PHYSICAL CHEMISTRY IN BRIEF

CHAP. 10: TRANSPORT PROCESSES [CONTENTS] 335

Solution

We use equation (10.5) rewritten into the form

dQ=−S λ

dT

dz

dτ

and integrate

Q=−Sλ

dT

dz

τ.

We estimate the temperature gradient, i.e. the derivative of temperature with respect to location

dT

dz

=. ∆T

∆z

=

263. 15 − 298. 15

0. 20

=− 175 K m−^1

and substitute

Q=− 4 × 2. 41 × 10 −^2 ×(−175)×3600 = 60732J.

Note: When solving the above example we neglected the heat conduction through the

glass and through the glass-air interface. We assumed that the windows were perfectly

sealed, and that the air between the two windows was motionless. We also neglected the

heat transfer by convection. This assumption was not fully justified. The cold air at the

outer window went down while the warm air at the inside window went up. This air flow

could significantly affect the result, i.e. increase the heat transfer.

10.2.4 Fourier-Kirchhoff law.

For the dependence of temperature on timeτ and location z we use the following partial
differential equation
∂T
∂τ

=

λ

c Cpm

∂^2 T

∂z^2

, [p] (10.6)

wherecis the molar density (see2.1.1) andCpmis the molar isobaric heat capacity (see3.2.4).
If we want to solve this equation we need to know the initial conditions (forτ= 0) and boundary
conditions (forz= 0). The solution yields temperature as a function of timeτand locationz.
Specific cases are solved within the course in Chemical Engineering.