Asme Viii Div. 1- 3 Statichead

Cairo Inspection Company Eng. Ibrahim Eldesoky

ASNT –NDT-- LEVEL III

ASME AUTHORIZED INSPECTOR

API AUTHORIZED INSPECTOR

API 510

Hydrostatic head pressure

What is hydrostatic head pressure? Let’s 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 the greater 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

2 Cairo Inspection Company (CIC) Www. CIC-Egypt.com Eng. Ibrahim Eldesoky

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. Let’s have a look at a graphic of a water tower that will detail this principle.

Hydrostatic Head of Water A Common Thing


Hydrostatic Head of a Water Tower 140’ x 0.433 = 60.6 psig and 100’ x 0.433 = 43.3 psig


What would be the hydrostatic head pressure if a gage were inserted into the side of the tower at the 110’ elevation when the tower was completely full? Hint, the darker area is exerting the pressure.

Class Quiz Hydrostatic Head of Water


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

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

Solution Hydrostatic Head of Water


The hydrostatic head of water is equal to 0.433 psi per vertical foot above the 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.

Hydrostatic Head of Water Basic Principle


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

Hydrostatic Head of Water

0 psi

43.3

psi

100 Feet


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

143.3

psi

100 Feet


What would be the pressure at the bottom if an external pressure of 235 psi were applied ?

•Class Quiz Hydrostatic Head of Water

235

psi

?

100 Feet


235 + 43.3 = 278.3 psi

Solution

235

psi

278.3

psi

100 Feet


From these simple water tower and pressure vessel examples the 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?



Case 1: To determine hydrostatic head based on an elevation from a stated problem it must be understood that elevations are normally taken from the ground level to a vessel’s very top. You must subtract the Given elevation from the Total elevation 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 head pressure of water at the 11 foot elevation which is located at the bottom of the top shell course?

•Hydrostatic Head of Water


Remember it is the number of vertical feet above the given elevation in question which causes the hydrostatic head at that point. To find the hydrostatic head you must subtract the elevation of the Given point from the Total elevation 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


Case 2: Hydrostatic head at a point in a vessel must be added to the pressure used (normally vessel MAWP) when calculating the required thickness of the vessel component at that elevation. Example: Determine the required thickness of the shell course in Case 1. The vessel's MAWP (Always measured at the 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"



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

-


Case 3 You must subtract hydrostatic head pressure when determining the MAWP of a vessel. If given a vessel of multiple parts and the MAWP for each of the parts, the MAWP of the entire vessel is determined by subtracting the hydrostatic head pressure at the bottom of each part to find the part which limits the MAWP of the vessel. Example: A vessel has an elevation of 40 feet including a 4 foot base. The engineer has calculated the following part’s MAWP to the bottom of each part based on each part's minimum thickness and corroded diameter. Determine the MAWP of the vessel as measured at the top.



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



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



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



The final step in determining the MAWP of the vessel at its top is to subtract the hydrostatic head of water from each of the calculated Part MAWPs. 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



Therefore the bottom of the middle shell course’s MAWP limits the pressure at the top and, determines the MAWP of the vessel.


The MAWP of the vessel is 400.125 psi


One thing to remember is this pressure is static. In our example the if the applied external pressure at the top were raised above 400.125 psi, then down at the 16.5’ elevation the gage would exceed that shell course’s MAWP of 410.3.




What would be the pressure at 16.5’ if the top read 410 psi instead of 400.125 ?


Solution Hydrostatic Head of Water

410 + 10.175 = 420.175

Since our part is only good for 410.3 we have now exceeded this shell course’s MAWP. Not a good thing!



One last example using a vessel which is horizontal, just to reinforce the concept that it is the Vertical Height that must be considered. The 6.928 psi total H.H. must be considered at the bottom when calculating the sump head.



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


Solution

Bottom of top chamber 3 x 0.433 = 1.299 psi Bottom of main shell 13 x 0.433 = 5.629 psi Total H.H. = 6.928 psi


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 radius is the depth of the hemispherical head

Depth of a Hemi and Ellipsoidal and Hydrostatic Head of Water


An ellipsoidal head's I. D. will be the same as the shell's. The inside diameter of an ellipsoidal head is also its 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 the Minor Axis of an ellipse which is used to form the head.

Standard 2 to 1 Ellipsoidal Head


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.



Therefore, our 2 to 1 Ellipsoidal head has a depth of 14 inches. 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.


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



1. The depth of a 2 to 1 ellipsoidal head having a diameter of 64 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


1. The depth of a 2 to 1 ellipsoidal head having a diameter of 64 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 a vessel shell that has inside diameter of 96” with an internal fit up ? b. 48” (96”/2 = 48”)

Solution


Asme Viii Div. 1- 3 Statichead

Documents

Transcript of Asme Viii Div. 1- 3 Statichead