上海交通大学：《热力学 Thermodynamics（I）》课程教学资源（课件讲义）Lecture 32_Internally reversible processes, Closed system entropy balance

上岸充通大学 SHANGHAI JIAO TONG UNIVERSITY Engineering Thermodynamics I Lecture 32 Chapter 7 Entropy (Section 7.13) Spring,2017 福 Prof.,Dr.Yonghua HUANG 几nMA http://cc.sjtu.edu.cn/G2S/site/thermo.html 1日G

Engineering Thermodynamics I Lecture 32 Spring, 2017 Prof., Dr. Yonghua HUANG Chapter 7 Entropy (Section 7.13) http://cc.sjtu.edu.cn/G2S/site/thermo.html

△S~Q relationship Internally reversible process Here only condider closed system cases 6Q+→dS>0 entropy transfer 6g-→dS<0 accompanies heat transfer 00→S=0y adiabatic internally reversible process (δQ)mt=TdS rev 2 T(2)in=Tds rev Tds Isentropic process rev Internally reversible--- 上游充通大 Wednesday,Ap...-,. 2 SHANGHAI JIAO TONG UNIVERSITY

Wednesday, April 12, 2017 2 ∆S ~ Q relationship Internally reversible process Here only condider closed system cases 0 0 0 0 Q dS Q dS Q dS            entropy transfer accompanies heat transfer adiabatic internally reversible process Isentropic process Internally reversible

An example for illustrating All internal reversible processes: Net work (enclosed) TH TH Te 4 b a b a S S (a) (b) Carnot power cycle Carnot refrigeration cycle 1→2：adiabatic,internal reversible,Q12=0 2→3：Q23=∫23TdS=TH(S3-S2),area2-3-a-b-2 Weycle area1-2-3-41 3→4：adiabatic,internal reversible,Q34=0 7= Q23 area 2-3-a-b-2 4→1：Q41=J41TdS=T(S1-S4),area4-1-b-a-4 (Ta-Te)(S3-S2)Ti-Tc 1-工 T(S3-S2) Ta 上泽通大学 Wednesday,April 12,2017 3 SHANGHAI JLAO TONG UNIVERSITY

Wednesday, April 12, 2017 3 An example for illustrating Carnot refrigeration cycle 12: adiabatic, internal reversible, Q12=0 23: Q23 = ∫2 3 TdS=TH (S3 -S2 ), area 2-3-a-b-2 34: adiabatic, internal reversible , Q34=0 41:Q41 = ∫4 1 TdS=TL (S1 -S4 ), area 4-1-b-a-4 Carnot power cycle All internal reversible processes: Net work (enclosed) H C 3 2 H C C H 3 2 H H ( )( ) 1 ( ) T T S S T T T T S S T T        

Example 32.1 Known:water,initially sat.liq.100C;internally reversible heating process; final sat.vap. Find:W/m and Q/m T Water g 100°C System boundary g 100°C m S 上游充通大 Wednesday,April 12,2017 4 SHANGHAI JIAO TONG UNIVERSITY

Wednesday, April 12, 2017 4 Example 32.1 Known: water, initially sat. liq. 100˚C; internally reversible heating process;  final sat. vap. Find: W/m and Q/m

Solution Internal reversible process;const.T&const.p;closed system; Neglect△PE,△KE W m J pdo=plv:-v) (1.014bar)(1.673-1.0435×10-3) 1kJ = 103N·m =170 kJ/kg Tas-m Tds m =T(Sg-S)=(373.15K)(7.3549-1.3069)k/kgK=2257kkg or Sg-Sr= lighr2=TAs=T(s-5 )=hh Use table A-4 T m 上游充通大学 Wednesday,April 12,2017 5 SHANGHAI JLAO TONG UNIVERSITY

Wednesday, April 12, 2017 5 Solution Internal reversible process; const. T &const. p; closed system; Neglect ∆PE, ∆KE or g f g f ( ) Q T s T s s h h m       Use table A-4