Internal Flow Through Pipes And Ducts.pdf

Module18: Internal FlowFlow and Heat Transfer through Pipes and Ducts Introduction to Internal Flow • Characteristics follow I&D, Chapter 8 – primary difference from external flow is the presence of an opposing, confining surface that constrains the boundary layer growth – entry (entrance) length exists (B.L. is developing) – fully developed region eventually forms r0 ≡ R ref. Incropera & DeWitt, Chap. 8 Hydrodynamic Boundary Layer Development Terms and Notation • Critical Reynolds number Re D,c = umD ≈ 2300 ν onset of turbulence • Hydrodynamic entry length Laminar: x fd ≈ 0.05 Re D Turbulent: 10 ≤ D x fd ≤ 60 D • Mean velocity, um such that & = ρu m A c m  constant for steady incompressible flow • Mean temperature Tm (to be defined later) General Assumptions We will assume: – steady.1 µm continuum breaks down for – very low pressure – very small dimensions Careful using for microchannels with gases ! .constant properties – Kn (= λ/L) << 1 (continuum) for air @ STP λ ≈ 0. Newtonian. incompressible. Laminar Flow ∂u 1 ∂ (rv) + = 0 continuity ∂x r ∂r ∂u ⎞ ∂p µ ∂ ⎛ ∂u ⎞ ⎛ ∂u ρ⎜ u +v + ⎟=− ⎜ r ⎟ x − momentum ∂r ⎠ ∂x r ∂r ⎝ ∂r ⎠ ⎝ ∂x ∂v ⎞ ∂p µ ∂ ⎛ ∂v ⎞ ⎛ ∂v ρ⎜ u +v = − + ⎟ ⎜ r ⎟ r-momentum ∂r ⎠ ∂r r ∂r ⎝ ∂r ⎠ ⎝ ∂x u(x.Hydrodynamics . r) = uo(r) Needs numerical solution … Assume axisymmetry .s: ∂u =0 ∂r r =0 u(x = 0. R) = 0 v(x.C. R) = 0 Entrance Region: B. Laminar Flow •Fully Developed Region – Poiseuille Flow (parallel flow): ∂u v=0 ⇒ ∂x = 0 ⇒ u(r) • v-momentum equation yields ∂p = 0 ⇒ p = p( x) only ∂r •x-momentum reduces to balance between pressure & shear forces: dp µ d ⎛ du ⎞ = ⎜r ⎟ dx r dr ⎝ dr ⎠ •B.s u(R) = 0 ∂u ∂r =0 r =0 .C.Hydrodynamics . C.• Integrate twice and apply B.s to get µ d ⎛ du ⎞ dp ∴ ⎜r ⎟ = r dr ⎝ dr ⎠ dx • Mean velocity • Velocity distribution • Velocity profile can integrate easily since dp/dx is independent of r 2 ⎡ 1 ⎛ dp ⎞ 2 ⎛ r ⎞ ⎤ u(r) = − ⎜ ⎟ R ⎢1 − ⎜ ⎟ ⎥ 4µ ⎝ dx ⎠ ⎣ ⎝ R ⎠ ⎦ & m = um = 2 ρπR u(r) um ∫ R 0 ρu(2πr)dr ρπR 2 ⎡ ⎛r ⎞ ⎤ = 2 ⎢1 − ⎜ ⎟ ⎥ ⎣ ⎝R⎠ ⎦ 2 R 2 dp =− 8µ dx Notice dimensionless velocity distribution not a functions of Re – why? . Fully Developed Laminar Flow Pressure Drop : Expressed in terms of the Moody (or Darcy) friction factor ⎛ dp ⎞ −⎜ ⎟D dx f≡ ⎝ 2 ⎠ ρu m / 2 ρu 2m ∆pfd = f L 2D D = 2R f= 64 Re D • fully developed • laminar . 20 (ReD > 20000) . but f is not.ref.316 ReD-0. Chap.184 ReD-0. For smooth surfaces. Incropera & DeWitt.25 (ReD < 20000) = 0. and the pressure drop is very sensitive to roughness (unlike in laminar flow). due to funny nondimensionaliza tion For turbulent flow the analysis is not as simple as above. fturb = 0. 8 Note: dP/dx is constant. Thermal Considerations .Laminar Flow Characteristics: ref. Chap. 8 r0 ≡ R . Incropera & DeWitt. Terms and Notation x fd. x) is different depending on whether Ts or q″ is a constant • Example: for engine oil (Pr ≈ 6000). a tube length of 54 m!! (δt never reaches the centerline in pipes of reasonable length) .t ≈ 0. the entrance length is nearly independent of Pr in turbulent flow.e.t = 0.05 Re Pr = 5455 tube diameters! i.01)/(550 x 10-6) = 18 (laminar) xfd. with Lt / D ~ 10) • The shape of the fully developed profile T(r.. um = 1 m/s. ν = 550 x 10-6 m2/s: ReD = (1) (0.05 ReD Pr •Thermal entrance length D (unlike in laminar flow. say D = 1 cm. mass flow rate • For a circular cross section. with constant-property flow 2 Tm = umR 2 ∫ T(r) u(r) r dr .Bulk Mean Temperature • Bulk mean temperature: rate of thermal energy transport & v Tm = E& t = mC ∫ ρuC v TdA c Ac Tm ∫ = Ac ρuC v TdA c & v mC Weighted w.t.r. local h is independent of x ! That is. props.Thermally Fully Developed Flow • Define a dimensionless temperature Ts − T(x. Nux is independent of x Can this happen without hydrodynamically fullydeveloped flow? . r) Ts − Tm (x) • The relative shape of the temperature profile no longer changes if field is “fully developed” ∂ ⎡ Ts − T ⎤ =0 ⎢ ⎥ ∂x ⎣ Ts − Tm ⎦ q ′′s = h(Ts − Tm ) = k −∂T/∂r r =R ∂ ⎡ Ts − T ⎤ = = constant ≠ f (x) ⎢ ⎥ Ts − Tm ∂r ⎣ Ts − Tm ⎦ r =R ∂T ⇒ h = constant ∂r r = R k In thermally f..d. flow with const. Energy Balance Temperature Distribution .Energy Balance Energy Balance: E& in = E& out specific volume 1 v= ρ d(C v Tm + pv) ⎤ ⎡& & & dq conv + m(C v Tm + pv) − ⎢ m(C v Tm + pv) + m dx ⎥ = 0 dx ⎣ ⎦ thermal energy flow work . Energy Balance (con’td) d(C v Tm + pv) ⎤ ⎡ & v Tm + pv) − ⎢ m(C & v Tm + pv) + m & dq conv + m(C dx ⎥ = 0 dx ⎣ ⎦ Perfect gas ⇒ pv= RTm . also dq conv = q′′s (x)Pdx d(C v + R)Tm & q′′s (x)Pdx − m dx = 0 dx But C v + R = Cp and d ( Cp Tm ) = di Perimeter For constant properties: dTm dx For a circular pipe P = πD dTm πDq′′s (x) = & p dx mC & p q′′s (x)P = mC What if it’s not a perfect gas ? Neglect pressure work. set Cp =Cv to get the same result. . r) Also.Constant Heat Flux Boundary Conditions dTm πDq′′s = = constant & p dx mC Furthermore. recall θ= =constant with x Tm − Ts Since Tm − Ts =constant. q′′s = h ( Ts − Tm ) ⇒ ( Ts − Tm ) = constant dTs dTm ⇒ = = constant dx dx Ts -T(x. = = ∂x dx dx All temperatures rise at the same rate axially! .r) is also constant ∂T dTs dTm Thus. Ts -T(x. Thus : d ( Ts − Tm ) dx = πD h ( Ts − Tm ) & p mC ( Ts − Tm ) = ( Ts − Tm )inlet exp(− πD hx) & p mC Careful! h is constant only in fully-developed region! Bulk temperature varies exponentially! Furthermore.r) =constant with x Tm − Ts Ts − T dTm dTm ∂T 1 ∂T 1 = ⇒ = = constant ∂x Ts − Tm dx Ts − T ∂x Ts − Tm dx All temperatures tend towards Ts exponentially with x! . q′′s (x) = h ( Ts − Tm ) .since θ= Ts -T(x.Constant Temperature Boundary Conditions dTm πDq′′s (x) = = not constant & p dx mC However. Axial Temperature Variation Is bulk temperature variation linear (or exponential) through the FD region? What about the temperature at a point (x.r)? . Profile. v=0) u • Constant Surface Flux Temperature → q′′s = constant = h(Ts − Tm ) ∂T dTm → = ∂x dx See Sec.4.d. 8. vel.1 I&D for solution ∂T α ∂ ⎛ ∂T ⎞ = ⎟ ⎜r ∂x r ∂r ⎝ ∂r ⎠ Constant Surface dTs =0 dx ∂T Ts − T dTm → = ∂x Ts − Tm dx → .Hydrodynamically and Thermally Fully Developed Flow Solution Energy Equation: (f. Solution for Constant Heat Flux BC • LHS known •Integrate twice to obtain ∂T α ∂ ⎛ ∂T ⎞ = u ⎟ ⎜r ∂x r ∂r ⎝ ∂r ⎠ 4 2 2u m R 2 dTm ⎡ 3 1 ⎛ r ⎞ 1 ⎛ r ⎞ ⎤ T(r) = Ts − ⎢ + ⎜ ⎟ − ⎜ ⎟ ⎥ α dx ⎢⎣16 16 ⎝ R ⎠ 4 ⎝ R ⎠ ⎥⎦ Is this known? Since both bc are Neumann-type. What is the physical meaning of this? . temperature can only be determined up to an additive constant. Nusselt Number Steps: 1. Hence find h. Find k ∂T ∂r ∂T ∂r = h (Ts − Tm ) r=R from temperature solution r=R 3.36 for q′′s = constant k Notice Nusselt number not a function of Re or Pr! . Find bulk temperature: 2 Tm = umR 2 ∫ T(r) u(r) r dr Known 4. Recall 2. and thus NuD hD Nu D = = 4. Solution for Constant Temperature BC • Solution is a bit more complicated because LHS is not constant.66 Ts = constant . • Solution obtained numerically • Can show that Nu D = 3. i + x & p mC q = q′′s P ⋅ L .outlet).inlet.i ) balance: q conv = mC dTm q′′s P P • Also = = h(Ts − Tm ) & p mC & p dx mC q′′s = constant Ts=constant ⎡ Px Ts − Tm (x) = exp ⎢ − & p Ts − Tm.o − Tm. o .i ⎢⎣ mC q = h A s ∆Tlm ⎤ h⎥ ⎥⎦ As = P ⋅ L ∆To − ∆Ti ∆Tlm ≡ ln(∆To / ∆Ti ) q′′s P Tm (x) = Tm.Other Useful Relationships • For the entire tube (i . overall energy & p (Tm. Developing Flow Terminology • “Thermal entry length problem” – Flow is fully developed. temperature is not • “Combined entry length problem” – Both flow and temperature are developing • “Unheated starting length” – There is an insulated length of duct at the entrance so that the flow has a chance to develop while the temperature does not – Synonymous with “Thermal entry length” T=Ts T=Ts . 5) – Colburn relation for friction factor for smooth circular tubes (Eq. 8. 8.use Table 8.assumes only thermal entry length. evaluate properties at mean temperature defined as average between inlet and outlet • – turbulent flow (section 8. 8.4.61) correlations for Nusselt number • – non-circular tubes (section 8. Dittus Boelter (Eq.58 I&D). 8.57 I&D) .2) – Hansen formula (Eq.for combined entry length.56 I&D) . where P is the wetted perimeter • – concentric tubes (section 8.59) and Seider-Tate (Eq.Convection Correlations • Refer to the course text to find correlations for NuD for: • – entry region (section 8. 8.7) . less accurate.6) – (Laminar flow . for constant surface temperature – Seider-Tate formula (Eq.1) – Turbulent flow. use correlations for circular tubes with hydraulic diameter: Dh = 4Ac/P. Entry Length in Circular Pipes • • • • NuD at x = 0 = ? Why the difference between thermal and combined entry lengths? Notice that curves are independent of Re. Pr if x axis is scaled as shown Graetz number = Re Pr/(x/D) (some texts use inverse) . •Solution: First. •Find the heat transfer coefficient & exit temperature.99 10 = × 6. The tube is 2 m long and water velocity = 1 m/s.634 W/mK 1* 0.36 × 10−7 turbulent .0254 4 Re D = 3.Example: •Water at 280 K enters a 1-inch diameter tube kept at a constant surface temperature of 360 K.16 .009 ⎝ ⎠ Pr = 4. Try 350 K T = T + T = 350 + 280 = 315 K mi m mo 2 3 µ 631 × 10−6 1 10 ⎛ −7 2 3 ⎞ ν= = = 6.36 × 10 m /s ⎜Q ρ = = = 991 kg / m ⎟ ρ 991 v 1. 2 k = 0. estimate the exit temperature to evaluate properties. Q Ts > Tm ) = 195 ∴ h = 4867 W/m 2 K • Use Eq.7 > 10 D 0.i m C p ⎠ ⎝ for Ts = constant . • Dittus-Boelter correlation (8.d.4 (n = 0. (8. 8.4 for heating.8 ( Pr ) 0.41 would be for qs” = constant) • ⎛ Px h ⎞ Ts − Tm (x) = exp ⎜ − ⎟ & Ts − Tm .43) to get exit temperature (Eq.0254 ⇒ f .L 2 = = 78.023(Re D ) 0.60) Nu D = 0. o πD 2 & = ρ um m 4 P = πD.3685 & Cp m ∴ 360−Tm. not = done here 2 . PL h 4L h 4L = = St & Cp m D ρ u m Cp D where St = Stanton # 195 St = = = 0.o = 304.L = Tm . Tm (x) = Tm .@ x = L.16 Nu PL h = − 0.6+ 280 exercise.0017 4 Re Pr 3.99×10 ×4.692 360− 280 not 350K as assumed! ⇒ Tm.o = 0.6 K now recalculate with new T m 304. 8.44) q conv = h As ∆ Tlm ∆ Tlm for Ts = constant 24. we can calculate this as: (Eq.6 ∆Ti −∆To = = ⎛ ∆Ti ⎞ ⎛ 360− 280 ⎞ ln ⎜ ln ⎜ ⎟ ⎟ ⎝ 360−304. πD 2 = ρ Cp u m ∆ Tm 4 & p ∆ Tm Q = mC π 4 = 991×1× (0.6) = 51622 W ∆Tm • Alternatively.• Total heat transfer rate.0254)2 × 4179 × (24.6 ⎠ ⎝ ∆To ⎠ • giving q = 51853 W Same answer!! (with slight calculation difference) .