# Homework Help: Continuity Equation

1. Nov 11, 2007

### Roger Wilco

1. The problem statement, all variables and given/known data
The inside diameters of the larger portions of the horizontal pipe as shown in the image (attached) are 2.50 cm. Water flows to the right at a rate of 1.80*10^4 m^3/s. What is the diameter of the constriction.

2. Relevant equations Continuity equation Rate of Volume flow=constant.

3. The attempt at a solution
I got a proportion worked down to
$$\frac{V_i}{V_o}=\frac{D_o^2}{D_i^2}$$

where V is velocity and the subscripts i and o denote in and out respectively.

I am having trouble figuring out how to utilize the two heights of the fluids in the vertical tubes.

RW

Edit: I am now considering using Bernoulli's equation to take into account pressure. Sound better?

Last edited: Nov 11, 2007
2. Nov 11, 2007

### rl.bhat

Using equation of continuity find the velocity of flow in larger tube. And using Bernoulli's equation find the velocity at constriction

3. Nov 11, 2007

### Roger Wilco

There is also pressure in the larger tube (left hand side) though, so wouldn't it be Bernoulli's only? I must admit, I have been eating cookies for the last 20 minutes....let me have another go here. :)

RW

4. Nov 11, 2007

### rl.bhat

Before asking next question you should have found the velocity of flow in the larger tube. Now a1v1 = rate of flow of water. From this find v1.
According to Bernoulli's equation P1 + 1/2dv1^2 = P2 + 1/2dv2^2. where d is the density of water. From this find v2.

5. Nov 12, 2007

### Roger Wilco

So, do I need to even consider the flow at the left-most portion of the tube (where the 10cm vertical pipe is)?

Or can I just look at it where h= 5 cm and the rightmost portion of the tube?

I have no trouble sorting out the algebra. My trouble is with which parts of the tube I actually need to condider.

RW

6. Nov 12, 2007

### Roger Wilco

Wow. This one is really giving me trouble if I choose to use all three sections of the tube. Maybe I can just limit it to the constricted portion and the right-most portions?

Last edited: Nov 12, 2007
7. Nov 12, 2007

### Roger Wilco

This isn't working out at all right now.

So i have so far from continuity equation:

Subscript A for rightmost portion of tube and B for the constricted portion under the 5cm vertical tube.

$$R_v=constant$$

$$\Rightarrow R_v=V_AA_A$$

$$\Rightarrow \pi r^2_AV_A=R_v$$

$$\Rightarrow V_A=\frac{1.8*10^{-4}}{(.025)^2\pi}=9.167*10^{-2} \frac{m}{s}$$

From Bernoulli:

$$p_B+.5\rho V_B^2+\rho gy_B=p_A+.5\rho V_A^2+\rho gy_A$$ p=0, rho is factor of all

$$\Rightarrow V_B=\sqrt{V_A^2-2gy_B}=DNE$$ The quantity I get under the radical is NEGATIVE.

Do I need to include the 10cm height? If so how?

I reall need some help on this one

8. Nov 12, 2007

### Roger Wilco

So I used this where P1= rho*g*y1 and P2=rho*g*y2 (i.e., Bernoulli's Equation)
Then instead of calculating V, I used $$R_v=AV$$

$$\Rightarrow V=\frac{R_v}{A}$$

$$\Rightarrow V= \frac{R_v}{\pi(.5 D)^2}$$
RW

Last edited: Nov 12, 2007
9. Nov 12, 2007

### rl.bhat

You have made 2 mistakes. 1) 2.5 cm is the diameter and not the radius. 2) PA is not equal to PB. PA = rho*g*0.10 and PB = rho*g*0.05. Since tube is horizontal rho*g*yA = rho*g*yB. Now try it again.

10. Nov 12, 2007

### Roger Wilco

I don't want to seem ungrateful, but I think you should reread my response again before assuming that I have made more errors. It is even more confusing when you say that I am incorrect when I am not.

As you can see from above quote: P1=rho*g*y1 and P2=rho*g*y2
which is exactly what you wrote.

And nowhere did I indicate that I planned on using 2.5cm as the Radius. I would be plugging in V=(R_v)/[pi*(D/2)^2]

Other than the miscommunication, I suspect that this is the correct approach.

RW

Last edited: Nov 12, 2007
11. Nov 12, 2007

### rl.bhat

I posted my responce when I received this:
Do I need to include the 10cm height? If so how? Both of us were wrighting the responce simultaneously. 7.24 and 7.32. When I was camposing and sending the responce I might have received your responce, which I don't know how to see simultaneously. Any way I am sorry.

Last edited: Nov 12, 2007
12. Nov 12, 2007

### Roger Wilco

Don't be I was just trying to be sure that we were on the same page. I think I can wrap this one up now Thank you for your help rl. bhat!

RW