up a system curve for the combined system (or even just solving the flow rates for a single instance) would require the solution of multiple simultaneous equations. For this system, it would be relatively simple, but what if we start having complex systems? Drawing up system curves simplifies the matter greatly.

COMBINED SYSTEM CURVES: Let us consider each system curve individually:

COMBINED SYSTEM CURVES (PARALLEL): If we combine the two systems BC & BD, we find that the volume flow will increase as the sum between the two curves, while the head remains the same.

COMBINED SYSTEM CURVES (SERIES): If we now combine the two systems (BC & BD) + AB, we find that the head will increase as the sum between the two curves, while the volume flow remains the same. This provides us with a final system curve as follows:

EXAMPLE 5: Using the system curves from the previous slides: If we have a series of pumps that can supply liquid at a head of 3000 m, how much liquid can we expect to receive at both points C and D respectively? From the final system curve, we find that at 3000 m head, the system supplies 145,91 m³/h. From the system curve BD & BC, we find that at a flow rate of 145,91 m³/h, the system requires a head of 1174 m. From system curve BD, we find that at a head of 1174 m, the system supplies a total of 73,91 m³/h to point D. From system curve BC, we find that at a head of 1174 m, the system supplies a total of 72 m³/h to point C.

SELECTING A PUMP: POSITIVE DISPLACEMENT VS KINETIC

Positive Displacement Pump. • A positive displacement pump operates by

forcing liquid through a confined space through mechanical means.

• A positive displacement pump delivers a constant volume of liquid based solely on the operating speed and size of the pump.

• The head to be supplied by the pump does not greatly affect the volume supplied, but directly influences the power requirement.

• A positive displacement pump has reduced flow with lower viscosity fluids.

• Typical Positive Displacement Pumps: • Reciprocating pumps (piston, plunger

etc.) • Power pumps. • Steam pumps. • Rotary pumps (gear, lobe screw etc.)

Kinetic Pump. • A kinetic pump operates on the principle of

kinetic/centrifugal energy. Pressure is generated by increasing the velocity of the liquid.

• A kinetic pump supplies a varying amount of fluid based on the operating speed and size of the pump and the pressure required.

• The power required to operate a kinetic pump does not vary as greatly as with a positive displacement pump as the flow decreases at higher pressures.

• A kinetic pump has reduces flow with higher viscosity fluids.

• Typical Kinetic Pumps: • End suction • In-line pumps • Submersible pumps • Double suction pumps • Axial flow pumps

SELECTING A PUMP: POSITIVE DISPLACEMENT PUMPS

Positive Displacement

Pumps

Reciprocating

Steam

Power

Controlled Volume

Plunger

Piston

Diaphragm Blow Case

Rotary

Vane Flexible Member

Lobe Gear Circumferential

Piston Radial Piston

Screw

SELECTING A PUMP: KINETIC PUMPS

Kinetic Pumps

Centrifugal

Overhung Impeller

End Suction In-line

Impeller Between Bearings

Axial Split Case

Radial Split Case

Turbine

Deep Well Turbine

Barrel Pump Axial Flow

Regenerative Turbine

Special

Reversible Centrifugal

Rotating Casing

PUMP CURVE:

QH – Pump Delivery

ETA – Efficiency

P – Pump Power

NPSH – Net Positive Suction Head

Source: http://www.bidgeepumps.com.au

QH:

Source: http://www.calpeda.com

QH CONTINUED:

Source: http://www.pumpscout.com

QH CONTINUED:

Source: http://www.pumpindustry.com.au

PUMP POWER:

Source: http://www.splitflowpumps.com

Pump power can be calculated as:

𝑃 =𝑄. 𝑕. 𝑔. 𝜌

𝐸𝑓𝑓

• Always remember units. • This does not include motor eff.

NET POSITIVE SUCTION HEAD: Always remember: Pumps do not suck!

𝑆𝑡𝑎𝑡𝑖𝑐 𝐻𝑒𝑎𝑑 + 𝐿𝑖𝑞𝑢𝑖𝑑 𝐻𝑒𝑎𝑑 > 𝑁𝑃𝑆𝐻 OR 𝑆𝑡𝑎𝑡𝑖𝑐 𝐻𝑒𝑎𝑑 + 𝐿𝑖𝑞𝑢𝑖𝑑 𝐻𝑒𝑎𝑑 − 𝑁𝑃𝑆𝐻 > 0

𝑺𝒕𝒂𝒕𝒊𝒄 𝑯𝒆𝒂𝒅:

Any pressure that acts on the source of the liquid being pumped.

𝑆𝑡𝑎𝑡𝑖𝑐 𝐻𝑒𝑎𝑑 = 𝐴𝑡𝑚𝑜𝑠𝑝𝑕𝑒𝑟𝑖𝑐 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 + 𝐺𝑢𝑎𝑔𝑒 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒.

This value will always be positive.

𝑳𝒊𝒒𝒖𝒊𝒅 𝑯𝒆𝒂𝒅 :

The difference in pressure due to the height and velocity of the liquid.

𝐿𝑖𝑞𝑢𝑖𝑑 𝐻𝑒𝑎𝑑 = 𝑕 −𝑣2

2𝑔− hVP − hf

The value of 𝑕 will be positive if the inlet of the pump is located below the liquid level and negative if the pump inlet is located above the liquid level.

PUMP COMBINATIONS:

PUMP COMBINATIONS EXAMPLE: Propose the following system curve:

If we use the first pump from the previous page, how much volume flow can we expect?

PUMP COMBINATIONS EXAMPLE:

Answer: 141 m³/h at a head of 380 m.


If the flow for the system we just analysed is 141,42 m³/h and the head is 380 m, what would the expected flow and head be for: a) Two pumps in parallel. b) Two pumps in series.

a) If two identical pumps are connected in parallel, the flow is doubled while the head remains the same: Therefore the flow is 282,82 m³/h and the head is 380 m.

b) If two identical pumps are connected in series, the head is doubled, while the flow is remains the same: Therefore the flow is 141,42 m³/h and the head is 760 m.

We must ALWAYS consider the system curve when evaluating the pumping characteristics.


a) For two pumps in parallel: Q = 161,89 m³/h; H = 460,7 m. b) For two pumps in series: Q = 187,61 m³/h; H = 557,6 m.

AFFINITY LAWS: Consider we have a pump that is delivering either too much or too little flow through our system. In order to alter it we can: a) Change the system to obtain a new system curve. b) Change the pump. c) Change the speed/impellor size of the pump.

Specific to Option C: if we use the same pump casing and impellor design but only change the speed or impellor size: take note the following: 𝑄1

𝑄2

=𝑛1

𝑛2

𝐷1

𝐷2

𝐻1

𝐻2

=𝑛1

𝑛2

2 𝐷1

𝐷2

2

𝑃1

𝑃2

=𝑛1

𝑛2

3 𝐷1

𝐷2

3

Note: The most basic assumption in these equations is that the pump is still operating at the same efficiency. Once you have calculated the new operating points it is important that you double check the new efficiency of the pump.

PIPE FLEXIBILITY CALCULATIONS:

Source: http://www.grantadesign.com

STRESS INTENSIFICATION/FLEXIBILITY FACTORS:

PIPE FLEXIBILITY EXAMPLE: Consider the following example:

Point A is fixed. Point B is supported vertically. Point C has a load of 5000 N acting downwards. Pipe is 4” schedule 40 stainless steel throughout. In order to simplify the calculation, we are going to ignore weight.

A B

C

PIPE FLEXIBILITY EXAMPLE (CONTINUED): Starting Point End Point Starting X Starting Y ΔX ΔY D t I L E

Point A Point B 0 0 0,8 0 114,3 6,0198 3,01044E-06 0,8 2,10E+11

Point B 10 0,8 0 0,27855 0 114,3 6,0198 3,01044E-06 0,27855 2,1E+11

10 20 1,07855 0 0,17145 0,17145 114,3 6,0198 3,01044E-06 0,26931303 2,1E+11

20 30 1,25 0,17145 0 0,1571 114,3 6,0198 3,01044E-06 0,1571 2,1E+11

30 40 1,25 0,32855 0,17145 0,17145 114,3 6,0198 3,01044E-06 0,26931303 2,1E+11

40 Point C 1,42145 0,5 0,32855 0 114,3 6,0198 3,01044E-06 0,32855 2,1E+11

Starting Point End Point FF (k) EP FX EP FY EP M dX dY dθ SP dX SP dY SP dθ TdX TdY Tdθ

Point A Point B

Point B 10

10 20

20 30

30 40

40 Point C


Point A Point B 0 0 0,8 0 114,3 6,0198 3,01044E-06 0,8 2,10E+11

Point B 10 0,8 0 0,27855 0 114,3 6,0198 3,01044E-06 0,27855 2,1E+11

10 20 1,07855 0 0,17145 0,17145 114,3 6,0198 3,01044E-06 0,26931303 2,1E+11

20 30 1,25 0,17145 0 0,1571 114,3 6,0198 3,01044E-06 0,1571 2,1E+11

30 40 1,25 0,32855 0,17145 0,17145 114,3 6,0198 3,01044E-06 0,26931303 2,1E+11

40 Point C 1,42145 0,5 0,32855 0 114,3 6,0198 3,01044E-06 0,32855 2,1E+11


Point A Point B 1 0 5000 4750

Point B 10 1 0 5000 3357,25

10 20 4,686 0 5000 2500

20 30 1 0 5000 2500

30 40 4,686 0 5000 1642,75

40 Point C 1 0 5000 0


Point A Point B 0 0 0,8 0 114,3 6,0198 3,01044E-06 0,8 2,10E+11

Point B 10 0,8 0 0,27855 0 114,3 6,0198 3,01044E-06 0,27855 2,1E+11

10 20 1,07855 0 0,17145 0,17145 114,3 6,0198 3,01044E-06 0,26931303 2,1E+11

20 30 1,25 0,17145 0 0,1571 114,3 6,0198 3,01044E-06 0,1571 2,1E+11

30 40 1,25 0,32855 0,17145 0,17145 114,3 6,0198 3,01044E-06 0,26931303 2,1E+11

40 Point C 1,42145 0,5 0,32855 0 114,3 6,0198 3,01044E-06 0,32855 2,1E+11


Point A Point B 1 0 5000 4750 0 0 0,001503

Point B 10 1 0 5000 3357,25 0,000263 0 0,001786

10 20 4,686 0 5000 2500 7,14E-05 5,81E-05 0,005535

20 30 1 0 5000 2500 0 4,88E-05 0,000621

30 40 4,686 0 5000 1642,75 5,15E-05 3,82E-05 0,003824

40 Point C 1 0 5000 0 9,35E-05 0 0,000427


Point A Point B 0 0 0,8 0 114,3 6,0198 3,01044E-06 0,8 2,10E+11

Point B 10 0,8 0 0,27855 0 114,3 6,0198 3,01044E-06 0,27855 2,1E+11

10 20 1,07855 0 0,17145 0,17145 114,3 6,0198 3,01044E-06 0,26931303 2,1E+11

20 30 1,25 0,17145 0 0,1571 114,3 6,0198 3,01044E-06 0,1571 2,1E+11

30 40 1,25 0,32855 0,17145 0,17145 114,3 6,0198 3,01044E-06 0,26931303 2,1E+11

40 Point C 1,42145 0,5 0,32855 0 114,3 6,0198 3,01044E-06 0,32855 2,1E+11


Point A Point B 1 0 5000 4750 0 0 0,001503 0 0 0 0 0 0,001503

Point B 10 1 0 5000 3357,25 0,000263 0 0,001786 0 0 0,001503 0,000396 0 0,003289

10 20 4,686 0 5000 2500 7,14E-05 5,81E-05 0,005535 0,000396 0 0,003289 0,000647 0,000238 0,008824

20 30 1 0 5000 2500 0 4,88E-05 0,000621 0,000647 0,000238 0,008824 0,000647 0,000728 0,009445

30 40 4,686 0 5000 1642,75 5,15E-05 3,82E-05 0,003824 0,000647 0,000728 0,009445 0,001214 0,001281 0,013269

40 Point C 1 0 5000 0 9,35E-05 0 0,000427 0,001214 0,001281 0,013269 0,002695 0,001281 0,013696

IMPORTANT TO NOTE: • Pipe flexibility assumes small displacements. Changes in load direction due to

deformation are not considered. • Pressure effects on the system are not accounted for.

• Elbows do not experience deformation due to pressure. • End points do not experience hydrodynamic loads. • Displacement (longitudinal growth) due to pressure is not accounted for. • Poisson’s effects due to pressure is not accounted for. • Change in inertia due to hoop stresses are not accounted for.

• The piping is considered uniform, and no deformation or ovalling is accounted for. • The piping flexibility model is based on simple beam theory with empirical values to

adjust certain factors such as flexibility. The process usually results in multiple simultaneous equations being solved in matrix form and then solved in a very similar fashion to Finite Element Analysis. Important to note however: THIS IS NOT A FINITE ELEMENT ANALYSIS.

• SIF’s and FF’s for components such as elbows and Tee’s are estimated values as provided by the Code. There are software programs capable of determining more accurate values by means of Finite Element Analysis. This is not however the norm.

FIXED SUPPORT VS REAL LIFE: Fully Fixed

If we use the same example as before, and add a temperature of 150°C without any other loads applied with the two ends fully fixed:

Using FEA, we find that the resultant axial load on the pipe is 24 000 N.

If we use exactly the same pipe and the same temperature, but only change the end conditions to a very stiff spring:

By allowing only 1mm of displacement (less than 1%), we immediately reduce the resultant force by 33%.

Stiff Spring

Unless the piping is therefore actually cast into stone, it is often worth considering the supports as stiff springs rather than fixed immovable objects, and where possible, it would be even more beneficial to consider modelling the actual structure as part of the analysis.

FIXED SUPPORT VS REAL LIFE CONTINUED: The same principle of course is also true when we need to restrain a certain section of the model. Consider the following case: We have a very fragile endpoint which cannot sustain a very high load. As such we fix two points near the connection with lateral supports. This immediately resolves the model and provides a satisfactory answer. In actuality, the supports are not entirely fixed and loads are still being transferred. Once again, using stiff springs would have highlighted the problem.

ASME DESIGN CODES:

• B31.1 - Power Piping • B31.2 – Fuel Gas Piping (Discontinued) • B31.3 – Process Piping • B31.4 – Pipeline Transportation Systems for Liquid Hydrocarbons And Other Liquids • B31.5 – Refrigeration Piping And Heat Transfer Components • B31.8 – Gas Transmission And Distribution Systems • B31.9 – Building Services Piping • B31.11 – Slurry Transportation Piping Systems • B31.12 – Hydrogen Piping And Pipelines

CODE CALCULATIONS (B31.1): Wall Thickness Sustained Loads

Occasional Loads

Expansion Loads Design Pressure


Occasional Loads Expansion Loads


Occasional Loads Expansion Loads

DYNAMIC ANALYSIS: We have considered the static analyses for piping and we are now confident to evaluate any piping system for pressure loading, weight, forced displacement, thermal growth and even wind. But what is often times far more dangerous is the dynamic loading on piping, mostly because it is not as easy to judge what the behaviour of a piping system will be purely by visual inspection and a basic understanding of science. There are many sources of dynamic loading including but not limited to: • Vibrating equipment. • Turbulent flows in the piping. • Seismic activity. • Sudden changes in flow velocity/direction. • Two phase flow. We are going to look at the sources and effects of these loads and how to analyse our piping system. However: because most of the mathematics for dynamic loading is being performed by software, we are not going to delve too deep into theory and mathematics. We will only try to achieve a rudimentary understanding of what happens in the system and how we can contain or eliminate the effects of dynamic loading. Detail theory for dynamic loading will be available in separate advanced courses.

DYNAMIC LOADING AND RESPONSE:

Let’s for one moment, consider a very basic dynamic load: a child being pushed by a parent on a swing and some pieces of information we can gather from it. • The swing will move on its own back and forth without being pushed. Even though the movement

can be small and undetectable, it will still occur naturally. • When the swing is being pushed in synchronisation with its natural movement, the height will

increase with each subsequent push. • If we discontinue pushing, the swing will (for the time being) keep on swinging at its current height

without assistance. • After a long period of time the swing will return to a state of “rest” at which the movement is again

minutely small.

NATURAL FREQUENCY: If we consider a basic system with a single mass attached to the end of a spring:

The natural frequency of the system can be obtained by the following formula:

𝜔𝑛² =𝑘

𝑚

With 𝜔𝑛 being the natural frequency or rate of oscillation given in radians/sec, 𝑘 is the stiffness of the spring and 𝑚 is the mass of the system.

Given this very basic equation and without having to perform detail calculations we can already derive the following information: • Piping that is more stiff (due to thicker walls, shorter spans and lower

temperatures) will have higher natural frequencies. • Piping that have less mass (due to smaller diameters, less dense fluids or less dead

loads from insulation etc.) will also have higher natural frequencies.

MODAL ANALYSIS (SIMPLE BEAMS):

MODAL ANALYSIS (SIMPLE BEAMS):

Source: Piping Vibration Analysis by J.C. Wachel

MODAL ANALYSIS (SIMPLE BEAMS): The figure below shows the first 4 modal shapes for a 2 dimensional cantilevered beam:

Source: http://www.wikipedia.org

A piping system has an infinite amount of modal shapes at varying frequencies. As a general rule of thumb, it is usually the lower frequencies which lead to disastrous results. As such only the first few modal shapes are generally used and frequencies above a certain threshold are ignored. It is the responsibility of the engineer performing the analysis (and not the software) to determine the cut-off threshold for frequencies to be evaluated based on the loading conditions and mass participation factors.

MODAL ANALYSIS (EIGENVALUES): Consider the following “complex” system with two masses:

Source: http://lpsa.swarthmore.edu

If we draw the free body diagrams for each mass and set up the equations of motion:

Rearranging the data into matrix form yields the following:

simplified slightly to produce:

MODAL ANALYSIS (EIGENVALUES):


If we now determine the eigenvalues from the matrix given in the following form:

where

Then we arrive at the following equation:

The stiffness matrix becomes a lot more complex if we consider a three dimensional piping system using beam theory to determine values for k. We are not (as part of this course) going to derive all the possible equations of motion using beam theory. We are going to suffice with the following information: • Using basic beam theory to determine the stiffness of a piping system comprised

of multiple interconnecting elements, we can create a stiffness matrix K for any given piping system.

• By solving the eigenvalues for the resulting stiffness matrix, we can extract as many eigenvalues from the matrix as we have degrees of freedom.

• These eigenvalues represents the natural frequencies of the piping system being evaluated.

• We do not need to extract all the natural frequencies, only the frequencies which are influenced by the excitation due to dynamic loads.

and

MODAL ANALYSIS (EVALUATION):


Obtaining the results from a modal analysis does not provide any quantifiable results as to the actual response of a piping system, in as much as it does not provide the expected stresses, forces or displacements for any loads imposed on the system. A modal analysis only provides insight into the characteristics of a system by supplying the natural frequencies of the system as well as the resulting mode shape for each frequency. The modal analysis is however the first step towards determining the actual response of the piping system. Further dynamic analyses uses the natural frequencies of the system, along with the mode shapes to determine the resulting excitation(s) due to imposed loads, whether these loads are due to harmonic, impulse or random vibration. The results from the modal analysis will however provide the engineer with insight to the response of the system and may negate any further analysis completely if properly investigated. As an example: We have a compressor operating at 5 Hz feeding a high pressure gas pipe. During the modal analysis, the lowest natural frequency of the piping system was found to be at 23 Hz. There should be no reason to investigate the response of the system due to the compressor as the compressor is unlikely to excite any of the natural frequencies of the system.

FORCED VIBRATION ANALYSIS: When we have a continuous cyclic load on a structure, we create forced vibration. Should this cyclic load be anywhere near the natural frequency of the structure, the structure will respond drastically. This is as a result of resonance. For a simple spring-mass-damper system, the steady state response of the system can be shown to be:

X=𝐹

𝑘 1−𝑟2 2+ 2𝜁𝑟 2 where 𝑟 =

𝑓

𝑓𝑛

Plotting the response of the spring-mass-damper for various values of 𝑓 and 𝜁, we can see clearly that as 𝑟 nears 1 the system has extreme responses. When 𝑟 starts deviating from 1, particularly at values exceeding 40%, the response of the system is much more subdued.

HARMONIC ANALYSIS: In the case of a real life scenario, there are however many natural frequencies for any given system. In order to simplify the situation, all we need to achieve is to calculate the natural frequencies for any given system (eigenvalues), determine the response of the system for each natural frequency, and then surmise them together. The plot below shows the response of a system with 3 natural frequencies: (2,3 rad/s; 5,4 rad/s; 14 rad/s)

HARMONIC ANALYSIS: Having the natural frequencies of a system is however not enough. We also need to consider the mode shape to determine the response of the system:

Looking again at the mode shape for a simple beam. Even though a natural frequency exists for mode shape 3, it will not be possible to excite that mode shape should we have a cyclic load located at L = 0,5. We will be able to excite the frequency if we locate the cyclic load at L = 0,3 and L = 0,7, but the biggest result will be obtained should we locate the cyclic load at L = 1,0.

HARMONIC ANALYSIS: If we now consider the same beam, but create 10 different intervals (nodes) we will find a combined response for each of the 4 mode shapes at each location:

-2,5

-2

-1,5

-1

-0,5

0

0,5

1

1,5

2

2,5

0 10 20 30 40 50

L=0,1

L=0,2

L=0,3

L=0,4

L=0,5

L=0,6

L=0,7

L=0,8

L=0,9

L=1

HARMONIC ANALYSIS: Note that in the previous example we added the 4 mode shapes together as a basic sum:

Δ𝑦𝑡𝑜𝑡 = Δ𝑦𝜔1 +Δ𝑦𝜔2 + Δ𝑦𝜔2 + Δ𝑦𝜔4

As the mode shapes given for a beam are generally non-dimensional (and therefore without direction) it would be more accurate to consider absolute values when determining the system’s modal response. It is also possible to consider the SRSS combination. Let us consider just the final node in our previous example where L=1:

HARMONIC ANALYSIS: Note that in the examples thus far, for simplification, we have assumed that in the

equation: X=𝐹

𝑘 1−𝑟2 2+ 2𝜁𝑟 2 F/k has been a constant value.

In reality, this is rarely the case. If we assume F = 1, and only evaluate the different values of k at each node, we find that the peak values for each natural frequency will alter and our true actual frequency response factors for a given node in a system might look more like this:

MASS PARTICIPATION FACTOR: The mass participation factor of a system for any given mode shape is a measure of how much mass is participating in that particular mode shape. The simplest means to understand it is to ask: how much percentage of the weight can translate how far in that given mode shape. If most of the mass remains stationary, then the mass participation for that mode shape will be low. If most of the mass is active in the given mode shape, then the mass participation for that mode shape will be high. The question always needs to be asked: I have a certain mass participation factor in my results, is it enough? Consider the following two scenario’s: In the left image, you would expect the mass participation factor to be close to 95% as most of the mass is mobile. In the right image, obtaining a mass participation factor higher than 40% would be near impossible.

RESPONSE SPECTRUM ANALYSIS: In the case of an earthquake however, the force applied to the system is not as clear. It is however possible to determine the acceleration of an earthquake and through the simple transformation of F=ma, we can calculate the force at each node by multiplying the acceleration with the mass at that node. The figure below shows a standard input curve used during a seismic analysis.

TIME-HISTORY ANALYSIS: A time-history analysis has the advantage where the solution can be determined as a function of time (and not of frequency) and the response of the system can be plotted for each time step. This is typically used for input that is not harmonic and can often times be as a result of a single impulse load.

WATER HAMMER: If we suddenly open or close a valve in a system or suddenly start/stop a pump, the system suddenly experiences a shift in energy balance. All the kinetic energy in the system needs to come to a halt. This results in a force being exerted on the piping components proportional to the mass of the liquid and the change of velocity and inversely proportional to the time required to open/close a valve.

Amount of mass changing velocity for pipe section.

𝐹 =𝑚∆𝑣

∆𝑡

WATER HAMMER: In the case where the system changes instantaneously, not all the process medium reacts at once. There is a certain delay from one particle to the next resulting a “shockwave” running through the system. The pressure of this wave is calculated by the Joukowsky equation:

∆𝑃 = 𝜌. 𝑎. Δ𝑣 Note that this pressure will be acting on the end elbow for as long as the shockwave is still in the leg. The rise time will be equal to the mean length of the elbow divided by 𝑎.

The time it takes for a pressure wave to travel up and down the entire section of pipe (from valve to reservoir) is referred to as the reflection time. The shock wave needs not to be considered if the change in velocity exceeds the reflection time.

PIPE FLEXIBILITY ANALYSIS SOFTWARE:

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Piping Presentation