Question

The hose shown below (point 1) has a radius of 1.25 cm, and the nozzle at the end (point 2) has a radius of 0.45 cm. Point 1 is on the ground and point 2 is 80 cm above the ground. If water is flowing out of the nozzle at 2 L/s,

1. What is the flow rate in units of m³/s?
2. What is the pressure within the hose?

Please show all math steps and how you set up the formulas to get the answers. I don't have the illustration but I don't think it's needed with the info available in the question. Thanks!

Answer 1

flow rate has been given to us = 2 L/s  

A)

now convert Liter to m³

1 m³ = 1000 L

1 L = 1/ 1000 m³

so 2 L = 2/1000 m³ =2*10^-3 m³

so flow rate = 2*10^-3 m³ /s

B)

equation of continuity

A1*V1 = A2*V2 = 2*10^-3

r1 = 1.25 cm = 0.0125 m

V1 = 0.002/ A1 = 0.002/( pi*0.0125^2) =4.0743 m/s

and  

r2= 0.45 cm = 0.0045 m

V2 = 0.002/A2 = 0.002 / ( pi*0.0045^2) =31.4380 m/s

density of water =ρ = 1000 kg/m³

NOW apply Bernoulli’s equation:

P1+, pu} + Pgy1 = P2 + 5 puả + Pgy2.

y1 = 0 m ( on ground)

y2 = 80 cm = 0.80 above ground

P1 + 0.5*1000*4.0743^2 + 1000*9.8*0 = P2 + 0.5*1000*31.438^2 + 1000*9.8*0.80

from here

P1-P2

= 0.5*1000*31.438^2 + 1000*9.8*0.80 - [ 0.5*1000*4.0743^2 + 1000*9.8*0 ]

= 493713.961755 Pa

= 4.937*10⁵ pa answer

I hope this helps you

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