Determine Inside Diameter of Horizontal Pipe Constriction

[zetabug]

Homework Statement

The inside diameters of the larger portions of the horizontal pipe in Figure P9.40 are 2.80 cm. Water flows to the right at a rate of 2.00 10-4 m3/s. Determine the inside diameter of the constriction. Answer in (cm)

Here is the picture http://www.webassign.net/sf5/p9_40.gif"

Could you guys please show me how to do this? Not just the answer I want to be able to grasp the concept. Thanks

Homework Equations

I'd Assume
1/2pv^2 + pgy = 1/2 pv^2 + pgy
Has something to do with it, but I don't really know where to start.

The Attempt at a Solution

Honestly I have no idea where to start. How does the info given relate to anything?
From the above equation I can get
1/2(1000)(2e-4)^2 + 1000 *9.8*.1?= 1/2 1000 (2e-4)^2? + 1000 *9.8* .25?

But this doesn't use a lot of the info. Any help would be appreciated.

 

[Doc Al]

That equation is Bernoulli's equation, but you are missing a term for pressure. (Look it up!) Hint: The heights of the fluid in those vertical segments will tell you the pressure in the fluid at those points.

Read this: http://hyperphysics.phy-astr.gsu.edu/hbase/pber.html"

 

[zetabug]

How exactly does the heights give me the pressure? I though pressure=F/Area?
And also is the velocity constant for the entire pipe?

 

[Doc Al]

How exactly does the heights give me the pressure?

Realize that the fluid in the vertical sections is not moving--so pressure just depends on depth.

I though pressure=F/Area?

That's still true.

And also is the velocity constant for the entire pipe?

If the velocity were constant throughout the pipe, you'd have some explaining to do. That would mean that more fluid flowed through the wide section of pipe per second than flowed through the narrow section--but that can't happen, right? (The amount of fluid flowing per second must be the same everywhere--otherwise where is the water going?) Hint: The volume flow rate is given by area*speed.