Water flows steadily from an open tank into a pipe as shown in the figure. The e
ID: 2149912 • Letter: W
Question
Water flows steadily from an open tank into a pipe as shown in the figure. The elevation of the top of the tank is 10.4 , and the elevation at the pipe is 2.70 . The initial cross-sectional area of the pipe (at point 2) is 6.90?10?2 ; and at where the water is discharged from the pipe (at point 3), it is 7.00?10?3 . The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe.
Assuming that Bernoulli's equation applies, compute the volume of water that flows across the exit of the pipe in the time interval 1.20 .
Explanation / Answer
Bernoullis from point 1 to 3
g h1 = (1/2) v3^2 + g h2
9.8 * 10.4 = (1/2) v3^2 + 9.8*2.7
v3 = 12.285 m/s is the speed exiting at point 3
I think your post says the area is 7.00 x 10^-3. I can only guess because you didn't post the information properly. No one can answer your question properly without guessing if you don't provide the necessary information.
Anyway..
flow rate = speed * area = 12.285 * 0.007 = 0.0860 m^3/s
Again, the last line of your post says "in the time interval 1.20" It doesnt says seconds, minutes, hours... so I have to guess.
If it is 1.20 seconds
volume = 0.0860 meter cubed = 86.0 liters
If it is 1.20 minutes
volume = 5.16 meter cubed = 5160 liters
If it is 1.20 hours
volume = 309.6 meter cubed = 309600 liters
You'll have to figure out which one you need.
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