Answer:
7.26 W/m^2 K ( average heat transfer coefficient )
2052.72 W ( convective heat transfer per unit length of pipe )
Step-by-step explanation:
Data given for this problem
Outer diameter ( O.D ) D = 30 CM = 0.3 m
outer surface temperature Ts = 600K = 327⁰c
atmospheric temperature ( air temp ) Ta= 300k = 27⁰c
assuming ; steady state conditions and atmospheric pressure = 1 atm
properties of air at film temperature which is = 177⁰c = ( (Ts +Ta)/2)
thermal conductivity k = 0.03625 W/m K
Kinematic viscosity , v = 3.17645 * 10^-5 m^2/s
Prandtl number, Pr = 0.6995
β = 1/450 = 2.22 * 10^-3
Rayleigh number (Rad) =
substituting the given values
Rayleigh number = 122.29 * 10^6
Nu =
![[0.6 + (0.387Rad^(1)/(16) )/([1+(0.559/Pr)^(9)/(16)]^(8)/(27) ) ]^2](https://img.qammunity.org/2021/formulas/engineering/college/bsdvzr6sqkzy8hw0r76bf24rxlyj1uv906.png)
substituting all the given values
Nu = 60.05
A) calculate the average heat transfer coefficient
H avg =

= (0.03625 / 0.3 ) * 60.05
= 7.26 W/m^2 K
B) calculate convective heat transfer per unit length of pipe
Q = H avg * (
Dl )(Ts - Ta)
assuming l = 1 m
Q = 7.26 * (
* 0.3 * 1 )(327 -27 ) = 2052.72 W