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Practical Use of Entropy Principle

If the process is more irreversible, the more will be the increase of entropy of the universe. The increase of entropy quantifies the extent of irreversibility of the process. The entropy principle is a quantitative statement of 2nd law. We illustrate some applications of entropy principle in the following.

 

8.11.1 Heat Transfer Through a Finite Temperature Difference

Fig. 8.14 shows two reservoirs A and B with T1 & T2 temperatures respectively. Let T1 > T2, so heat (Q) will flow from A to B. So for the isolated system comprising of two reservoirs A & B and the rod, we can write

ΔS univ = ΔSA + ΔSB + ΔSC

= Q / T1 + Q / T2 + -Q + Q / T

= Q ( T1 – T2 / T1 T2 + 0

= Q ( T1 – T2 / T1 T2

Since T1 > T2, ΔS univ > 0 and the process is possible and irreversible. If T1 = T2, ΔS univ = 0, the process is reversible.

If T 2 > T 1, ΔS univ < 0 and the process is impossible.

 

8.11.2 Mixing of two Fluids

Subsystem A               Subsystem B

mass m1                           m2

specific heat c1                 c2

temperature T1                 T2.

There are two subsystems A and B as shown in Fig. 8.15. When the partition is removed, the two fluids mix together and at equilibrium let T1 be the final temperature. So, T2 < T1< T1. As there is no heat transfer through adiabatic wall, we have

 

Heat lost by A= Heat gained by B

m1c1 (T1- T1) = m2c2 (TF T2)

T f = m1 c1 T1 + m2 c2 T2 / m1 c1 + m2 c2

 

The change of entropy for subsystem A is

ΔS1 < 0 , since T f < T1.

 

The change of entropy for subsystem B is

T f > T2 . ΔS1, > 0

ΔS univ = ΔS1 + ΔS2

= m1 c1 In T f / T1 + m2 c2 T f / T2

 

In actual calculation the RHS of equation (8.21) is positive,

Thus ΔS univ > 0 and the mixing process

is irreversible. For integration the irreversible path is replaced by reversible paths.

For                              m1 = m2 = m (say), and c1 = c2 = c ( say)

 

ΔS univ = mc In Tf2 / T1 T2

TF = m1 c1 T1 + m1 c2 T2 / m1 c1 + m2 c2 =  T1 + T2 / 2

Thus                  ΔS univ = 2 mc In ( T1 + T2 / 2√ T1 T2)

We have          ΔS univ > 0 this is possible when

T1 + T2 > 2 √ T1 T2

T1 + T2 / 2 > √ T1 T2.

 

This is already known that arithmetic mean of any two numbers is greater than the geometric mean. The above statement can be proved geometrically as

r  = T1 + T2 / 2

r21 = T1 T2

r1 = T1 T2

r > r1

T1 + T2 / 2 > T1 T2 proved.

 

The reverse of the statement i.e., if

T1 + T2 / 2 > T1 T2 the ΔS univ > 0

 

8.11.3 Maximum Work Out (from two Fixed Temperatures) by an Engine Working Reversibly

 ΔS univ = 0

ΔS1 + ΔS2 = 0

or                      In T f / T1 + In T f / T2 = 0

or                            In Tf2 / T1 T2 = 0 = In 1

T f = T1 T2

 

When the system reaches TF the work done by the engine during that period is given by

W max= Heat lost by source- Heat gained by sink

= Q1 – Q2

= mc p ( T1 – TF) – mc p ( TF – T2)

= mc p [ ( T1 + T2) – 2 T1 T2]

= mc p ( T1 – T2)2

 

c p is the specific heat at constant pressure in (KJ/kg K).

 

 

8.11.4 External Mechanical Irreversibility

(i) Isothermal dissipation of work

Let I current flows through a resistor in

contact with reservoir. At steady state,

dU = 0,

Q=dU+W

- Q= 0+(-W)

dU = 0 for isothermal process

Q =W

ΔS surr = Q / T = W / T

At steady state ΔS system = 0

ΔS univ = ΔS sys + ΔS surr

= O + ΔS surr

= ΔS surr = W / T

ΔS univ > 0 as W > 0.

 

Thus there is increase of entropy of the universe from this irreversible process.

 

(ii) Adiabatic dissipation of work

Let W is the work transfer to the system, according to

1st law,

or                     dQ = dU  + dW

0 = dU  + dW

Or             U2 – U1 = – ʃ d W = – W

ΔS univ = ΔS surr + ΔS sys

= 0 + mc p in T f / Ti

where c P is the specific heat at constant pressure.