Continuity Equation For a fixed, non-deforming control volume conservation of mass requires the time rate of change of mass inside the control volume plus net rate of mass flow through the control surface must be zero at ∫or+」∫m For steady flow, time rate of change of mass in control volume is 0 「m;lhnm-∑mn=0 Where 77z mass flow rate in and out of control volume in kg/s
Continuity Equation For a fixed, non-deforming control volume, conservation of mass requires the time rate of change of mass inside the control volume, plus net rate of mass flow through the control surface must be zero For steady flow, time rate of change of mass in control volume is 0 Where = mass flow rate in and out of control volume in kg/s + = 0 CS n CV dV v dA t 0 = out − i n = CS vn dA m m m
Continuity Equation Control volume p Stream tube For uniform normal flow across a control surface of area a 77Z =PAV=pQ Mass inflow into cv= mass outflow from cv In terms of mean density and velocity p1A1Ⅴ1=p2A2V For incompressible fluid, p is constant AI VI=A 2V2=Q Q= volume flow rate along stream tu be
Continuity Equation For uniform normal flow across a control surface of area A : = AV = Q Mass inflow into CV = Mass outflow from CV In terms of mean density and velocity : 1 A1 V1 = 2 A2 V2 For incompressible fluid, is constant A1 V1 = A2 V2 = Q Q = volume flow rate along stream tube Control volume Stream tube 2 , A2 , V2 1 A1 V1 m
Continuity equation A1,V1 A. V For steady flow, conservation of mass requires ∑mm=∑ For incompressible fluid, p constant ∑Qin=∑Q out AIVI+a2v2=A3V3+A4V4 V is uniform and perpendicular to a
Continuity Equation For steady flow, conservation of mass requires : For incompressible fluid, constant : Qin = Qout A1V1 + A2V2 = A3V3 + A4V4 V is uniform and perpendicular to A A1 , V1 A2 , V2 A3 , V3 A4 , V4 mout =min
Continuity equation Summary pply continuity equation to a Control volume ΣQ:n=ΣQ out ∑A:V.=AV In n out out Assumptions p is constant Flow is steady Ⅴ is uniform across a V is perpendicular to A
Continuity Equation Summary : Apply continuity equation to a Control Volume: Qin = Qout AinVin = AoutVout Assumptions : • is constant • Flow is steady • V is uniform across A • V is perpendicular to A
Example Circular plate, 0.3 m diam 0. 005 m thick -0 005 m diam B.m/ CVI 0.1 m diam Find Q, va vc? Apply continuity equation to CVi: A B VB 兀0.12/4V=0.0052/4x3 VA=0.0075m/s pply continuity equation to cv2 ACⅤc=ABVB 2(0.15)x0.005XVc=兀0.0052/4x3 VC=0.0125m/s
Example Find Q, VA, VC ? Apply continuity equation to CV1 : AA VA = AB VB 0.12 /4 VA = 0.0052 /4 x 3 VA = 0.0075 m/s Apply continuity equation to CV2 : AC VC = AB VB 2 (0.15) x 0.005 x VC = 0.0052 /4 x 3 VC = 0.0125 m/s Q VA VC 0.1 m diam 0.005 m diam Circular plate, 0.3 m diam VB=3 m/s 0.005 m thick CV1 CV2