Water Flow and Tank Pressure

P Atm works at the discharge at P2 as well.

Recalculating!!!!

my opinion is the water can still flow into tank 2 without any problem
This seem a ludicrous amount of overthink.
The water will flow as long as the water in A is above the outlet...
Havn't you ever noticed that water just does that?

If in doubt, make a model out of a couple of plastic cups and drinking straws (the bendy sort).
Or am I missing some unwritten query about flow rate

Del

snai85

As Del points out, you hardly need to prove this by applied science!

But apart from that, your formulas aren't quite right.

You say P₁= P_G+ P_atm = ρhg + P_atm but then ignore P_atm when evaluating P₁ (of course there was no need to bring it in, as it applies to P₂ as well and drops out when you subtract)

Also you say h = 3 ft but calculate as if it were 3 m.

Cheers.......Codey

Yes sir, thanks. I do realize about I shall neglect Patm because of the subtraction between this two pressure values. Anyway thank you sir. What about the new case that I've posted, is it ok? I am worried because of the elevation of the tanks is not always the same. At the same time, I supposed to close those two tanks to make sure the liquid will not flow out, but I worried if it effect the pressure at the small tank. What do you think about that?

It is the same old fault as made by many.

Note that the pressure in a U tube with big and small branches will be equal on equal distances from the top (without water flowing).

Ignoring friction the pressure that must be used for calculating the flow at P2 can only be the pressure at the corresponding height in the tank.

Note that the static pressure measured at P2 will be zero (measured against P atm)

A lot of people are confusing the lifting force of a bottle jack with pressure while the pressures are actually equal. and the force = P x Area)

And nobody saw the unit (kpa) the heights are 3Km and 2 Km not m of ft

I assumed by 30 000 he meant 30.000 kPa, but it does seem a waste of a lot of 0s!

Codey

snai85

Congratulations you are catching CR4 out here.

To see if it was your intention please give us your definitions

of pressure and force.

You may deserve a quick GA or you should go and stand in the corner.

He is only 24 years

it is ok ,friend

but for more accuracy,please consider the pressure drop in the connecting pipe lines between the two tanks.

I am really appreciate with all of your comments. I do realize there is a lot of silly mistakes I've made during calculating this. Actually, from my point of view for this case, it will just fine as long the water at tank A is not below than outlet level. This is the actual case that I am focus on. This tank would be install in a vessel. One of the thing that I am really concern right now is the ship is not in stable condition due to the sea waves. Please view at the diagram below.(click at the diagram for a better view)

Calculation Example

If the parameters are as below;

Patm = 101 kPa h1 = 4 m
ρ = 950 kg/m3 h2 = 1 m
g = 9.81 ms-2 L1= 0.3m
μ = 0.06 L2= 0.2m
QE = 299 L/hr L3= 0.1m
r = 12.5mm L = L1 + L2 + L6 = 0.6m

Obtaining Pressure P1

P1 = Patm + ρh1g
P1 = 101 kPa + 950 (4) (9.81)
P1 = 138278 Pa
P1 = 138.278 kPa

Obtaining Pressure P2

P2 = Patm + ρh2g
P2 = 101 kPa + 950 (4) (9.81)
P2 = 128958.5 Pa
P2 = 128.9585 kPa

Obtaining Flowrate at Tank 2 Inlet, Q2

Q2 = (P1-P2) x πr4 / 8μL

Q2 = 0.000715084 / 0.288

Q2 = 0.002482932 m3s-1



Flow Difference
Note that QE = 299 L/hr = 8.305556e-005 m3s-1

Δ Q = Q2 - QE

Δ Q = 0.002399876 m3s-1

**Would you advise me what will effect the flow or the pressure if I close the second tank. Thanks

Sorry, the above parameters is not alligned as I type just before I post it. Thanks

Patm = 101 kPa.........................h1 = 4 m

ρ = 950 kg/m3 .........................h2 = 1 m

g = 9.81 ms-2 ..........................L1= 0.3m

μ = 0.06 ..................................L2= 0.2m

QE = 299 L/hr ...........................L3= 0.1m

Please refer to the following example derived at my post 6 of CR4 Thread Selecting Pipe Size.

Again I say overthink mostly caused by using silly units.
If you measure the pressure as say 'inches of water' or the metric equivalent the (static) pressure at the outlet is simply the height difference between the outlet and surface of the water in rank A.
Any text book or a quick google will turn 'inches H₂O' into any unit you want.

Del

I have a related question to add to this thread. During a storm just before christmas we had water pouring over a wall into our garden. The water at the top of the wall was about 3" / 7.5cm high. This wall is about 10m long. At what rate would the water be flowing? It is imposible to be accurate in this because it is a dry stone wall with rough stones on top but I would like to know roughly how much water passed through my garden in the two days it flowed.

regards

Chas

Dear friend,

If water level i.e. h = 3ft = 300mm (approx.),

The pressure at bott om = 0.03 Kg/cm. Sq.

= 30 mmwg (approx).

Since P2 is at 2 ft height,

Pressure at P2 = (300 - 200)/10 000 mmwg

= 10 mmwg

Please convert the figures in other units as per your requirement.

And, if your smaller tank is not sealed at top your water is going to flow out.

If you seal the smaller tank you can not fill it completely since an air pocket will form at the top.

With regards,

R S Sahni

Dear friend,

Sorry for mistakes in calculations. Please:

Read 30 mmwg as 300 mmwg, and

and 10 mmwg as 100 mmwg.

Thanks.

R S Sahni

Dear R S Sahni,

Thanks for your advice. by the way, can you please advise me at the case

I have posted at comment number 14 (which I have replied at comment number 9). There is some typo error which I have highlighted below comment 14.

I am really confused right now is it okay to use that formula or i shall use the Darcy–Weisbach equation? I am confused to identify the flow either laminar or turbulence. Please advise me. Your kind attention is highly appreciated. thanks