You have an asbestos cement covered pipe with k = 0.208 W/m. k, it has an inside
ID: 1719133 • Letter: Y
Question
You have an asbestos cement covered pipe with k = 0.208 W/m. k, it has an inside radius of 0.5 in., and external heat-transfer coefficient of 1.5 Btu/h. ft^2. degree F, an inside temperature of 250 degree F and outside temperature of 70 degree F. Determine the critical radius in cm. Determine the equation in cm. Make a table of r_0 versus q/L and plot (not by hand) q/L (Btu/h. ft) versus r_0 (in) for the total range = r_1 to r_0 = 1.5 in. Let r_0 in inches be 0.5, 0.6, 0.7, 0.8, 0.9, 0.961, 1.0, 1.1, 1.2, 1.3, 1.4, 1.5. Determine the maximum q/L from the graph.Explanation / Answer
a) Critical radius = Rcrit= ?
The critical radius is ration between thermal conductivity and convectioanl heat transfer coefficient
Rcrit = k / h
= 0.208 W/mK / 0.26 W / m2 K
= 0.8m
ii) q / L = k A dT
q / L = 0.208(pi r02) (394.21 -294.26)
q / L = 0.208(3.14)(0.0127)2 (394.21 -294.26)
= 0.010 J / m
iii) A lot of values are possible for the graph , the basic equation for the graph will be given above
q / L = k A dT
q / L = k pi dT ro2
calcuating the values of kpi dT = x
so , altering the value of ro we will see q / L would also be changed. Please keep q / L on y axis and ro should be kept on x axis. Please draw graph by assigning different values to the ro for which you will find q /L.
iv) Maximum q / L from graph will be the peak value of q / L, which is known as maximma. where change in slope become zero.........
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