Lesson 04 StaticHead_New2

7/27/2019 Lesson 04 StaticHead_New2

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API 510 Preparatory Class

Lesson 4

Hydrostatic Head Pressure


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What is hydrostatic head pressure? Lets examine the words

to better understand the meaning of hydrostatic.

Hydro meaning liquid

Static meaning unchanging.Pressure is a force exerted over an area.

Which of leads us to the following;

It is a pressure that is generated by the weight of the liquid

due to gravity. The taller the height of a liquid column thegreater the force, which is expressed as pounds per square

inch (psi) for our purposes. The Hydro (liquid) of interest on

the exam is water, since it is the primary liquid we use for

Hydrostatic testing. Other liquids can be and are used.

Hydrostatic Head of Water

Overview


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The hydrostatic head of water is part of our everyday lives.

For example the water tower that supplies your home uses

the principle of Hydrostatic Head or gravity to push the

water into your home and out of your faucets. Lets have a

look at a graphic of a water tower that will detail thisprinciple.


A Common Thing


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Hydrostatic Head of a Water Tower140 x 0.433 = 60.6 psig and 100 x 0.433 = 43.3 psig


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What would be the hydrostatic head pressure if a gage were

inserted into the side of the tower at the 110 elevation whenthe tower was completely full? Hint, the darker area is

exerting the pressure.

Class Quiz



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The height of the water above the point causes the pressure.

140 - 110 = 30 therefore 30 x 0.433 = 12.99 psi

Solution



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The hydrostatic head of water is equal to 0.433 psi per

vertical footabovethe point where the pressure will

measured. For example the hydrostatic head of water at a

point in a vessel with 10 feet of water above it is calculated

by multiplying 10 x 0.433 psi.

10 x 0.433 = 4.33 psi

The 4.33 psi is being exerted totally by the weight of the

water. No other external pressure having been applied. If an

external source of pressure is applied it would be added to

the hydrostatic head pressure of the water at any given point

in the vessel. More on this later.


Basic Principle


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Now for a pressure vessel. No external pressure, filled with

water only. 0 psi at top, the bottom is 100 x 0.433 = 43.3 psi


0 psi

43.3psi

100 Feet


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External pressure of 100 psi is now applied resulting in a gage

pressure at the bottom of 143.3 psi. The 43.3 psi is static,

never changing.


100

psi

143.3psi

100 Feet


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What would be the pressure at the bottom if an external

pressure of 235 psi were applied ?

Class Quiz


235

psi

?

100 Feet


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235 + 43.3 = 278.3 psi

Solution

235

psi

278.3psi

100 Feet


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From these simple water tower and pressure vessel examplesthe following can be understood and applied to a pressure

vessel. For a pressure vessel the MAWP is always measured

at the top of a vessel in its normal operating position. Here are

the issues on the exam that must be understood to work H.H.problems that might be given.

Case 1: How do you determine hydrostatic head based on a

given elevation?

Case 2: When do you add the hydrostatic head pressure in

vessel calculations?

Case 3: When do you subtract the hydrostatic head pressure in

vessel calculations?



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Case 1:To determinehydrostatic head based on an

elevationfrom a stated problem it must be understood that

elevations are normally taken from the ground level to a

vessels very top. You mustsubtractthe Givenelevationfrom theTotalelevation to determine vertical feet of

hydrostatic head above the given elevation.

Example: A vessel has an elevation of 18 feet and is

mounted on a 3 foot base. What is the hydrostatic headpressure of water at the 11 foot elevation which is located at

the bottom of the top shell course?



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Remember it is the number ofvertical feet abovethe given

elevation in question which causes the hydrostatic head atthat point. To find the hydrostatic head you mustsubtractthe

elevation of theGivenpoint from theTotalelevation given for

the vessel.

18' feet total

-11' desired point

7' total hydrostatic head

Hydrostatic head pressure at 11' elevation is:7 x 0.433psi = 3.03 psi

Static Head of Water


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Case 2: Hydrostaticheadat a point in a vessel must be

addedtothe pressure used (normally vessel MAWP) whencalculating the required thickness of the vessel component

at that elevation.

Example: Determine the requiredthicknessof the shell

course inCase 1. The vessel's MAWP (Always measured atthe top in the normal operating position) is 100 psi. The

following variables apply:

Givens:

t = ? Circumferential stress from UG-27(c)(1)P = 100 psi + Hydrostatic Head

S = 15,000 psi

E = 1.0

R = 20"



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Since the bottom of this shell course is at the 11 foot

elevation the pressure it will see is 100 psi + the hydrostatic

head.

100 + 3.03 = 103.03 psi

Also our basic formula becomes;


"1379.18.14938

20606

)03.1036.0()0.1000,15(

2003.103

xXx

xt

.).(6.0

.).(

HHPSE

RHHPt

-


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Case 3 You mustsubtracthydrostatic head pressure when

determining the MAWPof a vessel. If given a vessel of

multiple parts and the MAWPfor each of the parts, the

MAWPof theentire vesselis determined by subtracting the

hydrostatic head pressure at the bottom of each part to findthe part which limits the MAWPof the vessel.

Example: A vessel has an elevation of 40 feet including a 4

foot base. The engineer has calculated the following partsMAWPto the bottom of each part based on each part's

minimum thickness and corroded diameter. Determine the

MAWPof the vessel as measured at the top.



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Calculated Part MAWP at the bottom of:

Top Shell Course 28' Elev. 406.5 psi

Middle Shell Course 16.5' Elev. 410.3 psi

Bottom Shell Course 4' Elev. 422.8 psi

Bottom of top shell course:

40.0' elev.-28.0' elev.

12.0' of hydrostatic head



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12' x 0.433 psi = 5.196 psi of Static

Bottom of the middle shell course:

40.0' elev.

-16.5' elev.23.5' of hydrostatic head

23.5' x 0.433 psi = 10.175 psi of

Hdrostatic Head



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Bottom of bottom shell course:

40.0' elev.

-4.0' elev.

36.0' of hydrostatic head

36' x 0.433 psi = 15.588 psi of

Hydrostatic Head



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The final step in determining the MAWPof the vessel at its

top is to subtract the hydrostatic head of water from each of

the calculated PartMAWPs. The lowest pressure will be the

maximum gauge pressure permitted at the top of the vessel.

Bottom of top shell course 406.5 - 5.196 = 401.3 psi

Bottom of mid shell course 410.3 - 10.175 = 400.125 psi

Bottom of btm. shell course 422.8 - 15.588 = 407.212 psi



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Therefore the bottom of the middle shell courses MAWP

limits the pressure at the top and, determines the MAWP of

the vessel.


The MAWP of the vessel is 400.125 psi


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One thing to remember is this pressure is static. In our

example the if the applied external pressure at the top wereraised above 400.125 psi, then down at the 16.5 elevation

the gage would exceed that shell courses MAWP of 410.3.


Cl Q i


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Class Quiz


What would be the pressure at 16.5 if the top read 410 psiinstead of 400.125 ?

S l ti


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Solution


410 + 10.175 = 420.175Since our part is only good for 410.3 we have now exceeded

this shell courses MAWP. Not a good thing!


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One last example using a vessel which is horizontal, just to

reinforce the concept that it is the Vertical Height that mustbe considered. The 6.928 psi total H.H. must be considered

at the bottom when calculating the sump head.

Cl Q i


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Class Quiz


What would be the hydrostatic pressure exerted at eachpoint in the vessel below?


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Solution

Bottom of top chamber 3 x 0.433 = 1.299 psi

Bottom of main shell 13 x 0.433 = 5.629 psiTotal H.H. = 6.928 psi


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One final thing the determination of H.H. for two formed

heads, Hemispherical and Ellipsoidal.

Hemispherical Head

For this example we will use a hemispherical head that has

an inside diameter of 48 inches which means it has a radius

of 24 inches. The radiusisthe depth of the hemispherical

head

Depth of a Hemi and Ellipsoidal and



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An ellipsoidal head's I. D. will be the same as the shell's. The

inside diameter of an ellipsoidal head isalsoits major axis.

This fact is the basis of finding the depth of a 2 to 1 ellipsoidal

head. Notice that we are strictly talking about 2 to 1 ellipsoidal

heads. The 2 to 1 refers to the ratio of the Major Axis to theMinor Axis of an ellipse which is used to form the head.

Standard 2 to 1 Ellipsoidal Head


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Of course only half of the Minor Axis is used for the head.

2 to 1 Ellipsoidal Head

Now add the 2 inch flange to the dish.


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Therefore, our 2 to 1 Ellipsoidal head has a depth of 14inches. Hint: To find the depth of a 2 to 1 ellipsoidal head

divide the major axis by 4. In our example 48/4 = 12 then

add the 2 flange.


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Ellipsoidal

Converting to feet: 18" divided by 12 =

1.5' x 0.433 psi = 0.6495 psi

Hemispherical

Converting to feet. 32" divided by 12 =

2.666' x 0.433 psi = 1.1543 psi


Cl Q i


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1. The depth of a 2 to 1 ellipsoidal head having a diameter of64 inches and a 1-1/2 flange is;

a. 33 -1/2

b. 16 -1/2

c. 17-1/2

2. What is the depth of a hemispherical head attached to a

vessel shell that has inside diameter of 96 with an internal

fit up ?

a. 96

b. 48

c. 32

Class Quiz

Depth of Heads


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1. The depth of a 2 to 1 ellipsoidal head having a diameter of64 inches and a 1-1/2 flange is;

c. 17-1/2 (64/4 = 16 + 1-1/2 + 17-1/2)

2. What is the depth of a hemispherical head attached to avessel shell that has inside diameter of 96 with an internal

fit up ?

b. 48 (96/2 = 48)

Solution


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This one is over..

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Lesson 04 StaticHead_New2