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ME3560 – Fluid Mechanics Summer 2016 ME 3560 Fluid Mechanics Chapter V. Finite control Volume Analysis 1 Chapter V. Finite control Volume Analysis ME3560 – Fluid Mechanics Summer 2016 5.1 Conservation of Mass–Continuity Equation • Reynolds Transport Theorem states that: The rate of change of a property in a control system is equal to the property accumulated within the control volume plus the rate at which the property is leaving the control volume. r dB ∂ = ρ b dV + ∫ ρ b (V ⋅ nˆ ) dA ∫ dt Sys ∂ t cv cs •The previous equation applied to the mass of a control system (B = m and b = 1) results in: r ∂ ˆ 0= ρ d V + ρ ( V ⋅ n ) dA ∫ ∫ ∂ t cv cs 2 Chapter V. Finite control Volume Analysis Summer 2016 ME3560 – Fluid Mechanics r ∂ ρ dV + ∫ ρ (V ⋅ nˆ ) dA = 0 ∫ ∂ t cv cs • This equation is also know as continuity equation. •The first term represents the mass accumulated within the CV. •The second term represents the amount of mass per unit time crossing the boundaries of the control surface (flux of mass) •If the process occurs under steady state conditions r ∫ ρ (V ⋅ nˆ ) dA = 0 m& = ρVA cs •If the density of the fluid is constant (incompressible flow): ∫ r (V ⋅ nˆ ) dA = 0 cs 3 Chapter V. Finite control Volume Analysis Summer 2016 ME3560 – Fluid Mechanics 5.2 Newton's Second Law: The Linear Momentum r dB ∂ = ρ b dV + ∫ ρ b (V ⋅ nˆ ) dA ∫ dt Sys ∂ t cv cs • To determine the r Linear Momentum Equation: B = mV and b = V r r r d (mV ) ∂ ˆ = ρ V d V + ρ V ( V ⋅ n ) dA ∫ ∫ dt Sys ∂ t cv cs • Additionally, we know that Newton’srSecond Law states that r d (mV ) = ∑F dt Sys Chapter V. Finite control Volume Analysis 4 ME3560 – Fluid Mechanics Summer 2016 • Thus, the conservation of momentum equation can be presented as r r r r ∂ ˆ ρ V d V + ρ V ( V ⋅ n ) dA = ∑ F ∫ ∫ ∂ t cv cs • That is: The amount of momentum accumulated per unit time in the CV plus the net rate of momentum flowing through the CS is equal to the forces acting on the CV. •In general the right hand side of the previous expression can be expressed as: r r r ∑ F = ∑ FSurface + ∑ F Body 5 Chapter V. Finite control Volume Analysis Summer 2016 ME3560 – Fluid Mechanics r r r ∑ F = ∑ FSurface + ∑ F Body • ΣFsurface are external forces acting on the CV. • ΣFbody is the component of the weight in the corresponding direction. In general the coordinate system will be selected such that the components of the weight in two of the three directions are zero. • ΣFsurface – Net force due to Normal Stress (pressure) r Fpressure = ˆ p n ∫CS dAr • If p = uniform =p0 Fpressure = CV ∫ pnˆdA = 0 CS •Gauss Theorem ∫ Φn̂ds = ∫ ∇ΦdV • If Φ=Const then∇Φ=0 S V Chapter V. Finite control Volume Analysis 6 ME3560 – Fluid Mechanics Summer 2016 • In conclusion, the net force due to a uniform pressure on the CV is zero. Therefore, it is necessary to use “gage pressures” when applying the momentum equation. •The momentum equation is a vector equation which can be written as: r ∂ ρ u dV + ∫ ρ u (V ⋅ nˆ ) dA = ∑ Fx ∫ ∂ t cv cs r ∂ ρ v dV + ∫ ρ v (V ⋅ nˆ ) dA = ∑ Fy ∫ ∂ t cv cs r ∂ ˆ ρ w d V + ρ w ( V ⋅ n ) dA = ∑ Fz ∫ ∫ ∂ t cv cs 7 Chapter V. Finite control Volume Analysis ME3560 – Fluid Mechanics Summer 2016 Selection of Control Volume •A horizontal jet of water exits a nozzle with a uniform speed of 10 ft/s, strikes a vane, and is turned through an angle θ. • Determine the anchoring force needed to hold the vane stationary if gravity and viscous effects are negligible 8 Chapter V. Finite control Volume Analysis Summer 2016 ME3560 – Fluid Mechanics •A horizontal jet of water exits a nozzle with a uniform speed of 10 ft/s, strikes a vane, and is turned through an angle θ. • Determine the friction between the fluid and the vane. Fy Fx 9 Chapter V. Finite control Volume Analysis Summer 2016 ME3560 – Fluid Mechanics 5.3 First Law of Thermodynamics–The Energy Equation ∆Esys = Q + W Time rate of total energy in the system = dE dt Net rate of heat added to the system •E = Total Energy = U + P.E. + K.E. = Q& + W& sys + Net rate of work added to the system P.E. = ½ mV2 K.E. = mgz •B = E and b = e = u + ½ V2 + gz •Units for E are J, cal, BTU •Units of e are J/kg, BTU/slug, lb⋅ft/slug 10 Chapter V. Finite control Volume Analysis Summer 2016 ME3560 – Fluid Mechanics •Applying Reynolds Transport Theorem: r ∂ ˆ Q& + W& = ρ e d V + ρ e ( V ⋅ n )dA ∫ ∫ ∂ t CV Cs • Heat and Work added to the control volume are positive. •In the most general situation W& = W& Shaft + W& pressure + W& Shear W& = Tω Shaft W& pressure r p r = − ∫ pV ⋅ nˆdA = − ρ ∫ V ⋅ nˆ dA CS CS ρ 11 Chapter V. Finite control Volume Analysis ME3560 – Fluid Mechanics W& Shear r = FTangential ⋅ V Summer 2016 • The work due to shear stress can be eliminated by selecting the CV on solid surfaces or setting the inlets and exits perpendicular to the velocity such that τ⋅V = 0. •Most of the time the effect of viscous shear stresses are grouped with other terms to determine the “losses”. • Thus, the energy equation can be written as ( p V r ∂ & + W& ˆ e ρ d V + u + + + gz ρ V ⋅ n dA = Q Shaft ∫ ∫ ∂ t cv ρ 2 cv 2 12 Chapter V. Finite control Volume Analysis Summer 2016 ME3560 – Fluid Mechanics • Realizing that h = u + p/ρ 2 r ∂ V & + W& ˆ e ρ d V + h + + gz ρ V ⋅ n dA = Q Shaft ∫ ∫ ∂ t cv 2 cv •If u, p, V and z are all assumed to be uniformly distributed over the flow cross-sectional areas involved: ( p V2 r ˆ + + + ⋅ u gz V n dA = ρ ∫cv ρ 2 ( p V2 u + + + gz m& ∑ ρ 2 flow out ( p V2 − ∑ u + + + gz m& ρ 2 flow in 13 Chapter V. Finite control Volume Analysis Summer 2016 ME3560 – Fluid Mechanics ( p V ∂ & & eρdV = Q + WShaft + ∑ u + + + gz m& ∫ ρ 2 ∂ t cv flow 2 in ( p V − ∑ u + + + gz m& ρ 2 flow 2 out ( p V u + + + gz m& ∑ ρ 2 flow 2 •For steady state processes: out ( p V2 − ∑ u + + + gz m& = Q& + W& Shaft ρ 2 flow 14 in Chapter V. Finite control Volume Analysis Summer 2016 ME3560 – Fluid Mechanics •By dividing this equation by the mass flow rate: ( p V2 ( p V2 u + + + gz − ∑ u + + + gz = q& + w& ∑ ρ 2 ρ 2 flow flow out in •For only one inlet and one exit: pout − pin ρ + −V ( ( + g ( zout − zin ) = q& + w& − uout + uin 2 2 out 2 in V •Applying the previous equation to a situation where there is not shaft work done by or on the CV. pout − pin ρ + 2 out V −V ( ( + g ( zout − zin ) = q& − uout + uin 2 2 in 15 Chapter V. Finite control Volume Analysis ME3560 – Fluid Mechanics Summer 2016 2 Vout pin Vin2 + + gzout = + + gzin − loss ρ 2 ρ 2 ( ( loss = uout − uin − q& pout 2 Vin2 pout Vout + + gzin = + + gzout + loss ρ 2 ρ 2 pin •In general, if the device analyzed has work (shaft) interactions: 2 Vin2 pout Vout + + zin + hshaft = + + zout + hL γ 2g γ 2g pin 16 Chapter V. Finite control Volume Analysis