How Does Fluid Dynamics Affect Pressure and Flow in Constricted Pipes?

[Jeann25]

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

 

[franznietzsche]

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

Looks fine.

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

Yes that is right. The important principle is conservation of mass, you have to have the same mass go in one end that goes out the other end (we're assuming that water is incompressible). So the formula would be:

$$\Delta V_1 = \Delta V_2$$

The volume of water going out side 1 must be the same as the volume going in side 2.

So

$$A_1 v_1 t \rho = A_2 v_2 t \rho$$

Since $$t$$ and $$\rho$$ are the same at both ends of the pipe they cancel and you get

$$A_1 v_1 = A_2 v_2$$

Just like you used. So in fewer words, yes that is correct.

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

The gauge pressure should be the pressure at the left end of the pipe minus one atmosphere, I believe. Maybe someone else knows for sure.