1. Entropy Change Property diagrams (T-s and h-s diagrams) –From the definition of the entropy, it is known that  Q=TdS during a reversible process. –Hence the total heat transfer during such a process is given by Q reversible =  TdS Therefore, it is useful to consider the T-S diagram for a reversible process involving heat transfer On a T-S diagram, the area under the process curve represents the heat transfer for a reversible process. T S

2. Example Show a Carnot cycle on a T-S diagram and identify the heat transfer at both the high and low temperatures, and the work output from the cycle. S T THTH TLTL S 1 =S 4 S 2 =S 3 1 2 3 4 AB 1-2, reversible isothermal heat transfer Q H =  TdS = T H (S 2 -S 1 ) area 1-2-B-A 2-3, reversible, adiabatic expansion isentropic process, S=constant (S 2 =S 3 ) 3-4, reversible isothermal heat transfer Q L =  TdS = T L (S 4 -S 3 ), area 3-4-A-B 4-1, reversible, adiabatic compression isentropic process, S 1 =S 4 Net work W net = Q H - Q L, is the area enclosed by 1-2-3-4, the shaded area QHQH QLQL W out W in

3. Mollier Diagram Enthalpy-entropy diagram, h-s diagram: it is valuable in analyzing steady-flow devices such as turbines, compressors, etc.  h: change of enthalpy from energy balance (from the first law of thermodynamics)  s: change of entropy from the second law ( a measure of the irreversibilities during an adiabatic process) ss hh h s

4. TdS (Gibbs) Equations Since, the area under the T-s line is equal to the heat transfer for a reversible process, it would be useful to have a relationship between Temperature and Entropy to obtain the heat transfer,T. Such a relationship exists for a closed system containing a pure, compressible substance undergoing a reversible process dU =  Q rev -  W rev = TdS - PdV => TdS = dU + PdV or Tds = du + pdv ( per unit mass) This is the well-known Gibbs equation Alternatively, eliminate du by using the definition of enthalpy, h = u + pv => dh = du + pdv + vdp, => du + pdv = dh – vdp Tds = du + pdv Tds = dh – vdp TdS

5. Comments Regarding theTdS Equations Equations relate the entropy change of a system to other properties, such as enthalpy (h), internal energy (u), pressure and volume. Hence, even though T-ds relations were derived for a reversible process, they are valid for any process, reversible or irreversible. They are applicable for closed or open systems. They are valid for all pure substances, single or multi-phase.

6. Example – Application of Tds relations Consider steam undergoing a phase transition from liquid to vapor at a constant temperature of 20°C. Determine the entropy change s fg =s g -s f using the Gibbs equations and compare the value to that read directly from the thermodynamic table. From table A-4: T=20°C, P = 2.338 kPa, v f = 0.001002(m 3 /kg), v g =57.79(m3/kg), u f =83.9(kJ/kg), u g =2402.9(kJ/kg) Substituting into the Tds relation: s fg = (1/293)(2402.9-83.9) + (2.338/293)(57.79-0.001002) = 8.375(kJ/kg K) Compares favorably with the tabulated value s fg =8.3715(kJ/kg K) 1/T u g - u f v g - v f P/T

7. Entropy Change of an Incompressible Substance (YAC:6-8) For most liquids and all solids, the density does not change appreciably as pressure changes, hence dv  0. Gibbs equation states that Tds = du+pdv  Tds = du where du = CdT, for an incompressible substance and C p =C v =C is a function of temperature only. Therefore, ds = du/T = C dT/T Specific heats for some common liquids and solids can be found in thermodynamic tables such as Table A-14 to A-18 Note: According to the above equation, an isothermal process for a pure incompressible substance is isentropic.

8. Example – Entropy Change of Incomp. Substances A 1-kg metal bar, initially at 1000 K, is removed from an oven and quenched by immersing in a closed tank containing 20 kg of water, initially at 300 K. Assume both substances are incompressible and C water = 4(kJ/kg K), C metal = 0.4(kJ/kg K). Neglect heat transfer between the tank and its surroundings. (a) Determine the final temperature of the metal bar, (b) entropy generation during the process. T m =1000 K, m m =1kg, c m =0.4 kJ/kg K T w =300 K, m w =20 kg, c w =4 kJ/kg K

9. Solution The total entropy of the system increases, thus satisfy the second law

10. Entropy Change of Ideal Gases (YAC:6-9) From Gibbs’ equations, the change of entropy of an ideal gas can be expressed as: For an ideal gas, u=u(T) and h=h(T), du=c v (T)dT and dh=c p (T)dT and Pv = RT

11. Entropy Change of Ideal Gases – Special Cases The dependence of specific heats on temperature makes the integration more complex. However, for certain cases, one can make simplifying assumptions. Case 1: If specific heats are assumed constant, integration is simplified: Note: The above is strictly true for monatomic gases, e.g.He, Ar It is fairly accurate if the temperature difference is small. Case 2: Calculate the specific heat at an average temperature, T avg, and assume it to be constant. This also allows the specific heat to be taken out of the integral Note: This approximation is generally fairly accurate if the temperature difference is not too large, usually good if  T < few hundred degrees. Strictly speaking, one should look at the temp. dependence of specific heats for the particular substance to evaluate the validity of this approximation.

12. Isentropic Processes for Ideal Gases If a process is isentropic (that is adiabatic and reversible), ds = 0, s 1 =s 2, then it can be shown that: The above are referred to as the: first, second and third, respectively, isentropic relations for Ideal Gases (assuming constant specific heats). They can also be written as: Tv k-1 = constant First isentropic relation TP (1-k)/k = constant Second isentropic relation Pv k = constant Third isentropic relation

13. Example Air is compressed from an initial state of 100 kPa and 300 K to 500 kPa and 360 K. Determine the entropy change using constant c p =1.003 (kJ/kg K) Negative entropy due to heat loss to the surroundings