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Forced-Convection-Internal_Compatibility_Mode_

# Forced-Convection-Internal_Compatibility_Mode_ - Forced...

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Unformatted text preview: Forced Forced Convection in Internal Flows The The General Thermal Problem Fully Developed Heat Transfer Laminar Flow and Heat Transfer Overview • The general thermal problem – Integration of the general energy for radially lumped flows lumped flows. – Average heat transfer coefficients • Fully developed heat transfer – Limiting heat heat transfer coefficients • Laminar flow and heat transfer in tubes – Developing Nusselt number – Fully developed Nusselt number Structure Structure of the problem Fluid Flow Hydraulic Diameter Developing vs. fully developed flow de flo Laminar flow vs. turbulent flow Heat Transfer Developing vs. fully developed heat transfer Constant wall temperatures vs. constant wall heat flux Fluid flow - Dh Hydraulic diameter, Dh or De D 4 A 4π D / 4 Dh = = =D πD P 2 (1) W L Dh = 4A 4( LW ) 2 LW = = P 2( L + W ) L + W (2) Laminar Laminar vs. turbulent flow For tube flow, the Reynolds criterion is that the flow becomes turbulent at becomes turbulent at Re D ≈ 2300 (3) For fully turbulent flow, ReD ~ 5,000 - 10,000 For channels and non-circular cross sections, use the hydraulic diameter to determine the flow regime. Developing and fully developed flow x=0 x=L u Le Developing flow Fully developed flow Entry Entry length, Le Entry length relations are developed from experimental studies: ⎛ Le ⎜ ⎝D ⎛ Le ⎜ ⎝D Re D ⎞ ⎟≈ 20 ⎠ lam (4) ⎞ ⎟ > 10 , Re D > 2300 ⎠ turb (5) Th The general thermal th problem... problem... Th The general thermal th problem... problem... The general thermal problem qw or TW hi ho x=0 Inlet x=L Outlet u , T Bi u , T Bo The The general thermal problem What is of interest is the change in the average, or bulk temperature of the fluid, the rate of heat transfer, and the pressure drop along the length of the pipe. The bulk temperature is defined in terms of the enthalpy distribution across the area of the tube. TB ∫ ρC p udA = ∫ ρC p TudA A A The general thermal problem hi q ′′ w or Tw ho x=0 Inlet x=L Outlet u , T mi u , T mo The energy balance is written for the fluid as the system of interest. The The general thermal problem Lumped analysis (Steady state): The energy balance for the fluid in the tube as the system of interest is: & 0 = q ′′ Aw + m (hi − ho ) w & 0 = q ′′ Aw + m C p (Tmi − Tmo ) w Note: h is the specific enthalpy of the fluid at the bulk temperature, TB(x). (13) The general thermal problem The energy balance can be interpreted in the following manner: Energy transferred to the fluid at the wall - Enthalpy change in the fluid =0 (14) (TBo q ′′ Aw − TBi ) = w & mC P (15) The The general thermal problem heat transfer coefficients defined Look at the bulk temperature of the fluid from x = 0 to x = L with the tube wall at constant temperature. TW = Constant TB(x) X 0 L Heat transfer coefficient defined The average heat transfer coefficient is defined in terms of a a mean temperature difference between the wall and the fluid temperatures q ′′ h= w Δ Tm (29.16) TW = Constant ΔTBm TB(x) X 0 L Heat Heat transfer coefficient defined TW = Constant ΔΤBm TB(x) X 0 L ΔTBm, is not defined in any specific way at this point. The concept that there exists a mean temperature difference that can define the heat transfer coefficient is correct and well grounded in both the physics of the transfer process and in mathematical fundamentals. Heat transfer coefficient defined & Qw (T Bo − T Bi ) = & mC P & Qw h= AΔTBm & m C P (T Bo − T Bi ) h= A Δ T Bm (15) (16) (17) Heat Heat transfer coefficients defined & mC P (TBo − TBi ) h= AΔTBm (17) Thi This relation holds for either type of wall boundary condition, condition, i.e., constant temperature or constant heat flux. The The general thermal problem Reconsider the energy balance, but now consider a dx length of the tube. Inlet Outlet u , T mi Ý m Tm ( x ) x x+Δx u , T mo Ý m Tm ( x + Δx ) q w = h x (πD Δ x )(Tw − Tm ) or q w = specified The general thermal problem Applying the steady state energy balance to the fluid element and using the Mean Value Theorem, we have & 0 = m C p [T B ( x ) − T B ( x + Δ x ) ] + q ′′ (π D ) Δ x w ⎡ dT B ⎤ & − m C P ⎢− w ⎥ = q ′′ (π D ) ⎣ dx ⎦ (1.3) The The general thermal problem Equation (30.3) can be rewritten as ⎡ dT B ⎤ q ′′ (π D ) =w ⎢ dx ⎥ & mC P ⎦ ⎣ The boundary condition is, (1.3a) TB (0) = TBi (1.4) Integration of the axially lumped energy balance ⎛ πD TB ( x ) = ⎜ ⎜ mC ⎝& P ⎞ ⎟ ∫ q ′′ dx + C ⎟w ⎠ (1.5) The integral on the r.h.s. is to be evaluated depending on the type of thermal boundary condition appliedat the tube wall. Thermal Thermal boundary conditions Heated wall at either a specified temperature or heat flux. & m, TBi τw TBo Wall shear stress Fully developed x = Le q ′′ ( x ), h ( x ) w Heat transfer at the wall transfer at the wall Fully developed X x = Lte Constant wall heat flux ⎛ πD TBm ( x ) = ⎜ ⎜ mC ⎝& P ⎞ ⎟ q ′′ dx + C ⎟w ⎠ ∫ (1.5) ⎛ πDq ′′ w TBm ( x ) − TBi = ⎜ ⎜ mC ⎝& P ⎞ ⎟x ⎟ ⎠ (1.6) Constant Constant wall heat flux ⎛ πD TB ( x ) − TBi = ⎜ ⎜ mC ⎝& P TB ⎞ ⎟x ⎟ ⎠ TB (1.6) TBi X Fully developed heat transfer... transfer... Fully Fully developed heat transfer The The local heat transfer coefficient is defined as by h ( x ) (T w ( x ) − T m ( x ) ) = q ′′ w q ′′ w h( x) = Tw − Tm Differentiate the Eq. (41.7) to get (1.7) dh − q ′′ ⎡d w (Tw − TB )⎤ = ⎥ dx (Tw − TB )2 ⎢ dx ⎣ ⎦ (1.8) Fully developed heat transfer When h(x) is constant, the heat transfer coefficient is said to be fully developed, that is the criterion for fully developed heat transfer is dh ⎡ d ⎤ = ⎢ (Tw − TB )⎥ = 0 dx ⎣ dx ⎦ (1.9) There are two important implications of this criterion, both of which are supported by experiment and more advanced analysis. Fully Fully developed heat transfer Linear variation of wall temperature in fully developed heat transfer for a constant heat flux at the wall. T Thermal entry region TBi Tw TB Tw-TB = Constant Constant Fully Developed Lte X Fully developed heat transfer The heat transfer coefficient as a function of axial distance: h(x) h(x) is infinite at x = 0 and drops off to an asymptotic value for large x. (Confirmed by experiment and analysis.) Thermal entry region Asymptotic value X Lte Fully Fully developed Nusselt number Define a Nusselt number in terms of the di diameter of the tube: th hD Nu D = kf For non-circular channels, use the hydraulic diameter, Dh. non- Temperature field in duct flow pipe ∂T ⎞ ∂ 2T 1 ⎛ ∂ ⎛ ∂T ⎞ ⎞ ⎛ ∂T +v ρC p ⎜ u ⎟⎟ + k 2 ⎟ = k ⎜ ⎜r ⎜ ⎟ ∂x ∂r ⎠ ∂x r ⎝ ∂r ⎝ ∂r ⎠ ⎠ ⎝ Neglect axial conduction ∂ 2T ∂ 2T >> 2 → 0 ∂r 2 ∂x For fully developed laminar flow v=0 v=0 ∂T k ∂ ⎛ ∂T ⎞ = ⎜r ⎟ ∂x r ∂r ⎝ ∂r ⎠ ∂T α ∂ ∂T (r ) u = ∂x r ∂r ∂r ρC p u Laminar Duct Flow Define Define mixed mean temperature 1o Tm = u (r )T (r )2πrdr AV ∫ 0 1 1 ∫ udA = A ∫ 2πrudr A T −T θ= w Tw − Tm um = r Thermal layer is fully developed when θ is independent of x for any set of B.C. θ is f(r) only Laminar Duct Flow ∂θ = 0 and ∂x r θ = θ ( ) only ro h= = A = − k ∂ ( Tw − T ) Tw − Tm ro ∂r ro Tw − Tm k ∂θ ro ∂ r ro q but θ ≠ θ(x) => h ≠ h(x) and h is defined at the wall r=ro => h r=r =constant ( uniform for fully developed flow) Laminar Duct Flow ∂θ =0⇒ ∂x ∂ ⎛ Tw − T ⎞ ⎜ ⎟=0⇒ ∂x ⎜ Tw − Tm ⎟ ⎝ ⎠ 1 ⎛ dTw dT ⎞ d⎛ 1 ⎞ ⎟=0⇒ − ⎜ ⎟ + (Tw − T ) ⎜ Tw − Tm ⎝ dx dx ⎠ dx ⎜ Tw − Tm ⎟ ⎝ ⎠ dTw dT ⎛ Tw − T ⎞⎛ dTw dTm ⎞ ⎟⎜ − −⎜ − ⎟=0⇒ ⎜ dx dx ⎝ Tw − Tm ⎟⎝ dx dx ⎠ ⎠ dT dT ∂T = (1 − θ ) w + θ m dx dx ∂x Laminar Duct Flow Two cases: cases: A- Constant heat flux (qw/A)=constant ⎛ qw ⎞ ⎜ A⎟ ⎝ ⎠ = h = const. Tw − Tm ⎛ qw ⎞ ⎜ A ⎟ −k ∂θ ⎝ ⎠= = const. Tw − Tm ro ∂r ro ⇒ Tw − Tm = const. or dTw dTm = = uniform dx dx (2) Laminar Duct Flow – Constant Flux Fully Fully Developed ∂T ∂Tw dTb = = ∂x dx dx Energy Equation ⎛ 1 ∂ ⎛ ∂T ⎞ ⎞ ∂T u b = α⎜ ⎜ r ∂r ⎜ r ∂r ⎟ ⎟ ⎟ ∂x ⎝ ⎠⎠ ⎝ ⎡ ⎛ r ⎞2 ⎤ u = 2⎢1 − ⎜ ⎟ ⎥ ⎜⎟ V ⎢ ⎝ ro ⎠ ⎦ ⎥ ⎣ r = ro T = Tw ( x) r =0 ∂T =0 ∂r symmetry ⎡ dT B ⎤ q ′′ (π D ) w ⎢ dx ⎥ = m C &P ⎣ ⎦ ⇒ q′′ 2π rο dx = ρ C p u avπ rο2 w ∂Tb dx ∂x 2q ′′ ∂T w = b = const LLL ( 2) ρ C p u av rο ∂x ⇒ bulk temp. increases linearly ∂T also at any r , = const ∂x ∂T ∂Tb ∂Tw or = = = const , independen t of r ∂x ∂x ∂x 2 q′′ ~ w T= x + T (r ) ρ C p u av rο 1 ∂ ⎛ ∂T ⎜r ur ∂r ⎝ ∂r with B.C : 1 ∂Tb ⎞ 1 ∂T = const = ⎟= α ∂x ⎠ α ∂x ⎡ ⎛r u = uο ⎢1 − ⎜ ⎜ ⎢ ⎝ rο ⎣ ⎞ ⎟ ⎟ ⎠ 2 ∂T ⎧ ⎪at r = 0, ∂r = 0 symmetry ⎪ ⎨ ∂T ⎪K = qw ⎪ ∂r r = rο ⎩ ⎤ ⎥ ⎥ ⎦ ∂ ⎛ ∂T ⎜r ∂r ⎝ ∂r 2 ⎞ uο ∂Tb ⎛ r ⎞ r ⎜1 − 2 ⎟ ⎟= ⎠ α ∂x ⎜ rο ⎟ ⎝ ⎠ ∂T 1 ∂Tb ⎡ r 2 r 4 ⎤ r = uο ⎢ − ⎥ + C1 ∂r α ∂x ⎣ 2 4 rο2 ⎦ 1 ∂Tb ⎡ r 2 r4 ⎤ ⇒T = uο − + C1 ln r + C 2 α ∂x ⎢ 4 16 rο2 ⎥ ⎣ ⎦ Apply B.C (1) : C1 = 0 Apply B.C ( 2) : T = TCL at r = 0 ⇒ C 2 = TC 1 ∂Tb uο rο2 T − TC = α ∂x 4 ⎡⎛ r ⎢⎜ ⎜ ⎢⎝ rο ⎣ 2 4 ⎞ 1⎛ r ⎞ ⎤ ⎟− ⎜ ⎟⎥ ⎟ 4⎜r ⎟ ⎝ ο⎠ ⎥ ⎠ ⎦ rο Tb = T = Then ∫ ρ 2π rdr uC T p 0 ρ C p u avπ r 2 7 uο rο2 ∂Tb Tb = Tc + LL (3) 96 α ∂x Also for wall temp 3 ∂Tb uο rο2 Tw( x ) = Tc ( x) + LL ( 4) 16 ∂x α 2q′′ 4q′′ ∂T w w = = where ∂x ρ C p u av rο ρ C p uο rο T − TC = Tb = Tc + 1 ∂Tb uο rο2 α ∂x 4 ⎡⎛ r ⎢⎜ ⎜ ⎢⎝ rο ⎣ ⎞ 1⎛ r ⎟− ⎜ ⎟ 4⎜r ⎠ ⎝ο 2 ⎞ ⎟ ⎟ ⎠ 4 ⎤ ⎥ ⎥ ⎦ ⎛ ∂T ⎞ q = KA⎜ ⎟ ⎝ ∂r ⎠ r =rο = hA(Tw − Tb ) K ⇒h= h= ∂T ∂r r = rο 7 uο rο2 ∂Tb LL (3) 96 α ∂x Also for wall temp Tw( x) = Tc ( x) + 3 ∂Tb uο rο2 LL (4) 16 ∂x α 2q′′ 4q′′ ∂T w w where = = ∂x ρ C p u av rο ρ C p uο rο Tw − Tb u sin g eqns 3 and 4 to calculate Tw − Tb 24 K 48 K = 11 rο 11 d hd = 4.364 K Nud = Laminar Duct Flow Case Case 2: B- Constant temperature Tw =const. =const. ∂T dT dT = (1 − θ ) w + θ m dx ∂x dx ∂Tw =0 ∂x ∂T ∂T = θ m L (1) ∂x ∂x ∂T α ∂ ⎛ r∂T ⎞ ⇒ uθ m = ⎜ ⎟ r ∂r ⎝ ∂r ⎠ ∂x ⎛ T − T ⎞ ∂Tw (r ) α ∂ ⎛ ∂T ⎞ ⇒ u⎜ w ⎟ ⎜ T − T ⎟ ∂x = r ∂r ⎜ r ∂r ⎟ ⎝ ⎠ ⎝ w m⎠ θ= Tw − T Tw − Tm Fully developed Nusselt number For fully developed heat transfer in fully developed laminar flow with a constant wall laminar flow with constant wall heat flux, Nu D , FD = 4.36 (1.10) Constant Constant wall temperature q ′′ (π D ) πD ⎡ dT B ⎤ = h (T w − T B =w ⎢ dx ⎥ & & mC P mC P ⎣ ⎦ ) T B ( 0 ) = T Bi T (1.11) (1.11a) Tw TB X Constant wall temperature Eq. (1.11) is rearranged. πD q ′′ (π D ) ⎡ dT B ⎤ w h (T w − T B = ⎢ dx ⎥ = m C &P & mC P ⎣ ⎦ ) 1 πD ⎡ dT B ⎤ ⎢ dx ⎥ = m C h ( x ) &P Tw − TB ⎣ ⎦ ⎡ T − TB ⎤ πD ln ⎢ w =− ∫ hdx ⎥ & mC p ⎣ Tw − TBi ⎦ (1.12) Fully Fully developed Nusselt number The local heat transfer coefficient is given by: q ′′ ( x ) w h= Tw − T B ( x ) (1.13) For fully developed heat transfer, h = constant., or dh d ⎛ q ′′ w ⎜ = dx dx ⎜ Tw − TB ⎝ ⎞ ⎟=0 ⎟ ⎠ (1.14) Laminar Duct Flow Iterate for solution, use θ(r) of qw= const, integrate to get new θ(r) and so on. Nu =3.66 =3 q = h(Tw − Tm ) A dT q 2πro = ρcvπro3 m dx A dT h(Tw − Tm ) = (const ) m dx dT c1 m − Tm = 0 ⇒ Tm ≈ e −c1x dx Fully Fully developed Nusselt number Based on experiment and analysis, we have for fully developed heat transfer in fully developed fully developed heat transfer in fully developed laminar flow with a constant wall temperature, Nu D , FD = 3.66 (1.15) K&N Forced Forced Convection in Internal Flows - 2 Thermally Developing Laminar Flows Turbulent Convection Structure of the problem Fluid Flow Hydraulic Diameter Developing vs. fully developed flow de flo Laminar flow vs. turbulent flow Heat Transfer Developing vs. fully developed heat transfer Constant wall temperatures vs. constant wall heat flux Thermally Thermally developing laminar flow with constant wall temperature Hausen’s Correlation Nu D , avg = 3 .66 + F ( = 3 .66 + 1 + 0 .04 [(D L ) Re D Pr ] D , Re, Pr) L 0 .0668 ( D L ) Re D Pr 2/3 Restrictions: (1) Constant wall temperature (2) Developing velocity and temperature profiles (32.7) Thermally developing laminar flow with constant wall temperature Seider-Tate Correlation Nu D ,avg ⎛ Re Pr ⎞ = 1.86⎜ D ⎟ ⎟ ⎜L D⎠ ⎝ 1/ 3 ⎛μ ⎜ ⎜μ ⎝w ⎞ ⎟ ⎟ ⎠ 0.14 (32.8) Restrictions: (1) Constant wall temperature (2) Developing flow and heat transfer in laminar flow (3) 0.48 < Pr < 16700 (4) 0.0044 <(μ/μw) < 9.76 K&N Developing Nusselt numbers for laminar tube flow for laminar tube flow. K&N K&N Turbulent flow Correlations for heat transfer are expressed in the form: Nu D = C Re m Pr n D (32.9) This form of correlation is obtained from either the Buckingham Pi theorem, the defining differential equations, or the approximate analysis of the boundary layer flow and heat transfer problem. The Chilton-Colburn analogy also holds for pipe flows, i.e., f = St Pr 2 / 3 8 f = 0.183 Re D −1/ 4 (32.10) The DittusThe Dittus-Boelter equation Most internal flows are handled by the Dittus-Boelter Equation: Nu D = 0.023 Re 0.8 Pr n D (32.11) Restrictions: (1) 0.7 < Pr < 160 (2) Re > 10,000 (3) L/D > 10 L/D 10 (4) n = 0.4 for heating the fluid (5) n = 0.3 for cooling the fluid (6) Estimated accuracy: 10% - 25% in Nu (7) Properties are evaluated at the average fluid temperature The Seider-Tate equation SeiderWhen the wall temperature and mean fluid temperatures exhibit a large difference such that fluid properties are highly variable, the Seider-Tate equation is used. ⎛μ Nu D = 0.027 Re 0.8 Pr 1 / 3 ⎜ D ⎜μ ⎝w ⎞ ⎟ ⎟ ⎠ 0.14 (32.12) Restrictions: (1) 0.7 < Pr < 16,700 (2) Re > 10,000 (3) L/D > 10 (4) Estimated accuracy: 10% - 25% in Nu (5) Properties evaluated at the mean fluid temperature. Liquid Liquid metals in turbulent flow with constant wall temperature Several correlations exist to handle liquid metals. The correlations recommend by the test are: recommend by the test are: 0 .8 D D Nu = 5.0 + 0.025 (Re Pr) (32.13) Restrictions: (1) Constant wall temperature (2) RePr > 100 (3) 0.003 < Pr < 0.05 (4) Fully developed turbulent flow Liquid metals in turbulent flow with constant wall heat flux Nu D = 4.82 + 0.0185(Re D Pr)0.827 (32.14) Restrictions: (1) Constant wall heat flux (2) 100 < Re < 9.05 x 105 (3) 100 < RePr <10,000 100 (3) 0.003 < Pr < 0.05 (4) Fully developed turbulent flow Key Key terms and concepts Fully developed Nusselt number Laminar flow: Constant wall temperature Laminar flow: Constant wall heat flux Fully developed heat transfer coefficient Fully developed temperature profile Fully developed velocity profile Tube flow with constant heat flux at the wall Thermal entry region Tube flow with constant temperature at the wall ...
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