Asked by Angie

I went to my teacher numerous times. I'm still not using the right formulas or equations. Please help.

A commonly used rule of thumb is that average velocity in a pipe should be about 1 m/s or less for "thin" (viscosity about water). If a pipe needs to deliver 6,000m^3 of water a day, what diameter is required. My teacher added flow area multiplied by the velocity equals the volumetric flow rate. The answer should be in cm.
Answered by Steve
if the diameter is d meters, the cross-section of the pipe has area pi/4 d^2 m^2

In one second, at 1 m/s, then, pi/4 d^2 m^3/s of water will flow through the pipe

now just plug in the numbers:

6000m^3/day * 1day/86400s = pi/4 d^2 m^3/s
d = 1/3 √(5/2pi) = 0.297m
Answered by Angie
When I calculate what you did, I don't get the same answer.
Answered by Steve
well, I may have botched it. What calculations do you have?
Answered by Angie
(6000)(1/86400)*(1/3)sqrt(5/2Pi)
Answered by Steve
you did not read carefully above: if you solve for d in the equation

6000m^3/day * 1day/86400s = pi/4 d^2 m^3/s

you get (ignoring the units)

6000/86400 = pi/4 d^2
d^2 = 5/(18pi)
Hmmm. I did have a typo above, but I still get
d = .297m = 29.7cm
Answered by Angie
Where does the 5 come from?
Answered by Steve
6000/86400 = 60/864 = 5/72

5/72 = pi/4 d^2
d^2 = 5/72 * 4/pi = 5/(18pi)

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