You could try load cells. There have been some previous threads in this forum.

Not knowing the nature of the business and how critical the volume measure is on a time bases. There laser measuring devices that could accurately take a measure and transfer that measure to a program to give real time volume of your storage containers.

OK. From first principles, and assuming the base of the silo is flat and circular for the moment and that the fill point is in the centre:

• Let the diameter of the silo be D
• Let the natural angle of repose of the material be Θ
• Let the height of the cone be H₂
• If 2H₂/D < cos(Θ) then the cone is not touching the sides of the silo. From the actual H₂ calculate D', being the actual diameter of the base of the cone. If D/2H₂ ≈ cos(Θ) then the cone is touching the sides of the silo and is standing upon a cylinder of material which is itself standing upon the flat base of the silo.
• Using either D or D' calculate the volume of material in the cone V₂ (1/3 base area times H₂).
• If the cone is standing on a cylinder, let the height be H₁. Calculate the volume in the cylinder V₁ using diameter D (base area times H₁). [If H₁ is zero then V₁ is zero.]
• Add V₁ and V₂ together to obtain total volume of material V.

Does that look a bit like what is going on in the worksheet?

Is there any allowance in the worksheet for slump as the material flows out?

And what is the new calculation method proposed?

The way my method works (dang, why can't I attach an Excel sheet here?) is taking the physical tank data;

1. diameter
2. cylinder height
3. roof height
4. bottom height
5. bottom bottom diameter (can be zero)

For each gauge location the following values are used / calculated;

1. distance from edge
2. length of gauge pipe
3. in-roof space
4. in-bottom space
5. maximum gauge length (cylinder height + 2. + 3. + 4.)

For the NORMAL calculation I take a "top" and an "edge" gauge measurement and (from the height distance relative to the cylinder top and the horizontal distance between them) calculate the product cone angle. By extending the cone I get the product height at the tank edge and the product cone height. The tank contents, then, are the product cone volume + the cylinder under the product level + the bottom volume. Amusingly this even works for a negative product angle; the product cone becomes a negative value then.

Things get tricky with non normal conditions;

• product level under the cylinder level (using the bottom angle - product angle intersection)
• product level > cylinder height (using product angle - roof angle intersection)

I've got the distinct impression that I'm missing something however?

Hi, if you still want to know the formula in your file than hack it. I mean there must be many tiny softwares that can help you by pass the protection of an Excel file. Try some Googling.

For storage tanks, we have to calibrate them from time to time (may be 5 years intervals), since its shape (roundness) changed by time due to buckling. It must to adapt the measured capacity with the actual capacity.

There are a contractors who measure and collect a lot of data about the ovality occurred to the cylindrical portion of tank and make the required formula.

Finding volume in partially filled tanks

Using an HP-97 desk calculator, you

can easily determine the volume

of liquid in a horizontal tank

if you know the tank dimensions

and the height of the liquid.

Ing. Enniàio Santi, Monda S.p.A., Italy

ð Horizontal cylindrical tanks are common in the chemical process industries. The level of the liquid contained in these tanks is usually indicated by a gage glass, but finding the liquid's volume calls for some tedious calculations, particularly if the tanks have

dished heads. - -

A method of calculating such volumes was given in an article in the June 11, 1973 issue of Chemical Engineering, on p. 130. This technique has been pro­grammed for the HP-97 desk calculator to give quick, accurate answers. As a bonus, the program also calcu­lates volumes of partially filled spherical tanks and of vertical tanks.

The formulas used are:

A_s = (R_c)² cos^-1 [(R_c-h)/R_c]-( R_c- h)(2 R_ch- h²)^1/2 (1)

V_dh =(2p/3)[( R_c)³ -B³]+p[R_cA² -BC²] +

A{(R_c)²[sin^-1(B/R_c)-(p/2)] + BD} —

2/3{(p/2)[3(R_d)² -(R_c)²]R_c-

[2(R_d)² + C² ]B sin^-1(A/C) .+

(A/2)[4(R_d)² -(R_c)²] sin^-1 (D/R_c)-

2(R_d)³ sin^-1 (R_dD/R_CC)-½ABD}. (2)

Note that in the 1973 article, the last two terms of Eq. (2) were incorrectly shown as being positive rather than negative. For Eq. (1) and (2), V_dh = volume of the two dished heads; h = liquid level; R_d = radius of the cyl­inder; R_d = radius of the dished heads; A = [(R_d)² -(R_c)²]^1/2 ; B = R_c - h; C = [(R_d)² - (R_c - h)²]^1/2

D = [(R_c)² - (R_c – h)²]^1/2.

V_h = (ph²/3)(3R_d)²-h) (3)

where: V_h = volume of a spherical segment to level h;

R_d= radius of the: sphere; h=R_d[(R_d)²-(R_c)²]^1/2 =R_d–A.

The combination of the three formulas provides the volume of: a horizontal tank with dished heads- Fig. 1a; a horizontal tank with hemispherical heads- Fig. 1b; a horizontal tank with flat heads-Fig. 1c; a spherical tank- Fig. 1d; and a vertical tank- Fig. 1e.

(text continues on p. 147)

The program first calculates the volume of the two heads and adds the volume of the cylinder—both at level h. For the two heads, if the tank is more than half full, the value of the headspace is calculated and then subtracted from the whole volume of the two heads.

In all cases, if the dimensions are entered in meters, the volume in liters will be obtained by pressing Key A.

If the dimensions are entered in feet, pressing Key B will provide the volume in gallons. (N.B. Before enter­ing program, initialize calculator to radians.)

Horizontal tank with dished heads (Fig. 1a).

Key in:

Example

Liquid level h Enter ­ l.00T

Cylinder radius R_c Enter ­ 2.00Z

Dished-heads radius R_d Enter ­ 6.00Y

Cylinder length L 3.00X

Press key A—~ volume 7,924.97L

Horizontal tank with hemispherical heads (Fig. 1b).

This is a special case in which R_d =R_c

Key in:

Liquid level h Enter ­ l.00T

Cylinder radius R_c Enter ­ 0.50Z

Dished-heads radius R_d Enter ­ 0.50Y

Cylinder length L 2.00X

Press key A—~ volume 2,094.40L

Horizontal tank with flat heads (Fig. 1c).

The same as Fig. 2, but here R_d =0.

Key in:

Spherical tank (Fig. 1d).

The same as before except that the cylindrical part has zero length (L) and the radius R_c is equal to R_d.

Key in:

Liquid level h Enter ­ 2.00T

Cylinder radius R_c Enter ­ 1.00Z

Dished-heads radius R_d Enter ­ 1.00Y

Cylinder length L 0.00X

Press key A—~ volume 4,188.79L

Vertical cylindrical tank (Fig. 1e).

In this case the height of the liquid becomes L; hi is 2R_c and R_d is zero.

Key in:

Diameter of the tank h=2 R_c Enter ­

Radius of the cylinder R_c Enter ­

Radius of the dished heads R_d = 0 Enter ­

Liquid level Press key L

Press key A ® volume

Roy V. Hughson, Editor

The author

Ermimio Santi work for Monda S.p.A., 37058 Sanguinetto (VR),VialeRoma 9, Italy, as production manager in a food plant. His experience has been in food engineering and food processing. He holds a degree in chemical engineering from the University of Padua, is a licensed professional engineer and is registered in the Italian Ordine Professionale degli Ingegneri.

I've used these equations to calculate the incremental (decimal fraction of an inch) volume of several vertical and horizontal storages in an Excel Work Book.

lwegrzyn2000@yhaoo.com

II

Using an HP-97 desk calculator, you

can easily determine the volume

of liquid in a horizontal tank

if you know the tank dimensions

and the height of the liquid.

Ing. Enniàio Santi, Monda S.p.A., Italy

ð Horizontal cylindrical tanks are common in the chemical process industries. The level of the liquid contained in these tanks is usually indicated by a gage glass, but finding the liquid's volume calls for some tedious calculations, particularly if the tanks have

dished heads. - -

A method of calculating such volumes was given in an article in the June 11, 1973 issue of Chemical Engineering, on p. 130. This technique has been pro­grammed for the HP-97 desk calculator to give quick, accurate answers. As a bonus, the program also calcu­lates volumes of partially filled spherical tanks and of vertical tanks.

The formulas used are:

A_s = (R_c)² cos^-1 [(R_c-h)/R_c]-( R_c- h)(2 R_ch- h²)^1/2 (1)

V_dh =(2p/3)[( R_c)³ -B³]+p[R_cA² -BC²] +

A{(R_c)²[sin^-1(B/R_c)-(p/2)] + BD} —

2/3{(p/2)[3(R_d)² -(R_c)²]R_c-

[2(R_d)² + C² ]B sin^-1(A/C) .+

(A/2)[4(R_d)² -(R_c)²] sin^-1 (D/R_c)-

2(R_d)³ sin^-1 (R_dD/R_CC)-½ABD}. (2)

Note that in the 1973 article, the last two terms of Eq. (2) were incorrectly shown as being positive rather than negative. For Eq. (1) and (2), V_dh = volume of the two dished heads; h = liquid level; R_d = radius of the cyl­inder; R_d = radius of the dished heads; A = [(R_d)² -(R_c)²]^1/2 ; B = R_c - h; C = [(R_d)² - (R_c - h)²]^1/2

D = [(R_c)² - (R_c – h)²]^1/2.

V_h = (ph²/3)(3R_d)²-h) (3)

where: V_h = volume of a spherical segment to level h;

R_d= radius of the: sphere; h=R_d[(R_d)²-(R_c)²]^1/2 =R_d–A.

The combination of the three formulas provides the volume of: a horizontal tank with dished heads- Fig. 1a; a horizontal tank with hemispherical heads- Fig. 1b; a horizontal tank with flat heads-Fig. 1c; a spherical tank- Fig. 1d; and a vertical tank- Fig. 1e.

(text continues on p. 147)

The program first calculates the volume of the two heads and adds the volume of the cylinder—both at level h. For the two heads, if the tank is more than half full, the value of the headspace is calculated and then subtracted from the whole volume of the two heads.

In all cases, if the dimensions are entered in meters, the volume in liters will be obtained by pressing Key A.

If the dimensions are entered in feet, pressing Key B will provide the volume in gallons. (N.B. Before enter­ing program, initialize calculator to radians.)

Horizontal tank with dished heads (Fig. 1a).

Key in:

Example

Liquid level h Enter ­ l.00T

Cylinder radius R_c Enter ­ 2.00Z

Dished-heads radius R_d Enter ­ 6.00Y

Cylinder length L 3.00X

Press key A—~ volume 7,924.97L

Horizontal tank with hemispherical heads (Fig. 1b).

This is a special case in which R_d =R_c

Key in:

Liquid level h Enter ­ l.00T

Cylinder radius R_c Enter ­ 0.50Z

Dished-heads radius R_d Enter ­ 0.50Y

Cylinder length L 2.00X

Press key A—~ volume 2,094.40L

Horizontal tank with flat heads (Fig. 1c).

The same as Fig. 2, but here R_d =0.

Key in:

Spherical tank (Fig. 1d).

The same as before except that the cylindrical part has zero length (L) and the radius R_c is equal to R_d.

Key in:

Liquid level h Enter ­ 2.00T

Cylinder radius R_c Enter ­ 1.00Z

Dished-heads radius R_d Enter ­ 1.00Y

Cylinder length L 0.00X

Press key A—~ volume 4,188.79L

Vertical cylindrical tank (Fig. 1e).

In this case the height of the liquid becomes L; hi is 2R_c and R_d is zero.

Key in:

Diameter of the tank h=2 R_c Enter ­

Radius of the cylinder R_c Enter ­

Radius of the dished heads R_d = 0 Enter ­

Liquid level Press key L

Press key A ® volume

Roy V. Hughson, Editor

The author

Ermimio Santi work for Monda S.p.A., 37058 Sanguinetto (VR),VialeRoma 9, Italy, as production manager in a food plant. His experience has been in food engineering and food processing. He holds a degree in chemical engineering from the University of Padua, is a licensed professional engineer and is registered in the Italian Ordine Professionale degli Ingegneri.

This is what I used in an Excel spreadsheet to calculate the partial volumes of various vertical and horizontal (dished head) storage tanks. This system of equations was in service for over twenty (20) years until the plant closed

Larry A. Wegrzyn

Thank you for your feedback (Ad C. Lockhorst here) BUT apparently most respondents quite failed to realize that my question is about a tank used for SOLIDS (well, pelletized material) rather than liquids. For liquids it would be rather easier to produce a calculation method because only ONE gauge would be needed and the calculation would only switch if the measured level was below or above the cylindrical part of the tank.

With solids, however, at least TWO gauges are needed as the product will have an angle (positive or negative): the product level therefore has to be calculated. Even worse is the fact that this calculated level is only valid if it is within the cylindrical section of the tank (otherwise the top / bottom wall intersection with the product angle line has to be found).

I THINK that the method I finally worked out (one base method, five special case methods) is acceptable. If anyone would be kind enough to review it for me (5 pages of text (DOC file) plus a simple calculator program) ..... let me know via alockhorst@chello.nl please. If my method is okay I'll put it up here (after its checked).