Second Law of Thermodynamics &

Entropy

Annexure

• A reversible engine operates between temperatures T1 and T (T1 > T). The

energy rejected from this engine is received by a second reversible engine

at the same temperature T. The second engine rejects energy at

temperature T2 (T2 < T).

Show that:

(a) Temperature T is the arithmetic mean of temperatures T1 and T2 if the

engines produce the same amount of work output

(b) Temperature T is the geometric mean of temperatures T1 and T2 if

the engines have the same cycle efficiencies

W1

Q3

Q2

Q2

Q1

T

T1

T2

W2

a. If he produce the same amount of work

Then, W1 = W2 Or 𝜂Q1 = 𝜂Q2a,

"#" - 1=1- "$""#%"$" =2

T="#%"$$

(1- ""#)("#" )Q2=(1-

"$" )Q2

b. If their efficiency is same then

1- ""#=1- "$"

T=√𝑇1𝑇2

As T is positive, so negative sign neglected

T is geometric mean of T1 and T2

We know that +#

"#= +$

"

Or+#

"#Q1 = "#" Q2

Entropy Generation

• Entropy generation Sgen is always a positive quantity or zero (reversible process)

• Its value depends on the process, thus it is not a property of a system

• The entropy of an isolated system during a process always increases, or in the limiting case of a

reversible process remains constant (it never decreases),this is known as the increase of entropy

principle.

• The entropy change of a system or its surroundings can be negative; but entropy generation

cannot

6-9

ST

M

L

Approximation of Heat Absorbed

The curve LM is obtained by heating ! Kg of gas from initial state L to final state MLet the temperature during heating increases from T1 to T2.Heat absorbed by the gas will be given by the area under curve LM

Heat absorbed = Area under the curve LM = (s1 – s2) . (T1+T2)$

Heat absorbed approximately equals the product of change in entropy and mean absolute temperature

T2

T1

S2S1

Q

ENTROPY CHANGES FOR A CLOSED SYSTEM

• General Case for Change of Entropy of a Gas- Let 1 kg of gas at a pressure p1, volume v1, absolute temperature T1 and entropy s1, be heated such that its

final pressure, volume, absolute temperature and entropy are p2, v2, T2 and s2 respectively

6-9

S

T

S2

S1

T2

T1

S2S1

Q

• Heating a Gas at Constant Volume

- Let 1 kg of gas be heated at constant volume and let the change in entropy and absolute

temperature be from s1 to s2 and T1 to T2

6-9

S

T

S2

S1

T2

T1

S2S1

Q

• Heating a Gas at Constant Pressure

- Let 1 kg of gas be heated at constant pressure, so that its absolute temperature changes from

T1 to T2 and entropy s1 to s2

• Isothermal Process

- Entropy changes from s1 to s2 when gas absorbs heat during expansion

- The heat taken by the gas is given by the area under the line 1-2 which also represents the work

done during expansion.

- In other words, Q = W

An isothermal expansion at constant temperature T

S

T

21T2

T1

S2S1

T = Constant

• Adiabatic Process

- Entropy s1 = s2 when gas absorbs adiabatic process

- There is no change in entropy and hence it is known as isentropic process- It is a vertical line (1-2) and therefore area under this line is nil ; hence heat supplied or

rejected and entropy change is zero

Adiabatic Process

S

T

2

1T2

T1

S1 = S2

• Polytropic Process 6-9

S

T

2

1

S2S1

Q

Formulae for Change of Entropy for Different Processes for Closed Systems

Entropy Relations to Other Thermodynamic Properties Consider a closed system undergoing a reversible process during which there are heat and workinteractions. Then applying the first law to the process, considering unit mass of the system, we canwrite dq = dw + du

Substituting dq = T.ds and dw = p.dv

we can write T.ds = p.dv + du …(1)The enthalpy is given by h = u+pv

The differential enthalpy is dh = du + p.dv + v.dpp.dv +du = dh –v .dp

Substituting this value in equation T.ds = dh – v. dp …(2)

• Equation 1 and 2 are know as fundamental thermodynamic relations combining first law and second

law of thermodynamics into single equation

• Derived for a reversible process they are applicable to irreversible processes also, as they

express relation between the properties and therefore are independent of the path followed by

the process

Change of Entropy for an Open System • In an open system, as compared with closed system, there is additional change of entropy due to

the mass crossing the boundaries of the system

• The net change of entropy of a system due to mass transport is equal to the difference between the product of the mass and its specific entropy at the inlet and at the outlet of the system

• Therefore, the total change of entropy of the system during a small interval is given by

𝑑𝑆 ≥𝑑𝑄𝑇𝑜

+2𝑠𝑖. 𝑑𝑚𝑖 −2𝑠𝑜. 𝑑𝑚𝑜

Generalform𝑑𝑆 ≥ 8+"9+∑ 𝑠. 𝑑𝑚

To – temperature of surroundings

Subscripts i and o refer to inlet and outlet conditions

𝑑𝑆 ≥ 𝑑𝑄𝑇𝑜

+ 2𝑠. 𝑑𝑚

𝑑𝑆𝑑𝑡 ≥

1𝑇𝑜𝑑𝑄𝑑𝑡 +2𝑠.

𝑑𝑚𝑑𝑡

0≥ #"9

8+8<+∑𝑠. 8=

8<

1𝑇𝑜Ǭ + 2𝑠. ṁ ≤ 0

2𝑠.ṁ ≤ 0

2𝑠.ṁ = 0