(41) Bulk Cylindrical Tank

[Thomas Klemm]

(10-14-2018 10:15 PM)Dieter Wrote:  The paper uses a "different" formula for the slant bottom section and assumes a spherical dome top with a given radius (which is different from the calculated radius).

This is something in the paper that I don't understand: Why would they determine the radius R graphically or by measurement when it can be calculated?

(10-15-2018 12:33 PM)Dieter Wrote:  2. If the dome top section is actually spherical there is exactly one solution: the radius R must be 90".

From your excellent sketch I concluded:

\(
\begin{align*}
a^2+r^2=R^2=(a+v_0)^2&=a^2+2av_0+v_0^2 \\
r^2&=2av_0+v_0^2
\end{align*}
\)

And thus:

\(2a=\frac{r^2-v_0^2}{v_0}\)

Let \(x\) be the radius of the surface in the dome on level \(v\).
Then:

\(
\begin{align*}
x^2 &= R^2-(a+v)^2 \\
&= R^2-a^2-2av-v^2 \\
&= r^2-(2a+v)v
\end{align*}
\)

And thus:

\(\left ( \frac{x}{r} \right )^2=1-\left ( \frac{2a}{r} + \frac{v}{r} \right )\frac{v}{r}\)

But since:

\(\frac{2a}{r}=\frac{r^2-v_0^2}{rv_0}=\frac{r}{v_0}-\frac{v_0}{r}\)

We end up with:

\(\left ( \frac{x}{r} \right )^2=1-\left ( \frac{r}{v_0}-\frac{v_0}{r} + \frac{v}{r} \right )\frac{v}{r}\)

Here's the modified program for the HP-42S:

Code:

00 { 72-Byte Prgm }
01▸LBL "AREA"
02 MVAR "t"
03 RCL "t"
04 RCL "h0"
05 X<Y?
06 GTO 00
07 ÷
08 2
09 ×
10 1
11 -
12 +/-
13 ACOS
14 LASTX
15 RCL ST Y
16 SIN
17 ×
18 -
19 RTN
20▸LBL 00
21 -
22 RCL "d"
23 X<Y?
24 GTO 01
25 PI
26 RTN
27▸LBL 01
28 -
29 RCL÷ "r"
30 RCL "r"
31 RCL÷ "v0"
32 ENTER
33 1/X
34 -
35 RCL+ ST Y
36 ×
37 1
38 X<>Y
39 -
40 PI
41 ×
42 END

Of course the constant \(\frac{r}{v_0}-\frac{v_0}{r}\) could be stored in another variable instead of calculating it again and again:

Code:

30 RCL "r"
31 RCL÷ "v0"
32 ENTER
33 1/X
34 -

Examples:

195 XEQ "TANK"
13,318.1905

198 XEQ "TANK"
13,509.0116

228 XEQ "TANK"
14,488.2093


Here I used \(10^{-12}\) for ACC. Do these results agree with your calculations?

Kind regards
Thomas

[Albert Chan]

(10-15-2018 12:33 PM)Dieter Wrote:  Take a look at this sketch which should be quite exactly to scale:



Red solid line: elliptical dome top
Blue dotted line: spherical dome top

Very nice sketch. Thanks !

To get the 1943 gallon dome, I assumed the author was not stupid.
There must be a reason he measured R of 78" (not calculated 90")

If what he was doing is do get a fit with same cross section area, it all make sense.
With R=90", blue dotted line always under-estimated true dome volume.

With R=78", blue-dotted line will go pass the Dome Top.
This over-estimated area should cancelled out the lower under-estimated part.

Also, 6" liquid volume is better approximated as cylinder of 6" high. (Red line almost vertical)

For Elliptical Dome Top, I assumed bottom part of dome is a cyclinder, top part as elliptical dome.
Matching both tank height and tank volume. The revised dimensions is

Height = (slant height = 12) + (cylinder height = 190.67) + (elliptical dome height = 25.33)

With this revised numbers, the tank should match better to actual shape ...

Quote:Now imagine a radius of 75,22". That's not much more than the tank radius r, so the dome top section would be much higher than v₀ = 36" ...

That R were for a different dome shape, Spherical Segment Dome Top.
The dome is actually flat at v0 = 36", with radius = sqrt(2320) ~ 48.17"

You can confirm with this (set a1=72, a2=48.17, h=36): https://rechneronline.de/pi/spherical-segment.php

[Dieter]

(10-15-2018 04:31 PM)Thomas Klemm Wrote:  This is something in the paper that I don't understand: Why would they determine the radius R graphically or by measurement when it can be calculated?

You take the words right out of my mouth. ;-)

(10-15-2018 04:31 PM)Thomas Klemm Wrote:  We end up with:

\(\left ( \frac{x}{r} \right )^2=1-\left ( \frac{r}{v_0}-\frac{v_0}{r} + \frac{v}{r} \right )\frac{v}{r}\)

Fine.

(10-15-2018 04:31 PM)Thomas Klemm Wrote:  195 XEQ "TANK"
13,318.1905

198 XEQ "TANK"
13,509.0116

228 XEQ "TANK"
14,488.2093

Here I used \(10^{-12}\) for ACC. Do these results agree with your calculations?

Yes, these are exactly my results for a spherical dome top. Both with the classic formula, determining R first, as well as with yours where R is not explicitely calculated.

Dieter

[Dieter]

(10-15-2018 05:52 PM)Albert Chan Wrote:  To get the 1943 gallon dome, I assumed the author was not stupid.
There must be a reason he measured R of 78" (not calculated 90")

So the 78" have been measured, and not calculated.

(10-15-2018 05:52 PM)Albert Chan Wrote:  If what he was doing is do get a fit with same cross section area, it all make sense.
With R=90", blue dotted line always under-estimated true dome volume.

How do you know if the blue dotted line (R=90) underestimates the true dome volume? How do you know the true dome value in the first place?

But maybe this is another argument for an elliptical dome top. ;-)

(10-15-2018 05:52 PM)Albert Chan Wrote:  With R=78", blue-dotted line will go pass the Dome Top.
This over-estimated area should cancelled out the lower under-estimated part.

Here is another picture:



Red line: elliptical dome top
Blue line: spherical dome top, R=90"
Green line: spherical dome top, R=78"

I don't think that things cancel out here. Even if the liquid level in the dome is limited to v₀. "Mr. Chan, I am not convinced". ;-)

(10-15-2018 05:52 PM)Albert Chan Wrote:  With this revised numbers, the tank should match better to actual shape ...

Since noone knows the actual shape we cannot conclude this.
All we know is that we know nothing. #-)

Dieter

[Dieter]

(10-15-2018 05:52 PM)Albert Chan Wrote:  There must be a reason he measured R of 78" (not calculated 90")

I just came across another problem. The formula for the spherical dome top include R and Vo as parameters. We have already seen that R=78 does not match Vo=36. But if you do the calculation with these two contradicting parameters, as suggested in the paper and shown in the results in table 25-3, you get volumes that are less (!) than those with R=90 and V0=36. This is because you do not (!) calculate the volume within the green line in the sketch above. To calculate the latter Vo would have to be adjusted to 48".

Here are some results for different ombinations of R and Vo.

Code:

level     R=90 Vo=36     R=78 Vo=36     R=78 Vo=48     Ellipsoid
----------------------------------------------------------------
 198        13509          13444          13521          13533
 208        14035          13889          14118          14167
 218        14370          14175          14591          14628
 228        14488          14276          14911          14805

I wonder what the figures in the R=78 and Vo=36 column mean. They are the lowest of all possible parameter combinations.

Dieter

[Albert Chan]

(10-15-2018 07:36 PM)Dieter Wrote:  Since noone knows the actual shape we cannot conclude this.
All we know is that we know nothing. #-)

Well put ...

To make everything consistent, the dome have to be very flat on top, but possible.
(even flatter than ellipsoidal dome, the red line, in post 44)

The problem with the setup is that graphically fitted R is then thrown away.
The maximum v is 36", so the carefully fitted volume is wasted ...
Worst, this wasted piece affected calculated dome volume (badly under-estimated).

I like the ellipsoidal dome setup, since the shape is adjustable.
(If v0 is allowed to change, like what I did in post 36)

[Dieter]

(10-15-2018 09:56 PM)Albert Chan Wrote:  The problem with the setup is that graphically fitted R is then thrown away.

The advantage is that no cumbersome and inaccurate graphically fitted R is required.

(10-15-2018 09:56 PM)Albert Chan Wrote:  The maximum v is 36", so the carefully fitted volume is wasted ...
Worst, this wasted piece affected calculated dome volume (badly under-estimated).

Apropos "carefully fitted": what about this idea?

1. Assume an elliptical dome top. But do not enter its height Vo.
2. Instead enter a max. allowed liquid level (this can be the hatch height, or a limit set by the manufacturer) and the corresponding max. volume.
   Calculate Vo from this.
   

In our example the tank may have an allowed liquid level of 215" with a volume of 14500 gallons. For an elliptical dome top this means Vo = 34,8942".

This way we get:

Code:

level    volume
------------------
 192"    13113 gal
 198"    13532 gal
 205"    13988 gal
 210"    14270 gal
 215"    14500 gal

How do you like that?

Dieter

[Albert Chan]

(10-16-2018 08:57 PM)Dieter Wrote:  Apropos "carefully fitted": what about this idea?

1. Assume an elliptical dome top. But do not enter its height Vo.
2. Instead enter a max. allowed liquid level (this can be the hatch height, or a limit set by the manufacturer) and the corresponding max. volume.
   Calculate Vo from this.
   

Good idea. Now, we only need 4 inputs, instead of 5:

1. tank radius r
2. slant bottom height h0
3. total height H
4. total volume V

With above, we can calculate elliposoid dome height v0, cyclinder height h1:

v0 = 3 * (H - h0/2 - V/(Pi r²))
h1 = H - h0 - v0

Edit: I would avoid using (in, gallon) units, and use (in, in³) instead.
This way, the code also work in other units, say (ft, ft³), (cm, cm³), (m, m³), ...

[SlideRule]

My last post on this thread, enjoy!

extract from volume 5, issue 4 (April 2016) International Journal of Research in Engineering and Technology Semi-Elliptical Head on a Vertical Tank & Torispherical Head in Vertical Tank.

BEST!
SlideRule
edit: [ missing ('s & )'s restored ]

[Thomas Klemm]

Here are some parentheses: (, (, (, ), ), )
Just sprinkle them here and there in the document where they are missing.

Cheers
Thomas

[rprosperi]

(10-18-2018 02:10 AM)Thomas Klemm Wrote:  Here are some parentheses: (, (, (, ), ), )
Just sprinkle them here and there in the document where they are missing.

LOL!!

I thought it was me at first (or maybe a bad scan?)!

Is that some kind of 'special' notation, or just an incredibly bad job of proof-reading?

[SlideRule]

(10-18-2018 01:11 PM)rprosperi Wrote:  … thought it was me …at first … maybe a bad scan?) … 'special' notation … bad job of proof-reading?

One [ ( ] was missing in the original, the others were dropped in the extract process. A first, for me, using Acrobat XI Pro. I'ld like to blame it on my very recent cataract surgery but most likely just negligent proof reading.

BEST!
SlideRule

[SlideRule]

On second thought, just one more post: an extract from volume I of a Structural Analysis Manual from General Dynamics / Convair & Space Systems Division
[attachment=6506]

BEST!
SlideRule