A horizontal stainless steel (conductivity, k = 15 W/mK) circular pipe of outsid

Question

A horizontal stainless steel (conductivity, k = 15 W/mK) circular pipe of outside and inside
diameters of 0.50 m and 0.45 m, respectively, carries fluid (convection coefficient, h = 200
W/m2K) and passes through a large room whose air and solid surfaces are at 20oC. Given that the
outside of the pipe is un-insulated, with asurface temperature of 80oC, and has a convection
coefficient of 20 W/m2K and an emissivity of 0.9, what is the total heat transfer rate (Watts)
from the pipe per metre length and the temperature of the fluid in the pipe? Note: Stefan
Boltzmann constant is 5.67 x 10-8 W/m2K4.

Explanation / Answer

let,
t1 be the inner surface temperature of the pipe
t2 be the outer surface temperature = 80 degC = 353K
t3 be the surface of ambient atmosphere = 20 degC = 293K

h be the convection co-eff of outer atmosphere

r1 be inner radius = 0.225m

r2 be outer radius = 0.25m

so total lost per unit length = heat lost due to convection + heat lost by radiation

heat lost be convection (per unit length) = h**r2*(t2-t3) = 20**0.25*60 = 942.48 W/m

heat lost by radiation = (emmisivity)*()*(t24-t34) = 0.9*(5.67*10-8)*(3534-2934) = 416.27W/m

net heat lost = 1358.75 W/m

as far as conduction is concerned, you will need the temp of the fluid
if you have the temperature of the fluid,

that is what i have assumed to be t1

if you get the the value of t1, the conduction loss will be [2k(t1-t1)/ln(r2/r1)] which can be added to above answer to get final heat lost

all values except t1 are known and no data except convection co-eff of fluid are known

plz note that the heat that is conducted by the pipe is equal to the heat lost to env by convection and radiation. heat is only transferred to outer surface via pipe by conduction and as heat reaches the outer surface it is simultaneously lost by conv and radiation.

so by calculating the combined loss of radiation and convection you are indirectly calculating heat conducted from fluid to outer surface. you need to know heat lost to env and conduction just occurs in the pipe and not in the outer environment. i guess this is why fluid temp is not given and need not be known. hence the answer remains quite the same i guess.

hope this helps :)

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