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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

1 Answer

4 votes

Answer:

h = 10,349.06 W/m^2 K

Step-by-step explanation:

Given data:

Inner diameter = 3.0 cm

flow rate = 2 L/s

water temperature 30 degree celcius


Q = A* V


2* 10^(-3) m^3 = (\pi)/(4) * (3* 10^(-2))^2 * velocity


V = (20* 4)/(9* \pi) = 2.83 m/s


Re = (\rho* V* D)/(\mu)

at 30 degree celcius
= \mu = 0.798* 10^(-3)Pa-s , K  = 0.6154


Re = (10^3* 2.83* 3* 10^(-2))/(0.798* 10^(-3))

Re = 106390

So ,this is turbulent flow


Nu = (hL)/(k) = 0.0029* Re^(0.8)* Pr^(0.3)


Pr= (\mu Cp)/(K) = (0.798* 10^(-3) * 4180)/(0.615) = 5.419


(h* 0.03)/(0.615)  = 0.0029* (1.061* 10^5)^(0.8)* 5.419^(0.3)

SOLVING FOR H

WE GET

h = 10,349.06 W/m^2 K

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