For only one inlet and one outlet According to continuity Fx ∑F d (mv rm(vout-vin) Fy=n(2y-vly) 2-out,1一 F==m(V2=-V1=) Example: A fixed control volume of a streamtube in steady flow has a uniform inlet (u,Al, v1 )and a uniform exit (2, A2, V2). Find the net force on the control volume Solution: XF=ri(v2-vi)m=pAiv=p,a2v )=ni(v2-V1 cos 0) EF,=i(v-Vi)=-mvisin8 O
m m m out = = in For only one inlet and one outlet According to continuity ) ( ) ( s out in d mV F m V V dt = = − 2－out, 1－ in Example: A fixed control volume of a streamtube in steady flow has a uniform inlet (r1 ,A1,V1 )and a uniform exit (r2 ,A2,V2) . Find the net force on the control volume. V 1 1 V 2 2 o x y F m V V x x x = − ( ) 2 1 F m V V y y y = − ( ) 2 1 F m V V z z z = − ( ) 2 1 o x y z 2 1 Solution: = − F m( ) V V 2 1 ( ) = − Fx m V V x x 2 1 = − m( cos ) V V 2 1 ( ) = − F y m V V y y = −mV1 sin m = = 1 2 A A 1 2 V V 1 2
Example Given p1=P2=4.19×105M m=785k8 10cm, d2=&cm 98 Neglect the weight of the fluid. Find the force on the water by the elbow pipe Solution select coordinate control volume ∑F=mi(V2-1) Fx=m(v2x-Vix=my Fsx-p2A2=mv2 X Fx=P2A2+m278.52×4 +p2=3696N 998ad2
Given 5 1 2 4.19 10 2 p p N m = = 78.5Kg m s = d d 1 2 = = 10 , 8 cm cm 998 3 Kg m = Neglect the weight of the fluid. Find the force on the water by the elbow pipe. Example: 1 2 1 2 Solution: x y o select coordinate ,control volume 2 1 = − F m( ) V V 2 1 ( ) F x m V V x x • = − mV 2 • = F A sx p2 2 mV 2 • − = F A sx p2 2 mV 2 • = + 2 2 2 2 2 2 78.5 4 998 4 d p d = + = 3696N
In the like manner Fs+p1A1=m(0-V1)=-m1 5934N Fs=-P1A1-ml1=-4642N 0=1g2=14147° F Find the force to fix the elbow ex Solution: coordinate Net force on the control volume 2Fx=P,AL-P(Ar-A2)-p2A2+Fe PoAr-PAr-(p,-P,)AL +F Where Fex is the force on the cv by pipe, on elbow) ∑Fx=Fex-(P2-p)A2 In like manner 2Fr=Fey+(p1-pa)at Surface force: (1) Forces exposed by cutting though solid bodies which protrude into the surface (2)Pressure,viscous stress
2 2 F F N s sy = + = Fsx 5934 1 1 1 1 (0 ) F A sy p m m V V • • + = − = − F A sy p1 1 mV1 = -4642N • = − − In the like manner 1 sy sx F tg F − = =141.47 Find the force to fix the elbow. Solution: coordinate, CV Net force on the control volume: 2 2 2 x L R ex ( ) = − − − + F A A A A F p p p a a Where Fex is the force on the CV by pipe,( on elbow) 2 L R ( ) = − − − p p p p a a a A A AL + Fex 1 2 x y o 2 2 ( ) x ex a = − − F F A p p In like manner 1 1 ( ) y ey a = + − F F A p p Fex Surface force: (1) Forces exposed by cutting though solid bodies which protrude into the surface.(2)Pressure,viscous stress
a fixed vane turns a water jet of area a through an angle e without changing its velocity magnitude The flow is steady pressure pa is everywhere, and friction on the vane is negligible. Find the force F applied to vane Fx=mv(cos 8-1) F,=mv sin e
A fixed vane turns a water jet of area A through an angle without changing its velocity magnitude. The flow is steady, pressure pa is everywhere, and friction on the vane is negligible. Find the force F applied to vane. V F V (cos 1) F x = − mV F y = mV sin
a water jet of velocity vi impinges normal to a flat plate which moves to the right at velocity vc. Find the force required to keep the plate moving at constant velocity and the power delivered to the cart if the jet density is 1000kg/m3 the jet area is 3cm 2 and Vj=20m/s, Vc=15m/s Neglect the weight of the jet and plate, and assume steady flow with respect to the moving plate with the jet splitting into an equal upward and downward half-jet 7.5N
A water jet of velocity Vj impinges normal to a flat plate which moves to the right at velocity Vc. Find the force required to keep the plate moving at constant velocity and the power delivered to the cart if the jet density is 1000kg/m3 the jet area is 3cm2, and Vj=20m/s,Vc=15m/s V j V c x Neglect the weight of the jet and plate,and assume steady flow with respect to the moving plate with the jet splitting into an equal upward and downward half-jet. −7.5N