# 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

Apr 23, 2018

see below

#### Explanation:

$R e = \frac{\rho \times d \times v}{\mu}$
where $\rho$ is density of water = $1 \frac{g}{c m} ^ 3$
d is the diameter = 3 cm
v= velocity
$\mu$ is dynamic viscosity $= {10}^{- 3} K \frac{g}{m \times s} = {10}^{- 2} \frac{g}{c m \times s}$

if Re must be less 2000 you have
$v = \frac{R e \times \mu}{\rho \times d} = \frac{2000 \times {10}^{- 2} \frac{g}{c m \times s}}{3 c m \times 1 \frac{g}{c m} ^ 3} = 6.7 \frac{c m}{s}$

the section of pipe is
$S = \pi \times {r}^{2} = 7 c {m}^{2}$
so you have a flow (F=V/t) in the tube of $F = v \times S = 6.7 \frac{c m}{s} \times 7 c {m}^{2} = 47 {\left(c m\right)}^{3} / s = 0.000047 {m}^{3} / s$

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