HEAT TRANSFER chAPTER 8 Internal flow 们au Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 1 HEAT TRANSFER CHAPTER 8 Internal flow
Internal flow Heat Transfer Where we’ ve been. Introduction to internal flow, boundary layer growth, entry effects Inviscid flow region Boundary layer region ro Hydrodynamic entrance region Fully developed region Where were going Developing heat transfer coefficient relationships and correlations for internal flow Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 2 Internal Flow Heat Transfer Where we’ve been …… • Introduction to internal flow, boundary layer growth, entry effects Where we’re going: • Developing heat transfer coefficient relationships and correlations for internal flow ro
Internal flow Heat Transfer KEY POINTS THIS LECTURE Energy balance for internal flow in a tube Temperature and heat transfer relations for two cases Constant surface heat flux Constant surface temperature Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 3 Internal Flow Heat Transfer KEY POINTS THIS LECTURE • Energy balance for internal flow in a tube • Temperature and heat transfer relations for two cases: – Constant surface heat flux – Constant surface temperature
Basic concepts-thermal considerations 1. The mean temperature viscid flow regio Boundary layer region l Hydrodynamic entrance region Fully developed region fd力 For the internal energy puc,l4.andE,≡ mc T So SJ puc TdA puc,TdA C nc pumAc For incompressible flow in a circular tube 2 Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 4 Basic concepts—thermal considerations • 1. The mean temperature • For the internal energy • So • For incompressible flow in a circular tube, ro = Ac t v TdAc E uc t v Tm E m c and m c v A v c v A v c m u A c uc TdA mc uc TdA T c c = = = 0 0 2 0 2 r m m uTrdr u r T
Basic concepts-thermal considerations 2. Newtons law of cooling qy=h(T。-Tmn Not constant Here, the mean t plays the same role as the free stream T for external flows 3. Fully developed conditions because of heat transfer, T(r) is continuousl changing with X can fully developed conditions be reached?? For thermally fully developed 0|T(x)-7(r,x) 0 OxLT。(x)-Tm(x) fa, t Although the temperature profile t(r) changes with x, the relative shape of the profile no longer changes Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 5 Basic concepts—thermal considerations • 2. Newton’s law of cooling • Here, the mean T plays the same role as the free stream for external flows. • 3. Fully developed conditions because of heat transfer, T(r) is continuously changing with x. can fully developed conditions be reached?? • For thermally fully developed • Although the temperature profile T(r) changes with x, the relative shape of the profile no longer changes. ( ) qs = h Ts −Tm T Not constant! 0 ( ) ( ) ( ) ( , ) , = − − m f d t s s T x T x T x T r x x
Basic concepts-thermal considerations In the thermally fully developed flow of a fluid with constant properties, the local convection coefficient is a constant, independent ofx h≠f(x) For the special case of uniform surface heat flux aT d Axial t Independent of gradient fa, t dx radial location fa, t cons tan t For the case of constant surface temperature Depends on the aT (T-T)dmn radial coordinate Ox\d, (T。-Tn)ax T=cons tan t Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 6 Basic concepts—thermal considerations • In the thermally fully developed flow of a fluid with constant properties, the local convection coefficient is a constant, independent of x. • For the special case of uniform surface heat flux • For the case of constant surface temperature h f (x) fd t m fd t dx dT x T , , = q cons t s = tan Axial T gradient Independent of radial location f d t m s m s f d t dx dT T T T T x T , , ( ) ( ) − − = T cons t s = tan Depends on the radial coordinate
Example: Velocity and temperature profiles for laminar flow in a tube of radius o10mm have the form ()=0.-(/) 7(r)=3448+750(/)-18.8(/rn with units of m/s and K, respectively. Determine the corresponding value of the mean(or bulk) temperature,, at this axial position KNOWN: Velocity and temperature profiles for laminar flow in a tube of radius ro=10mm FIND: Mean(or bulk) temperature, Tm, at this axial position. SCHEMATIC u(n,7(0 Fluid ASSUMPTIONS: (1) Laminar incompressible flow, (2)Constant properties Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 7 Example: Velocity and temperature profiles for laminar flow in a tube of radius have the form with units of m/s and K, respectively. Determine the corresponding value of the mean (or bulk) temperature, , at this axial position. ro =10mm Tm ( ) ( ) ( ) 2 4 2 ( ) 344.8 75.0 / 18.8 / ( ) 0.11 / o o o T r r r r r u r r r = + − = −
ANALYSIS: The prescribed velocity and temperature profiles, (m/s and K, respectively)are u(r)=0.11(rn) T(r)=3448+75.0(rr02-18.8(r (1,2) For incompressible flow with constant c, in a circular tube, from Eq. 8. 27, the mean temperature and un the mean velocity, from Eq. 8.8 are, respectively 25u()r()r u(r).rdr Substituting the velocity profile, Eq (1), into Eq (4)and integrating, find Substituting the profiles and um into Eq. (3), find 4{38()+088(+080) Tn=417240+1875-3131820+1250-235}=367K The velocity and temperature profiles appear as shown below. Do the values of um and Tm found above with th ive profiles as you thought? Is the fluid being heated or cooled? 008 E> 004 002 0 002040608 002040608 Radial coordinate riro Radial coordinate r/ro Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 8
Review: Energy balance Analysis E convection rnc t p m, inadl out.aa out, adv=mc T dx cony conv X X+dx Energy Balance Change in energy in the control volume energy input- work out energy in by advection energy carried out by advection (flow) OI P·ax+mc nnc T cony p m,x p m,x+dx 0 conT P dx- nnc dT. cony h(Ts-tm) m c Two Special Cases: 1. Constant surface heat flux 2. Constant surface temperature Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 9 Review: Energy Balance Analysis x x + dx Ein,adv Eout,adv Econvection Energy Balance Change in energy in the control volume = energy input – work out + energy in by advection – energy carried out by advection (flow) or Two Special Cases: 1. Constant surface heat flux 2. Constant surface temperature E q P dx q dx E m c T E m c T conv conv conv out adv p m x dx in adv p m x = = = = + , , , , ( ) 0 0 , , s m p p m conv conv p m conv p m x p m x d x h T T m c P m c q P d x d T q P d x m c d T q P d x m c T m c T = − = = − = + − +
Case 1: Constant Surface Heat Flux Example application Electrical heater element around a pipe Recognize that q≠∫(x) From the energy balance equation "P Tm(x)=Tmi+ if s constant Entrance region Fully developed regi T,G) If g"is a known function (T-In) ofx instead must T.Gx grate the above to obtain Tm.x q Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 10 Case 1: Constant Surface Heat Flux • Example application – Electrical heater element around a pipe • Recognize that: • From the energy balance equation • if q” is constant If q” is a known function of x instead, must integrate the above to obtain Tm,x q f (x) s ( ) T m,i x m c q P T x p s m = +
