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Chapter 4 Continuity, Energy, and Momentum Equations

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Ch 4. Continuity, Energy, and Momentum Equation Chapter 4 Continuity, Energy, and Momentum Equations 4.1 Conservation of Matter in Homogeneous Fluids • Conservation of matter in homogeneous (single species) fluid → continuity equation positive dA1 - perpendicular to the control surface negative 4.1.1 Finite control volume method-arbitrary control volume - Although control volume remains fixed, mass of fluid originally enclosed (regions A+B) occupies the volume within the dashed line (regions B+C). Since mass m is conserved: mmm m (4.1) ABBtttdt Ctdt m m mmBBtdt t A t C tdt (4.2) dt dt •LHS of Eq. (4.2) = time rate of change of mass in the original control volume in the limit mmBB m tdt t B dV (4.3) dt t t CV 4−1 Ch 4. Continuity, Energy, and Momentum Equation where dV = volume element •RHS of Eq. (4.2) = net flux of matter through the control surface = flux in – flux out = qdA qdA nn12 where q = component of velocity vector normal to the surface of CV q cos n dV qdAnn12 qdA (4.4) CV CS CS t ※ Flux (= mass/time) is due to velocity of the flow. • Vector form dV qdA (4.5) t CV CS where dA = vector differential area pointing in the outward direction over an enclosed control surface qdAq dA cos positive for an outflow from cv, 90 negative for inflow into cv, 90 180 If fluid continues to occupy the entire control volume at subsequent times → time independent 4−2 Ch 4. Continuity, Energy, and Momentum Equation LHS: dV dV (4.5a) ttCV CV Eq. (4.4) becomes dV q dA 0 CV CS t (4.6) → General form of continuity equation - integral form [Re] Differential form Use Gauss divergence theorem F dV FdAi VA xi Transform surface integral of Eq. (4.6) into volume integral qdA q dV CS CV Then, Eq. (4.6) becomes qdV 0 CV t (4.6a) Eq. (4.6a) holds for any volume only if the integrand vanishes at every point. q 0 t (4.6b) → Differential form ◈ Simplified form of continuity equation (1) For a steady flow of a compressible fluid 4−3 Ch 4. Continuity, Energy, and Momentum Equation dv 0 CV t Therefore, Eq. (4.6) becomes qdA0 (4.7) CS (2) For incompressible fluid (for both steady and unsteady conditions) d const. 0, 0 tdt Therefore, Eq. (4.6) becomes qdA0 (4.8) CS [Cf] Non-homogeneous fluid mixture → conservation of mass equations for the individual species → advection - diffusion equation cq = conservation of mass equation 0 tx c + mass flux equation due to advection and diffusion qucD x c c uc D 0 txx cuc c D txxx 4−4 Ch 4. Continuity, Energy, and Momentum Equation 4.1.2 Stream - tube control volume analysis for steady flow There is no flow across the longitudinal boundary ∴ Eq. (4.7) becomes qdA qdA qdA 0 (4.9) 11 1 2 2 2 qdA const. If density = const. q11 dA q 2 dA 2 dQ (4.10) where dQ = volume rate of flow (1) For flow in conduit with variable density qdA V → average velocity A 4−5 Ch 4. Continuity, Energy, and Momentum Equation dQ → average density Q 111VA 2 VA 22 (4.11) (2) For a branching conduit qdA0 11qdA 1 2 qdA 2 2 33 qdA 3 0 AA A 12 3 111VA 2 VA 22 333 VA 4−6 Ch 4. Continuity, Energy, and Momentum Equation [Appendix 4.1] Equation of Continuity → Infinitesimal (differential) control volume method At the centroid of the control volume: ,,,uvw Rate of mass flux across the surface perpendicular to x is u dx flux in u dydz x 2 u dx flux out u dydz x 2 u net flux = flux in flux out dxdydz x v net mass flux across the surface perpendicular to y dydxdz y w net mass flux across the surface perpendicular to z dzdxdy z dxdydz Time rate of change of mass inside the c.v. = t 4−7 Ch 4. Continuity, Energy, and Momentum Equation Time rate of change of mass inside = The sum of three net rates dxdydz u v w dxdydz txyz By taking limit dV dxdydz uvw qdivq tx y z q 0 (A1) t → general point (differential) form of Continuity Equation uvw [Re] qdivq xyz By the way, qq q Thus, (A1) becomes qq 0 (A2) t 1) For incompressible fluid d 0 ( const.) dt d q 0 tdt 4−8 Ch 4. Continuity, Energy, and Momentum Equation Therefore Eq. (A2) becomes qq 00 (A3) In scalar form, uvw 0 (A4) xyz → Continuity Eq. for 3D incompressible fluid For 2D incompressible fluid, uv 0 xy 2) For steady flow, 0 t Thus, (A1) becomes qq q 0 4−9 Ch 4. Continuity, Energy, and Momentum Equation 4.2 The General Energy Equation 4.2.1 The 1st law of thermodynamics The 1st law of thermodynamics: combine continuity and conservation of energy → energy equation – property of a system: location, velocity, pressure, temperature, mass, volume – state of a system: condition as identified through properties of the system The difference between the heat added to a system of masses and the work done by the system depends only on the initial and final states of the system (→ change in energy). → Conservation of energy Q W E QWdE (4.14) where Q = heat added to the system from surroundings W = work done by the system on its surroundings dE = increase in energy of the system Consider time rate of change QWdE (4.15) dt dt dt (1) Work Wpressure = work of normal stresses acting on the system boundary Wshear = work of tangential stresses done at the system boundary on adjacent external fluid in motion 4−10 Ch 4. Continuity, Energy, and Momentum Equation Wshaft = shaft work done on a rotating element in the system (2) Energy Consider e = energy per unit mass = E mass eu = internal energy associated with fluid temperature = u ep = potential energy per unit mass = gh where h = local elevation of the fluid q2 e = kinetic energy per unit mass = q 2 p u enthalpy q2 ee e e ugh (4.16) upq 2 • Internal energy = activity of the molecules comprising the substance = force existing between the molecules ~ depend on temperature and change in phase 4.2.2 General energy equation QWdE (4.15) dt dt dt 4−11 Ch 4. Continuity, Energy, and Momentum Equation Consider work done W WW W pressure shaft shear dt dt dt dt (4.15a) dA2 q 2 W pressure = net rate at which work of pressure is done by the fluid on the surroundings dt = work fluxout – work fluxin = p qdA CS 2 p = pressure acting on the surroundings = FA F/ L Positive for outflow into CV qdA Negative for inflow qdAQ L3 / t 4−12 Ch 4. Continuity, Energy, and Momentum Equation FL3 p qdA 2 FL// t E t L t Thus, (4.15a) becomes W Wshaft W p qdA shear CS dt dt dt (4.15b) Consider energy change dE = total rate change of stored energy dt = net rate of energy flux through C.V. (@energytt t energy @) tt + time rate of change inside C.V. = e qdA edV CS CV t (4.15c) e E/; mass qdA mass / time eEt qdA / Substituting (4.15b) and (4.15c) into Eq. (4.15) yields Q Wshaft W shear p qdA dt dt dt CS e qdA edV CSt CV QWW shaft shear dt dt dt 4−13 Ch 4. Continuity, Energy, and Momentum Equation p e qdA edV (4.17) CS t CV Assume potential energy eghp (due to gravitational field of the earth) q2 Then eu gh 2 Then, Eq. (4.17) becomes Q W W shaft shear dt dt dt pq2 ugh qdA edV (4.17) CS CV 2 t ◈ Application: generalized apparatus At boundaries normal to flow lines → no shear → Wshear 0 4−14 Ch 4. Continuity, Energy, and Momentum Equation 2 Q Wshaft pq ugh qdA edV (4.19) CS CV dt dt 2 t For steady motion, 2 Q Wshaft pq ugh qdA (4.20) CS dt dt 2 ◈ Effect of friction ~ This effect is accounted for implicitly. ~ This results in a degradation of mechanical energy into heat which may be transferred away (Q , heat transfer), or may cause a temperature change → modification of internal energy. → Thus, Eq. (4.20) can be applied to both viscous fluids and non-viscous fluids (ideal frictionless processes). 4.2.3 1 D Steady flow equations For flow through conduits, properties are uniform normal to the flow direction. → one - dimensional flow 1 2 V1 V2 Integrated form of Eq. (4.20) = ②-① 22 Q Wshaft pV pV ugh QQ ugh dt dt 22②① 4−15 Ch 4. Continuity, Energy, and Momentum Equation V 2 where average kinetic energy per unit mass 2 Section1: qdA Q mass flow rate into CV 1 Section2: qdA Q mass flow from CV 2 Divide by Q (mass/time) 22 heat transfer Wshaft pV pV ughugh mass mass 22②① Divide by g 22 heat transfer Wshaft up V up V hh weight weight gggg22②① (4.21) ◈ Energy Equation for 1-D steady flow: Eq. (4.21) ~ use average values for p,,, huand V at each flow section ~ use Ke (energy correction coeff.) to account for non-uniform velocity distribution over flow cross section 1 mV 2 KVQ22 qdQ ---- kinetic energy/time e 22 2 t qdQ2 K 2 1 e VQ2 2 4−16 Ch 4. Continuity, Energy, and Momentum Equation W 22 heat transfershaft pVpV u21 u hKee hK weight weight22gg②① g (4.23) Ke = 2, for laminar flow (parabolic velocity distribution) 1.06, for turbulent flow (smooth pipe) For a fluid of uniform density p V22 p VW heat transfer u u 1122hK hK shaft 21 12ee1222g g weight weight g (4.24) → unit: m (energy per unit weight) For viscous fluid; heat transfer u u 21H L12 weight g → loss of mechanical energy ~ irreversible in liquid Then, Eq.
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  • 1 the Continuity Equation 2 the Heat Equation

    1 the Continuity Equation 2 the Heat Equation

    1 The Continuity Equation Imagine a fluid flowing in a region R of the plane in a time dependent fashion. At each point (x; y) R2 it has a velocity v = v (x; y; t) at time t. Let ρ = ρ(x; y; t) be the density of the fluid 2 −! −! at (x; y) at time t. Let P be any point in the interior of R and let Dr be the closed disk of radius r > 0 and center P . The mass of fluid inside Dr at any time t is ρ dxdy: ZZDr If matter is neither created nor destroyed inside Dr, the rate of decrease of this quantity is equal to the flux of the vector field −!J = ρ−!v across Cr, the positively oriented boundary of Dr. We therefore have d ρ dxdy = −!J −!N ds; dt − · ZZDr ZCr where N~ is the outer normal and ds is the element of arc length. Notice that the minus sign is needed since positive flux at time t represents loss of total mass at that time. Also observe that the amount of fluid transported across a small piece ds of the boundary of Dr at time t is ρ−!v −!N ds. Differentiating under the integral sign on the left-hand side and using the flux form of Green’s· Theorem on the right-hand side, we get @ρ dxdy = −!J dxdy: @t − r · ZZDr ZZDr Gathering terms on the left-hand side, we get @ρ ( + −!J ) dxdy = 0: @t r · ZZDr If the integrand was not zero at P it would be different from zero on Dr for some sufficiently small r and hence the integral would not be zero which is not the case.