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 = 473

Ts2 = 1000+273= 1273

E/A=(Ts14 Ts24 )

= 146kW/m2

EXAMPLE2:

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 K

Calculate total heat loss / unit length

Sol

Conv.q = h As T= h ( x d x l)(60-25)= 6.5x ( x 0.05x1)(60-25) = 35.72W

Rad.q= A(Ts14 Ts24 )= 0.7 x ( x d x l) x 5.67 x 10 -8 ((60+273)4-(25+273)4)

q= 33.72W

totalQ=35.72+33.72=69.44W

OVERALL HEAT TRANSFER COFF.:

qci= hiA(Ti-Ta) = (Ti-Ta)

Rci

qk = kA(Ta-Tb) = (Ta-Tb)

l Rk

qco = hoA(Tb-To) = (Tb-To)

Rco

qci = qk = qco = q

Now 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 Rco

Ti To = q (Rci + Rk + Rco)

q = (Ti To)

(Rci + Rk + Rco)

By Newtons law of cooling

q = 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)

Rci

Rk

Rco

1/ hiA

l/kA

1/ 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 wall

qci/A= hi(Ti-Ta) =10(1000-800)=2000Wm2

From wall to atmos.

q/A = ho(Tb-To) = 5(Tb - 25) = 2000

Tb = 425 oC

Eq for conduction

q/A = k(Ta - Tb)/ l = 2000

l = 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,

q

ci

q

k

q

co

Ti

Ta

Tb

To

Ti

Ta

Tb

To