Home work: P190-346 P191-350 P192-3.54 P192-p358
Home work: P190-p3.46 P191-p3.50 P192-p3.54 P192-p3.58
Derive the thrust(Etj)equation for the jet engine air drag is neglect Solution Coordinate, CV F inlet: uniform V,p, outlet outer v p v p inner Ve p e 2F=(mnout-(mv)in i,,=rn m 1)=(m+m1)-mV m: mass flux of fuel i m me left paA ight-paAee-pe v p ∑F Fx balance with thrust 2F=Fx+P A0o'-p,Aee-P, Aee=Fx+P,Aee-p Aee Fx+(p.-pAee=meve-mv x=meve-mv+(pe-p)Aee m≤0.02m R=-Fx=-[meve-mv+(p-p)Aee R-m(e-v)+(p-pu)Aeel
Derive the thrust(推力) equation for the jet engine. air drag is neglect inlet : , a uniform V p outlet outer V pa inner V e pe Solution: = − F mV mV ( ) ( ) out in = − m V mV e e ( ) e = + − m m V mV f mf : mass flux of fuel o o e e e e m o o V a p V a p F x right ' left − − p p a e A A e e e ' ' o o pa A Balance with thrust = + Fx p p a e A A ee ee − Fx V e e p me V a p Fx e e ( ) Fx m V mV = − + − p p e a Aee R Fx m V mV = − = − − + − [ ) ] e e (p p e a Aee R m V V − − + − [ ( ) ) ] e (p p e a Aee ( ) Fx + − p p a e Aee = − m V mV e e ' ' ' o o = F Fx A + − − p p p a a e A A e ee e Coordinate, CV m m e m m f 0.02 moo mee moo = mee
Example: In a ground test of a jet engine pa=1.0133×105N/m2Ae=0.1543m2Pe=1.141×105N/m2 Ve=542m/s, i=43 4Kg/s. Find the thrust force Solution: coordinate R≈-[mi(Ve-1)+(P。-p)Aa =0 R=-lmve+(Pe- pa)ael 25493N F16R=65.38KN
Example: In a ground test of a jet engine, pa=1.0133×105N/m2 ,Ae=0.1543m2,Pe=1.141×105N/m2, Ve=542m/s, . m Kg s = 43.4 / Find the thrust force. V e e p me a p Solution: [ ( ) ] 25493 R mV P P A e e a e N = − + − = − V = 0 F16 R=65.38KN R m V V − − + − [ ( ) ) ] e (p p e a Aee x coordinate
Exampl e A rocket moving straight up. Let the initial mass be moand assume a 个v() steady exhaust mass flow and exhaust velocity ve relative to the rocket If the flow pattern within the rocket motor is steady and air drag is neglect Derive the differential equation of vertical rocket motion v(tand integrate using the initial condition V=O at t=o Solution: coordinate The Cv enclose the rocket, cuts through p the exit jet, and accelerates upward at rocket speed v(t)
A rocket moving straight up. Let the initial mass be M0 ,and assume a steady exhaust mass flow and exhaust velocity ve relative to the rocket. If the flow pattern within the rocket motor is steady and air drag is neglect. Derive the differential equation of vertical rocket motion v(t) and integrate using the initial condition v=0 at t=0 . Example: ve , pe Ae mf v t( ) Solution: The CV enclose the rocket,cuts through the exit jet,and accelerates upward at rocket speed v(t). coordinate z v(t)
Z-momentum equation ∑F=(mv)an-(mv)m m v)-(-m4y) >F=-PA+P(A-A)+Pa -mo dt (P - - -m g vedt-gdt m(t)=Mo-n 0 dv=mfveJo Mo-m gl.dt p v(t) ve ne Mo
Z-momentum equation: = F mv mv ( ) ( ) out in − e f dv mg m m v dt − − = − ( ) 0 m t M m t = − f f e m dv v dt gdt m = − 0 0 0 0 v t t e f f dt dv m v g dt M m t = − − 0 ( ) ln(1 ) f e m t v t v gt M = − − − mA mA ( ) ( ) = − − − − m v m v m v f e A A = −m vf e −P A a ( ) + − P A A a e +P A e e −mg dv m dt = F − ( ) e a e dv P P A mg m dt = − − − a e if P P ve , pe Ae mf v t( ) v(t) z A