1-Dfow:< Tis only the function of s.Φ=Φ(s) (d)mn=(Bdm)m=(pAds)im=(BpAVdt) In the like manner (d)out =(Bpavdtout 2 tt+at dd +-,[(dd)o-(a)m dt -+lBpADout-BpAl)in RTT For stead docy dd y (BpADout-BpADin flow dt
( ) ( ) ( ) ( ) d dm Ads AVdt = = = in in in in In the like manner ( ) ( ) d AVdt = out out 1 [( ) ( ) ] s cv out in d d d d dt dt dt = + − [( ) ( ) ] cv out in d AV AV dt = + − s 1-D flow : is only the function of s . = ( )s For steady flow : 0 d cv dt = ( ) ( ) s out in d AV AV dt = − t+dt t+dt t t ds R T T
If there are several one-D inlets and outlets: dd ∑(月41)m-∑(月pA Steady, 1-D only in inlets and outlets, no matter how the flow is within the Cv 33 Conservation of mass(质量守恒) (Continuity Equation) O=m B=dm/dm ∑(A1om-(OA)m=0 Σ(P,A1)=E(P,A)m∑(m)m=∑(m2) out Mass fluⅹ(质量流量m)
If there are several one-D inlets and outlets : ( ) ( ) s out i i i i i i i i in i i d AV AV dt = − Steady , 1-D only in inlets and outlets, no matter how the flow is within the CV . 3.3 Conservation of mass (质量守恒) (Continuity Equation) f=m =dm/dm=1 ( ) ( ) 0 s out in i i i i i i i i dm A A V V dt = − = ( ) ( ) out in i i i i i i i i AV V = A ( ) ( ) i in i out i i m m = Mass flux (质量流量 m )
For incompressible flow ∑(A)m=∑(A)mQ=A1 Volume flu体积流量 If only one inlet and one outlet A11=A22 -Leonardo da vinci in 1500 妻口瀑布是我国著名的第二大瀑布。两百多來宽的黄炣河面,突然紧缩 为50米左右,跌入30多米的形蛱谷。入妻之水,奔腾咆,势如奔马,浪 声震天,声闻十里。“黄河之水天上来”之惊心动魄的景观
For incompressible flow: ( ) ( ) out in i i i i i i A A V V = i i Qi = AV Volume flux 体积流量 A A 1 2 V V 1 2 = -------Leonardo da Vinci in 1500 If only one inlet and one outlet 壶口瀑布是我国著名的第二大瀑布。两百多米宽的黄河河面，突然紧缩 为50米左右，跌入30多米的壶形峡谷。入壶之水，奔腾咆哮，势如奔马，浪 声震天，声闻十里。 “黄河之水天上来”之惊心动魄的景观
Exampl l Ajet engine working at design condition at the inlet of the nozzle p1=205×10Nm2T1=865K,Ⅵ1=288m/s,A1=0.19m3; At the outlet p,=1.143 x10N/m T2=766K, A2=0.1538m Please find the mass flux and velocity at the outlet Given R=287. 4 J/k, K gas constant Solution rm= pAv Rn AV-P,AV 45.1kg/s R71 According to the conservation of mass m=14V1=P2A12→P41 RT RT 2=565.1m/s Homework P185P3 12 P189P336
Example: A jet engine working at design condition. At the inlet of the nozzle At the outlet Please find the mass flux and velocity at the outlet. Given gas constant 5 2 1 p N m = 2.05 10 / T1 =865K，V1=288 m/s，A1=0.19㎡； 5 2 2 p N m = 1.143 10 / T2 =766K，A2=0.1538㎡ R=287.4 J/kg.K。 Solution 1 1 1 1 45.1 / p AV kg s RT m AV = = = p AV RT = 1 1 1 2 2 2 1 2 p AV p A V RT RT = According to the conservation of mass m AV A V = = 1 1 1 2 2 2 1 1 2 2 1 2 2 1 565.1 / A p T V V m s A p T = = Homework: P185 P3.12, P189P3.36
34 The Linear Momentum Equation(动量方程) Newton's Second Law dop dt ∑(BpA)m-∑(B,A) steady RtT g-my (linear -momentum) g flux dmk B v momentum perunit mass d(mv) ∑(p,A1)m-∑(pA)m=∑(mV)-∑(mT)m Newton's d(my) ∑(mV1)om-2(mv)m second law ∑ m;: Momentum flux(动量流量) ∑ F: Net force on the system or CV(体系或控制体受到的合外力
= mV ( ) linear momentum − dmV V dm = = momentum perunit mass ( ) ( ) ( ) s out in i i i i i i i i i i d mV A A V V dt = − 3.4 The Linear Momentum Equation (动量方程) （ Newton’s Second Law ） ( ) ( ) out in i i i i i i = − m m V V Newton’s second law d mV ( )s F dt = ( ) ( ) out in i i i i i i = − m m V V F :Net force on the system or CV (体系或控制体受到的合外力) miV i ：Momentum flux (动量流量) ( ) ( ) s out in i i i i i i i i d A A V V dt = − 1-D in & out steady RTT flux