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
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.
7 answers
When I calculate what you did, I don't get the same answer.
well, I may have botched it. What calculations do you have?
(6000)(1/86400)*(1/3)sqrt(5/2Pi)
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
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
Where does the 5 come from?
6000/86400 = 60/864 = 5/72
5/72 = pi/4 d^2
d^2 = 5/72 * 4/pi = 5/(18pi)
forget your algebra skills?
5/72 = pi/4 d^2
d^2 = 5/72 * 4/pi = 5/(18pi)
forget your algebra skills?