Amount - Water

[Guest]

What have you tried to do since.............
Did you look up Bernouli equations.........
Energy is constant.....cannot be created or distroyed.
Pressure energy + velocity energy + static head is constant throughout
So pressure is 60 psi.....we can convert this to equivalent static head
The pressure will be converted to velocity energy as the water flows.
from that we get the velocity.............

[Guest]

'What have you tried to do since............. I don't know how.
Did you look up Bernouli equations......... Yes- don't understand.

"Pressure energy + velocity energy + static head is constant throughout
So pressure is 60 psi.....we can convert this to equivalent static head" No idea what you mean.

[Guest]

This is a maths forum
Why are you asking questions that seem so difficult to you. Subjects you must not be studying
I would have thought you would ask question relating to your studies and as such would have some idea but just needed help.
It seems this question is totally out of your depth.
It is like if my car does not work properly and I want a mechanic to fix it. If all I tell him is "the car is not working" it is not much help to the mechanic. I need to tell him more precisely what the problem is. Like "it does not go into 3rd gear". Then the mechanic knows the engine is OK and everything else is OK so all he has to fix is the gearbox.

In your case you just say you" cannot do it". What is it you cannot do?

"did you use the formula for area of a circle pi x d x d /4" pi x r^2.....Are you querying the use of this formula

Pi x d x d / 4 is the same as pi x r^2 so why do you query it...???

r = d /2 So r^2 = (d/2)x(d/2) = (d^2)/4

Did you find anything out about flow through pipes, holes, orifices etc.

[Guest]

"In your case you just say you" cannot do it". What is it you cannot do?" I am confused how to solve.

"Did you use the formula for area of a circle pi x d x d /4" pi x r^2.....Are you querying the use of this formula

Pi x d x d / 4 is the same as pi x r^2 so why do you query it...??? Was not sure if right formula.


did you find anything out about flow through pipes, holes, orifices etc." Yes - don't understand the formulas.

I was reading about water conservation and saw a table with different size diameter holes with water wasted over a 3 month period. I was just curious how to calculate.

Sorry I posted the question and wasted your time.

[Guest]

Please post some of the figures from you table for different sizes of pipes etc over 3 months period so I can see if my results compare

[Guest]

"Please post some of the figures from you table for different sizes of pipes etc over 3 months period so I can see if my results compare"


1/16 " - 9,850 cu. ft. 280 cu. meters

3/16 " 89,031 " " 2,521 " "

[Guest]

Additional: 1/8" - 296,000 gallons.

[Guest]

Theoretical solution................
1/4" hole in a tank leaks a continuous flow for 3 months. Find the volume of water lost.
The pressure in the tank is 60 psi.

dia in mm is 0.25 x 25.4 = 6.35 mm

area of flow is 3.142 x 6.35 x 6.35 / 4 = 31.673 sq mm = 0.3167 sq cms

1 psi = 2.31 feet of water guage = 2.31 x 0.3048 = 0.704 metre water guage

Pressure of water (h) = 60 x 0.704 = 42.25 metres water guage

velocity of flow = Sqroot(2xgxh) = sqroot( 2 x 9.81 x 42.25) = 28.79 m/s = 2879 cms/s


Time is 3 months (take 3x30 days) = 90 x 24 x 60 x 60 = 7776000 seconds

Quantity of water = Area x Velocity x Time = 0.3167 x 2879 x 60 x 60 / 1000 = 3282 litres per hour.

Quantity of water = Area x Velocity x Time = 0.3167 x 2879 x 7776000 / 1000 = 7089995 litres water

Equals........ = 7089995 / 1000 = 7090 cu. metres

The above figures assume full flow and no friction or turbulance effects

Normally a coefficient of discharge of approx 0.8 to about 0.95 would be taken

So 7090 x 0.85 = 6000 cu metres approx OR 6000000 / 4.4 = 1363636 gallons

divide that by 4 gives 340909 gallons for a 1/8" hole

which is roughly same as your table value for 1/8" hole

...........................Simple........................

[Guest]

Thanks.

"velocity of flow = Sqroot(2xgxh) = sqroot( 2 x 9.81 x 42.25) = 28.79 m/s = 2879 cms/s"


I have a question:

What is the "g" ?

Also, what if the opening was a 1/4 " square ? Would the amount be calculated basically the same ?

[Guest]

I used the metric system of units because I am familiar with them and didn't need to look up any conversions.
I worked out the area in sq centimetres............
I worked out the velocity in centimetres per second
I multiplied these together to give cubic centimetres per second. There are 1000 cubic centimetres in 1 litre and 1000 litres in a cubic metre.
The velocity is worked from the change from potential energy of the pressure to the kinetic energy of the velocity of the flow.
The formula I used needed the pressure converted to an equivalent metres head of the liquid flowing (water). that is what height of column of water would exert a pressure of 60 PSI.
Pressure(N/m^2) = density(Kg/m^2) x Acceleration due to gravity(m/s^2) x Head or height of water(m).
But I just used the conversion that 1 PSI = 2.31 feet of water OR 0.704 metres water.
g is the acceleration due to gravity = 9.81 metres per second squared (m/s^2).

Pressure energy = (density(kg/m^3) x gravity(g) x head(h)) and this is converted to kinetic energy ) = 0.5 x mass(kgs) x (velocity(m/s))^2
Density is 1000 kg/m^3, So for per unit flowrate the above becomes g x h = 0.5 x (velocity(v))^2

OR velocity(v) = sqroot(2 x g x h) metres per second (m/s)

And Area of flow x velocity give the quantity of flow per second and multiply by time to get total.

Formula for velocity and flowrates through holes and pipes are what is known as emperical formulae....that is derived from experiment or testing.....sometimes using models. Normally there is a coefficient of velocity or coefficient of discharge applied to allow for the effect of sharp edged holes or rounded edges or streamline flow or turbulant flow.

Similarily you ask about a square hole the methods of calculation would be the same, you would consider the area of the square hole and apply a coefficient maybe 0.7 in this case. The flow from a square hole will not be square and will not have a constant velocity across it.

[Guest]

Thanks for the detailed explanation, patience, and understanding.

[Guest]

I have a follow-up question:

The city where I reside recently had a spill of partially treated wastewater from the municipal water treatment plant. The spill was from the mechanical failure of a 4" check valve. Over a 24 hour period the spill was determined to be 4.4 million gallons.

My question: Would the amount of wastewater spilled by calculated in the same manner?

[Guest]

ARISTOTLE said........What we have to learn to do we learn by doing............

You should work some of these thing out yourself..............

based on the earlier calculation we got...........for a 1/4" hole.........
3282 litres per hour = 746 gallons per hour

= 17902 gallons in 24 hours

Based on same pressure and velocity but the diameter is now increased by a factor of 16 to 4 inches
So area will be up by factor od 16 x 16 = 256 times so flowrate is 256 time the 1/4" hole.

256 x 17902 = 4582912 gallons in 24 hours = 4.58 million gallons......not a bad guess..????

You should read up on ARISTOTLE and take his advice.................

...................................Simple........................

[Guest]

Thanks.

"You should read up on ARISTOTLE and take his advice................."

I will - Thanks again.