Amount - Water

Please reply to my question. Thanks.

Somewhere in the problem they should have led you to an equation that looked something like this

m=C*A*Sqrt(2*rho*g*DelP)

where m is flow rate lbm/s
C is a dimensionless flow orifice coefficient between 1 and 0
A is area of opening
rho is fluid density (lbm/ft^3)
DelP is pressure drop (lbf/ft^2)

Dimensions on numbers are important as they provide units for the answer.

A is a circular hole 1/4 inch in diameter so
A(ft^2)=pi*r^2
r=1/8in
so A(ft^2)=3.04x10^4 ft^2
(using scientific notation because this is a small number)

C=1 (worst case)

g is acceleration of gravity (32.2 lbm ft/ lbf s^2)

rho=60 lbm/ft^3
Good approximation for water

DelP=60psi=60 lbf/in^2= 8640 lbf/ft^2

plug and chug

m=0.55 lbm/s

This is a flow rate per second. Now for the second part.--3 months

30 day months==3 months is 90 days

1 day is 24 hours with 3600 seconds per hour--so

T=3 months = 3*30*24*3600=7.8 million (10^6) seconds

Total loss is Q

and

Q= 0.54 lbm/s * 7.8*10^6 s=4.2 million pounds of water.

Q at about 8 pounds per gallon is about a half million gallons

BTW really large amounts of water is given in acre feet--that is an acre of area (43560 ft^2) covered by one foot of depth this is about 350,000 (3.5x10^5) gallons.

So a half million gallons is 1.4 acre feet of water.

Rap

Thanks. I found the information in a chart about amounts of water wasted with different diameter holes. No equation was provided. I was interested in how the amount was calculated.

I am not totally clear on your solution.

Bernoulli's equation ----an early 19th century crunch of energy and the flow of fluids---through and around solids.

Rap

Thanks.

I have a follow-up question - If the opening was rectangular, would the calculations be basically the same?

The actual fluid dynamics is accounted for in the coefficient C of the above orafice flow equation. This C is somewhere between 0 (zero) and 1.

When I crunched the numbers I stated that I used 1.0 , this would be the worst case. If I used this assumption on any shaped hole--the answer would be the same for any shape having the same cross sectional area .

So the answer is yess if the area if the rectangle is the same as the area of the circle.

Rap

Thanks again.

I have a related follow-up question:

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

Would that spill be determined the same way? I don't know the PSI.

Lets look at what we know and ask if it makes sense.

OK start with 4.4E+6 gallons in 24 hours--how many gallons per second is that?

4.4Eg gallons/24 hrs*1 he/3600 sec = 50 gal/s Good flow but believable.

OK what is the cross sectional of an open 4" gate valve....a gate valve is guillotine across a 4" diameter opening.

A=pi*d^2/4 = 3.14159*4*4/4 in^2 = 12.5 in^2

A gallon is 231 cubic inches--so the velocity of 50 gallons per second divided by the opening area.

V=50 gal/s *231 in^3/gal * 1/12.5 in^2 = 925 in/s * 1 ft/in = 77 ft/s

Now I happen to know that 88 ft/s is 60 mph so this waste water was exiting the open valve at close to 52.5 mph

That's a pretty good clip for close to waste water--it was under a pretty good head (head is another way to express pressure--15 pound per square inches is equivalent to 32 feet of water head.

The beauty of Bernoulli is that his equation id a relation--a relation between kinetic (velocity) and potential (elevation) energy.

OK Potential Energy (PE) is given as PE=m*H*g where m is mass H is height and g is acceleration of gravity (32.2 ft/s^2)

Kinetic Energy (KE) is given as KE=1/2*m*V^2 where m is mass and V is velocity in ft/s.

So

m*H*g=1/2*m*V^2

Divide through by m and g

So

H=1/2*V^2/g = 1/2*(77 ft/s)^2/32.2 ft/s^2 = 92 ft That a hell of a standpipe.

As 32 feet is 15 psi, 92 ft is 43 psi

That's a pretty good pressure---I'd be suspicious-- this was more than an open 4" check valve, there was a running pump behind it.

Think of it this way--the typical backyard inground swimming pool is 60,000 gallons--4.4 million gallons is 70 swimming pools full---that almost 3 pools per hour.

Rap

Thanks. I know what you are saying but I am not totally clear.