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s2008m [1.1K]
3 years ago
8

Calculate the convective heat-transfer coefficient for water flowing in a round pipe with an inner diameter of 3.0 cm. The water

flow rate is 2 L/s and the water temperature is 30 °C
Engineering
1 answer:
olasank [31]3 years ago
6 0

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

Q = A\times V

2\times 10^{-3} m^3 = \frac{\pi}{4} \times (3\times 10^{-2})^2 \times velocity

V = \frac{20\times 4}{9\times \pi} = 2.83 m/s

Re = \frac{\rho\times V\times D}{\mu}

at 30 degree celcius = \mu = 0.798\times 10^{-3}Pa-s , K  = 0.6154

Re = \frac{10^3\times 2.83\times 3\times 10^{-2}}{0.798\times 10^{-3}}

Re = 106390

So ,this is turbulent flow

Nu = \frac{hL}{k} = 0.0029\times Re^{0.8}\times Pr^{0.3}

Pr= \frac{\mu Cp}{K} = \frac{0.798\times 10^{-3} \times 4180}{0.615} = 5.419

\frac{h\times 0.03}{0.615}  = 0.0029\times (1.061\times 10^5)^{0.8}\times 5.419^{0.3}

SOLVING FOR H

WE GET

h = 10,349.06 W/m^2 K

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