1. ME 200 L24: Definition of Entropy & Entropy as a Property 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 https://engineering.purdue.edu/ME200/ ThermoMentor © Program Launched 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 is a Property ►Entropy is easier to understand if thought of as a property analogous to specific volume. It is defined as:

3. Entropy Change Calculations Q H =1000 kJ, T H =500 K and Q C =600 kJ at T C : (a) 200 K, (b) 300 K, (c) 400 K. Find if each heat transfer is reversible find entropy change for (the material that makes up) the reservoir. Solution: Use the given Q H, Q C and T H, T C values to find the entropy change for the reservoirs:

4. Examples: Entropy Change for the Reservoirs (Property of materials making up the reservoirs) (a) Entropy can decrease and increase! Net entropy change can be zero! (b) (c) Net entropy change can be negative! However, can’t continue that process in a cyclic manner! Since by definition reservoir properties are fixed, these changes must be compensated by reverse actions!

5. Entropy Change using Tables 1 5 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. Adiabatic diffuser with given pressure gain leads to decrease in entropy. In reality, this diffuser design will not function!

6. 6 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

7. Entropy Change using Tables 2 7 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.

8. 8 State 4 State 3 T-s Diagram and Demonstration of Throttle Action; h-s diagram State 3 State 4