Answer:
h = 10,349.06 W/m^2 K
Explanation:
Given data:
Inner diameter = 3.0 cm
flow rate  = 2 L/s
water temperature 30 degree celcius
[tex]Q = A\times V[/tex]
[tex]2\times 10^{-3} m^3 = \frac{\pi}{4} \times (3\times 10^{-2})^2 \times velocity[/tex]
[tex]V = \frac{20\times 4}{9\times \pi} = 2.83 m/s[/tex]
[tex]Re = \frac{\rho\times V\times D}{\mu}[/tex]
at 30 degree celcius [tex]= \mu = 0.798\times 10^{-3}Pa-s , K Â = 0.6154[/tex]
[tex]Re = \frac{10^3\times 2.83\times 3\times 10^{-2}}{0.798\times 10^{-3}}[/tex]
Re = 106390
So ,this is turbulent flow
[tex]Nu = \frac{hL}{k} = 0.0029\times Re^{0.8}\times Pr^{0.3}[/tex]
[tex]Pr= \frac{\mu Cp}{K} = \frac{0.798\times 10^{-3} \times 4180}{0.615} = 5.419[/tex]
[tex]\frac{h\times 0.03}{0.615} Â = 0.0029\times (1.061\times 10^5)^{0.8}\times 5.419^{0.3}[/tex]
SOLVING FOR H
WE GET
h = 10,349.06 W/m^2 K