Download Irreversible heating of a Bar

Entropy change in irreversible heating of a metal bar
Consider a metal bar of cross sectional area A that is initially at temperature T0 . The bar
runs from x = 0 to x = L. At time t = 0, the temperature of the cross section at x = 0 is
raised to the temperature T1 (and held at this temperature), while the remaining surfaces
of the bar are insulated. This is an example of an irreversible transient heat transfer
process in which the temperature distribution within the bar will be non-uniform, running
monotonically from higher temperatures near x = 0, to lower temperatures near x = L.
The temperature at all points within the bar (except at the end x = 0) will be increasing
with time. Even though the process within the system takes place under irreversible
conditions, is it possible to get a handle on the entropy change during this irreversible
process? Lets see how.
Because we are dealing with a metal bar, the work done by the system on the
surroundings is negligible (since the volume change is negligible). The rate of heat
transfer into the bar at x = 0 is given by:
dQ
 T 
 kA 
dt
 x  x0
(1)
where k is the thermal conductivity of the metal. Because there is no work done, the rate
of change of internal energy is given by:
dU dQ

dt
dt
(2)
where
L
U  Cv A T  T0 dx
(3)
0
This assumes that the heat capacity at constant volume is approximately equal to the heat
capacity at constant pressure, and is independent of temperature. It also assumes
negligible variations in density  .
At each location along the bar, the transient energy balance equation is given by
T
 2T
CV
k 2
t
x
If we take the partial derivative of the quantity
(4)
1 T
k
with respect to x, we obtain:
T x
  1 T 
k  T  1   2T 

k





  k
x  T x 
T 2  dt  T  x 2 
2
(5)
If we substitute Eqn. 5 into the differential energy balance equation (Eqn. 4), we obtain:
CV T
T
t

  1 T  k  T 
 k



x  T x  T 2  dt 
2
(6)
The left hand side of this equation, for this problem, can be recognized as the rate of
change of entropy per unit volume ds / dt , where
ds 
CV dT
(7)
T
Therefore,
s   1 T  k  T 
  k



t x  T x  T 2  dt 
2
(8)
If we multiply Eqn. 8 by the cross sectional area A, and integrate between x = 0 and x =
L, we obtain:
2
dS
A  T 
 1 T 
  k    kA 
 dx
dt
T1  x  x0
T

x


0
L
(9)
where S is the total entropy of the bar.
Substituting Eqn. 1 into Eqn. 9 then yields:
2
dS 1 dQ
 1 T 

 kA 
 dx
dt T1 dt
T x 
0
L
(10)
The second term on the right hand side of Eqn. 10 is positive definite. This means that,
for this irreversible process, in which there are temperature gradients within the material,
the rate of change of entropy is greater than dQ divided by the temperature at which the
heat enters the system T1 . This gives you an idea of what happens with an irreversible
process.
At any time t after placing the bar in contact with the reservoir, the change in entropy of
the system, relative to the initial state, can be precisely calculated as
2
Q
 1 T 
S 
 kA  
 dxdt
T1
T

x


0 0
t L
(11)
This shows that, even though the process taking place in the system is irreversible, one
can precisely calculate the running history of the change in entropy of the system under
these irreversible conditions. Therefore, at least in this case, the system does not have to
be at thermodynamic equilibrium in order to determine its change in entropy. In Eqn. 11,
the heat added Q and the double integral on the right hand side can be calculated from
the solution to the transient heat conduction equation (Eqn. 4), which is well-known (see
Transport Phenomena by Bird, Stewart, and Lightfoot).
For the present problem, the change in entropy for the surroundings, comprised of a huge
heat reservoir at T1 , is just:
S surroundings   Q
(12)
T1
Therefore, the total change in entropy for the combination of system and surroundings
(i.e., the universe) as a result of this process is:
S  S surroundings
2
 1 T 
 kA  
 dxdt
T x 
0 0
t L
(13)
Note that, for our reversible process, this will be positive and monotonically increasing
with time because of the positive definiteness of the integrand. Eqn. 13 provides us with
a relationship for precisely calculating the increase in the entropy of the universe as a
result of our process, by making use of the solution to the transient heat conduction
equation.
Eqn. 10 provides an example of the so-called Cauchy inequality applied to our system:
dS 
dQ
T
(14)
If heat enters a closed system through several different portions of the system boundary,
and the temperatures at these various locations of entry all differ from one another, then
Cauchy’s inequality takes the more general form:
dS  
dQ j
Tj
(15)
where dQ j is the heat flow through the j’th portion of the boundary, and T j is the
temperature at that portion of the boundary.