One of the new programs in the software library (Thanks, Don, for pointing them out to us) is a small HP-42S program that calculates the volume of a horizontal cylindrical tank, given its current fluid level, by Ken Delsnider:
http://www.hpmuseum.org/software/42tankv.htm
Perhaps the following could be a companion program to it, but some discussion is required. Years ago at work my chief, also an electrical engineer, gave me a book on Fluid Mechanics and asked me to try to solve this problem. I had not been an outstanding student in that discipline (quite the contrary!), but after an hour or so I managed to find a formula (by following a solved example for a vertical tank in the book). I tested it with a 20-liter water bottle and a small piece of a plastic pen body as a nozzle and it worked. I guess it may work for diesel oil as well (that was the fluid in the original problem) given its low viscosity, but I don't know what discrepancy one could expect when using the formula below. The constant in line 21 has to be recalculated for British units.
Formula
t = 4*L*(sqrt((D - h2)^3) - sqrt((D - h1)^3))/(3*S*c*sqrt(2*g))where
t = time do drain from upper to lower level [s]
L = length of the tank [m]
D = diameter of the tank [m]
h1 = upper level of fluid [m]
h2 = lower level of fluid [m]
S = cross-sectional area of the nozzle [m^2]
g = acceleration of gravity [m/s^2]
c = nozzle coefficient (dimensionless)
Gerson.
---------------------------------------------------------------------------------------------(*) The nozzle constant is a dimensionless constant
TIME TO DRAIN A HORIZONTAL CYLINDRICAL TANK FROM AN INITIAL LEVEL TO A FINAL LEVEL OF STORAGE00 { 103-Byte Prgm }
01>LBL "T2MT"
02 MVAR "DIA"
03 MVAR "HT"
04 MVAR "HTF"
05 MVAR "LEN"
06 MVAR "NDIA"
07 MVAR "NC"
08 MVAR "T"
09 RCL "DIA"
10 RCL- "HTF"
11 3
12 Y^X
13 SQRT
14 RCL "DIA"
15 RCL- "HT"
16 3
17 Y^X
18 SQRT
19 -
20 RCL× "LEN"
21 383330.627104 ; (8/15)*10^7/(pi*sqrt(2*9.80665))
22 ×
23 RCL "NDIA"
24 X^2
25 RCL× "NC"
26 ÷
27 RCL- "T"
28 ENDDIA = Diameter [m]
HT = Initial level of storage [m]
HTF = Final level of storage [m]
LEN = Length of tank [m]
NDIA = Nozzle diameter [mm]
NC = Nozzle constant (*)
T = Time [s]
related to the ratio of the length and diameter
of the nozzle, according to the following table:
| NCThe formula doesn't take the fluid viscosity into account. This works for water and other low viscosity fluids.
-------------
l<d | 0.62
l=2d | 0.82
l=3d | 0.82
l=12d | 0.76
l=24d | 0.73
l=48d | 0.63
l=60d | 0.60
l=100d | 0.50
Example:
Given the following data, calculate the time to empty a horizontal cylindrical water tank.
DIA = Diameter 1.488 m
HT = Initial level of storage 0.177 m
HTF = Final level of storage 0.000 m
LEN = Length of tank 2.988 m
NDIA = Nozzle diameter 50.800 mm
NC = Nozzle constant (*) 0.620
Shift SOLVER T2MT
1.488 DIA
0.177 HT
0 HFT
2.988 LEN
50.8 NDIA
0.62 NC
\/ T --> 224.808926919 seconds3600 / Shift CONVERT ->HMS --> 00h 03m 45s
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