Bernoulli's equation limitations?

[madchemist]

A firehose must be able to shoot water to the top of a building 35 m tall when aimed straight up. Water enters the hose at a steady rate of 0.500 m3 / s and shoots out of a round nozzle. (a) What is the maximum diameter this nozzle can have? (b) If the diameter of the hose is 10 cm, what is the pressure in the hose 1.00 m below the nozzle?

(Vyf)^2 = (Vyi)^2 + 2ay(yf-yi)
J = AV
A = (pie)r^2
P1 + 0.5pv1^2 + pgh1 = P2 + 0.5pv2^2 + pgh2

Found that maximum diameter in part (a) was 15.6 cm. This means the nozzle diameter is more than 50% larger than the hose diameter. When I applied Bernoulli's equation, I came up with a negative number for the pressure in part (b) like this... P1 = (1.01*10^5) + 500(26.19^2 - 63.66^2) + 9800 = -1572539.75 pa. I'm not sure how to explain it, but I think it has something to do with the fact that the water ejected from the narrow hose would remain at the same velocity, because the is no force causing it to conform to the larger dimensions of the nozzle. Therefore, the pressure would be simply Patm + pgh = 110800 pa?

 

[nasu]

1. What speed MUST the water have at the nozzle in order to reach 35 m height?
The answer has nothing to do with Bernoulli's law.

2. What diameter of the nozzle will provide this speed?
Again, nothing to do with Bernoulli. (you'll get something definitely smaller than the diameter of the hose.)
Hint: continuity equation

3. For part b) you need to use Bernoulli's law.

 

[madchemist]

1. What speed MUST the water have at the nozzle in order to reach 35 m height?
The answer has nothing to do with Bernoulli's law.

2. What diameter of the nozzle will provide this speed?
Again, nothing to do with Bernoulli. (you'll get something definitely smaller than the diameter of the hose.)
Hint: continuity equation

3. For part b) you need to use Bernoulli's law.

I agree to "1". I also agree to "2" that the continuity equation is useful here. I disagree however that the diameter of the nozzle is smaller than the diameter of the hose. I got 15.6 cm for the nozzle and 10 cm for the hose is given. I know it's a funny looking hose, but who am I to argue with the problem statement?
I agree to "3" but as previously stated, Bernoulli's law gives negative Pascals.

 

[madchemist]

I got Vi = 26.19 m/s.

 

[nasu]

I've got the same value.
I think you are right, I did not do the complete calculation before.
Now I did the hole think and I get as you said, around 16 cm for diameter.
Sorry for your time.
It seems that the hose becomes larger at the nozzle (unless there is some mistake in the number for the flow rate).

If the water slows down in the nozzle, then the pressure should increase, compared to the region 1 m below where it flows faster (smaller diameter). It makes sense to have a negative difference.

 

[madchemist]

But isn't there no such thing as NEGATIVE pressure? I think the water ejected from the hose would remain at the same velocity (I calculated it as 63.66 m/s) as it passed by the nozzle (assuming the nozzle isn't very long). In other words there would be no change in velocity, so 0.5pv^2 drops out of Bernoulli's law and we're left with 1.01*10^5 pascals (for the atmosphere) plus the hydrostatic pressure, i.e. pgh. Anyone agree with me?

 

[madchemist]

Is anyone out there?

 

[Doc Al]

But isn't there no such thing as NEGATIVE pressure?

A fluid cannot support tension, only compression. So you're right, the pressure cannot be negative.

I think that the person who devised this problem just made up the numbers without doing a reality check.

 

[madchemist]

Thanks Doc Al and nasu! This problem is, in a sense, "solved".