Thermodynamics and Heat Engine

LESSON - 17 PRINCIPLE OF INCREASE OF ENTROPY AND NUMERICAL PROBLEMS

17.1.1. Closed System Consider the process shown in Fig. 17.1. in which a quantity of heat δQ is transferred from the surroundings at temperature To to the system at temperature T, and let the work done by the system during this process be δW.  For this process entropy change for a system can be written as,      dSsystem ≥ For this process entropy change for a surrounding can be written as,

Fig. 17.1. Entropy change for the system plus surroundings.

dS_surroundings ≥       [ because for the surrounding δQ is negative]

The total change of entropy is therefore

dS_system +  dS_surroundings ≥     

                                                                                                                            ……………………………….(17.1)

For irreversible process,

Since T_o > T , the heat transfer is from the surroundings to the system, both and the quantity  are positive and we conclude  that

The total change of entropy from eqt. (17.1),    dS_system +  dS_surroundings >   0                                                    …….. (17.2)

If T > T_o , the heat transfer is from the system to the surroundings, both and the quantity  are negative, thus yielding the same result

The total change of entropy from eqt. (17.1),     dS_system +  dS_surroundings >   0                                                     ……. (17.2)

For reversible process,

Since T ≈ T_o  , the heat transfer is from the surroundings to the system or system to the surroundings, we conclude  that

The total change of entropy,                 dS_system +  dS_surroundings =   0                                                         ……………..(17.3)

Thus, from equation (17.2) and equation (17.3) we can conclude that for all possible processes that a system in a given surroundings can undergo,

The total change of entropy,                dS_system +  dS_surroundings ≥   0                                                          ………………… (17.4)

where, the equality holds for reversible processes and the inequality for irreversible processes.     

Since all macroscopic processes that occur are to some degree irreversible, there is always a net increase in the entropy of the system plus surroundings (entropy of universe).

17.1.2. For isolated system

If a system is isolated from its surroundings (or simply an adiabatic closed system), the heat transfer is zero, and equation (17.1) reduces to

dS _{isolated system}  ≥   0                                                                                                …………………………….. (17.5)

The equation (17.5) can be expressed as the entropy of an isolated system increases during an irreversible process and in cases of a reversible process it remains constant.

Note that in the absence of any heat transfer, entropy change is due to irreversibilities only, and their effect is always to increase entropy.

 Thus from equations (17.4) and (17.5) we can conclude that the entropy of a Universe (system and its surroundings) or an isolated system will continually increase, or, in the limiting cases of a reversible process remains constant. There is no way in which the entropy of an isolated system or the entropy of a system and its surroundings can decrease. This is known as the increase of entropy principle for closed/isolated system.

 Irreversible or spontaneous processes can occur only in that direction for which the entropy of the universe or that of an isolated system, increases. These processes cannot occur in the direction of decreasing entropy.

17.1.3. Open System  Refer Fig. 17.2. For control volume (system), the entropy rate balance (refer  eqt. 16.3) can be written as,                                                                       ....................….. (17.6)   For surroundings, the entropy rate balance (refer eqt. 16.3) can be written as,

Fig. 17.2.  Entropy change for a control volume plus surroundings.

                ......................…...(17.7)                [because for the surrounding δQ is negative]

 Add (17.6) and (17.7)

(dS/dt)_system + (dS/dt)_surroundings  ≥  

  For  irreversible process

Since T_o > T , the heat transfer is from the surroundings to the system, and both   and the quantity   are positive and we conclude  that

The total change of entropy,         (dS/dt)_system + (dS/dt)_surroundings  > 0                          …….(17.8)

If T > T_o , the heat transfer is from the system to the surroundings, and both   and the quantity  are negative, thus yielding the same result

The total change of entropy,         (dS/dt)_system + (dS/dt)_surroundings  > 0                      ……….(17.8)

 Therefore when heat transfer takes place between system and surroundings, entropy of the system plus its surroundings (entropy of universe) increases.

 For reversible process, 

 Since,  T ≈ T_o    the heat transfer is from the surroundings to the system or system to the surroundings, we conclude  that

The total change of entropy,       dS/dt_system + dS/dt_surroundings   = 0                   …………..…(17.9)

        The above equation suggests that entropy of the system plus surroundings (entropy of universe) is constant in reversible process.

Thus, from equation (17.8) and equation (17.9) we can conclude that for all possible processes that a system in a given surroundings can undergo,

The total change of entropy,                dS/dt_system + dS/dt_surroundings ≥   0                 ………… (17.10)

where, the equality holds for reversible processes and the inequality for irreversible processes.     

The equation (17.10) can be expressed as the entropy of a system and its surroundings will continually increase, or, in the limiting cases of a reversible process remains constant. There is no way in which the entropy of a system and its surroundings can decrease. This is known as the increase of entropy principle for open system.

Example 17.1: A heat exchanger uses 5000kg/h of water to cool oil from 150° to 50°C. The rate of flow of oil is 2500 kg/h. The average specific heat of oil and water are2.5 kJ/kgK and 4.1868kJ/kgK, respectively. The water enters the heat exchanger at 21°C.

Determine:

 (i) Change in the entropy due to heat exchange process.

(ii) The amount of work obtained if cooling of oil is done by using the heat to run a reversible engine with sink temperature of 21°C.

Solution:

Given:  m_water = 5000kg/h     C_{P, water} = 4.1868kJ/kgK;      T_{i, water} = 21°C

               m_oil = 2500 kg/h         C_{P, oil} = 2.5 kJ/kgK;               T_{2, oil} = 150° C;        T_{3, oil} = 50° C;     

 

 

 

 

                                                   Heat exchanger

(i)     Determine Change in the entropydue to heat exchange process:

Formula: The net entropy change as result of heat exchange process in the heat exchanger

  = The change of entropy of oil + The change of entropy of water

  = dS_oil + dS_water

Finding unknown, dS_oil and dS_water;

dS_oil = m  C_P,oil  ln    = 2500 x 2.5 x   = -1687.5 kJ/K

dS_water = m_waterC_{P, water} ln

Finding unknown,T_{e, water};

Heat lost by oil = Heat gain by water = Q

m_oilC_{P, oil} (T_{2, oil} – T_{3, oil}) = m_waterC_{P, water} (T_{e, water} – T_{i, water})  

2500 x 2.5 x (150 - 50) = 5000 x 4.1868 (T_{e, water}  - 21)

T_{e, water} = 51°C

Therefore, dS_water   = 5000 x 4.1868 x ln  = 2034 kJ/K

Answer: The net entropy change as result of heat exchange processin the heat exchanger

= dS_oil + dS_water= -1687.5 + 2034 = 346.5 kJ/K

(ii)   Determine the amount of work  (W) obtained if cooling of oil is done by using the heat to run a reversible engine with sink temperature of 21°C:

 

       Heat engine                                                                                                 T-s diagram

Formula: The amount of work that could have been obtained by the use of reversible engine,

           W  = Q_H - Q_L= area‘1-2-3-4’– area ‘1-6-5-4’

Finding unknown,Q_H  and    Q_L;

Q_H= m_oil C_oil dT_oil= 2500 x 2.5 x (150-50) = 625000 kJ/h

Q_L = area ‘1-6-5-4’ under line 5-6

      = T_sink dS_oil   = (21 +273) x 1687.5 = 496125 kJ/h

Answer:  Thus, work achievable by reversible engine,

                   W = Q_H - Q_L = 625000 - 496125 = 128875 kJ/h = 35.8 kW

 Example 17.2: 3 Kg of water at 80 °C is mixed with 4 kg of water at 15 °C in an insulated system. Calculate the change of entropy due to mixing process

Solution:

 Given: (m₁)_{cold water} = 4 kg;     (t₁)_{cold water} = 15 °C;      (t₂)_{Hot water} = 80 °C ;    (m₂)_{Hot water} = 3 kg

Determine Change in the entropy due to mixingprocess:

Formula: Net change in entropy, ΔS = S₂ - S₁

                                                                    

Finding unknown,temperature after mixing (t);

Heat gained by cold water = Heat lost by hot water

[m₁. C_p,w. (t- t₁)]_{cold water}  =  [m₂ . C_p,w. (t– t₂)]_{hot water}

4 .C_p,w. (t – 15)  = 3 . C_p,w. (80 – t)

4 (t – 15) = 3 (80 – t)

t = = 42.85°C

Answer: Net change in entropy, ΔS = S₂ - S₁

        

                 = 4 C_p,w 0.0923 + 3 C_p,w (− 0.1112)

                 = 0.3692  C_p,w − 0.3336  C_p,w = 0.0356  C_p,w = 0.0356 x 4.187 = 0.1491 kJ/K

Example 17.3:  In an air turbine the air ex pands from 7 bar and 460 °C  to 1.012 bar  and 160°C. The heat loss from the turbine can be assumed to be negligible. 

(i)     Show that the process is irreversible

(ii)   Calculate the change of entropy per kg of air.

Solution:

Given: p₁ = 7 bar;   T₁ = 460 + 273 = 733K    p₂ = 1.012 bar;   T₂’ = 160+273 = 433K    Q = 0.

(i)     Prove that the process is irreversible

Since the heat loss is negligible, the process in air turbine is adiabatic.

If the process is reversible adiabatic, the outlet temperature (T₂) of the air turbine is calculated by following equation:

                                                                                               = 421.6 – 273 = 148.6 °C

But the actual temperature at the outlet of the air turbine [T₂’ = 160°C (160+273 = 433 K)] at the pressure of 1.012 bar is more than outlet temperature (T₂) of the air turbine for reversible process.

As per T-s diagram there is increase in entropy even heat transfer is zero.

Hence proved that the process irreversible.

 

 

 

 

 

 

 

 

(ii)   Determine the change of entropy per kg of air

Formula: The change in entropy (s₂’- s₁) = (s₂’- s₂)

Finding unknown,(s₂’- s₂);

The change in entropy for a reversible constant pressure process 2 to 2’,

 (s₂’ - s₁) =   =  0.02681 kJ/kg K

Answer: The change in entropy (s₂′- s₁) = (s₂′- s₂) =  0.02681 kJ/kg K

                 i.e. Increase of entropy.