pressure surges in pipelines

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CHAPTER 4 PRESSURE SURGES IN PIPELINES 4.1 FEATURES OF PRESSURE SURGE In the past, water engineers were facing problems with repeated pipe bursts, failing seals etc. They termed the cause of failure as "Water Hammer". Now this phenomenon is termed as a surge. Surge is a phenomenon occurring during hydraulic transient and is due to pressure waves caused due to the change in flow velocity in the pipe line. This pressure wave will be reflected from the end of the pipe line towards the pump. The main causes for a surge to develop are; * Opening or closing of valves. * Starting and stopping of pumps. * Change in pump speed. * Pipe fittings. * Turbulence flow. * Cavitation. * Air bubbles. Out of the above, the worst situation would be the sudden stoppage of a pump due to a power failure. Then the pumping capacity is lost and a negative pressure wave occurs just next to the pump in the pipeline and it travels along the pipeline and is reflected at the discharge end so that it is changed into a positive pressure wave which returns along the pipeline. For short pipelines, the pressure drop near the pump is usually small and the subsequent surge effect caused is relatively small. In the case of long pipelines, these effects are much greater. 70

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Page 1: PRESSURE SURGES IN PIPELINES

CHAPTER 4

PRESSURE SURGES IN PIPELINES

4.1 FEATURES OF PRESSURE SURGE

In the past, water engineers were facing problems with repeated pipe bursts, failing seals etc. They termed the cause o f failure as "Water Hammer". Now this phenomenon is termed as a surge. Surge is a phenomenon occurring during hydraulic transient and is due to pressure waves caused due to the change in flow velocity in the pipe line. This pressure wave will be reflected from the end o f the pipe line towards the pump.

The main causes for a surge to develop are;

* Opening or closing of valves.

* Starting and stopping of pumps.

* Change in pump speed.

* Pipe fittings.

* Turbulence flow.

* Cavitation.

* Air bubbles.

Out of the above, the worst situation would be the sudden stoppage of a pump due to a power failure. Then the pumping capacity is lost and a negative pressure wave occurs just next to the pump in the pipeline and it travels along the pipeline and is reflected at the discharge end so that it is changed into a positive pressure wave which returns along the pipeline. For short pipelines, the pressure drop near the pump is usually small and the subsequent surge effect caused is relatively small. In the case of long pipelines, these effects are much greater.

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Page 2: PRESSURE SURGES IN PIPELINES

The factors that govern the surge when a pump is suddenly stopped are;

* Flow velocity in the pipeline.

* Moment of Inertia of the rotating parts (Pump & Motor).

* Length and pipeline profile.

* Material of pipeline.

* Pump and valve characteristics.

4.2 PROPAGATION SPEED OF PRESSURE WAVE

The speed o f pressure wave propagation in water pipelines and can be calculated by using the following equation;

a = 1425 /V [1 + ( K / E x D/e)] m/s

Where;

a = Propagation o f speed o f pressure wave m/s

K = Bulk Modulus o f liquid (2 .07 x 1 0 8 kgf/m 2 )

E = Youngs Modulus for pipe material kgf/m )

D = Internal Diameter o f Pipe (m)

e = Pipe Thickness (m)

(Source : Pumping Station Engineering Hand Book 1991)

Graphs are available to obtain the wave velocities in different pipes when D/e ratio is known,

(please see Annexes 4.1 and 4 .2 ) .

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Page 3: PRESSURE SURGES IN PIPELINES

4.3 RAPID CHANGES IN FLOW VELOCITY AND PRESSURE

The first major publication on pressure transients in pipelines was by Joukowski. His expression for maximum/minimum possible pressure head above/below following a sudden stoppage o f flow is well known; namely Joukowski's law (some-times called Allievi formula) is expressed as;

AH = ± a / g x A V

Where,

AH = Pressure rise/fall (m)

a = Wave velocity (m/s) » , v

i •

AV = Change in fluid velocity (m/s)

g = Acceleration due to gravity (m/ s 2 ) ',<.- ...

4.4 REFLECTION OF PRESSURE WAVE IN THE ABSENCE OF FRICTION

The following sequence o f events occur on a pumping system after an instantaneous power failure. The events are shown graphically in pages 7 4 and 75 .

1. The forward flow will immediately cease and the check valve will close. The head will fall by av/g at the check valve and travel along the pipe at a velocity "a" reducing the downstream pressure after a time L/a. L = length o f pipeline, a = Wave velocity.

2. After a time L/a reflected pressure wave returns at a velocity "a". The pressure at the reservoir will remain H° and cause a flow out o f the reservoir towards the pump. At the end o f this reflection the time period is 2L/a .

3. After a time 2 L / a the reverse flow having encountered a close valve pressure rise occurs o f av/g and the head rise will travel along the pipe at a velocity "a" reaching the reservoir at a time 3 L/a.

4 . After time 3 L / a the flow is into the reservoir and the head in the system returns to H 0 . Thus at a time 4 L / a the original conditions are established and a complete cycle o f changes o f pressure and velocity occurs in the pipe in a time 4L/a .

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Page 4: PRESSURE SURGES IN PIPELINES

Theoretically this cycle is repeated indefinitely, but friction causes the oscillations to die away.

5. The "pipe line period" time t is 2L/a.

I f the check valve closes within a time 2L/a the full head rise/fall av/g has time to develop and the closure is regarded as rapid.

If the closure takes longer the head rise/fall is less than av/g.

To determine maximum and minimum heads in such cases a numerical, graphical methods or computer programmes must be used.

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Page 5: PRESSURE SURGES IN PIPELINES

H 0~aV / g wave moving downstream velocity

y a

V = 0

H,

(1) between time = 0 and time = L/a

Ho^aV wave moving upstream

V = o

L 2L. (2) between time = /a and time = /a

H,

wave moving downstream

V = 0

HQ

2L 3L (3 ) between time =• /a and time = /a

"Si

L wave moving upstream H Q+aV r g

<

\ V = 0

3 L 4 L ( 4 ) between time = /a and time » /a

PIPELINE CONDITIONS FOR FOUR TIME INTERVALS AFTER PUMP FAILURE

Page 6: PRESSURE SURGES IN PIPELINES

PRESSURE (HEAD)

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Bergeron developed a graphical method originated by Schneider which enable many engineering problems to be solved graphically. This method requires pump characteristic curves as well as other system information. Other authors from various countries have done studies on this subject and now computer soft wear are available to handle complex situations. What is dealt in this chapter is an elementary treatment o f the subject related to water industry.

4.5 PUMP TORQUE AND SPEED

The decrease in pump rotational speed after a sudden power failure depends on the inertia effect o f the rotating parts o f pump and motor and the rotating torque o f the pump.

In obtaining graphical solutions to maximum and minimum heads after a sudden power failure, two independent parameters are used. These are,

(1 ) P the pipeline constant given by;

(2) . K) x ( 2 L / a ) a constant which includes the effect o f pump and motor inertia and pressure wave travel time in the pipeline.

Where,

H = Total Head ( m )

P = a v / 2 g H

g

a

V

L Length o f pipeline (m)

Change in velocity (m/s)

Wave velocity (m/s)

Acceleration due to gravity (m/s 2 )

4 5 0 x g x w x H x Q / (7 t 2 GD 2 r |pN 2 )

Where,

Q = Flow rate ( m 3 / s )

G D 2 = Pump and Motor inertia ( K g m 2 )

n p = Pump efficiency (%)

N = Pump Speed ( R P M )

= 22 /7

w = Specific weight o f water ( K g / m 3 )

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4. 6 SURGE ANALYSIS CHARTS

Pressure changes in pipelines after a sudden power failure can be found from charts by using the above mentioned numerical values.

X - Axis gives the values Ki x 2 L / a

For different values o f 2P ,

Y-Axis gives the up surge/down surge as a percentage o f the pumping head.

Figures over leaf give the upsurges and down surges at pump and mid length o f the pipe line.

(Source: Parmakian 1963)

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Page 9: PRESSURE SURGES IN PIPELINES

WATERHAMMER IN PUMP DISCHARGE LINES a LOOR-I

•OJ .04 .OS 06 .06 .10

VALI

OOWNSUHOC AT PUMP

>I00 {

* -3 * i £ 3 as 10 VALUES OF K - Q .

0 OOWNSURGE AT MIOLENOTH

WATERHAMMER IN PUMP DISCHARGE LINES

Mr

VALUES OF K

U P S U R G E AT PUMP

VALUES OF K

UPSURGE AT MIDLENGTH

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4.7 SAMPLE CALCULATION

Flow rate = 5 0 m 7 H r

HeadH = 36 m (inclusive of a static head of 13m)

Speed = 2 9 0 0 R P M

Motor Power = 7.5 kW

Pump Efficiency = 0 .65

Internal diameter o f pipe = 140 mm

Pipe thickness = 10.4 mm

Length o f pipeline = 7 5 0 m

Velocity V = 50/ (n /4 x = 5 0 / ( 7 t / 4 x 0 . 1 4 x 0 . 1 4 x 3 6 0 0 )

= 0.9 m/s

From the graph wave velocity a = 4 8 0 m/s

Max surge pressure = ± av/g

- ± 4 8 0 x 0 . 9 / 9 . 8

= ± 4 4 m

Therefore without considering inertia effects the maximum and minimum surges will be 57 ( 1 3 + 4 4 ) and -8m ( 3 6 - 4 4 ) respectively.

Motor G D 2 = (0 .03 ~ 0 . 0 0 5 ) k W 1 4 x P ° 7 5

Where P = Number o f poles in the motor

Motor G D 2 = 0 .005 x 7 . 5 1 4 x 2 ° 7 5

= 0 . 1 4 k g m 2

Total G D 2 = ( 0 . 1 4 / 9 0 ) x 100 = 0 . 1 6 (assumed motor inertia is 9 0 % o f total inertia).

K , = 4 5 0 x g x w x H x Q/(TT G D 2 x n p x N 2 )

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Page 11: PRESSURE SURGES IN PIPELINES

4 5 0 x 9 . 8 1 x 1 0 0 0 x 3 6 x 5 0 / ( t T x 0 .16 x .65 x 2 9 9 0 x 2 9 9 0 x 3 6 0 0 )

= 0 . 2 5 6

av/g = 4 4

p = av/2gH

= 44 /2 x 3 6

= 0 .6

:.2p = 1 . 2

2L/a = 2 x 7 5 0 / 4 8 0

= 3 . 1 2 5 Sec.

K, x 2 L / a = 0 . 2 5 6 x 3 .125

= 0 . 7 9 9

Therefore from the charts;

Down surge at pump = 0 .82H = 2 9 . 5 m

Up surge at pump = 0 .43H = 1 5 . 5 m

Down surge at Mid Length = 0 .64H = 23 m

Up surge at Mid Length = 0 .25H = 9.0 m

over and below the total pumping head.

4 8 C H A R T S F O R G D 2

Generally moment o f Inertia o f the rotor of the motor G D 2 amounts to 9 0 % o f the total G D 2 and that o f the rotating parts o f the pump amounts to 10% at most. The rule o f the G D 2 o f the motor differs considerably by the type and brand o f the motor, but it can be roughly expressed by the following equation.

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Page 12: PRESSURE SURGES IN PIPELINES

G D 2 = (0 .013 - 0 . 0 0 5 ) k W l 4 x P (

Where;

kW = Power o f the motor in kW

P = Number o f poles o f the motor.

(Source : Kubota Pump Hand Book Vol I, 1 9 7 7 )

Out o f the range o f the above coefficient the larger part is for double squirrel cage type motors and smaller part is for wound rotor type motors. For the surge calculations it is safer when the above coefficient is smaller.

Annex 4.3 shows how the GD o f different motors vary with their capacity.

(Source : Pumping Station Engineering Hand Book).

4.9 CHANGES TO CODE OF PRACTICE CP 312

The curves for wave speeds in PVC pipes are based on standard data available from manufacturers published tables. Independent experimental data are not yet available on wave speeds. The principle properties o f U P V C depend upon both temperature and rate o f strain. If the operating temperature is significantly different from 2 0 deg. C, the appropriate value o f elastic modulus would have to be substituted.

Several failure o f PVC pipes have occurred, and in some cases fatigue arising from repeated surge effects is suspected to be the cause. Code o f Practice CP 3 1 2 has been revised with regard to U P V C Pipes and imposes restrictions on the range and frequency o f pressure variations resulting from surge.

The recent change to the code CP 3 1 2 include:

(1) . The total surge pressure variation from the minimum to the maximum should be limited to 5 0 % o f the maximum sustained working pressure of the pipe.

(2) . Class B PVC should not be used for situations where the surge pressure variation includes negative gauge (subatmospheric) pressures.

(3) . Pump switching, or valve movement, which generates surge should not be more frequent that 6 times per hour.

Nomograms given in Annex 4 .4 illustrate the ranges over which Class B and Class C P V C Pipes can be used.

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4 .10 SURGE PREVENTION MEASURES

The pressure transients following power failure to electric motor driven pumps are the usually the most extreme that a pumping system will experience. The different devices available for surge protection are as follows:

Fly-Wheel.

Pump bypass reflex valve.

Inline reflex valve.

Surge tank.

Automatic air release valve.

Discharge tank.

Air vessels.

The best method o f surge protection will depend on the hydraulic and physical

characteristics o f the system. The accompanying figure illustrates where the devices are

usually accommodated and the table summarises the ranges over which various devices are suitable.

8 2 .

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4 11 SUMMARY OF METHODS OF SURGE PROTECTION

The table below gives the different surge protection methods available for different system conditions and figure over leaf gives the pipe line profile illustrating various devices for surj protection.

Method of Protection (in approximate order

of increasing cost)

Required range

of variables

Remarks

Inertia o f pump M N 2 / w A L H > 0 . 0 1 Approximate only

Pump bypass reflux valve

av/gh » 1 Some water may also be drawn through pump

In-line reflux valve av/gh > 1 Normally used in conjunction with some other method of protection. Water column separation

possible

Surge tank h small Pipeline should be near hydraulic grade line so height o f tank is practical

Automatic release valve av/gH « 1

2 L / a > 5 sec.

Pipeline profile should be convex downwards. Water column separation likely

Discharge tanks av/gh > 1 h = pressure head at tank. Pipeline profile should be convex upwards.

Air vessel av/gH < 1 Pipeline profile preferable convex downwards.

Table 4.1 - Methods of Surge protections.

(Source : Stephensen 1979)

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Hydraulic grade line Surge

Pump ; with bypass

PIPELINE PROFILE ILLUSTRATING VARIOUS DEVICES FOR SURGE

' PROTECTION

Source Sfe^W^^w, )

Delivery reservoir

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4 .12 A I R VESSELS

When negative pressures are encountered, water is forced into the system by using air vessels. For rising mains which attain the highest point down stream o f the reservoir at a point three quarters o f its way along and provided the static head is not less than 15m Lupton proposed a set o f relations for air vessel dimensions to avoid column separation.

Total volume = ( S + 3 6 . 6 ) x L x A x V / 8 3 6 1 m 3

Air volume = S x L x A x V/8361 m 3

Where,

L = Total length o f pipe m

A = Pipe cross sectional area m 2

V = Steady state velocity m/s

S = Static head m

(Source : Unconfirmed).

If a high point occur near to the pumping station a larger vessel might be needed, when a smaller one would suffice for a uniformly rising main.

Now air vessels are available which isolates the air from being dissolved in the water by use o f a bladder. Otherwise a compressor will be needed to feed the vessel and reinject into the system the elastic energy lost by dissolution.

If no bladder is used a lot o f control and sensing devices will have to be used. The installation o f a bladder type surge vessel is simple but must be done with care.

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