- #1
Jeann25
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There is a drawing of a pipe, with the left end (diameter = 5.0 cm, v2) being wider than the right end (diameter = 3.0 cm, v1). Water flows through a horizontal pipe and then out into the atmosphere at a speed v1=15 m/s.
a. What volume of water flows into the atmosphere during a 10 minute period?
Here I used the equation ∆V=v2πr²∆t=(15 m/s)π(.015 m)²(600 s)=6.36 m³
b. In the left section of the pipe, what is the speed v2?
Can I just use A1V1=A2V2? And use a cross-sectional area?
(7.07 cm²)(15 m/s)/(19.6 cm²)=5.41 m/s
c. In the left section of pipe, what is the gauge pressure?
Would the gauge pressure be the difference between the p2 and p1(which is equal to atmospheric pressure=1 atm)? So could I use Bernoulli's Equation as: p2=1/2ρ(v1)²-1/2ρ(v2)²+p1
a. What volume of water flows into the atmosphere during a 10 minute period?
Here I used the equation ∆V=v2πr²∆t=(15 m/s)π(.015 m)²(600 s)=6.36 m³
b. In the left section of the pipe, what is the speed v2?
Can I just use A1V1=A2V2? And use a cross-sectional area?
(7.07 cm²)(15 m/s)/(19.6 cm²)=5.41 m/s
c. In the left section of pipe, what is the gauge pressure?
Would the gauge pressure be the difference between the p2 and p1(which is equal to atmospheric pressure=1 atm)? So could I use Bernoulli's Equation as: p2=1/2ρ(v1)²-1/2ρ(v2)²+p1