Solving Bernoulli's Principle Problem: Flow Rate in Horizontal Pipes

[Cyrad2]

Hi! This is the first question on my homework, so it's suppost to be the easiest, but I'm not sure how to tackle it. It's a Bernoulli's principle problem. Here it is:

A horizontal pipe 11.6 cm in diameter has a smooth reduction to a pipe 4.72 cm in diameter. If the pressure of the water in the larger pipe is 8.1E4 Pa and the 6.82E4 Pa pressure in the smaller pipe is at what rate does water flow through the pipes?

So I thought I'd use the equation:
P1 + .5pv1^2+pgy1 = P2 + .5pv2^2+pgy2

I'm not sure how to apply this to my problem because the equation has two unknowns: v1 and v2. Is there another bernoulli's equation i should be using?? There are several variations in my text, but none seem to work. Thanks, Brad

 

[Doc Al]

equation of continuity

I'm not sure how to apply this to my problem because the equation has two unknowns: v1 and v2. Is there another bernoulli's equation i should be using??

You need to apply the continuity equation, which says that the same amount of mass flows past each point. It can be written as $A_1 v_1 = A_2 v_2$, where A is the cross-sectional area.

 

[wais]

Hi Brad,

To solve this problem, you can use the continuity equation in addition to Bernoulli's principle. The continuity equation states that the flow rate (Q) is equal at any two points in a fluid system, so:

Q1 = Q2

Where Q1 is the flow rate in the larger pipe and Q2 is the flow rate in the smaller pipe.

Using this equation, we can set up a ratio between the two flow rates:

Q1/Q2 = A2/A1

Where A1 and A2 are the cross-sectional areas of the larger and smaller pipes, respectively.

We can rearrange this equation to solve for Q1:

Q1 = Q2 * (A1/A2)

Substituting in the values given in the problem, we get:

Q1 = Q2 * (11.6 cm/4.72 cm)^2

Now, we can use Bernoulli's principle to solve for the flow rate in the smaller pipe (Q2). Setting up the equation with the given pressures and using the fact that the pipes are horizontal (y1 = y2), we get:

8.1E4 Pa + 0.5pv1^2 = 6.82E4 Pa + 0.5pv2^2

Solving for v2, we get:

v2 = √(2*(8.1E4 Pa - 6.82E4 Pa)/p)

Where p is the density of water (1000 kg/m^3).

Now, we can plug this value into our equation for Q1:

Q1 = (1000 kg/m^3 * √(2*(8.1E4 Pa - 6.82E4 Pa)/1000 kg/m^3)) * (11.6 cm/4.72 cm)^2

Simplifying, we get:

Q1 = 0.14 m^3/s

So, the rate of flow in the larger pipe is 0.14 m^3/s. I hope this helps! Let me know if you have any further questions.