Question

A house with its own well has a pump in the basement with an output pipe of inner radius 6.4 mm. Assume that the pump can maintain a gauge pressure of 420 kPa in the output pipe. A showerhead on the second floor (7.5 m above the pump's output pipe) has 38 holes, each of radius 0.34 mm. The shower is on "full blast" and no other faucet in the house is open.


(a) Ignoring viscosity, with what speed does water leave the showerhead?
m/s

(b) With what speed does water move through the output pipe of the pump?
m/s


Sorry for reposting this, but i still don't understand it. Could someone please give a step for step equation or explaination on how to do this. Thank you.

Answer 1

Bernoulli's equation.

Pressurebase+1/2 rho*V1^2 + rho*gh1 =Pressureabove+1/2*18 rho*V2^2 + rho*gh2

let the base be h1=0 ; pressure above=0, pressure base= 420kPa
h2=7.5m
rho is density water.

one equation, two unknowns (V2, V1)

Equation of Continunity:
Area1*V1=Area1*V2

now you can solve the equation for V1, V2, just a tad of algebra :)

Answer 2

I got this

(420+(1/2)1000(V1^2)+1000(0)= 0+(1/2)18(1000)V2^2+1000(9.8)(7.5)

so it's 920 V1^2=82500 V2^2

but I don't understand how to get part a and b n how to solve for V1^2 and V2^2 and what does Equation of Continunity:
Area1*V1=Area1*V2 mean I mean what is the area and how can I get V1 and V2. Sorry, but am confused! Thank you.