# Minimum diameter fluid dynamics

1. May 24, 2010

### pat666

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

A fire-hose must be able to shoot water to the top of a building 35.0 m tall when aimed straight up. Water enters this hose at a steady rate of 0.500 m³.s¯¹ and shoots out of a round nozzle.

(a) What is the maximum diameter that this nozzle can have? (4 marks)

(b) If the only nozzle available is twice as great, what is the highest point that the water can reach?

2. Relevant equations
Vcylinder=pir^2h
gh=1/2v^2

3. The attempt at a solution
first i found the initial velocity required which was 26.21m/s. then 0.5m^3/s=pi*r^2h/s therefore h/s is the velocity, then i solved that and got 15.6cm......... For b i did basically the reverse and got 2.2m.... Can someone please check this.

2. May 24, 2010

### rock.freak667

A = cross-sectional area

so the cross sectional area of a cylinder is ?

hence 0.5=Av.

so what is v?

When you get that, your second equation will come into play now.

3. May 24, 2010

### pat666

ok so ive assumed the hose is a cylinder... the given flow rate is 0.5m^3 per second so therefore 0.5m^3/s is equal to (pi r^2*h/)s ... h is in meters so h/s is the velocity(m/s) as it exits the hose. therefore 0.5=pi*r^2*v.............. Im trying to solve this by logic and am unsure if its right. im guessing that you do not agree with me??

4. May 24, 2010

### rock.freak667

oh sorry, you solved it the other way. I didn't read your post through, sorry.

5. May 24, 2010

### pat666

ok cool thanks for that, just out of interest how do you solve these sort of problems?

6. May 24, 2010

### rock.freak667

Exactly how you solved it, except I would have done it in the reverse to how you did it. Just because I could get the diameter in terms of variables I know the value of so that rounding errors would be reduced.