LESSON - 15 ENTROPY, CLAUSIS INEQUALITY, ENTROPY A PROPERTY, ENTROPY CHANGE DURING PROCESS

 

15.1. INEQUALITY OF CLAUSIS

Let a reversible heat engine and an irreversible heat engine both operate between same two given temperature reservoirs (i.e. high temperature reservoir at T_H and low temperature at T_L). Let the heat transfer to both the engines be Q_H and heat rejected from both the engines be Q_L .

We know that for a reversible heat engine operating in a cycle,

Efficiency,                                              ………………….  (15.1)

For an irreversible engine operating in a cycle,

Efficiency,                                       …………………….. (15.2)

From equations (15.1) and (15.2) we find

                         or                 

or                                 or              

or                                            or             

Therefore, in differential form, for any cyclic process, reversible or irreversible

 

This is called inequality of Clausius. A cycle which violates this inequality is not possible to execute in practice.

                               (for a reversible cycle)    

and                        (for an irreversible cycle)

Problem 15.1: A heat engine receives 1000 kW of heat at constant temperature of 285°C. The heat is rejected at 5°C. The possible heat rejected is (a) 840 kW, (b) 492 kW, (c) 300kW. Comment on results.

Solution: Given: Heat receives, Qreceives = 1000 kW;              TH = 285+273 = 558 K;     TL = 5+273 = 278K Formula: Apply the clausius inequality in each case    The cycle is possible if it satisfy clausius inequality     i.e. cyclic   Comment on results. (a)    Given: Qrejected =  840kW         Answer:         The clausius inequality does not hold good, Hence, the cycle is not possible. (b)   Given: Qrejected = 492 kW        Answer:           The clausius inequality holds good, Hence, the cycle is possible

Fig. 15.1. Schematic of problem 15.1

(c)    Given: Q_rejected = 300 kW

        Answer:      

        The clausius inequality does not hold good, hence, the cycle is not possible.

15.2. ENTROPY A PROPERTY  ENTROPY CHANGE FOR A SYSTEM DURING A REVERSIBLE  PROCESS

Consider a system in state 1 undergoes a change of state to state 2 through path A. This process is reversible. Now it has the option to come back to state 1 by many processes (paths).  Say, B and C are two such paths. Processes through path B and C are also reversible. Thus the system has undergone two reversible cycles, i.e. 1-A-2-B-1and 1-A-2-C-1 as shown in Fig. 15.2. For reversible cycle through paths A and B

Fig. 15.2. Two reversible cycles.

                                        …………. (15.3)                          [  for reversible cycle, cyclic ]       

For reversible cycle through paths A and C

                                      …………. (15.4)                            [  for reversible cycle, cyclic ]                    

Subtracting Equation (15.4) from Equation (15.3), we get

                     

Since the quantity δQ/T is same for two different paths between two states, we conclude that this quantity is independent of path and is a function of the end states only, and is therefore a thermodynamic property. This property is called entropy, and is designated by S. It follows that, mathematically, entropy may be defined as a property of a substance in accordance with the relation

                 dS = (δQ/T)_rev                                                                                                                           ………………………..(15.5)

Considering mathematical and physical aspects, the entropy is defined as follows:

The amount of heat transported to a substance and the degree to which it is ordered are described by the property of state known as entropy. The higher the disorder, the greater is the increase in entropy.

The change in entropy of a system as it undergoes a change of state may be found by integrating equation (15.5)

            S₂ – S₁ =              or             ΔS =                    , kJ/k                     ………………………..(15.6)

The entropy per unit mass is

     s₂ – s₁ =                                  , kJ/kg k

The above equation suggests that entropy is conserved in a reversible process.

For reversible adiabatic process

δQ = 0

From equation (15.6), we have

ΔS =  0

This means change in entropy is zero for a reversible adiabatic process.

Temperature entropy diagram From Equation (15.5), the heat interaction for a reversible process is given by          (δQ)rev = TdS Total amount of heat interaction for a reversible process 1-2 is given by,   1Q2 = Therefore for a reversible process drawn with T-S co-ordinates shown in Fig. 15.3, the area beneath the curve represents the heat transfer during the process.

Fig. 15.3. T-S diagram of reversible process

15.3. ENTROPY CHANGE FOR A SYSTEM DURING AN IRREVERSIBLE PROCESS

Consider a system in state 1 (Fig. 15.4). It undergoes a change of state through a path A to state 2. This process is reversible. It can complete the cycle through reversible path B or irreversible path C.   For a reversible cycle through paths A and B   For reversible cycle, cyclic                                 ……… (15.7)

Fig. 15.4. Entropy change of a system during an irreversible process.

For an irreversible cycle through paths A and C

   For irreversible cycle, cyclic    

                                                                               ............……… (15.8)            

 Substracting Equation (15.8) from (15.7) we get

                                                                                   ……………………. (15.9)

 Since path B is reversible,

         

and entropy is a property      

     Therefore,          

       From the above two equations, we have

      

Using this in Equation (15.9), we have

           or                                                                           ……………………. (15.10)                                                   

The above equation suggests that entropy increases in an irreversible process.

For an irreversible adiabatic process

δQ = 0

From equation (15.10), we have

ΔS >  0

Note that for an irreversible adiabatic process, though heat transfer is zero but there is still an increase in entropy due to irreversibilities.