上海交通大学：《热力学 Thermodynamics（I）》课程教学资源（课件讲义）Lecture 16_Control volume analysis - mass conservation

上浒充通大￥ SHANGHAI JIAO TONG UNIVERSITY Engineering Thermodynamics I Lectures 16 Chapter 5 Mass and Energy Analysis of Control Volume Analysis Spring,2017 Prof.,Dr.Yonghua HUANG 强 AMMAMRA http://cc.sjtu.edu.cn/G2S/site/thermo.html 1日G

Engineering Thermodynamics I Lectures 16 Spring, 2017 Prof., Dr. Yonghua HUANG Chapter 5 Mass and Energy Analysis of Control Volume Analysis http://cc.sjtu.edu.cn/G2S/site/thermo.html

Closed system/Open system Closed system(CM) Open system (CV) T2,V2 D2 T State 1 ---> pT,v inlet exit > A Pr Ty Vi m Besides 0&W,energy accompanying State 2 mass as it enters or exits. Objective of Chap.5:develop illustrate the use of the CV forms of the conservation of mass and conservation of energy 上游究通大学 Mar/15,Wed,2017 2 SHANGHAI JLAO TONG UNIVERSITY

Mar/15, Wed, 2017 2 Closed system/Open system Closed system(CM) Open system (CV) State 1 State 2 p1 , T1 , v1 p2 , T2 , v2 m p1 , T1 , v1 p2 , T2 , v2 inlet exit Objective of Chap. 5: develop & illustrate the use of the CV forms of the conservation of mass and conservation of energy Q W Besides Q & W, energy accompanying mass as it enters or exits. pi , Ti , vi pe , Te , ve

Developing the mass rate balance Dashed line defines the control volume boundary mi nev(①) nev(t+△) Region i LRegion e m →m; Time t Time t+△t me→ Study: the fixed quantity of matter m as time elapses: mev(t)+mi=mev(t At)+me mev(t At)-moy(t)=mi-me me(t+△t)-nev(t) mi me △t △t△t 上游充通大 Mar/15,Wed,2017 3 SHANGHAI JIAO TONG UNIVERSITY

Mar/15, Wed, 2017 3 Developing the mass rate balance Study: the fixed quantity of matter m as time elapses: mi me m

conservation of mass principle min=50 kg Dashed line defines the control volume mi boundary One-inlet,one-exit control volume system mev(t) > me Inlet i mcv(t+△t)-mev(t) mi me Exit e △t △t △t At time t time rate of change of time rate of flow time rate of flow mass contained within of mass in across of mass out across the control volume at time t inlet i at time t exit e at time t Multi-streams dmev At time t: dt mi me → L Mass flow rate (SI units:kg/s) 上游充通大 Mar/15,Wed,2017 4 SHANGHAI JIAO TONG UNIVERSITY

Mar/15, Wed, 2017 4 conservation of mass principle At time t mcv(t) mi me At time t: Mass flow rate (SI units: kg/s) Multi-streams One-inlet, one-exit control volume system

Evaluating the mass flow rate ri velocitys、 time interval dm dA Volume Control volume (CV) of matter Incremental area Control surface(CS) amount of mass A:area crossing dA during =pVn△t)dA the time interval△t Divided by△t ≥(N.a) Muti-streams instantaneous rate Integration over A of mass flow =pVn dA 防= PVn dA across dA 上究大学 Mar/15,Wed,2017 5 SHANGHAI JIAO TONG UNIVERSITY

Mar/15, Wed, 2017 5 Evaluating the mass flow rate m velocity time interval : area Incremental area Divided by ∆t Integration over A Muti-streams

Different forms of mass flow rate balance One dimensional flow form assumption: The flow is normal to the boundary at inlet and outlet locations. All intensive properties,including velocity and density,are uniform with position over each inlet or exit area. 1 D flow: Area=A area,m2、 velocity,m/s Air V.T.v m=pAV or Air compressor density,kg/m3 Air mass flow rate, volumetric flow rate, kg/s m3/s -Ay-Σ dt 上游充通大学 Mar/15,Wed,2017 6 SHANGHAI JLAO TONG UNIVERSITY

Mar/15, Wed, 2017 6 Different forms of mass flow rate balance One dimensional flow form assumption: • The flow is normal to the boundary at inlet and outlet locations. • All intensive properties, including velocity and density, are uniform with position over each inlet or exit area. 1D flow: or velocity, m/s area, m2 density, kg/m3 mass flow rate, kg/s volumetric flow rate, m3/s

Steady state All properties are unchanging in time For a CV the total amount present at any instant remains constant ∑=∑m Steady sate今∑m=∑m】 ∑m=∑成芩Sady stae (T,p...might varying with t) every property is independent of time 上游充通大 Mar/15,Wed,2017 7 SHANGHAI JLAO TONG UNIVERSITY

Mar/15, Wed, 2017 7 Steady state All properties are unchanging in time For a CV the total amount present at any instant remains constant Steady state     Steady state (T, p… might varying with t) every property is independent of time 0

Example 16.1:Steady flow feedwater heater,Steady State (SS) b Find:Determine mz,m,and the velocity V2. A2=25cm2 T1=200C T2=40C P1=7 bar rin =40 kg/s Solution: (AV)3 0.06m3/s P2=7 bar m3= (1.108×10-3m3/kg) =54.15kg/s 3 个 Control volume boundary Saturated liquid P3 =7bar,@sat.liq.>Table A-5 P3 7 bar (AV)3=0.06m3/s =∑成一∑m: dt 0 given dm dt m1+m2一m3 ? m2=m3-m1=54.15-40=14.15kg/s 上游充通大 Mar/15,Wed,2017 8 SHANGHAI JIAO TONG UNIVERSITY

Mar/15, Wed, 2017 8 Example 16.1: Steady flow feedwater heater, Steady State (SS) Find: Determine , , and the velocity V2 . m2 m3 Solution: ? p3 =7bar, @sat. liq.  Table A-5 given

Solution cont. How to find V2? 14.15kg/s T=200C 1 A2=25cm2 T2=40C P1=7bar P2=7bar rin =40 kg/s V2=m22/A2 25cm2 Control volume boundary Saturated liquid P3=7bar (AV)3=0.06m3/s > State2:T2=40C----> Subcooled/Compressed liquid. P2 7 bar Approximation Tables/Graphs/Software U2≈U(T2) Table A-4:40°C =1.0078×10-3m3/kg (14.15kg/s)(1.0078×10-3m/kg)104cm2 V2= 5.7m/s 25cm2 1m2 上游充通大 Mar/15,Wed,2017 9 SHANGHAI JIAO TONG UNIVERSITY

Mar/15, Wed, 2017 9 Solution cont. ? State 2: Subcooled/Compressed liquid. Approximation Tables/Graphs/Software Table A-4: 40˚C How to find V2 ? V2 ?

Example 16.2 Transient problem open! Air Initially filled with water pulled out discharge plug near the bottom Water ·average velocity V=√2gh h is the height of water in the tank measured from the center of the hole (a variable) 121.92cmh01 g is the gravitational acceleration,=9.81m/s2 1.27cm Determine: h how long it will take for the water level in the tank to drop to ho/2 from the bottom? 0 Dtank 91.44cm Assumptions: [Textbook Example 5-2] 1.Water is an incompressible substance. 2.The distance between the bottom of the tank and the center of the hole is negligible compared to the total water height. Analysis: take the volume occupied by water as the CV---a variable CV an unsteady-flow problem since the properties within the CV change with time. 上降文通大学 Mar/15,Wed,2017 10 SHANGHAI JIAO TONG UNIVERSITY

Mar/15, Wed, 2017 10 Example 16.2 Transient problem 121.92cm 91.44 cm open 1.27 cm Determine: how long it will take for the water level in the tank to drop to h0/2 from the bottom? Assumptions: 1. Water is an incompressible substance. 2. The distance between the bottom of the tank and the center of the hole is negligible compared to the total water height. Analysis: • take the volume occupied by water as the CV--- a variable CV • an unsteady-flow problem since the properties within the CV change with time. [Textbook Example 5-2]