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KONVEKSI: ALIRAN INTERNAL Convection Heat Transfer: Internal Flow

Dr. Ir. I Made Astina, M.Eng. 1

Aim of this part 

Focus on problem of computing heat transfer rates from a surface in internal flow



Introduce evaluation methods of heat transfer coefficient in the internal flow for tube, either heating or cooling



Implementation of empirical equation to calculate heat transfer rate

2

Introduction • We will obtain convection coefficients for geometries involving internal flow, such as flow in tubes, non-circular duct  Recall Newton’s law of cooling:

qs"  h (Ts  T )

 For flow inside a tube we cannot define T  Must know how temperature evolves inside the pipe and find alternative expressions for calculating heat flux due to convection.

3

Flow Conditions for Internal Flow

du / dx  0 Onset of turbulent flow at

ReD 

um D  2300 

• Hydrodynamic entry length: x fd ,h / D  0.05 Re D – Laminar flow – Turbulent flow

x fd ,h / D  10 4

Mean Velocity Velocity inside a tube varies over the cross section. For every differential area dAc:

dm  u( r , x )dAc





 m  dm  u ( r, x )dAc A

A

Overall rate of mass transfer through a tube with cross section Ac:

m  um Ac

and

m um  Ac

Combining with the above Eq. :

um

where um is the mean (average) velocity

 

Ac

 u( r, x) dAc  Ac

2  2 ro



ro

u (r , x )rdr

0

 Can determine average temperature at any axial location (along the x-direction), from knowledge of the velocity profile 5

Velocity Profile in a Pipe For laminar flow of an incompressible, constant property fluid in the fully developed region of a circular tube (pipe):

1  dp  2   r u (r )     ro 1   4  dx    ro 

  

2

  

ro2 dp um   8 dx  r u( r )  2 1   um   ro 

  

2

   6

Thermal Considerations: Mean Temperature 

• •

Newton’s law of cooling inside a tube can written by considering a mean temperature, instead of T The internal energy per unit mass for a differential area is: dE 

q"x  h(TS  Tm )

dm cT (r , x )  (udAc )cT ( r , x )

Integrating over the entire cross section:



E  dE  A

•

 uc T (r, x)dA A



c

Overall rate of energy transfer:

E  m cTm Combining with the above Eq. :

E Tm   c m

and

T  m

where Tm is the mean (average) temperature

  uc T (r , x )dA Ac

v

c

m c



2   uT (r , x )rdr ur ro

2

m

0

o

7

Example 1 The velocity and temperature profiles for flow of a liquid metal through a circular tube at a particular axial location may be approximated as being uniform and parabolic respectively. That is, u(r)=C1 and T(r)-Ts=C2[1-(r/ro)2], where C1 and C2 are constants and Ts the temperature at the surface of the tube. What is the value of the Nusselt number, NuD at this location?

r

u ( r )  C1

o

T (r )  T  C [1  (r / r ) ] 2

s

2

8

o

Ans. NuD=8

Example 2 Velocity and temperature profiles for laminar flow in a tube of radius ro = 10 mm have the form U(r) = 0.1 [1+(r/ro)2] T(r) = 344.8 + 75.0(r/ro)2 – 18.8(r/ro)4 With units of m/s and K, respectively. Determine the corresponding value of the mean (or bulk) temperature, Tm, at this axial position and convective heat transfer coefficient.

9

Fully Developed Conditions (i)

? Can we claim that dT ( r )  0 ? dx • For internal flows, the temperature, T(r), as well as the mean temperature, Tm always vary in the x-direction, i.e.

dT ( r) dT  0, m  0 dx dx What is its difference with velocity behavior?

10

Fully Developed Conditions (ii) • Although T(r) changes with x, the relative shape of the temperature profile remains the same: Flow is thermally fully developed.

  Ts ( x )  T ( r, x)  0   x  Ts ( x)  Tm ( x)  fd ,t • A fully developed thermally region is possible, if one of two possible surface conditions exist: – Uniform temperature (Ts=constant) – Uniform heat flux (qx”=const) • Thermal Entry Length:

( x fd , t / D )lam  0.05 Re D Pr ( x fd ,t / D) turb  10 11

Fully Developed Conditions (iii) 

It can be proven that for fully developed conditions, the local convection coefficient is a constant, independent of x:

h  f (x )

12

Friction in Fully Developed Flow Non dimension

Lamiar flow

Large Reynolds and smooth surface Wide range of condition:

13

Mean temperature variation along a tube (i)

 –

We are still left with the problem of knowing how the mean temperature Tm(x), varies as a function of distance, so that we can use it in Newton’s law of cooling to estimate convection heat transfer. Consider an energy balance on a differential control volume inside the tube: Main contributions are due to internal energy changes [= m cTm ], convection heat transfer and flow work [=pv], needed to move fluid.

P=surface perimeter

The rate of convection heat transfer to the fluid is equal to the rate at which the fluid thermal energy increases, plus the net rate at which is work is done in moving the fluid through the control volume

dqconv  m d (cTm  p ) 14

Mean temperature variation along a tube (ii) Considering perfect gas, or incompressible liquid:  c p dTm dqconv  m By integrating:

qconv  m c p (Tm, o  Tm ,i )  qconv is related to mean temperatures at inlet and outlet. Combining the equations:

dT m q s" P P   h(Ts  Tm )  cp m  cp dx m

where P=surface perimeter =

pD for circular tube, width for flat plate

 Integration of this equation will result in an expression for the variation of Tm as a function of x. 15

Case 1: Constant Heat Flux qconv 

q"s A



q"s ( P  L)

where P=surface perimeter  p D for circular tube, = width for flat plate

qs"  const

Integrating equation:

Tm ( x )  Tm ,i

q s" P x   cp m

16

Example 3 A flat-plate solar collector is used to heat atmospheric air flowing through a rectangular channel. The bottom surface of the channel is well insulated, while the top surface is subjected to a uniform heat flux, which is due to the net effect of solar radiation absorption and heat exchange between the absorber and cover plates. For inlet conditions of mass flow rate = 0.1 kg/s and Tm,I = 40oC, what is the air outlet temperature, if L = 3 m, w = 1 m and the heat flux is 700 W/m2? The specific heat of air is cp = 1008 J/kg·K

Ans: Tm,o=60.8°C

17

Case 2: Constant Surface Temperature, Ts=constant dTm d (D T ) P   hDT dx dx m c p

From eq. Ts-Tm=DT:

Integrating from x to any downstream location:

 Px  Ts  Tm ( x)  exp  h  m c p  Ts  Tm,i   For the entire length of the tube:

 PL  Ts  Tm ,o D To  h  exp    DTi Ts  Tm ,i  m c p 

qconv  h As DTlm

where

As is the tube surface area, As=P·L=pDL

DTlm 

D To  D Ti ln(DTo / DTi ) 18

Example 3 Steam condensing on the outer surface of a thin-walled circular tube of 50 mm diameter and 6-m length maintains a uniform surface temperature of 100oC. Water flows through the tube at a mass flow rate of 0.25 kg/s and its inlet and outlet temperatures are Tm,1=15oC and Tm,o=57oC. What is the average convection coefficient associated with the water flow? Ts=100oC D = 50 mm

Tm,i = 15oC

Tm,o=57oC L=6 m Ans . h  756 W / m 2 K 19

Case 3: Uniform External Temperature



Replace Ts by T and h by U (the overall heat transfer coefficient, which includes contributions due to convection at the tube inner and outer surfaces, and due to conduction across the tube wall)

 U As  D To T  Tm, o   exp      DTi T  Tm ,i  mc p 

q  U As DTlm 20

Heat Transfer Correlations for Internal Flow Knowledge of heat transfer coefficient is needed for calculations shown in previous slides.  Correlations exist for various problems involving internal flow, including laminar and turbulent flow in circular and non-circular tubes and in annular flow.  For laminar flow we can derive h dependence theoretically  For turbulent flow we use empirical correlations  Recall from Chapters 6 and 7 general functional dependence

Nu  f (Re, Pr) 21

Laminar Flow in Circular Tubes Fully Developed Region We start from the energy equation, written for fully developed, flow in one direction and substitute known velocity profile for flow in tubes

u

T T    T    r  x  r r r  r 

where =0 and

  r 2  u( r )  2 1     um   ro    

For constant heat flux, the solution of the differential equation is:

11 q s" Tm ( x)  Ts ( x )   48 k

qs"  const

Combining with Newton’s law of cooling:



h

q"x  h (TS  Tm )

48 k / D  11 22

Laminar Flow in Circular Tubes • For cases involving uniform heat flux: Nu D 

hD  4.36 k

qs"  const

• For cases involving constant surface temperature:

NuD  3.66

Ts  const

23

Laminar Flow in Circular Tubes (i) Entry Region: Velocity and Temperature are functions of x •

Thermal entry length problem: Assumes the presence of fully developed velocity profile

•

Combined (thermal and velocity) entry length problem: Temperature and velocity profiles develop simultaneously

24

Laminar Flow in Circular Tubes (ii) For constant surface temperature condition: Thermal Entry Length case

Nu D  3.66 

0.0668( D / L) Re D Pr 1  0.04[( D / L ) ReD Pr]2 / 3

Combined Entry Length case 1/ 3

 Re Pr  Nu D  1.86 D   L/ D 

      s 

Ts  const 0.14

0.48  Pr  16,700   9.75 0.0044  s

All properties, except s evaluated at average value of mean temperature

Tm 

Tm, i  Tm,o 2

25

Turbulent Flow in Circular Tubes (i) For a smooth surface and fully turbulent conditions the Dittus–Boelter equation may be used for small to moderate temperature differences TsTm:

NuD 

0.023 Re D4 / 5

Pr

n

0.7  Pr  160 Re D  10,000 L / D  10

n=0.4 for heating (Ts>Tm) and 0.3 for cooling (Ts...