Page 29 - Modelling in Transport Phenomena A Conceptual Approach

10 CHAPTER 1. INTRODUCTION

Solution
For steady-state conditions without generation, the heat transfer rate is constant
and can be determined from Eq. (1.3-1) as
Heat transfer rate = (Heat flux),=, (Area),=,

Since the cross-sectional area of the cone is rD2/4, then

Heat transfer rate = (45) [r (oy]= 2.21

The value of the heat transfer rate is also 2.21 W at x = L. However, the heat flux
does depend on position and its value at x = L is
2.21
(Heat flu^)^=^ = = 1126 W/ m2
[T (0.05)2 /4]

Comment: Heat flux values are different from each other even though the heat
flow rate is constant. Therefore, it is important to specify the area upon which a
given heat flux is based when the area changes as a function of position.

1.4.2 Steady-State Transport With Generation
For this case Eq. (1.1-1) reduces to

Rate of ) = ( 02E&
of
( input Rateof cp )+( generation of cp ) (1.45)
Equation (1.45) can also be written in the form
JJ,,. (Inlet flux of Cp) d~ + JJLs, RdV = JLoue
(Outlet flux of cp) dA (1.46)

where R is the generation rate per unit volume. If the inlet and outlet fluxes
together with the generation rate are constant, then Q. (1.46) reduces to
( System ) = ( Outlet flux ) ( Outlet )

of cp volume of cp area
( InIet flux ) ( rri: ) +
(1.47)

Example 1.6 An exothermic chemical reaction takes place in a 20 cm thick slab
and the energy generation rate per unit volume is 1 x lo6 W/ m3. The steady-state
heat transfer rate into the slab at the left-hand side, i.e., at x = 0, is 280W.
Calculate the heat transfer rate to the surroundings from the right-hand side of the
slab, Le., at x = L. The surface area of each face is 40cm2.