# Diameter of a stream of water from a faucet

1. Feb 19, 2013

### sluo

1. The problem statement, all variables and given/known data

Consider a stream of water flowing from a kitchen faucet. At the mouth of the faucet, the diameter of the stream is 0.960 cm. The stream fills a 125 cm^3 container in 16.3 seconds. Find the diameter of the stream 13.0 cm below the mouth of the faucet.

2. Relevant equations

Equation of continuity: A_1v_1 = A_2v_2
Bernoulli's equation: P_1 + 0.5ρv_1^2 + ρgh_1 = P_2 + 0.5ρv_2^2 + ρgh_2

3. The attempt at a solution

Ok, so using the flow rate and the diameter at the mouth of the faucet we can solve for v_1. We can then write v_2 = (A_1v_1)/A_2 and plug this into Bernoulli's equation and solve for A_2 from which we can extract the diameter. My main question is what is the pressure difference? I assumed P_1=P_2 and from this I got an answer that is close to what the book gives, but not close enough for me to think I'm doing it right. Can anyone tell me what the pressure difference ought to be and why? Thanks!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 19, 2013

### haruspex

I would think that the main cause of a pressure difference from atmospheric would be from surface tension, but it doesn't sound like you should be taking that into account.

3. Feb 19, 2013

### rude man

I don't see why there should be a pressure difference. p = 1 at. at A1 and A2. Roundoff error? Converted 125 cm3 correctly into m3?

4. Feb 19, 2013

### danielakkerma

Since the water stream "descended" a little, a pressure difference will form due to the gain in velocity.
Remember Bernoulli's equation:
Suppose the initial velocity was v_1, and take the faucet's height to be at relative level 0; Then surely:
$$\frac{\rho {v_1}^2}{2}=\rho \frac{{v_2}^2}{2} - \rho gh$$
Then use continuity.
And you are given parameters to find out the initial velocity(v_1):Recall the relations between volume, speed, and area.
(Is the correct answer, by the way, 0.245 cm?)

5. Feb 19, 2013

### haruspex

This is in open air. Why, other than surface tension, would the pressure at any point differ from atmospheric?

6. Feb 19, 2013

### sluo

The book lists the correct answer as 0.247 cm. I keep getting 0.268 cm. It seems from your equation above, that you took P_1 = P_2 even though you said that there should be some difference. So is there a pressure difference or not?

7. Feb 19, 2013

### sluo

Here is my work:

$$A_1v_1 = 125 cm^3/ 16.3 s \\ v_1 = 125/(16.3 *\pi*0.48^2) = 10.6 cm/s$$

Next, write $$v_2 = A_1v_1/A_2$$ and plug into Bernoulli's equation. If we take $$P_1 = P_2$$, then the density cancels out and it simplifies to

$$2gh = v_2^2 - v_1^2 = \Bigg(\frac{A_1^2}{A_2^2} - 1\Bigg)v_1^2$$

From this we get

$$A_2^2 = \frac{A_1^2}{\frac{2gh}{v_1^2} + 1}$$

Plugging in values, we have

$$A_2^2 = \frac{\pi .48^2}{\frac{2*980*13}{10.6^2} + 1} = .00319$$

So

$$A_2 = .05698 = \pi r^2$$

Solving for r and multiplying by 2, I get 0.268 cm. The book says 0.247 cm is the correct answer.

8. Feb 19, 2013

### rude man

What gives here? You cite the Bernoulli equation without pressure appearing in it, thereby contradicting yourelf!

9. Feb 19, 2013

### rude man

I went thru the op's math & came up with the same answer. daniel?

Last edited: Feb 19, 2013
10. Feb 20, 2013

### haruspex

Try $A_2^2 = \frac{\left(\pi .48^2\right)^2}{\frac{2*980*13}{10.6^2} + 1}$
I think you got 2.68cm.

11. Feb 20, 2013

### danielakkerma

Sorry, I must have misunderstood...
There is obviously no external pressure variation.
For some reason I thought you were referring to the change in Dynamic pressure, which obviously exists.
Apologies all around!
And it seems you have done everything correctly.

12. Feb 20, 2013

### haruspex

Not so - see my previous post (#10)

13. Feb 20, 2013

### danielakkerma

I see what you mean, though through your method, one still gets 0.268 cm, which the OP stated is incorrect.
But, if you do this:
$$v_1 = \frac{V}{A_1 t} \approx 10.5947 \ \frac{cm}{s} \\ {v_2} = \sqrt{{v_1}^2+2gh} \approx 159.976 \ \frac{cm}{s} \\ A_2 = \frac{A_1 \ v_1}{v_2} \approx 0.0479367 \ {cm}^2 \\ d_2 = 2\sqrt{\frac{A_2}{\pi}} = 0.247 cm$$
Which is what the OP wanted to obtain...

14. Feb 20, 2013

### rude man

OK, so no roundoff error!

Last edited: Feb 20, 2013
15. Feb 20, 2013

### sluo

Thank you, yes I was forgetting to square the A_1. And the answer is .247cm, not 2.47cm. I kept everything in cm throughout. Thank you!!

16. Feb 20, 2013

### haruspex

It was never a round-off error; it was out by an order of magnitude.