Lesson 04 StaticHead_New2
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Transcript of 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.
Hydrostatic Head of Water
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
Hydrostatic Head of Water
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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
Hydrostatic Head of Water
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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.
Hydrostatic Head of Water
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
Hydrostatic Head of Water
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.
Hydrostatic Head of Water
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
Hydrostatic Head of Water
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?
Hydrostatic Head of Water
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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?
Hydrostatic Head of Water
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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"
Static Head of Water
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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;
Static Head of Water
"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.
Static Head of Water
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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
Static Head of Water
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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
Static Head of Water
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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
Static Head of Water
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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
Static Head of Water
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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.
Hydrostatic Head of Water
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.
Hydrostatic Head of Water
Cl Q i
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Class Quiz
Hydrostatic Head of Water
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
Hydrostatic Head of Water
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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Hydrostatic Head of Water
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
Hydrostatic Head of Water
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
Hydrostatic Head of Water
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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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2 to 1 Ellipsoidal Head
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
2 to 1 Ellipsoidal Head
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..