Sciencemadness Discussion Board - Physics brain teaser: storage tank filling time

[...]It may be complicated by the change pressure at the output of the pipe as the tank fills.[...]

This will lead to a differential equation. As the height of the water increases, the pressure at the pipe's entry point at the bottom of the tank increases from 1.3 to 1.8 bar. So, you can write the pressure difference between pump outlet and tank inlet as a simple function of h, where h is the height of water, already in the tank.

The time derivative of h can be written as a constant times the inlet flow, the constant depending on the radius of the tank.

aga - 5-3-2016 at 11:54

The way i see it, the water is leaving a 10cm pipe at 3 bar pressure and entering a 3m pipe at 1.3 bar pressure, just that the pipe going in is at the side and not in via the bottom.

Not found a forumula to convert that to flow rate yet.

wg48 - 5-3-2016 at 12:15

The way i see it, the water is leaving a 10cm pipe at 3 bar pressure and entering a 3m pipe at 1.3 bar pressure, just that the pipe going in is at the side and not in via the bottom.

Not found a forumula to convert that to flow rate yet.

You can use the venture formular as I suggested above but that would imply the rate is unaffected by the height of the water but that does not seem valid.

aga - 5-3-2016 at 12:24

With the venturi forumla i found, i get the velocity of the water in the tank being the square root of v₂2 - 3.4

still not found the formula to convert pipe diameter and pressure to flow rate or velocity.

[Edited on 5-3-2016 by aga]

wg48 - 5-3-2016 at 12:25

Quote: Originally posted by woelen

[...]It may be complicated by the change pressure at the output of the pipe as the tank fills.[...]

Yes I agree. But the weakness is the position of the inlet pipe is given as "near the bottom" it seems invalid to assume its at the bottom. The surface pressure could be used but I don't see how that can be valid either see my other post.

blogfast25 - 5-3-2016 at 12:43

Quote: Originally posted by woelen

I think I need the length of the pipe between the pump and the point where it enters the tank.

Not needed: all friction losses are ignored, as stated.

But the weakness is the position of the inlet pipe is given as "near the bottom" it seems invalid to assume its at the bottom. The surface pressure could be used but I don't see how that can be valid either see my other post.

Consider the water to come through a pipe in the bottom of the tank, if you want. It makes no difference to the solution of the problem. in my derivation.

[Edited on 5-3-2016 by blogfast25]

wg48 - 5-3-2016 at 12:44

With the venturi forumla i found, i get the velocity of the water in the tank being the square root of v₂2 - 3.4

still not found the formula to convert pipe diameter and pressure to flow rate or velocity.

[Edited on 5-3-2016 by aga]

You know know the ratio of v1 and v2 given by the ratios of the pipe to tank diameters.
Ok the maths is a bit more complicated than I originally thought

[Edited on 5-3-2016 by wg48]

wg48 - 5-3-2016 at 13:32

Ok so using the ratio of pressures and flow rates we can solve for the file time. I guess the water height effects how hard the pump works but the flow rate is fixed by the pressure ratios.

But I still have no justification for picking the water surface pressure and not the pressure at the height of the inlet pipe.

We know the rate of energy input given by the rate of the water rising in the tank but I don't see how that helps. Can it be used in a fuller continuity equation?

[Edited on 5-3-2016 by wg48]

[Edited on 5-3-2016 by wg48]

aga - 5-3-2016 at 13:38

From Bernoulli (googled) it seems that the velocity V = square root of (2P/d) where P is differential pressure and d is density (=1) so the velocity of the flow is sqrt(2P).

This means that the velocity of the water is sqrt(2*1.7) somethings, associated with pounds, inches and square bits.

Flow rate Q = VA, so sqrt(3.4)*pi*1.5*1.5 = 13.03 is the flow rate, in some unknown units involving metres, inches, pounds etc.

The tank's volume is pi*(3/2)2*5 = 35.35m3

So the tank fills in about 2.71 hours, based on some unknown poundy inchy cuby metre-like unit times hour-1

[Edited on 5-3-2016 by aga]

wg48 - 5-3-2016 at 14:33

From Bernoulli (googled) it seems that the velocity V = square root of (2P/d) where P is differential pressure and d is density (=1) so the velocity of the flow is sqrt(2P).

This means that the velocity of the water is sqrt(2*1.7) somethings, associated with pounds, inches and square bits.

Flow rate Q = VA, so sqrt(3.4)*pi*1.5*1.5 = 13.03 is the flow rate, in some unknown units involving metres, inches, pounds etc.

The tank's volume is pi*(3/2)2*5 = 35.35m3

So the tank fills in about 2.71 hours, based on some unknown poundy inchy cuby metre-like unit times hour-1

[Edited on 5-3-2016 by aga]

I checked the consistence of the units, that got my grey matter working LOL. So with the velocity in m/s density in kg/m**3 the pressure is in Newtons/m**2 . Yes the conversions are yucky especially with bars thrown in.

Ref: N/m**2 - kg/m**3. x (m/s)**2
Simplifying the right side results in kg/(m x s**2)
Multiplying both sides by m**2 results in N equals kg.m/s**2
Which is correct QED

[Edited on 5-3-2016 by wg48]

Fulmen - 5-3-2016 at 14:43

No, it is sufficient to know the pump delivers 3 bar consistently.

You're right, I've just never dealt with idealized problems before. I've calculated many similar problems, but always with some pump specifications (and with pipe losses to calculate). Bernoulli's should give the answer to the flow rate out of the pipe.

aga - 5-3-2016 at 14:44

Not sure what you mean wg48.

Glad your brain cells were mobilised, so what are the units ?

I certainly don't know.

[Edited on 5-3-2016 by aga]

wg48 - 5-3-2016 at 15:09

Not sure what you mean wg48.

Glad your brain cells we mobilised, so what are the units ?

I certainly don't know.

See my addendum

aga - 5-3-2016 at 15:11

what does ** mean ?

wg48 - 5-3-2016 at 15:21

what does ** mean ?

Raised to the power of ie x**2 is x squared, x**3 is x cubed

aga - 5-3-2016 at 15:31

x^3 would be more understandable.

Thanks for the explanation.

Edit:

So what are the units ?

[Edited on 5-3-2016 by aga]

blogfast25 - 5-3-2016 at 15:53

All: you're getting warm with Bernoulli.

Fulmen: I love so-called gedanken experiments. Much can be learned from them.

[Edited on 5-3-2016 by blogfast25]

aga - 5-3-2016 at 15:57

DOH !

So twelvety thingies is not the right answer ?

No wonder i never got a degree.

wg48 - 5-3-2016 at 15:58

x^3 would be more understandable.

Thanks for the explanation.

Edit:

So what are the units ?

[Edited on 5-3-2016 by aga]

I don't have that symbol on my phone key board.

Units means the units of a dimension ie meters are units of length. Newtons are units of force and so on.

Perhaps I answered the wrong question. Do you mean what units you must use in your calculation? If you do then you can can use what ever consistent units you want. I would use SI units ie meters, seconds, kilograms for mass. Newtons for forces. So bars needs to be converted to Newtons per square meter.

[Edited on 6-3-2016 by wg48]

blogfast25 - 5-3-2016 at 16:36

No wonder i never got a degree.

Yours is one from the University of Hard Knocks.

Don't knock it!

Metacelsus - 5-3-2016 at 16:47

The flow velocity at the top of the tank will be 1/3600 the flow velocity through the pipe, from the area ratio.

I've set up the Bernoulli equation:

1/2 * v² + 3*10⁵ N/m² / (1000 kg/m³) = 1/2 (v/3600)² + h * 9.81 m/s² + 1.3*10⁵ / (1000 kg/m³)

where v is the flow velocity at the bottom of the pipe, and h is the height of the water.

However, I'm getting some odd results with it (imaginary numbers). Did I miss anything?

[Edited on 3-6-2016 by Metacelsus]

wg48 - 5-3-2016 at 17:20

Quote: Originally posted by Metacelsus

The flow velocity at the top of the tank will be 1/3600 the flow velocity through the pipe, from the area ratio.

I've set up the Bernoulli equation:

1/2 * v² + 3*10⁵ N/m² / (1000 kg/m³) = 1/2 (v/3600)² + h * 9.81 m/s² + 1.3*10⁵ / (1000 kg/m³)

where v is the flow velocity at the bottom of the pipe, and h is the height of the water.

However, I'm getting some odd results with it (imaginary numbers). Did I miss anything?

[Edited on 3-6-2016 by Metacelsus]

The equation look wrong h x 9.81 needs to be multiply by the mass of one cubic meter of water perhaps your missing brackets or I am falling asleep.

wg48 - 5-3-2016 at 18:38

Having thought about the Bernoulli equation is only valid for the pressure at the height of the entry into the tank. As that is not given I can see no unique solutions. Obviously the height could be ignored but that would be a cheat.

I give up what's the solution then blogfast ???

blogfast25 - 5-3-2016 at 19:08

Having thought about the Bernoulli equation is only valid for the pressure at the height of the entry into the tank.

I give up what's the solution then blogfast ???

Not true at all. The form of Bernoulli's equation to be used is valid at any arbitrary point along the streamline. But you were right in stating one integration is needed to get a filling time.

You can use Bernoulli to prove Torricelli's Law, e.g. But you need to set it up properly.

I'll publish the solution tomorrow. It's very late here now.

So far Metacelsus has gotten the furthest. But he needs to develop a expression for the volumetric flow rate in function of height of water in the tank. Your criticism of his equation was wrong though.

woelen was also basically correct.

[Edited on 6-3-2016 by blogfast25]

deltaH - 5-3-2016 at 21:07

Nice problem blogfast. My 2.3h is the filling time at a constant dP. So the real answer will be larger. It would be interesting to see by how much.

Magpie - 5-3-2016 at 22:44

I converted all pressures to height of water in meters. So the energy balance (Bernoulli eqn) is:

h + 13.3 + (v^2)/2g = 30.6 (eqn 1)

g = acceleration of gravity, 9.81 m/s^2

(v^2)/2g is a loss term for a pipe entrance into a wide opening expressed as a height of fluid.

h is the height of water at any time t (in seconds) and v is the instantaneous average velocity of water in the small pipe in m/s.

V = dh/dt, where V is the instantaneous average velocity in the tank.

v = [(R/r)^2]V, where r and R are radii for the pipe and tank, respectively.

v = [(R/r)^2]dh/dt (eqn 2) (continuity eqn)

Substitute eqn 2 in eqn 1 to eliminate v. Then integrate to definite integrals between h=0 and h=5 and t=0 and t=T. Solve for T the total time.

My solution was T = 16.4 min.

--------------------------------------------------

I know this is not correct as I've ignored the kinetic energy of the incoming stream (v^2/2g) and taken some loss at the entrance which you said to ignore. But I'll leave it up for discussion.

[Edited on 6-3-2016 by Magpie]

[Edited on 6-3-2016 by Magpie]

[Edited on 6-3-2016 by Magpie]

[Edited on 7-3-2016 by Magpie]

wg48 - 6-3-2016 at 02:22

Looking at the problem again and the posts I realize I thought the venturi equation and baliouni equation where the same. Apparently the latter equation (which I was not familiar with) includes the effects of gravity. That explains some of the posts and blogfast,s comments.

But I am still lost how the entrance pipe can be anywhere other than at the bottom of the tank for a solution.

Fulmen - 6-3-2016 at 03:35

Actually the solution would be simpler if it were on top, then the back pressure would be constant. With the inlet on the bottom you have to calculate the increase in back pressure as the level rises. With the inlet anywhere between you'll have to calculate it in two steps, first with constant back pressure until it reaches the inlet, then with increasing pressure.

blogfast25 - 6-3-2016 at 06:56

@Magpie:

I like your approach a lot but don't understand where the numbers 13.3 and 30.6 come from. So I haven't checked the integration.

Edit: and I can confirm that plugging in the numbers into your DV, I get:

$$h+660550\dot{h}^2=17.3$$

And integrated between t=0, h=0 and t, h=H I get t = 17.7 min.

Edit 2: oooopsie, now I can see where these numbers come from and where you went a bit wrong.

[Edited on 6-3-2016 by blogfast25]

My solution:

blogfast25 - 6-3-2016 at 07:01

Start from Bernoulli's Equation. Applied here:

$$\frac{v_z^2}{2}+gz+\frac{p_z}{\rho}=\frac{v_0^2}{2}+gz_0+\frac{p_0}{\rho}$$

$$p_z=1.3\:\mathrm{bar}, p_0=3\:\mathrm{bar}, z_0=0$$

Call:

$$p_0-p_z=\Delta p$$

Then:

$$\frac12(v_0^2-v_z^2)=gz-\frac{\Delta p}{\rho}$$

Let:

$$Q$$

... be the volumetric throughput and as water is incompressible, with A the respective cross section:

$$Q=A_zv_z=A_0v_0$$

So:

$$v_z=\frac{Q}{A_z}, v_0=\frac{Q}{A_0}$$

With some reworking we get:

$$\frac12(v_0^2-v_z^2)=\frac{Q^2}{2} \times \frac{A_z^2-A_0^2}{A_z^2A_0^2}$$

Set:

$$\alpha^2=\frac{A_z^2-A_0^2}{A_z^2A_0^2}$$

So:

$$Q(z)=\frac{1}{\alpha}\sqrt{2\big(\frac{\Delta p}{\rho}-gz\big)}$$

So the volumetric throughput is a function of height, as expected. To calculate filling time, note that (with V volume):

$$dV=A_zdz=Q(v)dt$$

Integrate between z=0 and z=H to find total filling time:

$$A_z\int_0^H\frac{dz}{Q(v)}=t$$

$$A_z=\pi R_z^2$$

$$t=\frac{\alpha\pi R_z^2}{\sqrt{2}} \int_0^H\frac{dz}{\sqrt{\frac{\Delta p}{\rho}-gz}}$$

So then:

$$t=\frac{2\alpha \pi R_z^2}{\sqrt{2}g}\Big[\big(\frac{\Delta p}{\rho}\big)^{1/2} - \big(\frac{\Delta p}{\rho}-gH \Big)^{1/2}\Big]$$

Also:

$$R_0=0.05\:\mathrm{m}, R_z=3\:\mathrm{m}$$

$$\alpha=127.4\:\mathrm{m^{-2}}$$

$$\Delta p=(3-1.3) \times 10^5\:\mathrm{Pa}=1.7 \times 10^5\:\mathrm{Pa}$$

$$\rho=1000\:\mathrm{kg/m^3}$$

$$\frac{\Delta p}{\rho}=170\:\mathrm{m^2/s^2}$$

$$gH=49.05\:\mathrm{m^2/s^2}$$

So:

$$t \approx 1050\:\mathrm{s}=17.5\:\mathrm{min}$$

An interesting thing happens when:

$$gH>\frac{\Delta p}{\rho}$$

Then the equation is no longer valid because the square root of a negative number occurs.

$$H=\frac{\Delta p}{\rho g}$$

... is the maximum height to which the pump can still deliver (zero) volumetric throughput, due to hydrostatic pressure of the water column.

[Edited on 7-3-2016 by blogfast25]

aga - 6-3-2016 at 08:33

Wow ! Just Wow.

That's a fine display of mathsiness there bloggers.

Lost me totally at the integration step.

$$t \approx 166\:\mathrm{s}$$

Hey ! 166s = 2.7 somethings !

Incredible co-incidence.

[Edited on 6-3-2016 by aga]

blogfast25 - 6-3-2016 at 08:54

Wow ! Just Wow.

That's a fine display of mathsiness there bloggers.

Lost me totally at the integration step.

$$t \approx 166\:\mathrm{s}$$

Hey ! 166s = 2.7 somethings !

Incredible co-incidence.

Mathsiness or madness?

You want me to take you through the integration?

aga - 6-3-2016 at 09:05

Having an idea how do an integration would be wonderful.

blogfast25 - 6-3-2016 at 09:29

Well, for this specific integration:

$$t=\frac{\alpha\pi R_z^2}{\sqrt{2}} \int_0^H\frac{dz}{\sqrt{\frac{\Delta p}{\rho}-gz}}$$

$$t=fI$$

Where f is that composite factor upfront and I is the actual integral. This way we can concentrate on the integral:

$$I=\int_0^H\frac{dz}{\sqrt{\frac{\Delta p}{\rho}-gz}}=\int_0^H\big(\frac{\Delta p}{\rho}-gz\big)^{-1/2}dz$$

Make a substitution:

$$u=\frac{\Delta p}{\rho}-gz$$

Differentiate:

$$du=-gdz$$

$$dz=-\frac1g du$$

The integral now becomes:

$$=-\frac1g \int_0^H u^{-1/2}du$$

This is a tabled integral, so from memory or a table:

$$I=-\frac2g \Big[u^{1/2}\Big]_0^H$$

Substitute back:

$$I=-\frac2g \Big[\big(\frac{\Delta p}{\rho}-gz\big)^{1/2}\Big]_0^H$$

$$I=-\frac2g \Big[\big(\frac{\Delta p}{\rho}-gH\big)^{1/2}-\big(\frac{\Delta p}{\rho})^{1/2}\Big]$$

Invert the sign and multiply with f to get the expression for t in my solution.

[Edited on 7-3-2016 by blogfast25]

deltaH - 6-3-2016 at 10:46

I used this online venturi calculator because I'm a lazy shit: http://www.efunda.com/formulae/fluids/venturi_flowmeter.cfm#...

If you do the problem backwards, d=6m pipe constricting to 0.1m pipe and use a constant pressure drop of 1.7bar, I get a fill time of 975s using their answer of volumetric flow rate of 145l/s, which is much longer than your solution. Considering that I didn't even use it correctly the first time, I may well be making some stupid mistake again, but try it yourself and if not, perhaps there's a mistake somewhere (could be in their calculator of course).

Note, the constant pressure drop is just a quick check. The volumetric flow rate with an increasing head should take longer, but not faster, if I'm not mistaken.

[Edited on 6-3-2016 by deltaH]

blogfast25 - 6-3-2016 at 11:01

Quote: Originally posted by deltaH

I used this online venturi calculator because I'm a lazy shit: http://www.efunda.com/formulae/fluids/venturi_flowmeter.cfm#...

If you do the problem backwards, d=6m pipe constricting to 0.1m pipe and use a constant pressure drop of 1.7bar, I get a fill time of 975s using their answer of volumetric flow rate of 145l/s, which is much longer than your solution. Considering that I didn't even use it correctly the first time, I may well be making some stupid mistake again, but try it yourself and if not, perhaps there's a mistake somewhere (could be in their calculator of course).

Note, the constant pressure drop is just a quick check. The volumetric flow rate with an increasing head should take longer, but not faster, if I'm not mistaken.

[Edited on 6-3-2016 by deltaH]

Thanks delta, will look into it now.

Edit: yup. Spotted an error in alpha. Will correct ASAP.

[Edited on 6-3-2016 by blogfast25]

Magpie - 6-3-2016 at 11:37

@Magpie:
I like your approach a lot but don't understand where the numbers 13.3 and 30.6 come from.

One common form of the Bernoulli equation, which is an energy balance, expresses the energy as "height of fluid." So my pressure terms become p/density. Then 3bar = 3bar(10.2m/bar) = 30.6m. But you said you had figured that out.

I didn't get going on this problem until late last night but had my solution by 11PM. Then after going to bed I realized my errors and had to get back up and work on the problem. After going back to bed at 1PM I still had no satisfactory solution. So, thanks for keeping me up all night.

[Edited on 7-3-2016 by Magpie]

blogfast25 - 6-3-2016 at 11:52

deltaH was correct.

My value for alpha was way off, due to a silly mistake. 1050 s is in line with the 974 s for the venturi: it's slightly higher due to gravity (hydrostatic pressure).

[Edited on 6-3-2016 by blogfast25]

blogfast25 - 6-3-2016 at 11:57

Quote: Originally posted by Magpie

Funnily enough that's the form I was taught 30 years ago. But I'm a little rusty.

It kept me up all night too, as I kept finding niggly errors. SORRY!

deltaH - 6-3-2016 at 12:10

deltaH was correct.

My value for alpha was way off, due to a silly mistake. 1050 s is in line with the 974 s for the venturi: it's slightly higher due to gravity (hydrostatic pressure).

[Edited on 6-3-2016 by blogfast25]

HA, that's nothing, I initially didn't hit the calculate button and worked from the 'default' answer given

of 17l/s

Very cute problem though, thanks again!

[Edited on 6-3-2016 by deltaH]

blogfast25 - 6-3-2016 at 12:43

Quote: Originally posted by deltaH

Very cute problem though, thanks again!

Welcome!

aga - 6-3-2016 at 13:59

Just finished filling this newly-built water tank from a bit of 100mm dia pipe and a 3 bar pump.

Took 2.71 hours to fill.

Where Bernoulli ballsed up was in the calculation of how long it took to get the pump running again after some idiot wired it up wrong.

needs to add Ki^d

where K is the idiocy constant, i is the idiot and d is the idiot density.

This is/was an excellent brain teaser, and highly informative.

Post more !

[Edited on 6-3-2016 by aga]

blogfast25 - 6-3-2016 at 15:00

Just finished filling this newly-built water tank from a bit of 100mm dia pipe and a 3 bar pump.

Took 2.71 hours to fill.

Where Bernoulli ballsed up was in the calculation of how long it took to get the pump running again after some idiot wired it up wrong.

needs to add Ki^d

where K is the idiocy constant, i is the idiot and d is the idiot density.

This is/was an excellent brain teaser, and highly informative.

Post more !

[Edited on 6-3-2016 by aga]

I doubt that: Bernoulli was an excellent electrician, although they do say he didn't have the edge on Aristotle!

Maybe a basic course in calculus could be useful?

Ah, C-A-L-C-U-L-U-S!!! Succulent abracadabra, bish, bosh, bash magic of numbers!

aga - 6-3-2016 at 15:40

Some maths would be immensely helpful.

Perhaps the B&D Uni could offer such a course ?

Magpie - 6-3-2016 at 15:48

@blogfast:

I have a couple comments:

Start from Bernoulli's continuity equation. Applied here:

$$\frac{v_z^2}{2}+gz+\frac{p_z}{\rho}=\frac{v_0^2}{2}+gz_0+\frac{p_0}{\rho}$$

minor comment: The Bernoulli equation is an energy conservation equation, not a continuity equation. The equation that defines Q is the continuity equation.

So:

$$Q(z)=\frac{1}{\alpha}\sqrt{2\big(\frac{\Delta p}{\rho}-gz\big)}$$

It looks like you switched signs under the radical. Did I miss something?

Metacelsus - 6-3-2016 at 16:13

Yes, my equation also had the signs switched from what Blogfast wrote (but that was also the equation that gave me imaginary numbers).

Magpie - 6-3-2016 at 16:30

Quote: Originally posted by Metacelsus

Yes, my equation also had the signs switched from what Blogfast wrote (but that was also the equation that gave me imaginary numbers).

Yes, when I tried to correct my solution by removing my entrance loss term and adding the velocity term of the incoming stream I also ended up with a negative quantity under the radical.

blogfast25 - 6-3-2016 at 16:33

Quote: Originally posted by Magpie

So:

$$Q(z)=\frac{1}{\alpha}\sqrt{2\big(\frac{\Delta p}{\rho}-gz\big)}$$

It looks like you switched signs under the radical. Did I miss something?

You are the most apprehensive reader of that derivation, Magpie! :cool:

Kudos.

Yes, there's been a switcheroo of signs and let me explain why.

You are correct that this form of the Bernoulli equation:

$$\frac{v^2}{2}+gz+\frac{p}{\rho}=\text{Constant}$$

... if you multiply it with the mass flow:

$$\dot{m}$$

... is an energy (per unit of time) conservation equation and if we assume there no mass flow loss or increase and no loss or addition of energy, the equation represents transformations of energy (kinetic, potential, internal) between one point on the streamline and another.

However, kinetic energy is direction invariant because of the squares. Squares give the same (positive) outcome regardless whether the argument is positive or negative.

Here we have a situation where the velocity vector:

$$\vec{v_z}$$

points in the opposite direction of gravimetric acceleration:

$$\vec{g}$$

That's why a change in sign is needed because otherwise the expression for

$$Q(z)$$

makes no sense.

I noticed the problem a few days ago when modelling a sinking ship: with a hole in the bottom the ship takes on water due to outside pressure but hydrostatic pressure of water already taken on board actually slows down that flow. There too a change in sign in the argument of Q(z) is needed to represent reality.

[Edited on 7-3-2016 by blogfast25]

Magpie - 6-3-2016 at 17:05

I've never run across anything like this before when using the Bernoulli equation. I'm going to have to continue thinking about it for awhile.

Most textbook problems incorporate friction loss terms due to wall friction as well as "minor losses." Minor losses are due to ells, tees, and other fittings. Maybe that's why I've never seen this issue before.

Oscilllator - 6-3-2016 at 17:08

Wait, I don't get it. If the pressure in the tank is a constant 1.3 bar and the pressure outside is a constant 3bar, then surely there is a constant pressure differential of 1.7bar.

So we have a constant pressure differential from which it is easy to get a flow rate using the cross-sectional area, and then the fill time using the volume.

Where did I go wrong in this reasoning?

blogfast25 - 6-3-2016 at 17:23

Quote: Originally posted by Magpie

I've never run across anything like this before when using the Bernoulli equation. I'm going to have to continue thinking about it for awhile.

If you are, then think 'Hamiltonian' and how the first time derivative of it gives you the Equation of Motion (Newtonian).

Alternatively, determine the time needed for the tank to empty from full (take away the pump, of course!

), using Bernoulli. You'll see no sneaky sign switcheroo is needed then.

Another reason why you may never have encountered is the following. In my own Uni textbook I haven't found a single application of Bernoulli where flow wasn't somewhat downward or simply plain horizontal. In all these cases no sign change is needed to get a Q_v(z) equation.

[Edited on 7-3-2016 by blogfast25]

blogfast25 - 6-3-2016 at 17:28

Quote: Originally posted by Oscilllator

Wait, I don't get it. If the pressure in the tank is a constant 1.3 bar and the pressure outside is a constant 3bar, then surely there is a constant pressure differential of 1.7bar.

So we have a constant pressure differential from which it is easy to get a flow rate using the cross-sectional area, and then the fill time using the volume.

Where did I go wrong in this reasoning?

As the tank fills, the water already in the tank exerts hydrostatic pressure:

$$p_{hydr}= \rho g z$$

with z the height of the water.

The 'effective' pressure is thus:

$$\Delta p_{eff}= \Delta p - \rho g z$$

Flow rate into the tank thus slows down as it fills up.

[Edited on 7-3-2016 by blogfast25]

blogfast25 - 6-3-2016 at 19:10