LESSON - 16 CHANGE OF ENTROPY OF GASSES IN THERMODYNAMICS PROCESSES AND NUMERICAL PROBLEMS

16.1. TWO IMPORTANT ‘Tds’ THERMODYNAMIC RELATIONS

We have,    

or                                             

or                                            

In differential form, it can be written as   

or         TdS = dU + pdV                                                                           …………………..(16.1)

or         Tds = du + pdv       ( for unit mass )      

The above equation is very important since it establishes a relationship between all thermodynamic properties and does not involve path functions like work and heat. It is interesting, therefore, to note that equation δQ = dU + pdV   and  δQ = TdS  are true for reversible process but equation TdS = dU + pdV  is true for all processes reversible as well as irreversible. This equation is true for any two equilibrium states of a system.

Since H = U + pV

It follows that   dH = dU + pdV + Vdp

Substituting this relation in Equation (16.1)

Thus    TdS = dH - Vdp                                                                             ………………….(16.2) 

or         Tds = dh – vdp             (for unit mass)         

 Equation (16.1) and (16.2) are two important ‘TdS’ thermodynamic relations.

16.2. ENTROPY CHANGE IN A PROCESS FOR IDEAL GAS:

Various processes are shown on p-v and T-s diagrams in Fig. 16.1. The entropy change in these processes is determined as follows.

Fig. 16.1. Polytropic processes on P-v and T-s diagrams.

16.2.1. Constant volume process (Isochoric process) (Fig. 16.2):

       V1 = V2 = V            

From (16.1), we have     TdS = dU + pdV    

But   dV = 0 and  dU = m Cv dT;          

therefore    TdS = m Cv dT + 0     = m Cv dT

 

  or                    [ If Cv is constant for the gas]

                           S2 – S1 =  m Cv ln 

For unit mass,

                     s2 – s1 =  Cv ln 

 

 

Fig. 16.2. Constant volume process

16.2.2. Constant pressure process (Isobaric process) (Fig. 16.3):

P1 = P2 = P

From Equation (16.2), we have    

TdS = dH - Vdp    

But   dP = 0 and  dH = mCpdT; 

therefore      TdS = m Cp dT + 0   = m CpdT

     

or                   [ If Cp is constant for the gas]

     S2 – S1 =  m Cp ln  

For unit mass,

     s2 – s1 =   Cp ln    

 

Fig. 16.3 Constant pressure process

16.2.3. Constant temperature process (Isothermal process) (Fig. 16.4):

PV = C = mRT            or T1 = T2 = T

From Equation (16.1), we have

    TdS = dU + pdV    

But   dU = m Cv dT =0    as  dT = 0; 

therefore    TdS = 0 + pdV    = pdV

    

or        

Using perfect gas relation

, we can write

 or   S2 – S1 =  m R ln  

For unit mass,

or   s2 – s1 =   R ln  

       

 

Fig. 16.4. Constant temperature process

16.2.4. Reversible and an irreversible adiabatic process (isentropic process)(Fig. 16.5):

For reversible adiabatic process, we have,  


But   δQ = 0        (for adiabatic process)

Therefore,      S2 – S1 =  0       or     S1 =  S2 

Therefore a reversible adiabatic process is an isentropic process

For an irreversible adiabatic process, for unit mass, we have,  


But   δQ = 0        (for adiabatic process)

Therefore,      S2 – S1 >  0      or       S2 >  S1

              

                          Fig. 16.5. Adiabatic process

16.2.5. Polytropic process:

we have,  

But for a gas,  .   Using this in the above equation, we get

                                        

S2 – S1 = 

For unit mass,

s2 – s1 = 

16.2.6. Equal-internal energy process:

U1 = U2 = U

From Equation (16.1), we have     TdS = dU + pdV    

But   dU = 0;         therefore    TdS = 0 + pdV    = pdV

dS =         or                

     or  S2 – S1 =   

 For unit mass,

or   s2 – s1 = 

 16.2.7. Change in Entropy for an Equal Enthalpy Process

     H1 = H2 = H

From Equation (16.2), we have     TdS = dH - Vdp    

   But   dH = 0 ;       therefore      Tds = 0 - Vdp    = - Vdp

     dS =       or                                         

   or     S2 – S1 = = 

For unit mass,

       s2 – s1 =  

 16.3. CARNOT CYCLE ON T-s DIAGRAM

Let a heat engine operate on a Carnot cycle as discussed in ‘Fig. 14.11 of Lesson 14’. It works on four reversible processes as shown on T-S diagram in Fig. 16.6.

It is clear from the T-S diagram that,

The amount of heat transferred by the reservoir at temperature ‘TH’ to the working substance,

        1Q2  = T1(S2 – S1)

The amount of heat transferred by the working substance to the reservoir at temperature ‘TL’,

            3Q4 = T3 (S4 – S3) = T3 (S1 – S2).

or     3Q4 = - T3 (S2 – S1).


Fig. 16.6. Carnot cycle on T-S diagram.

Since the system undergoes a cycle, the net heat transfer should be equal to the net work done by the system.  

 or    Net work done by the system  = Net heat transfer by the system = 1Q2 + 3Q4

                                                                                                                      =  T1 (S2 – S1) + [- T3 (S2 – S1)]

                                                                                                        =  [T1 (S2 – S1) - T3 (S2 – S1)]

Thermal efficiency of the engine working on cycle =   

                                                                                           

             

16.4. SECOND LAW OF THERMODYNAMICS FOR A CONTROL VOLUME (C.V.)  

The second law of thermodynamics can be applied to an open system (C.V.) by a procedure similar to the application of First law for open system. Figure 16.6 shows an open system.

 Fig. 16.6. The general case for an open system

Case 1: For more than one inlet and one exit of control volume

 For a flow process as shown in Figure 16.6, the entropy rate balance gives

                                                                    ………. (16.3)

In the tabove equation the equality holds for a reversible process and the inequality for an irreversible process.

  and    is rate of entropy change within the system (C.V).

              and    rate of entropy transfer in and out of the system (C.V) accompanying mass flow.

              is rate of heat transfer during time dt at the location of boundary where the instantaneous surface temperature is T   

                 represents the accompanying rate of entropy transfer.   

                 sign for more than one inlet and one  exit of control volume

For the steady state steady flow (SSSF) process, which is defined as a process in which the conditions within the control volume do not vary with time.

i.e.                  and        

For steady state steady flow process, the change in entropy from Equation (16.3) can be given by the following equation

Case2: For one inlet and one exit of control volume

Many applications in engineering problems involve one inlet and one exit of control volume. For such cases the steady state and steady flow entropy rate balance is given by

 

Example 16.1. 5 kg of oxygen is heated in a reversible non-flow constant volume process from temperature of 60°C until the pressure is doubled. If CP =0.913 kJ/kgK, determine, (i) Final temperature. (ii) work done, (iii) change in internal energy. (iv) heat transferred, (v) change in enthalpy and (vi) change in entropy. Assume CV= 0.653 kJ/kgK,

Solution:

Given: m = 5 kg;   T1= 60 °C  = 60 + 273 = 333 K; p2 = 2p1    CP =0.913 kJ/kgK;  dV=0

 (i)     Determine the final temperature, T2;

Formula: For constant volume process, V1 = V2, we have

T2 =  T1

         Answer:    T2 =  x T1= 2 x 333 = 666 K or 393°C

 (ii)   Determine the work done, 1W2;

Formula:  work done, 1W2 = ∫pdV

      Answer:    1W2 = ∫pdV= 0 K or 0°C

 (iii) Determine the change in internal energy, (U2 - U1);

               Formula: U2- U1=  m CV (T2 -T1)

     Answer:    U2 - U1=  m CV (T2 -T1) = 5 x 0.653 x (666-333) = 1087 kJ

 (iv)  Determine the heat transfer, 1Q2;

Formula: 1Q2=  (U2- U1)+1W2

Answer:    1Q2=  (U2- U1)+1W2=  1087 + 0 = 1087 kJ

(v)    Determine the change in enthalpy, (H2 - H1);

Formula:  H2-H1= m CP(T2 - T1)

        Answer:    H2-H1= m CP(T2 - T1) =  5 x0.913 x (666 - 333) = 1520 kJ

(vi)  Determine the change in entropy, (S2 - S1);

Formula: S2-S1 = 

Answer:    = 2.263 kJ/K

Example 16.2: An insulated cylinder contains 25 kg of nitrogen. Its volume capacity is 5 m3. Paddle work is done on the gas by stirring it till the pressure in the vessel gets increased from 5 bar to 10 bar. Determine (i) change in internal energy. (ii) work done (iii) heat transferred and (iv) change in entropy. Assume Cp =1.04 kJ/kgK and CV = 0.7432 kJ/kgK for nitrogen.

Solution:

Given:   m = 25 kg;p1= 5 bar = 5 x102kN/m2 ;     p2= 10 bar = 10 x102 kN/m2

 V1 = V2 = 5m3and it is constant for both end states.

 R = Cp - CV = 1.04 - 0.7432 = 0.2968 kJ/kgK.         Insulated cylinder,  1Q2= 0

(i)     Determine the change in internal energy, (U2 - U1);

Formula: (U2 -U1) = m CV (T2-T1) = CV (mT2 - mT1)

Finding unknown,  mT1 and  mT2 ;

 The mass of the gas in the cylinder is given by

       

Therefore, mT1 =   =   8.423 x103 kgK

and            mT2 =  = 16.846 x103 kgK

         Answer:   U2 - U1=  CV(mT2 − mT1) = 0.7432 x (16.346 − 8.423)x103= 6260 kJ

 (ii)   Determine work done, 1W2 ;

Formula:  1Q2 = U2 − U1+ 1W2

                        therefore  1W2= − (U2 − U1) = − 6260 kJ

        Answer:   1W2= − (U2 − U1) = − 6260 kJ

 (iii) Determine heat transfer, 1Q2 ;

        Answer:   1Q2 = 0 kJ

(iv)  Determine the change in entropy, (S2 - S1);

Formula:    For constant volume process,  

Finding unknown, ;

For constant volume,   = 2

Answer:    S2-S1 =  = 12.878 kJ/kg K

Example 16.3: Nitrogen is heated in a steady flow process from 700 kPa and 27°Cto 620 kPa and927oC. Derive an expression for the change in entropy and calculate the change of entropy during the process if CP = 39.65 – 7.65x103/T + 1.5x106/T2 kJ/kg-mole-K and the value of R is 0.3026 kJ/kgK. Assume molecular weight of nitrogen as 28.016.

Solution:

Given:   p1= 700 kPa;          p2= 620 kPa;                            T1 = 27+273 = 300 K;       T2 = 927 + 273 = 1200 K

CP = 39.65 – 7.65x103/T + 1.5x106/T2 ;                 R=  0.3026 kJ/kgK;           Mol. wt. = 28.016 kg/mol

Determine the change in entropy:

Formula: We know   Tds = du + pdv = d(h – pv) + pdv = dh – vdp

or    

               

              

Answer: (S2-S1) 

    =  1.5215 kJ/kg K

Example 16.4: 5 kg of air is compressed in a reversible steady flow polytropic process from 100 kpa and 40°Cto 1000 kpa and during this process the law followed by the gas is pV1.25= C. Determine the change in entropy CV = 0.717 kJ/kgK , R = 0.287 kJ/kgK.

Solution:

Given:   m = 5 kg;     Ti =40 °C= 40 + 273 = 313 K;       pV1.25= C

Pi = 100 kPa  = 1 x 105 N/m2 ;Pe= 1000 kPa  = 1 x 106 N/m2; 

R =CP - CV    or    CP = R + CV      or    CP = 0.287 + 0.717 = 1.005 kJ/kgK

 Determine the change in entropy:

Formula: For unit mass,       Tds = dh – vdp

  TdS = dH – Vdp             or   

        or     

   

Finding unknown, Te ;

                   Equation of state,

Therefore,         

                                              or  Te =    = 496 K

Answer:     (Se-Si) =

                                        = − 0.9887 kJ/K

Example 16.5: Air is throttled from 1000 kPa at 27°Cto 100 kPa in a steady flow adiabatic process. Changes in kinetic and potential energies are negligible. Determine the change in entropy per kg of air flow. CP= l.004 kJ/kgK, R =0.287 kJ/kgK.

Solution:

Given: Ti=27 °C = 27 + 273 = 300 K;     Pi = 1000 kPa;        Pe= 100 kPa

              R =0.287 kJ/kgK       CP= l.004 kJ/kgK      ΔPE = 0     ΔKE = 0

Determine the change in entropy per kg of air flow:

Formula: In steady flow energy equation for this process (throttling), we have

   hi = he   or     he − hi  = 0 ;

or  CP(Te-Ti) =  0   or      Te = Ti

 Now we have,  Tds = dh – vdp = − vdp ;

or          

         

Answer:   (Se-Si) = = 0.6608 kJ/kgK

Example 16.6: A gas undergoes a non-flow process according to the law p = (0.16/V + 2.1) bar, where V is the volume in cubic meters. Initial volume is 0.6 m3 and the final volume is 0.2 m3. Determine the change in enthalpy during the process if 25 kJ of heat is rejected from the system.

Solution:

Given:   V1= 0.6 m3;            V2 = 0.2 m3;         1Q2 = - 25 KJ;        p = (0.16/V + 2.1) bar

 Determine the change in enthalpy(H2 – H1):

Formula:  Equation for energy balance for closed system is given by

         δQ  = dU + pdV

As    H =U+pV;       Therefore   δQ  = d(H – pV) + pdV

or     δQ  =dH – pdV - Vdp + pdV

or     δQ  =dH  − Vdp

Integrate, 

       1Q2  = (H2 – H1) −

or    (H2 – H1) = 1Q2 +

Finding unknown, ;

Now,   p =  bar;      p = x 105 N/m2

dp = − 

              

               = 17550 J = 17.55 kJ

Answer:    (H2 – H1)  = 1Q2 +  = −25 + 17.55 = − 7.45 kJ

Last modified: Tuesday, 10 September 2013, 5:29 AM