1. ME 200 L27: Control Volume Entropy Balance ME 200 L27: Control Volume Entropy Balance Kim See’s Office ME Gatewood Wing Room 2172 Please check your HW and Examination Grades on Blackboard Please pick up all graded Home Work and Examinations from Room 2172 before Friday, 4/4/14 https://engineering.purdue.edu/ME200/ ThermoMentor © Program Spring 2014 MWF 1030-1120 AM J. P. Gore gore@purdue.edu Gatewood Wing 3166, 765 494 0061 Office Hours: MWF 1130-1230 TAs: Robert Kapaku rkapaku@purdue.edu Dong Han han193@purdue.edurkapaku@purdue.eduhan193@purdue.edu

2. Entropy Balance Equation ►Like Internal Energy, Enthalpy, and Specific Volume, Entropy is an extensive property of a working substance. ►Analogous to and must apply simultaneously with the Conservation of Energy and ►Conservation of Mass No Internal Source of mass!

3. Substitute Entropy Definition into First Law ►Like Internal Energy, Enthalpy, and Specific Volume, Entropy is an extensive property of a working substance. ►Analogous to and must apply simultaneously with the Conservation of Energy and ►Conservation of Mass

4. Entropy Generation using Tables: Example 1 4 Given: Steam at 120 o C, 0.7 bar is pressurized through a diffuser to 1 bar, 160 o C and negligible velocity. Find: Find the change in entropy of steam in kJ/kg-K and comment on whether the diffuser can be adiabatic and the resulting impact. Assumptions: Change in PE neglected, No heat transfer, No work done other than flow work, Steady state, Steady flow, Mass is conserved.

5. 5 State 1: 0.7bar, 100 C State 2: 1 bar, 160 C T-s Diagram and Diffuser Action State 2: 1 bar, h2>h1, s2>S1 State 1: 0., 7bar, 100 C

6. On the T-s diagram drawn to scale State 1 and State 2 2 1

7. Entropy Generation Calculation: Example 2 7 Given: 0.5 kg/s of steam at 280 o C, 20 bar is expanded in a turbine to 1 bar in a constant entropy process. If the process was not a constant entropy process and resulted in saturated steam at 1 bar, find the decrease in work and increase in entropy in kW/K. Find: Find the work produced by the steam in kW and show the processes on a T-s diagram. If the process was not a constant entropy process and resulted in saturated steam at 1 bar, find the decrease in work and increase in entropy in kW/K. Assumptions: Change in PE neglected, No heat transfer, work done on Turbine shaft, Steady state, Steady flow, Mass is conserved. StP,Thsx i20,2802976.46.6828- es1,99.632423.26.68280.8883 e1,99.632675.57.35941.0000 x=(s-s f )/(s g -s f )=(6.6828-1.3026)/(7.3594-1.3026)=5.3802/6.0568=0.8883; h es = 417.46 +0.8883 (2258) = 2423.22

8. State 2a: Saturated State 1: 20 bar, 280 C On the T-s diagram drawn to scale State 1 and State 2 are close to each other as illustrated below. State 2s: Mixture

9. Entropy Generation: Example 3 9 Given: Consider R134 throttled from p3 =120 lbf/in 2 to p4 =40 lbf/in 2. Find: Find the change in entropy of R134. Assumptions: Change in KE, PE neglected, No heat transfer, No work done other than flow work, Steady state, Steady flow, Mass is conserved. Adiabatic throttle with a pressure loss and phase change lead to increase in Entropy while keeping Enthalpy constant. Entropy is generated by fluid friction Or viscosity in this case, in spite of the process being (externally) adiabatic.

10. 10 State 4 State 3 T-s Diagram and Demonstration of Throttle Action; h-s diagram State 3 State 4

11. Entropy Generation: Example 4 11 Given: Consider R134 condensed from saturated vapor (state 2) to saturated liquid at p3 =120 lbf/in 2. Find: Find the change in entropy generation rate in the process of condensing R134. Assumptions: Change in KE, PE neglected, Heat transfer to a sink at 90.54 o F- δ and heat transfer to sink at 80.54 o F. No work done other than flow work, Steady state, Steady flow, Mass is conserved.

12. 12 State 2 State 3 T-s Diagram and Demonstration of Condenser Action; h-s diagram State 2 State 3 Sink Temperature