Question

An uninsulated steam pipe passes through a room where the temperatures of the air and its walls are 38 degrees. The outer diameter of the pipe is 90 mm, the temperature and spreading rate of its surface are 200 degrees and 0.8, respectively. Calculate the radiating power of the surface and the radiation to the surface. Since the heat transfer coefficient with natural convection from the surface to the air is 25 w/m^2.K, what is the heat loss per unit length for the length of one pipe from the surface of the pipe? wwwwwwwwwwwwwwwwwwwwwwwwwwwam

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Answer 1

air and walls Given :- Temperature of Outer diameter of pipe (do)= 90 mm of room (Ta) 38°C 200°C. Jemperature of pipe's outer surface (Ts) = Em misivity (E). 08 convective heat transfer coefficient for pipe's surface (h) - a5 W/mk. To find :- er of the 1) Radiation falling unit (i) Radiative power the surface (E) to the surface (G) (T) Heat loss th from the pipe lengta P1% 8 = 8 heat loss rate from the pipe Length Pipe L the pipe. > Length Gradiation (G & connection Gradiation (E)

2 the heat by The above diagram depicts that pipe is closing processes i.e. & convection and radiation. And the pipe is gaining from roundings radiation through cheat the surr the The amount by is b t of of heat lost surface per unit area per unit time radiation is called. Emissive power (E). And the amount unit area per unit time on the surface of pipe is called Irradiation. (G). heat falling per

3 رام = 0.8 + - 5.67 Is Emissive power (E)= EFTst E lemissivity of surface Stefan-Boltzmann constant x108 W/m² surface temperature = 200°c - 200+ 273.15) K T = E = 5.67×108 x (0.8) (473. 173.15) + 2, 273.36 W/m² (ii) Irradiation (G) • Tsurroundings 473.15K E = 4 -8 Х G = o T G 25.678 10 (38 + 273-15) = 5.67 X10 x (311.15)4 = 531.44 W/m² 67

4 (iii) Total heat loss from the pipe heat loss due to convection + heat loss due to radiation convection t & radiation Q . ha(Ts- Ta) + E A (Ts 4- Ta*) surface here A area a do L of cylindrical pipe hla do ) (Ts - Ta) + E vlada) (Ts 4- Tat) Q5 W/m2K) (2.(09)(_)(200 - 3 8)) Km? + (0-8)(5.67x10-6)(a (0.09)(_)) ( 473-154 311-159) g = L [1,144.53 +522.3] w 1,666.83 watt /m. rlo :. heat loss unit per length from pipe 1,666.83 Watt/m. -