13.3 Flowand heat transferinamicro-channel微通道内流动换热问题Focus: Compared with previous examples, the focusof this example is about pressure-out boundarycondition and‘two-side-wall"boundary condition.6/37
13.3 Flow and heat transfer in a micro-channel 微通道内流动换热问题 Focus: Compared with previous examples, the focus of this example is about pressure-out boundary condition and ‘two-side-wall” boundary condition. 6/37
13.3 Single-phase fluid flow and heat transfer in microchannelwithrectangularribs(MC-RR)Known: Cold water at T-20 °C flows into the inletof a MC-RR with velocity u=0.1 m/s. The side wall ofMC-RR is heated with a uniform heat flux q = 30 W/cm?Assumption: (1) steady state, (2) laminar flow, (3)incompressible fluid, (4) constant fluid properties, (5)negligible radioactive and natural convective heattransferfrom the micro channel heat sink.7/37
7/37 13.3 Single-phase fluid flow and heat transfer in micro channel with rectangular ribs (MC-RR) Known：Cold water at Tf =20 ℃ flows into the inlet of a MC-RR with velocity u=0.1 m/s. The side wall of MC-RR is heated with a uniform heat flux q = 30 W/cm2 . Assumption: (1) steady state, (2) laminar flow, (3) incompressible fluid, (4) constant fluid properties, (5) negligible radioactive and natural convective heat transfer from the micro channel heat sink
LRR-WRLwLDLtFig.1 Computational domainTable.1 Geometrical parameters of MC-RRGeometricalLtWLRRWRLRParameters30.50.70.10.3Value/mm8/37
8/37 Fig.1 Computational domain Lt LR w LRR wR 1 2 LR Table .1 Geometrical parameters of MC-RR Geometrical Parameters W LRR WR LR Lt Value/mm 0.5 0.7 0.1 0.3 3
Find:Temperaturedistributionandpressuredistribution in the domain.Governing equations:duavContinuity equation::0ayaxMomentum equations:duau(a2uau1 apufuax0x2dy0y2PraxPf(a2vavava201 apMfuV十ax0x2dy2ayPfdypf9/37
9/37 Find: Temperature distribution and pressure distribution in the domain. Governing equations: Continuity equation: 𝜕𝑢 𝜕𝑥 + 𝜕𝑣 𝜕𝑦 = 0 Momentum equations: 𝑢 𝜕𝑢 𝜕𝑥 + 𝑣 𝜕𝑢 𝜕𝑦 = − 1 𝜌𝑓 𝜕𝑝 𝜕𝑥 + 𝜇𝑓 𝜌𝑓 𝜕 2𝑢 𝜕𝑥 2 + 𝜕 2𝑢 𝜕𝑦 2 𝑢 𝜕𝑣 𝜕𝑥 + 𝑣 𝜕𝑣 𝜕𝑦 = − 1 𝜌𝑓 𝜕𝑝 𝜕𝑦 + 𝜇𝑓 𝜌𝑓 𝜕 2𝑣 𝜕𝑥 2 + 𝜕 2𝑣 𝜕𝑦 2
Energyequation:(a2Tfa2Ta(pfCpruTp),a(pfCpfvfT)kaxay0x20y2where T, is the coolant's temperature, Cpr is fluidspecific heat and k,is fluid thermal conductivity.Energy equation for the solid region:(a2Ts,a2Ts0=ks0x2Qy2where T, is solid temperature and k, is solid thermalconductivity10/37
10/37 Energy equation: where Tf is the coolant’s temperature, cPf is fluid specific heat and kf is fluid thermal conductivity. Energy equation for the solid region: where Ts is solid temperature and ks is solid thermal conductivity 0 = 𝑘𝑠 𝜕 2𝑇𝑠 𝜕𝑥 2 + 𝜕 2𝑇𝑠 𝜕𝑦 2 𝜕(𝜌𝑓 𝑐𝑝𝑓𝑢𝑓𝑇𝒇 ) 𝜕𝑥 + 𝜕(𝜌𝑓 𝑐𝑝𝑓𝑣𝑓𝑇𝒇 ) 𝜕𝑦 =kf 𝜕 2𝑇𝑓 𝜕𝑥 2 + 𝜕 2𝑇𝒇 𝜕𝑦 2