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Area of Ellipsoidal Heads
- Thread starter Sharik
- Start date
V=D^3(a + 5.86 L/D)
V= Volume in gallons
D= Diameter in feet
L/D= Length of Cylinder section to diameter ratio
a = constant= 1.164 for ASME flanged and dished heads
1.909 for 2 to 1 Elliptical heads
3.92 for hemispherical heads
Here is formula for a cyclinderical vessel with 2 heads. It's an approximation but fairly accurate. You can work from this. I'll try to find my calculations for each head type.
V= Volume in gallons
D= Diameter in feet
L/D= Length of Cylinder section to diameter ratio
a = constant= 1.164 for ASME flanged and dished heads
1.909 for 2 to 1 Elliptical heads
3.92 for hemispherical heads
Here is formula for a cyclinderical vessel with 2 heads. It's an approximation but fairly accurate. You can work from this. I'll try to find my calculations for each head type.
Sorry I didn't read "Sharik's' post correctly as I transposed volume and area. Crossed neurons again.
This is a formula we used early on for the total volume of a vessel(in gallons) with different type heads base on the L/D of the cylinder.
Its origin was from a United States Steel handbook "Low Temperature and Cryogenic Steels"
Sorry
This is a formula we used early on for the total volume of a vessel(in gallons) with different type heads base on the L/D of the cylinder.
Its origin was from a United States Steel handbook "Low Temperature and Cryogenic Steels"
Sorry
- Thread starter
- #6
Volume is in US gallons.
Length and Diameter of cylinder are both in feet.
This formula came from the early 60's so there wasn't much use at the time of metric units except by chemists.
Where one uses a spreadsheet today we made tables of values and in the handbook I mentioned there are tables of different Diameters vs Gallons using different L/D curves.
I have some more information and will try to find it.
Length and Diameter of cylinder are both in feet.
This formula came from the early 60's so there wasn't much use at the time of metric units except by chemists.
Where one uses a spreadsheet today we made tables of values and in the handbook I mentioned there are tables of different Diameters vs Gallons using different L/D curves.
I have some more information and will try to find it.
- Thread starter
- #8
unclesyd,
Your formula is good but I need a formula for surface area of semi-elipsoidal 2:1 heads. I'm in a remote area and am trying to complete some API calculations for safety valve relieving capacity requirements. I'm too far away from my own resource material in my office.
Your formula is good but I need a formula for surface area of semi-elipsoidal 2:1 heads. I'm in a remote area and am trying to complete some API calculations for safety valve relieving capacity requirements. I'm too far away from my own resource material in my office.
Sharik
In my formula 'D' is in metres for both surface area and volume.
Excuse my ignorance but why is the surface area of a dished end required to safety valve relieving capacity requirements? - I would have thought volume would be reuired.......??
In my formula 'D' is in metres for both surface area and volume.
Excuse my ignorance but why is the surface area of a dished end required to safety valve relieving capacity requirements? - I would have thought volume would be reuired.......??
- Thread starter
- #10
- Thread starter
- #13
You got me, as my tables for head blanks run out at 144" tank diameter. I was going to give you a head blank size which would have worked, but on the conservative side.
For a 1.875" thick head
The head blank prior to forming for a 144" OD head is 188".
The head blank prior to forming for a 72" OD head is 99"
It works out that your head blank should have been between 198" and 200" Dia.
What I would do is pull a tape on your head and use that number or check and see if it approaches these numbers or use these numbers and be on the conservative side.
The loss on forming usually isn’t very much though will vary by manufacturer.
These number are from a series of tables that were used for cost estimation of a vessel in the old days.
Part of a class I had in 1967.
I'm still looking for the other information on heads that I had.
For a 1.875" thick head
The head blank prior to forming for a 144" OD head is 188".
The head blank prior to forming for a 72" OD head is 99"
It works out that your head blank should have been between 198" and 200" Dia.
What I would do is pull a tape on your head and use that number or check and see if it approaches these numbers or use these numbers and be on the conservative side.
The loss on forming usually isn’t very much though will vary by manufacturer.
These number are from a series of tables that were used for cost estimation of a vessel in the old days.
Part of a class I had in 1967.
I'm still looking for the other information on heads that I had.
You can calculate the blak diameter before forming from the spreadsheets available at this link.
Click on the 4 Head Spreadsheets link.
But the surface area of a formed head may be 10% more than that of the blank diameter before formation (because of thinning).
Regards,
Believe it or not : Had we trusted Archimedes and assigned him the work of lifting the earth(or any mass equivalent to that of earth on earth),with a lever of suitable length, it would have taken him 23 million million years to lift the earth by one centimeter, if he worked at the rate of 1 HP.
Click on the 4 Head Spreadsheets link.
But the surface area of a formed head may be 10% more than that of the blank diameter before formation (because of thinning).
Regards,
Believe it or not : Had we trusted Archimedes and assigned him the work of lifting the earth(or any mass equivalent to that of earth on earth),with a lever of suitable length, it would have taken him 23 million million years to lift the earth by one centimeter, if he worked at the rate of 1 HP.
Christine74
Mechanical
I worked it out (don't you guys remember calculus?). Ignoring the straight flange, the formula for the surface area of a 2:1 ellipsoidal head is:
A = [(1/4) * pi + (1/24) * sqrt(3) * ln(sqrt(3) + 2) ] * d^2
-Christine
A = [(1/4) * pi + (1/24) * sqrt(3) * ln(sqrt(3) + 2) ] * d^2
-Christine
Christine74,
the calculation of the length of an ellipse and of the area of an ellipsoid are known as being 'elliptic' (as the word says) problems: this means that closed form solutions do not exist for them, only approximate values.
You should revise your calculus exercise.
prex
Online tools for structural design
See if you want to use symbols on these fora
the calculation of the length of an ellipse and of the area of an ellipsoid are known as being 'elliptic' (as the word says) problems: this means that closed form solutions do not exist for them, only approximate values.
You should revise your calculus exercise.
prex
Online tools for structural design
See if you want to use symbols on these fora
Christine74
Mechanical
Actually, you're wrong. This problem can readily be solved by calculating the surface of revolution of an ellipse. If an ellipse has a diameter ratio of 2:1, a its formula can be written:
x ^ 2 / (d / 4) ^ 2 + y ^ 2 /(d / 2) ^ 2 = 1
or
y = sqrt (d ^ 2 / 4 - 4 * x ^ 2)
if you rotate the ellipse about the x-axis, and the surface can be found by integrating from 0 to d/4:
2 * pi * y * sqrt (1 + y' ^ 2)
where y' is the derivative of y.
-Christine
x ^ 2 / (d / 4) ^ 2 + y ^ 2 /(d / 2) ^ 2 = 1
or
y = sqrt (d ^ 2 / 4 - 4 * x ^ 2)
if you rotate the ellipse about the x-axis, and the surface can be found by integrating from 0 to d/4:
2 * pi * y * sqrt (1 + y' ^ 2)
where y' is the derivative of y.
-Christine
Christine74
Mechanical
BTW, I just noticed that the formula I typed yesterday should have read:
A = [(1/4) * pi + (1/24) * pi * sqrt(3) * ln(sqrt(3) + 2) ] * d^2
which works out to A = 1.084d^2
-Christine
A = [(1/4) * pi + (1/24) * pi * sqrt(3) * ln(sqrt(3) + 2) ] * d^2
which works out to A = 1.084d^2
-Christine
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