What is the speed and pressure of water exiting a cylindrical tunnel in a dam?

[chawkdee]

Homework Statement

A cylindrical water tunnel runs through a dam. The entrance to the tunnel has a radius of 1.30m and is located at a depth of 22.0m below the surface of the water. The other end of teh tunnel has a radius of 0.840m., is 46.0m below the water's surface, and is open to the air.
a) What is the speed of the water when it exits the tunnel?
b) Determine the absolute pressure of of the water just after it enters the tunnel

Homework Equations

v=Sq Rt (2gh)

P1 + (1/2 x density x v1 squared) + (density x g x h1) = P2 + (1/2 x density x v2 squared)+ (density x g x h2)

The Attempt at a Solution

I really don't know where to start as we don't know the velocity at the entrance.

v= the square root of 2 x 9.81 x (46-22) = square root 470 = 21.6 m\s

We can ignore density.

But we don't know P1 or P2 so I'm stumped! Please help me understand the question and if I'm using the correct equations.

 

[minger]

Let's start with the outlet, it's slightly easier.

The other end of teh tunnel has a radius of 0.840m., is 46.0m below the water's surface, and is open to the air.

Can you say anything now about the exit pressure?

Now the inlet. There is a critical piece of information that you're only thinking of in one way.

A cylindrical water tunnel runs through a dam. The entrance to the tunnel has a radius of 1.30m and is located at a depth of 22.0m below the surface of the water.

anything? ;)

 

[chawkdee]

Minger, thanks for your time.

Firstly, id the speed correct?

v= the square root of 2 x 9.81 x (46-22) = square root 470 = 21.6 m\s

The exit pressure:-

pressure of liquid = specific gravity of liquid x height of liquid xgravitational acceleration

sp gravity of water=1

h=46m

g=9.8

p=1 x 9.8 x 46

p= 450 ?

 

[chawkdee]

A cylindrical water tunnel runs through a dam. The entrance to the tunnel has a radius of 1.30m and is located at a depth of 22.0m below the surface of the water.

I can't figure out how I can find the Pressure if it is not open to the air.

 

[minger]

First of all, let's look at your calculation. Specific gravity is dimensionless as it's merely a ratio. So, we have
[tex]p = (1)(46 m)\left(9.8 \frac{m}{s^2}\right)[/tex]
This gets us:
[tex]p = 450 \frac{m^2}{s^2}[/tex]
I don't know where you come from, but m2/s2 is NOT a unit of pressure. First rule of engineering/physics/etc: UNITS UNITS UNITS!

Had you written that out, you would see that the proper formula for hydrostatic pressure is:
[tex]p = \rho g h[/tex]
Density, not specific gravity.

Now, at the outlet, the fluid releases to the atmosphere. That probably means atmospheric pressure, or 0 psig (gauge pressure). At the inlet, you can use the hydrostatic pressure formula I provided. With this, you have both elevations and pressures at both the inlet and outlet. At the velocity you can simply assume that the fluid is at rest since it's coming from basically a resevoir.

This leaves only outlet velocity to calculate for.

 

[chawkdee]

P1 + (1/2 x density x v1 squared) + (density x g x h1) = P2 + (1/2 x density x v2 squared)+ (density x g x h2)

So P2 is atmospheric pressure = 1 x 10*3 Pa
(density x g x h2)= 1x10*3 x 9.8 x 46 (do we need the 46 if it's open to the air?)
(density x g x h1)= 1x10*3 x 9.8 x 22
v1=0


If the above are correct, I can do it - thanks a lot!