TITLE.PM5

788 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th15-1.pm5

But, dt

dx

d

dx

tt

L

xt tt

L

= F −

HG

I

KJ

+

L

N

M

O

Q

P=

21 −

1 21

∴ Q = – kA() ()tt

L

kA t t

L

21 − = 12 − ...(15.24)

Eqn. (15.24) can be written as :

Q =

()

(/ )

()

()cond.

tt

LkA

tt

Rth

12 − = 12 − ...(15.25)

where, (Rth)cond. = Thermal resistance to heat conduction. Fig. 15.3 (b) shows the equivalent
thermal circuit for heat flow through the plane wall.
Let us now find out the condition when instead of space, weight is the main criterion for
selection of the insulation of a plane wall.
Thermal resistance (conduction) of the wall, (Rth)cond. =
L
kA
...(i)
Weight of the wall, W = ρ A L ...(ii)
Eliminating L from (i) and (ii), we get
W = ρA.(Rth)cond. kA = (ρ.k)A^2 .(Rth)cond. ...(15.26)
The eqn. (15.26) stipulates the condition that, for a specified thermal resistance, the lightest
insulation will be one which has the smallest product of density (ρ) and thermal conductivity (k).
15.2.5.2. Heat conduction through a composite wall
Refer Fig. 15.4 (a). Consider the transmission of heat through a composite wall consisting of
a number of slabs.
Let LA, LB, LC = Thicknesses of slabs A, B and C respectively (also called path lengths),
kA, kB, kC = Thermal conductivities of the slabs A, B and C respectively,
t 1 , t 4 (t 1 > t 4 ) = Temperatures at the wall surfaces 1 and 4 respectively, and
t 2 , t 3 = Temperatures at the interfaces 2 and 3 respectively.

Q t 1 t 2 t 3 t 4 Q

Rth–A Rth–B Rth–C

R=th–A k.ALA

A

R=th–B k.ALB

B

R=th–C k.ALC

C

,,

(b)

A B C

kA kB kC

t 1

t (^2) t 3
t 4
Interfaces
Temperature
profile
Q
(^1234) L
LA LB C
(a)
Q
Fig. 15.4. Steady state conduction through a composite wall.