1. Thermal energy is converted in to em waves: all body above absolute zero emit radiant energy via photons.2. photons can travel in space without media @ light3. when photon strike another surface they are absorbed /reflected/ transmitted but absorbed and converted in to heat in perfect black body.

Radiation rate equation is described by the STEFAN BOLTZMAN LAW: Total emission from a black body per unit area per unit time is proportional to forth power of absolute temperature of the body.

E=Ts4 (W/ m 2 )

Where: = emissivity, which depand on surface, finish and material ( = 1 is black body) = Steffan Boltzman constant = 5.67 x 10 -8W/m 2 K 4 .Ts = Absolute temperature of the surface (K)This eq does not give heat exchange, for heat exchange

E=A(Ts4 - Tsur4 )

Where: Tsur = Absolute temperature of surroundings. (K)EXAMPLE1:Two perfect black bodies surround each other such that all radiant energy of inner surface at 1000oC reaches outer surface at 200oC find net rate of heat transfer per unit area.Sol Ts1 = 200+273 = 473Ts2 = 1000+273= 1273E/A=(Ts14 Ts24 )= 146kW/m2EXAMPLE2:5 cm dia pipe at steady state temp 60 oC kept in a room of temp 25 oC , =0.7, h= 6.5 W/m 2 KCalculate total heat loss / unit lengthSol Conv.q = h As T= h ( x d x l)(60-25)= 6.5x ( x 0.05x1)(60-25) = 35.72WRad.q= A(Ts14 Ts24 )= 0.7 x ( x d x l) x 5.67 x 10 -8 ((60+273)4-(25+273)4)q= 33.72WtotalQ=35.72+33.72=69.44W

OVERALL HEAT TRANSFER COFF.:

qci= hiA(Ti-Ta) = (Ti-Ta) Rciqk = kA(Ta-Tb) = (Ta-Tb) l Rkqco = hoA(Tb-To) = (Tb-To) Rco

qci = qk = qco = qNow we can measure Ti and T o but not Ta and Tb so we eliminate them

(Ti-Ta) + (Ta-Tb) + (Tb-To) = q Rci + q Rk + q RcoTi To = q (Rci + Rk + Rco)q = (Ti To) (Rci + Rk + Rco)By Newtons law of coolingq = UAT = (Ti To) (Rci + Rk + Rco)UA= 1 . (Rci + Rk + Rco)

ELECTRICAL ANALOGY TO HEAT FLOW

V= I(R1+R2+R3)T= q (Rci + Rk + Rco)

RciRkRco

1/ hiAl/kA1/ hoA

Q. in furnace, combustion is at 1000oC and outside temp is 25oC, convection heat transfer coefficient between furnace and wall= 10W/m2 and wall to outside air 5 W/m2 , thermal conductivity of brick material of wall is k= 1.04 W/mK. Find thickness of wall if wall temperature should not exceed 800oC.

Heat from furnace to wallqci/A= hi(Ti-Ta) =10(1000-800)=2000Wm2 From wall to atmos.q/A = ho(Tb-To) = 5(Tb - 25) = 2000Tb = 425 oCEq for conductionq/A = k(Ta - Tb)/ l = 2000l = 0.195 m

USE OF HEAT TRANSFER CALCULATIONS IN DESIGN:Engineers have only two types of problems regarding HMT they either have to stop heat flow or to promote heat flow. Design of condenser, heat exchanger, air-conditioning (to calculate heat load as well as air-conditioning required). To calculate heat load of a building in civil engineering. Heat treatment of metals,