Advanced Transport Phenomena

1. The first system we consider is an electric wire
of circular cross section with radius R and
electrical conductivity k 1/Ω cm,
Through this wire there is an electric current
passing with
current density I A/cm2
The surface of the wire is maintained at
temperature To.

2. Objective
To find out the radial temperature distribution
with in the heated wire.
Assumption
The temperature rise in the wire is not so large
that the temperature dependence of either the
thermal or electrical conductivity need be
considered.

3. To
R
r
Δr
L

4. Energy balance
➢Rate of thermal energy in across cylindrical
surface at r (by conduction)
=heat flux * surface area =
➢Rate of thermal energy out at (r + Δr) across
cylindrical surface (by conduction)
=

5. ➢Heat flux = q= rate of heat transfer/area for
heat transfer (q= Q/A)
=> Q=qA
➢Rate of production of thermal energy by
electrical dissipation
=vol. * heat produced per unit vol.
=(2πrΔrL)
Energy balance

6. ➢ Where,
is the rate of heat production per unit
volume.
➢Actually the transmission of an electric current is
an irreversible process, and some electrical
energy is converted into heat (thermal energy).
➢The rate of heat production per unit volume is
given by the expression
Energy balance

7. ➢We know the steady state energy balance is,
{rate of thermal energy in} – {rate of thermal
energy out} + {rate of thermal energy
production}= 0
➢Putting values in above equation;
- + = 0
(2πrΔrL)
Energy balance

8. ➢ Or,
- = (2πrΔrL)
Dividing throughout by 2πΔrL,
= r
Taking limit when Δr →0
{ } = r
( )= r = r ….. (1)
➢ This is the differential equation for heat flux.
Energy balance

9. ➢Integrating (1) ;
=
= +
…. (2)
➢In order to find , we use boundary
conditions;

10. ➢B.C.1:
At r = 0 , becomes infinity from (2), but
is not infinity but a finite value, so = 0.
Putting =0 in (2), we have
…. (3)
➢This shows that heat flux varies linearly with r.

11. ➢According to Fourier’s law of heat conduction,
[putting in (3)]
=
.dr ….(4)
➢Integrating (4) assuming k to be a constant.
= -
T= - ….(5)

12. ➢To calculate we use boundary conditions
➢B.C.2:-
At r = R , T = So,
= + [ putting values of in (5)]
T= - + +
T - = -
❖ T - = {1- } ….(6)

13. Once heat flux and temperature
distribution are known, various
quantities about the system
may be found.
1. Maximum temperature rise
2. Average temperature rise
3. Heat flow rate at surface

14. ➢ The maximum temperature is at the center,
where r = 0,
so,
➢At r = 0, T = T max [Putting in (6)]
❖ T max - = {1 –0} =

15. ➢Average temperature rise can be found by
integrating the temperature rise over whole
surface divided by the whole surface area.
➢Replacing T by < T >, we can write average
temperature rise as,
<T> - =
=

16. =
=
➢ Putting values of T - from (6),
=

17. =
=
=
=

18. < T > - =
So,
< T > - =
➢ Which shows average temperature is half of the
maximum temperature.

19. ➢For the length L of the wire;
➢Heat flow rate at r = R = = ?
Now,
➢ Heat flux = heat rate/area
Or
➢ Heat rate = heat flux * area

20. = * (2πRL) [at r = R]
Or
= [from (3) where r = R]
❖ = π