Internal Flow: Heat Transfer in Pipes

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1 Internal Flow: Heat Transfer in Pipes V.Vuorinen Aalto University School of Engineering Heat and Mass Transfer Course, Autumn 2016 November 15 th 2016, Otaniemi

2 First about the exercise class 2 Application of boundary layer flow Idea to increase surface area and convect/radiate the heat away

3 Learning Objectives Convective heat transfer in laminar pipe flow, also, impact of turbulence Boundary layer transition and thermal entry length concepts Boundary conditions and their influence on Nusselt numbers (non-dimensional convection coefficient Turbulent boundary layer animation Turbulent pipe flow Laminar vs turbulent flow

4 Basic Setup Assume laminar inflow Temperature wall: either uniform T s or q ' ' U T When the non-dimensional temperature profile reaches a constant shape flow is said to be thermally fully developed How would this go for dimensional form of profile? Velocity profile

5 On entry lengths in laminar and turbulent conditions Re D = U D ν = U ρ D μ Re D, c 2300 Critical Re when turbulent Hydrodynamic entry length Laminar flow :(x/ D 0.05Re D Turbulent flow :10 <( x/ D<60 Thermal entry length LES(Vuorinen et al. 2013Re D Laminar flow :(x/ D 0.05Re D Pr Turbulent flow :10<( x/ D<60 DNS(Keskinen et al. 2016Re D 4000 Entry lengths depend on whether or not inflow is turbulent.

6 Mean velocity and velocity profile Mass flow rate (kg/s in a pipe with cross section A: ṁ=ρ u m A= A ρu (r, xda u m =mean velocity ρ u(r,xda A r =0 u m = =2 π ρ A R ρ r u(r, xdr = 2 R r=0 ρ π R 2 r u(r, xdr R 2 In steady state: Viscous terms = pressure Note: p =independent of x=const. x p x =μ(1/r r r u (r r u t +u u x +v u y = 1 ρ p x +ν 2 u x +ν 2 u 2 y 2 u(r= 1 4 μ ( p x R2 [1 (r / R 2 ]

7 Finally we get to the parabolic profile u m = R2 8 μ ( p x u(r/u m =2[1 (r /R 2 ] Note: for a pipe a channel flow profile would be parabolic as well with different max velocity

8 Pressure gradient and friction factor in fully developed flow Flow equipments cause pressure drop fan or pumping power needed Convenient tool: the Moody friction factor (dimensionless parameter Friction factor for laminar flow f = (dp/dxd ρ U 2 /2 Friction coefficient = 64 Re D C f =f /4 Friction factor for turbulent flow 1 f = 2.0 log[ e/d Re D f ] e=surface roughness Pressure drop between two points along x-axis Δ p=f ρu 2 2 D (x x 2 1 f =turbulent /laminar friction factor

9 Friction factor (Moody diagram This equation is plotted in Moody diagram 1 f = 2.0 log[ e/d Re D f ] Re= U D ν

10 Mean temperature T out q=ṁ c p (T out T in [q]=j /s=w R T m = 2 r=0 r u T dr u m R 2 T in Some surface heating options: 1 constant surface T 2 constant heat flux (electrical heating or uniform radiation from outside Increased internal energy of small fluid element δ m by δ E=δ m c p (T out T in δ E δt = δ m δ t c p (T out T in q=ṁ c p (T out T in Re= U D ν

11 Thermally fully developed conditions x T s (x T (r, x T s (x T m (x =0 The point: although the profile shape T(r,x changes the relative shape does not depend on x after a certain point thermally fully developed flow r T s ( x T (r, x T s (x T m (x r=r o = ( T (r, x/ r r =r o f (x T s (x T m (x Newton q s '' =h(t s T m q s '' =k T r r =r o h/ k f ( x Nu L =hl/k The point: in thermally fully developed conditions Local h is a constant independent of x

12 Example 8.1: Flow of liquid metal through circular tube u(r=c 1 and T (r T s =C 2 (1 (r /r o 2 Nu=? r o T m = 2 r=0 r u T dr u m R 2 u=u m =C 1 Needed in normalization T m =T s +C 2 /2 T s (x T (r, x x T s (x T m (x =0 q '' s =k T r = 2 C k /r 2 2 o q s '' =h(t s T m h=4 k /r o Nu=hD/k=8 Note: in thermally developed conditions Nu is constant! Note: we did not really need the numeric values of the liquid at all

13 Energy balance dq conv =ṁc p dt m =q s '' P dx P=surface perimeter of the tube ' ' P dt m dx = q s = P h(t ṁc p ṁc s T m p Depending on surface boundary conditions (constant T vs surface flux the mean temperature can be integrated. Case of constant surface heat flux: ' ' P dt m dx = q s f (x ṁc p q s ' ' P ṁc p =const. T m (x=t m, i + q ' ' s P x,q '' ṁ c s =const. p Linear profile

14 Axial temperatures on surface vs mean ' ' P dt m dx = q s = P h(t ṁc p ṁc s T m p Constant surface heat flux Constant surface temperature

15 Example 8.2: Cold water heating in a pipe from 20 C to 60 C Isolated outer walls D i =40 mm Heat generation inside the pipe wall T s =70C T in =20 C D i =20 mm Heat generation inside the pipe wall L=? ṁ=0.1 kg/s, T in =20 C T out =60 C q=10 6 W /m 3 Note: volumetric Note: q=ṁ c p (T out T in [q]=j /s=w ṁ c p (T out T in = q π /4 (D o 2 D i 2 L L=17.7 m

16 Temperature profile in fully developed region of pipe (Ch Recall convection-diffusion equation (i.e. just energy conservation T t +u T x +v T y =ν 2 T T x +ν 2 T 2 T y 2 Separate variables assuming T=T(r,x=R(rX(x T (r, x=t s (x 2u m R 2 α Transform to cylindrical coordinates (assume r,z-dependency u T x = α r r r T r u(r/u m =2[1 (r /R 2 ] For the constant surface heat flux condition an analytical 4 th degree polynomial solution reads: dt m dx [ (r/ R4 1 4 (r/ R2 ] Constant surface heat flux: Constant surface temperature: Nu D =4.36 Nu D =3.66

17 Thank you for your attention!

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