Question

Find the maximum amount of water that can flow through 3 cm i.d.pie per minutes without turbulence .Take the maximum Reynolds nomber for nonturbulent flow to be 2000?.for water at 20 c, viscosity=1*10^-3 pa.s Ans;0.0028 m^3

Answer 1

see below

Explanation:

`Re =(rho xx d xx v)/ mu`
where `rho` is density of water = `1 g/(cm)^3`
d is the diameter = 3 cm
v= velocity
`mu` is dynamic viscosity `= 10^(-3) Kg/(m xx s) = 10^(-2) g/(cm xx s)`

if Re must be less 2000 you have
`v= (Re xx mu)/(rho xx d)= (2000 xx 10^(-2) g/(cm xx s))/( 3 cm xx1 g/(cm)^3)=6.7(cm)/s`

the section of pipe is
`S= pi xx r^2 = 7 cm^2`
so you have a flow (F=V/t) in the tube of `F= v xx S = 6.7 (cm)/s xx 7 cm^2=47 (cm)^3/s= 0.000047 m^3/s`

you don't say for how many minutes water flow, but from the result you can have
`t= V/F= (0.0028 m^3)/(0.000047 m^3/s) = 59 s` = about 1 minute