# Minimum diameter fluid dynamics

## Homework Statement

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?

Vcylinder=pir^2h
gh=1/2v^2

## 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.

rock.freak667
Homework Helper

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.

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??

rock.freak667
Homework Helper
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??

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