Transcript HEAT PROCESSES - Czech Technical University in Prague

HP4
HEAT PROCESSES
Heat transfer
Mechanisms of heat transfer. Conduction, convection (heat transfer coefficients),
radiation (example: cooling cabinet). Fourier’s law of conduction, thermal resistance
(composed wall, cylinder). Unsteady heat transfer, penetration depth (derivation, small
experiment with gas lighter and copper wire). Biot number (example: boiling potatoes).
Convective heat transfer, heat transfer coefficient and thickness of thermal boundary
layer. Heat transfer in a circular pipe at laminar flow (derivation Leveque). Criteria: Nu,
Re, Pr, Pe, Gz. Heat transfer in turbulent flow, Moody’s diagram. Effects of variable
properties (Sieder Tate correction for temperature dependent viscosity, mixed and
Rudolf Žitný, Ústav procesní a
natural convection).
zpracovatelské
techniky ČVUT FS 2010
HP4
Mechanisms of heat transfer
There exist 3 basic mechanisms of heat transfer between
different bodies (or inside a continuous body)
Conduction in solids or stagnant fluids
Convection inside moving fluids, but first of all we shall
discuss heat transfer from flowing fluid to a solid wall
Radiation (electromagnetric waves) the only mechanism
of energy transfer in an empty space
Aim of analysis is to find out relationships between heat flows
(heat fluxes) and driving forces (temperature differences)
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Heat flux and conduction
General form of transport
Fourier equation
for temperature field T(t,x,y,z) in a solid or in a stagnant fluid
taking into account internal heat sources and an adiabatic
temperature increase during compression of gas.
T
cp

t
accumulation
of enthalpy
v p
T ( ) p
T t
   (T ) 
this term is zero for
incompressible liquid and
reduces to Dp/Dt for ideal gas.
conduction
heat flux
Benton
Q( R )
reaction and phase
changes enthalpies
and electric heat
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Heat flux and conduction
Energy balance of a closed system dq = du + dw (heat delivered to system equals
internal energy increase plus mechanical work done by system) tells nothing
about intensity of heat transfer at the surface of system, neither about
relationships between heat fluxes and driving forces (temperature gradients). This
problem is a subject of irreversible thermodynamics.

n
dA

q
Intenzity of heat transfer through element dA of
boundary is characterized by vector of heat flux

[W/m2] q
Direction and magnitude of heat flux is determined by
gradient of temperature and thermal conductivity of
media 

q  T
Fourier’s law of heat
conduction
Heat flow through boundary is projection of heat flux to the outer normal
 
nq
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Thermal conductivity 
Thermal and electrical conductivities are similar: they are large for metals (electron
conductivity) and small for organic materials. Temperature diffusivity a is closely

related with the thermal conductivity
a
Memorize some typical values:
Material
 [W/(m.K)]
a [m2/s]
Aluminium Al
200
80E-6
Carbon steel
50
14E-6
Stainless steel
15
4E-6
Glas
0.8
0.35E-6
Water
0.6
0.14E-6
Polyethylen
0.4
0.16E-6
Air
0.025
20E-6
c p
Thermal conductivity of nonmetals and gases increases with temperature (by
about 10% at heating by 100K), at liquids and metals  usually decreases.
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Conduction – Fourier equation
Distribution of temperatures and heat fluxes in a solid can be expressed in differential
form, based upon enthalpy balancing of infinitesimal volume dv
h

  q
t
Accumulation of
enthalpy in unit
control volume
qx q y qz
q


x y z
Divergence of heat fluxes (positive
if heat flows out from the control
volume at the point x,y,z)
Integrating this differential equation in a finite volume V the integral enthalpy balance
can be expressed in the following form using Gauss theorem


n
dA

q
V
Accumulation of
enthalpy in
volume V
h
dv    qdv    n qds
t
V
S
Heat transferred
through the
whole surface S
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Conduction – Fourier equation
Heat flux q as well as the enthalpy h can be expressed in terms of temperatures,
giving partial differential equation – Fourier equation
Internal heat source (e.g.
enthalpy change of a chemical
reaction or a volumetric heat
produced by passing electric
current or absorbed
microwaves)
h

  q  Qg
t
T
cp
  (T )  Qg
t
T

T

T

T
cp

(
)  (
)  (
)  Qg
t x
x
y
y
z
z
Thermal conductivity  need not be
a constant. It usually depends on
temperature, and for anisotropic
materials (e.g. wood) it depends
also on directions x,y,z – in this
case ij should be considered as the
second order tensor.
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Conduction - stationary
Let us consider special case: Solid homogeneous body (constant thermal conductivity
and without internal heat sources). Fourier equation for steady state reduces to the
Laplace equation for T(x,y,z)
2
2
2

T

T

T
0   2T 


x 2 y 2 z 2
 2T 1  T
0 2 
(r )
x r r r
1 
T
0  2 (r 2 )
r r
r
The same equation written in
cylindrical and spherical
coordinate system (assuming
axial symmetry)
Boundary conditions: at each point of surface must be prescribed either temperature
T or the heat flux (for example q=0 at an insulated surface).
Solution of T(x,y,z) can be found for simple geometries in an analytical form (see next
slide) or numerically (using finite difference method, finite element,…) for more
complicated geometry.
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Example temperature profile in a cylinder
Calculate radial temperature profile in a cylinder and sphere (fixed
temperatures T1 T2 at inner and outer surface)
R1
R2
cylinder
T
r
 c1 ,
r
r2
T
 c1 ,
r
Sphere (bubble)
T=c1 ln r  c2
T=-
c1
 c2
r
T2  T1
T1 ln R2  T2 ln R1
c1 
c2 
ln R2 / R1
ln R2 / R1
c1 
R1 R2
T R  T1 R1
(T2  T1 ) c2  2 2
R2  R1
R2  R1
Conduction – thermal resistance
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Knowing temperature field and thermal conductivity  it is possible to calculate heat
fluxes and total thermal power Q transferred between two surfaces with different (but
constant) temperatures T1 a T2
T1
T2
T1 T2
S2
R2
S
Q
S1
Q
T2
R1
h
T1 T2
R1
T1
h
h1 h2
RT 
1 h1 h2
(  )
S 1 2
Serial
RT [K/W] thermal resistance
L
L
RT 
h
S11  S 22
Parallel
RT 
ln R2 / R1
2L
Tube wall
RT 
T1  T2
RT
ln 2h / R
2L
Pipe burried under surface
In this way it is
possible to express
thermal resistance
of windows, walls,
heat transfer
surfaces …
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Conduction - nonstacionary
Time development of temperature field T(t,x,y,z) in a homogeneous solid body without
internal heat sources is described by Fourier equation
T
 2 T  2 T  2T
 a( 2  2  2 )
t
x
y
z
The coefficient of temperature diffusivity a=/cp is the ratio of
temperature conductivity and thermal inertia
with the boundary conditions of the same kind as in the steady state case and with
initial conditions (temperature distribution at time t=0).
This solution T(t,x,y,z) can be expressed for simple geometries in an analytical form
(heating brick, plate, cylinder, sphere) or numerically.
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Example Conduction - nonstationary
Typical application: How long must be a can held in a sterilizer so that all
microorganisms at the can center will be killed?
Simple answer: for infinitely long time. This is never possible to kill all of them, only some prescribed percents,
e.g. the number of Clostridium Botulinum spores should be reduced by 12 orders according to regulation.
Therefore only one of 1012 spores could survive the thermal treatment (not the viable form it would be much
easier). Number of survived pathogens depends upon the whole temperature history at a given place (in this
case in the center of can) and can be evaluated as an integral of temperature (see lecture aseptic processes).
Temperature field in a can is described by Fourier equation rewritten by using dimensionless temperature  that
is zero on the surface and 1 inside the can at time t=0
  2 1 

 2 
(r
),

x
r r
r
=at, where a is

T  Tw
T0  Tw
Initial temperature
temperature diffusivity
of meat in a can
wall temperature
There are many ways how to solve the problem:
Numerically (finite differences seems to be an obvious choice)
Using integral transforms (e.g. Laplace transform to transform differential equation to an algebraic one)
Using Fourier’s method of separated variables, see next slide
HP4
Example Conduction - nonstationary
Temperature field (=at,x,r) can be decomposed to the product X(,x)R(,r) of two
functions that fulfill the following two simplified equations
X
 X


x 2
R 1 
R

(r
)

r r
r
2
Axial temperature profile
Radial temperature profile
X and R satisfy zero
boundary condition (X=R=0 at
surface) and initial condition
(X=R=1)
It can be easily verified that the product (,x,r)= X(,x)R(,r) satisfies not only the
original Fourier equation but also the prescribed boundary (=0) and initial (=1)
conditions (this is the reason why it was necessary to introduce dimensionless temperature and
transformed time  = a t ).
Resulting equations can be solved again by the separation of variables
X ( , x)   xi Ti ( ) i ( x)
i
R( , r )   ri T i ( )i (r )
i
Sometimes (for very long time )
it is sufficient to approximate the
solution only by the first term in
these series
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Example Conduction - nonstationary
Resulting ORDINARY differential equations have the following solutions (the same
exponential term describes time profile, the sin-function axial variation of temperature
and the Bessel function J0 radial variation):
Ti ( )=e  i 
2
Ti ( )=e
 i2
 i ( x)  sin(  i x)
 i (r )  J 0 ( i r )
L
Eigenvalues i are determined from boundary conditions, e.q. sin(iL)=0.
Remark: It is possible to modify this solution for more complicated boundary
conditions with finite thermal resistance on the surface of can, see Baehr S. Wärme
und Stoff-übertragung, Springer, Berlin, 1994.
Coefficients xi,ri must be calculated so that the initial conditions (X=R=1) will be
satisfied (using special properties of eigenfunctions sin,J0, called orthogonality).
Seems too complicated? For more details look at the mentioned book (Baehr), or
wait to the next semester, where the same problem will be discussed in more details
during the course Numerical analysis of processes.
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Example Conduction - nonstationary
I wrote previous pages at home when I have only the Baehr’s book at
hand. Today (at work) I can add reference to the famous book (first
published at 1946!)
Carslaw H.S., Jaeger J.C.: Conduction of heat in solids, Clarendon Press,
Oxford, 2004 (previous problem is solved on pages 328-329)
Conduction - nonstacionary
HP4
Previous example was probably too complicated. However, there exists much simpler
and probably more important 1dimensional problem of heated halfspace, and the
step wise change of surface temperature as a boundary condition.
  2
 2,

x
Tw
t
T0
x
δ

d 2
d

2

0
2
d
d
x
2 
  1
2


2
 exp( )d  erfc(
0
x
)
2 at
T  T0

Tw  T0
This equation is the same as previously and therefore the same
method (separation of variables) can be used. The analyzed case
of heat transfer in a plate was complicated by the boundary
condition at x=L and now this requirement is shifted to infinity.
Therefore the solution in the vicinity of surface will be controlled
only by the boundary condition at x=0 (for short times) and a new
kind of solution, based upon similarity transformation can be
used. This transformation introduces suitable dimensionless
combination of time and coordinate , for example =/x2, =x2/,
=x/, recasting the partial differential equation (PDE) to an
ordinary differential equation (ODE). The simplest form of the
resulting ODE is obtained for =x/ (check other possibilities)
Erfc is complementary error
function (available also in
some pocket calculators)
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Conduction - nonstacionary
Erfc function describes temperature response to a unit step at surface (jump from
zero to a constant value 1). The case with prescribed time course of temperature at
surface Tw(t) can be solved by using the superposition principle and the response can
be expressed as a convolution integral.
t
Tw()
T (t , x)   Tw ( ) E (t   , x)d
0
T(t,x)


d
x
Time course Tw(t) can be substituted by short pulses
t
Temperature at a distance x is the sum of
responses to short pulses Tw()d
The function E(t,,x)=E(t-,x) is the impulse function (response at a distance x to a
temperature pulse of infinitely short duration but unit area – Dirac delta function).
The impulse response can be derived from derivative of the erfc function
x
x2
E (t , x) 
exp(
)
3/ 2
4at
2 at
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Theory of penetration depth
Still too complicated? Your pocket calculator is not equipped with the erffunction? Use the acceptable approximation by linear temperature profile

T
T
dx


a
|x  0
0 t
x
Tw

t
T0
T

Tdx  a w

t 0

t+t
x
T
 Tw
(
)a w
t 2

Integrate Fourier equations
(up to this step it is
accurate)
Approximate
temperature
profile by line
δ
+Δ
Result is ODE for thickness  as a function of time
  4at
Using the exact temperature
profile predicted by erffunction, the penetration
depth slightly differs =(at)
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Theory of penetration depth
=at penetration depth. Extremely simple and important result, it
gives us prediction how far the temperature change penetrates at the time t. This
estimate enables prediction of thermal and momentum boundary layers thickness
etc. The same formula can be used for calculation of penetration depth in
diffusion, replacing temperature diffusivity a by diffusion coefficient DA .
Wire Cu
=0.11 m
=398 W/m/K
=8930 kg/m3
Cp=386 J/kg/K

HP4
Example Continuation the can
Summary of previous results:
 For very short times and small penetration depth
the temperature profile can be always approximated
by erf-function or linear function (even for cylinder as
soon as the thickness <<R)
 For very long times the temperature profile can be
approximated only by the first term of Fourier’s
expansion, in case of the can by
  ce
 (  x2   r2 )
sin(  x x) J 0 (  r r )
R
L
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Convection
General form of transport Fourier Kirchhoff equation
T
v Dp
 c p (  u T )  T ( ) p
   q   :   Q( R)
t convection
T Dt conduction viscous
this term is zero for
incompressible liquid
heating
Benton
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Convection
Calculation of heat flux q from flowing fluid to a solid surface requires calculation of
temperature profile in the vicinity of surface (see previous Fourier Kirchhoff equation
but also for example temperature gradients in attached bubbles during boiling, all
details of thermal boundary layer,…).
Engineering approach simplifies the problem by introducing the idea of stagnant
homogeneous layer of fluid, having an equivalent thermal resistance (characterized
by the heat transfer coefficient  [W/(m2K)])
q   (T f  Tw )
T  f
q   (Tf  Tw )   f
 (Tf  Tw )
y 
Tf is temperature of fluid far from surface (behind the boundary of thermal boundary layer), Tw is wall
temperature. Thickness of stagnant boundary layer δ, f thermal conductivity of fluid.
Boiling
(bubbles)
d
A

n ,y

q

q  T
Outer flowthermal
boundary
layer
f


Tf
Tf

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Example heating sphere
Temperature distribution inside a solid sphere
Ts s  2 Ts
cp
 2
(r
)
t
r r
r
Boundary condition (convection)
Ts
s
  (T f  Ts )
r
It is correct only as soon
as the heat flux q or the
temperature is uniform
on the sphere surface
Heat flux calculated from
Fourier law inside the
sphere equals the flux in
fluid
Fourier equation can be integrated at the volume of body (sphere in this case)
 cp
V
Ts
dv   s 2Ts dv     qdv
t
V
V
The integrals can be evaluated by the mean value and by Gauss theorem,
assuming uniform flux at the surface
Mc p
Ts
  n qdS  n qS   (T f  Ts )S
t
S
HP4
Example heating sphere
For the case that the temperature inside the sphere is uniform (as soon as the
thermal conductivity s is very high) the mean temperature is identical with the
surface temperature
dTs
S

(T f  Ts )
dt
cp M
Ts  T f  (T0  T f )e  t /

cp M
S
This exponential solution works only for small values of Biot number
Bi 
D
 0.1 Thermal resistance of fluid >> thermal resistance of solid
s
HP4
Convection – Nu,Re,Pr
Heat transfer coefficient  depends upon the flow velocity (u), thermodynamic
parameters of fluid () and geometry (for example diameter of sphere or pipe D).
Value  is calculated from engineering correlation using dimensionless criteria
Nu 
Re 
D

uD

Pr 
a

Nusselt number (dimensionless , reciprocal thickness of boundary layer)
Reynolds number (dimensionless velocity, ratio of intertial and viscous forces)
Prandl number (property of fluid, ratio of viscosity and temperature diffusivity)
Rem:  is dynamic viscosity [Pa.s],  kinematic viscosity [m2/s], =/
And others
Pe=Re.Pr
Gz=Pe.D/L
De=Re√D/Dc
Péclet number
Graetz number (D-diameter, L-length of pipe)
Rayleigh
Dean number (coiled tube, Dc diameter of curvature)
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Convection in a pipe
Basic problem for heat transfer at internal flows: pipe (developed velocity
profile) and a constant wall temperature
Liquid flows in a pipe with the constant wall temperature Tw that is different than
the inlet temperature T0. Temperature profile depends upon distance from inlet and
upon radius r (only thin temperature boundary layer of fluid is heated). Heat flux
varies along the pipe even if the heat transfer coefficient  is constant, because
driving potential – temperature difference between wall and the bulk temperature
Tm depends upon the distance x. Tm is the so called mean calorific temperature
1
Tm ( x)   uTdA
m A
q( x)   ( x)(Tw  Tm ( x))
Heat flux from wall to bulk (
is related to the calorific
temperature as a
characteristic fluid
temperature at internal flows)
HP4
Convection in a pipe
Axial temperature profile Tm(x) follows from the enthalpy balance of system,
consisting of a short element of pipe dx :
0  m(h( x)  h( x  dx))   (Tw  Tm ( x)) Ddx
0  mcp (Tm ( x)  Tm ( x  dx))   (Tw  Tm ( x)) Ddx
Tw
T0
m c p
Q
Tm
dTm
 Ddx
Tw  Tm
D
Solution Tm(x) by integration
x
dx
mc p ln
Tw  Tm ( x)
  D x
Tw  T0
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Convection in a pipe
Previous integration is correct only if  and the wall temperature are constant.
 This doesn’t hold in laminar flow characterized by gradual development of thermal
boundary layer (at entry this layer is thin and therefore =/ is high,  decreases with
increasing distance). Typical correlation for laminar flow is Leveque formula
Nu 
D
D
 1.6 3 Re Pr

L
  is almost constant at turbulent flows characterized by fast development of
thermal boundary layer. Typical correlation (Dittus Boelter)
D
Nu 
 0.023Re0.8 Pr1/3

 More complicated are cases with mixed convection (effect of temperature
dependent density and gravity), variable viscosity and first of all influence of phase
changes (boiling/condensation).
Tw
general formula for variable wall
temperature and variable heat
transfer coefficient
Q
Toutlet
T0
D
Toutlet
mc p

T0
L
L
dTm
  D   dx
Tw  Tm
0
HP4
Convection Laminar Leveque
Leveque method is very important technique how to estimate thickness of
thermal boundary layer and the heat transfer coefficient  in many internal
flows (not only in circular pipes). This theory is applicable only for “short”
channels, in the region of developing temperature profile.
Tw

T0
y
x
umax
D
HP4
Convection Laminar Leveque
Linear approximation of
velocity profile in thermal
boundary layer
Linear approximation of
temperature profile in thermal
boundary layer
Tw

T0
y
umax
Linearized velocity profile
u( y) 
2umax D
8u
y  ky
D
Transit time of particle at
the distance  from wall
t
x
x
k
Thickness of boundary layer from penetration theory
x
   at   a
k
2
 
3
 axD
8u
2
D D
8
u
8
D
u 3 8 Du  D
D
Nux 
  D3
3

 c 3 Re Pr  c 3 Gz x


 aDx
 ax
  a x
x
Graetz number for
distance x from inlet
Gz=Re.Pr.D/x
HP4
Convection Laminar Leveque
Local Nux decreases with x. Mean value of Nu corresponding to the length of
pipe L is obtained by integration, so that the outlet temperature Toutlet can be
calculated from enthalpy balance
Heat capacity of
stream
Heat transfer
surface
T T
mc p ln w outlet   D   dx   DL ln
Tw  T0
0
L
1
 c 3
D
 3c 3
D
 3
 ln    dx 
Re
Pr
dx

Re
Pr

1.6
Gz

L0
DL0
x
D 2
L
D
L
Mean heat transfer
coefficient related to the
mean logarithmic
temperature difference
Remark: further on the index ln will be
frequently omitted, and Nu,  denotes
mean values along the whole channel.
L
Nu 
 ln D
 1.6 3 Gz

Graetz number
Gz=Re.Pr.D/L
HP4
Convection Laminar Leveque
Previous result holds only for parabolic velocity profile, corresponding to the
fully developed flow of a Newtonian liquid. For fluids with different velocity
profiles (e.g. profiles corresponding to power law liquids) the whole derivation
must be repeated – the idea of Leveque remains
Tw

T0
y
x
Linear approximation of power
law velocity profile in thermal
boundary layer
D
umax
cq const  0.975
Velocity profile for power law
liquid (n-flow index)
n1 

3n  1   2r  n 
u
u 1  

n 1   D 

1/3
 du

y

0
 dy

D
 ,
Nu 
 cD 

 ax 




cTwconst  0.8085
The same procedure can be applied to the
case with the constant wall temperature
and with a constant wall heat flux (only the
constant c differs)

1/3
D
 2(3n  1) D 
Nu 
 0.8085 
Pe 

n
x

This example demonstrates that
sometimes it is more important to know
a derivation than the final result.
HP4
Convection Laminar Hausen
Léveque solution is valid only at
 Laminar flow (Re<2300) and fully developed velocity profile
 In the region of thermal boundary layer development (“short” pipes, Gz>50)
Constant heat flux (therefore
increasing wall
temperature). Typical for the
counterflow heat
exchangers.
For a pipe of arbitrary length the Graetz solution in
form of series of eigenfunctions (see the Fourier
method of separation of variables) can be used. At a
very large distance from inlet the boundary layers
merge, transverse temperature profiles become
similar and the heat transfer coefficient approaches
limiting value (Nu=3.66 for the case of constant wall
temperature, Nu=4.36 for constant heat flux).
Hausen’s semiempirical correlation is a
blend of Léveque solution and
asymptotic values of Nu
Constant wall temperature
(maintained by boiling or
condensing steam)
Nu 
 ln D
0.0668Gz
 3.66 

1  0.04Gz 2/3
Prove that the Hausen formula reduces to Leveque for Gz
HP4
Example Graetz number=50
Thermal stabilization length
Hydraulic stabilization length
Maximum length for Leveque
Stabilization velocity for laminar/turbulent flow
D
D
LT  Re Pr
 Re Pr
Gz
50
Lhl  0.05 Re D
Water: kinematic viscosity 10-6 m2/s, Pr=6
Lht  1.36 4 Re D
Rohsenow
chapter 5.26
D [m]
U [m/s]
Re
Lh [m]
LT [m]
0.1
1
105
1.2
turbulent
0.05
1
50000
0.5
turbulent
0.01
1
10000
0.07
turbulent
0.01
0.1
1000
0.5
1.2
0.01
0.01
100
0.05
0.12
0.005
0.1
500
0.12
0.3
0.005
0.01
50
0.012
0.03
HP4
Mixed convection, Sieder Tate
Temperature dependendent properties of fluid are respected by correction
coefficients applied to a basic formula (Leveque, Hausen, …similar corrections
are applied in correlations for turbulent regime)
Temperature dependent viscosity results in changes of velocity profiles. In
case of heating the wall temperature is greater than the bulk temperature, and viscosity of liquid at wall
lowers. Velocity gradient at wall increases thus increasing heat transfer (look at the derivation of
Leveque formula modified for nonnewtonian velocity profiles). Reversaly, in case of cooling (greater
viscosity at wall) heat transfer coefficient is reduced. This effect is usually modeled by
Sieder Tate correction (ratio of viscosities at bulk and wall temperature).
Temperature dependent density combined with acceleration (gravity)
generate buoyancy driven secondary flows. Resulting effect depends upon orientation
(vertical or horizontal pipes should be distinguished). Intensity of natural convection (buoyancy) is
characterized by Grashoff number Gr)
Levequ
e
Mixed convection
(Grashoff)
1/3   
Nu  1, 618Gz1/3 1  0, 0111Gz1/3Gr 4/9  


 W
g  TD 3
Gr 
,
2

Sieder Tate
correction
0,14
,
HP4
Convection Turbulent flow
Boccioni
HP4
Convection Turbulent flow
Turbulent flow is characterised by the energy transport by turbulent eddies which is
more intensive than the molecular transport in laminar flows. Heat transfer
coefficient and the Nusselt number is greater in turbulent flows. Basic differences
between laminar and turbulent flows are:
Nu is proportional to
3
0.8
u in laminar flow, and u in turbulent flow.
Nu doesn’t depend upon the length of pipe in turbulent flows significantly (unlike
the case of laminar flows characterized by rapid decrease of Nu with the length L)
Nu doesn’t depend upon the shape of cross section in the turbulent flow regime (it
is possible to use the same correlations for eliptical, rectangular…cross sections
using the concept of equivalent diameter – this cannot be done in laminar flows)
The simplest correlation for hydraulically smooth pipe designed by Dittus Boelter
is frequently used (and should be memorized)
Nu  0.023Re0.8 Pr m
m=0.4 for heating
m=0.3 for cooling
Similar result follows from the Colburn analogy
Nu  0.023Re 0.8 Pr1 / 3
HP4
Convection Turbulent flow
 Dittus Boelter correlation assumes that Nu is independent of L. For very short
pipes (L/D<60) Hausen’s correlation can be applied
0 ,14
  D 2 / 3  3 / 4



 ,
Nu  0,0371     Re  180 Pr 0 ,42 
  L  
 W 
2300<Re<105
0,6<Pr<500


 Effect of wall roughness can be estimated from correlations based upon
analogies between momentum and heat transfer (Reynolds analogy, Colburn
analogy, Prandtl Taylor analogy). Results from hydraulics (pressure drop, friction
factor f) are used for heat transfer prediction. Example is correlation Pětuchov
(1970) recommended in VDI Warmeatlas
Friction factor
Nu 
 f Re Pr


8 .56  35 .92 Pr 2 / 3  1  f
  


 W 
104Re5,106
n
n=0,11
TW>T (heating)
n=0,25
TWT (cooling)
0,5Pr2000
0,08W /40.
HP4
Pressure drop, friction factor
Pressure drop is calculated from
Darcy Weissbach equation
Friction factor f
depends upon Re
and relative
roughness
p 
1
L
 f u 2
2
D
HP4
Turbulent boundary layer
Rougness of wall has an effect upon the pressure drop and heat transfer only
if the height of irregularities e (roughness) enters into the so called buffer
layer of turbulent flow. Smaller roughness hidden inside the laminar (viscous)
sublayer has no effect and the pipe can be considered as a perfectly smooth.
y
Buffer
layer
Laminar
sublayer
Dimensionless
distance from wall
Velocity profile
Friction
velocity
e-roughness
y 
u* y 

Thickness of laminar sublayer is at value y+=5
,
u* 
w
,

w 
dp D
dx 4
HP4
Example smooth pipe
Calculate maximum roughness at which the pipe D=0.1 m can be
considered as smooth at flow velocity of water u=1 m/s.
Re 
uD 

 10000
0.316
f  4
 0.0316
Re
p
1

 f  u 2  1600
L 2D
Blasius correlation for
friction factor (smooth
pipes)
L
e5  10
 25 m
Dp
Thickness of laminar
sublayer (y+=5)