Mercury is to be evaporated at 317°C in a furnace. The mercury flows through a 25.4 mm BWG No. 18 gauge 304 stainless-steel tube, that is placed in the center of the furnace. The furnace cross section, perpendicular to the tube axis, is a square 20 cm by 20 cm. The furnace is made of brick having an emissivity of 0.85, and its walls maintained uniformly at 977°C. If the convection heat transfer coefficient on the inside of the tube is 2.8 kW/(m2 K) and the emittance of the outer surface of the tube is 0.60, calculate the rate of heat transfer per meter of tube, neglecting convection within the furnace.

Question

GIVEN

Mercury flow through a tube in the center of a furnace

Mercury temperature (Tm) = 317°C = 590 K

Tube specification: 25.4 mm BWG no 18 gauge stainless steel

Furnace cross section is 20 cm x 20 cm = 0.2 m x 0.2 m

Furnace emissivity (Tube specification: 25.4 mm BWG no 18 gauge stainless steel

Furnace wall temperature (T2) = 977°C = 1250 K

Tube interior heat transfer coefficient (hci) = 2800 W/(m2 K)

Tube exterior emissivity (?1) = 0.602) = 0.85

FIND

The rate of heat transfer per meter of tube

ASSUMPTIONS

Steady state

Convection within the furnace is negligible

SKETCH



the Stephan-Boltzmann constant



for 25.4 mm 18 BWG tubes

Inside diameter (Di) = 22.9 mm= 0.0229 m

Outside diameter (Do) = 25.4 mm = 0.0254 m

for type 304 stainless steel,the thermal conductivity (ks) = 14.4 W/(m K)

Answer 1

The tube and furnace can be thought of as two infinitely long concentric gray cylinders. The rate of radiative heat transfer is





The thermal circuit for this problem is shown below



where

Rci = Convective thermal resistance inside the tube

Rk = Conductive thermal resistance of the tube wall

Rr = Radiative thermal resistance

qc = Convective heat transfer to the tube wall interior



qk = Conductive heat transfer through the tube wall



q1–2 = Radiative heat transfer



For steady state, these three heat transfer rates must be equal



Solving for Tw1 from the first part of the equation



Substituting this into the second part of the equation to solve for T1 yields





Checking the units, then eliminating them for clarity



By trial and error