Page 16 - Fundamentals of The Finite Element Method for Heat and Fluid Flow

8
Now, the energy balance for gas gives
dT g 4 INTRODUCTION
4
m g c pg = h f A g (T f − T g ) − h g A g (T g − T w ) + f A f σ(T − T ) (1.21)
f
g
dt
Wall:
Rise in internal energy of wall:
dT w
m w c pw (1.22)
dt
Radiation from filament to wall:
4
4
f σA f (T − T ) (1.23)
f w
Convection from wall to ambient:
h w A w (T w − T a ) (1.24)

Radiation from wall to ambient:
4
4
w σA w (T − T ) (1.25)
a
w
Energy balance for wall gives
dT w 4 4 4 4
m w c pw = h g A g (T g − T w ) + f σA f (T − T ) − h w A w (T w − T a ) − w σA w (T − T )
a
w
f
w
dt
(1.26)
where m g is the mass of the gas in the bulb; c pg , the specific heat of the gas; m w ,the mass
of the wall of the bulb; c pw , the specific heat of the wall; h f , the heat transfer coefficient
between the filament and the gas; h g , the heat transfer coefficient between the gas and wall;
h w , the heat transfer coefficient between the wall and ambient and is the emissivity. The
subscripts f, w, g and a respectively indicate filament, wall, gas and ambient.
Equations 1.21 and 1.26 are first-order nonlinear differential equations. The initial con-
ditions required are as follows:
At t = 0,

T g = T a and T w = T a (1.27)
The simultaneous solution of Equations 1.21 and 1.26, along with the above initial
condition results in the temperatures of the gas and wall as a function of time.

1.4.3 Systems with a relative motion and internal heat generation
The extrusion of plastics, drawing of wires and artificial fibres (optical fibre), suspended
electrical conductors of various shapes, continuous casting etc. can be treated alike.
In order to derive an energy balance for such a system, we consider a small differential
control volume of length, x, as shown in Figure 1.3. In this problem, the heat lost to