上海交通大学：《热力学 Thermodynamics（I）》课程教学资源（课件讲义）Lecture 34_Entropy balance to open systems

上游充通大学 SHANGHAI JIAO TONG UNIVERSITY Engineering Thermodynamics I Lecture 34 Chapter 7 Entropy (Section 7.13) Spring,4/26/2018 强 Prof.,Dr.Yonghua HUANG 几m是。 http://cc.sjtu.edu.cn/G2S/site/thermo.html 1日

Engineering Thermodynamics I Lecture 34 Spring, 4/26/2018 Prof., Dr. Yonghua HUANG Chapter 7 Entropy (Section 7.13) http://cc.sjtu.edu.cn/G2S/site/thermo.html

Entropy rate balance for CV systems Like mass and energy,transferred by streams Closed system +o T →open system special dSev = dt ∑m空m rate of rates of rate of entropy entropy entropy change transfer production 上游气通大粤 April 26,2018 2 SHANGHAI JLAO TONG UNIVERSITY

April 26, 2018 2 Entropy rate balance for CV systems Like mass and energy, transferred by streams Closed system  open system special

Integral form of entropy balance eq.for CV 号空包 Heat flux time rate of entropy Sev(t) transfer accompanying heat transfer 0 w-)+p）-sp1元 velocity 上游充通大 April 26,2018 3 SHANGHAI JIAO TONG UNIVERSITY

April 26, 2018 3 Integral form of entropy balance eq. for CV Heat flux velocity

CV at steady state Steady state: Mass balance: ∑m=∑m Energy balance: 0=a.-晚+公e+兰+)-Σ(a-兰+x) Entropy balance: 70 dS= "'dt 汽区 Not conserved!! These equations often must be solved simultaneously. irreversibilities 上游充通大 April 26,2018 4 SHANGHAI JIAO TONG UNIVERSITY

April 26, 2018 4 CV at steady state Steady state: Mass balance: Energy balance: Entropy balance: 0 These equations often must be solved simultaneously. Not conserved!! irreversibilities

1-inlet,1-oulet CV at steady state 0= T ri(s1 -s2)+fev +r Similar form as CM sys. m Entropy difference at exit and inlet Entropy transfer Entropy production accompanying heat transfer due to irreversibilities >,=,52-S=0 m 上游充通大 April 26,2018 5 SHANGHAI JLAO TONG UNIVERSITY

April 26, 2018 5 1-inlet, 1-oulet CV at steady state Entropy transfer accompanying heat transfer >, =, < 0 Entropy production due to irreversibilities ≥0 Similar form as CM sys. For adiabatic system: if reversible 0 (isentropic) Entropy difference at exit and inlet

Example 34.1 Entropy Production in a Steam Turbine Known: P1 =30 bar T =400°C Wey =540 kJ/kg steam expands through V1=160m/s a turbine at steady state. steam Find: Gev/m D>T2=100C Tb=350K- Saturated vapor V2=100m/s Sketch 30 bar 400°C Assumption: 1.CV,(1,1,1)steady state 2.Heat transfer at T 3.△PE neglected 100C 上游充通大学 April 26,2018 6 SHANGHAI JLAO TONG UNIVERSITY

April 26, 2018 6 Example 34.1 Entropy Production in a Steam Turbine Find: steam Known: steam expands through a turbine at steady state. CV  / m Assumption: 1. CV, (1,1,1) steady state 2. Heat transfer at Tb 3. ∆PE neglected Sketch ?

Solution 30 bar 400°C Steady state,(1,1,1) Mass balance: 100C 0=m1-2 Entropy rate balance: ris-mi2s2 +ev To 0 三 s+is)o T 上游究通大学 April 26,2018 7 SHANGHAI JLAO TONG UNIVERSITY

April 26, 2018 7 Solution Steady state, (1,1,1) Mass balance: Entropy rate balance:

Solution cont. 30 bar 400°C Tp +(S1-S2)+0cv 100°C Q./m 个 (S2 m Energy balance equation:∠？ -++（明 i TabA-6,(30bar,400C)→h1=3230.9k/kg; TabA-4,(100C)→h2=hg=2676.1k/kg -sn是+eos1--[7,@1(g）hlml 1kJ m kg =540-554.8-7.8=-22.6kJ/kg 上游充通大学 April 26,2018 8 SHANGHAI JLAO TONG UNIVERSITY

April 26, 2018 8 Solution cont. Energy balance equation: ?? Tab A-6, (30bar, 400˚C)h1 = 3230.9 kJ/kg; Tab A-4, (100˚C)h2=hg = 2676.1 kJ/kg ??

Solution cont. 30bar 400°C TabA-4,(100C)→s2=7.3549k/kg TabA-6,(30bar,400°C)→s1=6.9212k/kg-K; 100°C -22.6kJ/kg、 Qev/ri +(S2-S1) m Tp (-22.6kJkg) 350K +(7.3549-6.9212) = 0.0646+0.4337=0.4983kJ/kg·K 上游充通大 April 26,2018 9 SHANGHAI JIAO TONG UNIVERSITY

April 26, 2018 9 Solution cont. Tab A-6, (30bar, 400˚C)s1 = 6.9212kJ/kg-K; Tab A-4, (100˚C)s2=7.3549 kJ/kg

Example 34.2 Evaluating a performance claim a single stream of air An inventor claims: T1=21C P1=5 bars a device requiring no energy transfer by work or 2 Inlet heat transfer.Separate air T2=79C into hot and cold streams. P2=I bar Hot outlet Evaluate the inventor's claim, Cold outlet T3=-18C P3=I bar 60%mass Assumption: 1.Steady state 2.CV:Qcv=0,Wcv=0 3. ideal gas model for air,cp=1.0 kJ/kg-K 4.ignoring APE and AKE of the streams from inlet to exit. 上游充通大 April 26,2018 10 SHANGHAI JLAO TONG UNIVERSITY

April 26, 2018 10 Example 34.2 : Evaluating a performance claim An inventor claims: a device requiring no energy transfer by work or heat transfer. Separate air into hot and cold streams. Evaluate the inventor’s claim, a single stream of air 60% mass Assumption: 1. Steady state 2. CV: Qcv=0, Wcv=0 3. ideal gas model for air, cp=1.0 kJ/kg-K 4. ignoring ∆PE and ∆KE of the streams from inlet to exit