Can someone provide a formula to estimate the amount of time it would take to drain a small reservoir trap under atmospheric pressure? It holds about 9L and has a small 3/8" tube and I'm trying to find a formula that can give the amount of time it would take to drain it.
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Volume drainage calculation...
- Thread starter krugan
- Start date
quark
It seems the link offers results based on constant liquid levels, not falling as they would on draining. Am I right ?
European laboratory viscometers (Redwood, Saybolt, Engler) measure the time for a fluid to drain between two marks as a direct function of the fluid's viscosity. I wonder whether this concept is applicable to the question in hand.
Most every topic has been discussed at EngTips 
"vonlueke (Structural) 9 Jan 03 17:00
Could someone please advise me on how to calculate tank drainage time for a tank-full of still water in a large, relatively shallow, rectangular, open tank having a circular drain hole in the bottom (or direct me to the correct forum or thread)? (The eng-tips search function doesn't work, in case this was already covered.)"
"vonlueke (Structural) 9 Jan 03 17:00
Could someone please advise me on how to calculate tank drainage time for a tank-full of still water in a large, relatively shallow, rectangular, open tank having a circular drain hole in the bottom (or direct me to the correct forum or thread)? (The eng-tips search function doesn't work, in case this was already covered.)"
To quark, I've made so many mistakes in these forums that your kind remark is unwarranted.
![[blush]](/data/assets/smilies/blush.gif "[blush] [blush]")
Reconsidering..., your link could be used by changing liquid heads to find a relation between height and flowrate which, for a cylindrical vertical vessel of constant cross-section, would enable estimating time of drainage.
To krugan, there are complex formulas for cylindrical vertical vessels involving iterations and some assumptions. Even lab capillary viscometers use mean hydrostatic head in calculating the constants, in line with quark's usually practical and effective advice.
Question: otherwise, could you actually measure the draining time and then "build" backwards a relation between the draining time and the liquid height ?
trashcanman
Mechanical
If the container is handy, why not just fill it up get a stopwatch, pull the plug & time it?
btrueblood
Mechanical
The closed form solution, using the same parameters as the efunda site, would look like:
time to empty = 2*At*sqrt(h)/(C Ad sqrt(2*g))
where
At is the tank XC area (tank is assumed to have same area at every height "h")
h is initial height of fluid above discharge
C is discharge or flow coefficient
Ad is the discharge area
g is gravitational acceleration
Double checked that solution with a numerical simulation, and it checks to within a few percent. The derivation of the above formula uses seperation of variables. Start by writing
(1) dh/dt = -K (h)1/2
where K = CA(2g)1/2/At
eq. 1 can be rewritten
(2) dt = -k dh/(h)1/2
where k = 1/K
integrate both sides to get
(3) t = -2k h1/2 + A
(A is a constant of integration, which we can discard, since we really want to know the delta-time. Also, it bugged me that the answer is a negative number, thus verifying it with a numerical sim. I just discarded the negative sign in equation above.)
time to empty = 2*At*sqrt(h)/(C Ad sqrt(2*g))
where
At is the tank XC area (tank is assumed to have same area at every height "h")
h is initial height of fluid above discharge
C is discharge or flow coefficient
Ad is the discharge area
g is gravitational acceleration
Double checked that solution with a numerical simulation, and it checks to within a few percent. The derivation of the above formula uses seperation of variables. Start by writing
(1) dh/dt = -K (h)1/2
where K = CA(2g)1/2/At
eq. 1 can be rewritten
(2) dt = -k dh/(h)1/2
where k = 1/K
integrate both sides to get
(3) t = -2k h1/2 + A
(A is a constant of integration, which we can discard, since we really want to know the delta-time. Also, it bugged me that the answer is a negative number, thus verifying it with a numerical sim. I just discarded the negative sign in equation above.)
sailoday28
Mechanical
btrueblood (Mechanical)
where K = CA(2g)1/2/At What is A in this relation?
(3) t = -2k h1/2 + A You state A=0. What happens at t=0?
Regards
where K = CA(2g)1/2/At What is A in this relation?
(3) t = -2k h1/2 + A You state A=0. What happens at t=0?
Regards
btrueblood
Mechanical
Sailoday,
in the first equation "A" is the discharge area.
In equation 3, t=0 corresponds to a tank with zero initial height, i.e. it takes zero time to drain an empty tank. Thus, the constant of integration is taken to be zero. The final equation, as given earlier states that a tank with an initial height "h" takes time "t" to drain. Again, in going from eq. 3 to the earlier equation, I ignored the -sign.
Sorry about the mixup, also, it looks like the prior thread already covered this. I'd left the thread & a half-finished reply hanging on the browser before submitting (had other things to do) and didn't see CarlB's post before hitting send.
in the first equation "A" is the discharge area.
In equation 3, t=0 corresponds to a tank with zero initial height, i.e. it takes zero time to drain an empty tank. Thus, the constant of integration is taken to be zero. The final equation, as given earlier states that a tank with an initial height "h" takes time "t" to drain. Again, in going from eq. 3 to the earlier equation, I ignored the -sign.
Sorry about the mixup, also, it looks like the prior thread already covered this. I'd left the thread & a half-finished reply hanging on the browser before submitting (had other things to do) and didn't see CarlB's post before hitting send.
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