Assuming fully developed flow for the water, determine the required tube length
ID: 1717342 • Letter: A
Question
Assuming fully developed flow for the water, determine the required tube length per pass Is the fully developed assumption reasonable? A 1-kW commercial electrical heating rod, 8 mm in diameter and 0.3 m long is to be used in highly corrosive gaseous environment. Therefore, it is to be provided with a cylindrical shell of fireclay. The gas flows by at 120C, and h is 230 W/m^2.degreeC outside the shell. The surface of the heating rod cannot exceed 800degreeC. Derive the temperature distribution in the fireclay shell,. Find the maximum shell thickness and the outer temperature of the fireclay. The corrosive gas flow stops accidentally due to a malfunction in the process while the rod is still heating at steady-state. What would be the surface temperature of the rod if it is made of AISI304 steel? Will the rod melt? You are a blacksmith working for the Renaissance Festival and need to make a genuine mace (a weapon that consists of a 10 cm outside diameter ball with spikes on it, see figure). First you heat the mace to 600degreeC in the furnace, and then drop it into oil (T = 100 degree C. p = 1050 kg/m^3, C_0 = 1200 J.kg.K) to cool it Both the ball and the spikes are made of steel whose properties (k = 15.0 W/m.K. C_0 = 1050 J/ko K p = 4500 kq/m^3) are temperature insensitive. The spikes may be thought of as fins where each fins area is At= 0.01 m^2 and each fin covers 0/001 m^2 of the original ball's (spheres) surface The efficiency of these fins is 0.7 and the heat transfer coefficient into the oil is 5 W/m_.K. The total volume of the should be taken to be 0.00055 m' and ter characteristic length is 10 cm.Explanation / Answer
Heating rod diameter = 8mm = 0.08 m
Length of rod L = 0.3 m
Gas flow temperture = 120 0C
Heat transfer coefficient h = 230 w /m^2 oC
Resistance R = 1 / 2 x pie x h x r1
1/ 2 x 3.14 x 230 x 0.08
Heat transfer Q = T1 -T2 /R
Where T1 = initial temperture ,T2 = final temperature in the fire clay shell
1000 = 120-T2 /0.0023
T2 = tempertature in fire clay = 117.69oC
Q = T1 - T2 / 1 / 2x pie x k x L (ln r2 / r1 )
1000 = 800- 117.69 / 2 x 3.14 x 1.3x 0.3 (ln r2 /0.04)
r2 = 0.181 m
Shell thickness = 0.181 - 0.04
= 0.141m
= 141 mm
outer temperature of the fire clay
Q = T1 -T2 / R
R = 1/2X 3.14 X L (1 /h1r1 + 1/k1 (ln r2 /r1 )
1/2 x 3.14 x 0.3 (1 x 230 x 0.04 + 1 / 1.3(ln 0.181 / 0.04)
= 0.05769
Q = T1 -T2 / 0.05769
1000 = T1 -T2 / 0.05769
T1 -T2 = 57.69
800 - 57.69 = T2
T2 = 742 oC
If the steel rod is used then thermal conductivity of steel rod AISI304 K = 14.9 W/mK from data book then
Q = T1 - T2 /1/ 2 x 3.14 x 14.9 x 0.3 (ln r2 /r1 )
= 800 - 117.69 / 0.05269
= 12949 W
It will melt because heat transfer is so higher than 1000 W
R = 1 /2 X 3.14 X 14.9 X 0.3 (ln r2 / r1)