1. ME 200 L28: Control Mass Entropy Balance and Directionality of Processes ME 200 L28: Control Mass Entropy Balance and Directionality of Processes https://engineering.purdue.edu/ME200/ Spring 2014 MWF 0930-1020 AM Professor Wassgren Lecture by Robert Kapaku; slides adapted from Prof. Gore TAs: Robert Kapaku rkapaku@purdue.edu Dong Han han193@purdue.edurkapaku@purdue.eduhan193@purdue.edu

2. Control Volume Entropy Balance Illustrating an Impossible Process 2 Given: Steam at 100 o C, 1 bar is pressurized through a diffuser to 1.5 bars, 120 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. Equations: Starting with basic conservation equations from the equation sheet, we arrive at: Adiabatic diffuser with given pressure gain leads to decrease in entropy. In reality, this diffuser design will not function! The pressure gain will be less than what is assumed here.

3. 3 State 1: 1bar, 100 C State 2: 1.5 bar, 120 C Saturated State: 1.5b bar, 111 C T-s Diagram and Diffuser Action (This diffuser will not work!) State 2: 1.5 bar, h2>h1, s2<S1 State 1: 1bar, 100 C

4. State 2: 1.5 bar, 120 C State 1: 1bar, 100 C On the T-s diagram drawn to scale State 1 and State 2 are close to each other as illustrated below.

5. Entropy Balance Equation ►Control Mass equations result from recognizing that there can be no inflows and outflows of mass ►Analogous to and must apply simultaneously with the Conservation of Energy

6. Control Mass Entropy Generation: Example 1 6 Given: Saturated liquid water at 10 bar is heated in a piston-cylinder device while maintaining pressure until the volume increases by a factor of 10. Assume the boundary temperature is equal to the water temperature. Find: (a) Work done in a reversible process, (b) Heat transfer in a reversible process, and (c) entropy production in kJ/kg-K, if the work done is (90% of theoretical value). Assumptions: Change in KE, PE neglected, Control mass. Equations: StP barT, Cv, m 3 /kgv f, m 3 /kgv g, m 3 /kgx 110179.91.1273(10 -3 ) 0.19440 210179.91.1273(10 -2 )1.1273(10 -3 )0.19440.0525 Important: Know why these equations are simplified this way!

7. Control Mass Entropy Generation: Example 1 7 StP barT, Cu, kJ/kgu f, kJ/kgu g, kJ/kgx 110179.9761.68 2583.60 210179.9857.331761.682583.60.0525 StP barT, Ch, kJ/kgh f, kJ/kgh g, kJ/kgx 110179.9762.81 2778.10 210179.9868.613762.812778.10.0525

8. Control Mass Entropy Generation: Example 1 8 StP barT, Cs, kJ/kgKs f, kJ/kgKs g, kJ/kgKx 110179.92.1387 6.58630 210179.92.37222.13876.58630.0525 179.9+273

9. On the T-s diagram drawn to scale State 1 and State 2b 21

10. 10 12 On the p-v diagram drawn to scale State 1 and State 2

11. September 17th, 2010ME 20011 In-Class Example