heat transfer enhancement in pipe flow

A SEMINAR REPORT
Submitted by
Haidar Majeed Hachim
In partial fulfillment for the award of the degree
Of
MASTER OF POWER PLANT
IN
MECHANICAL ENGINEERING
Prof. Dr. veysel ozceyhan


 The need to increase the thermal performance of heat transfer
equipment (for instance heat exchangers) thereby effecting energy
,material and cost saving has led to the development and use of
many heat transfer enhancement techniques this seminar deals
with the analysis of heat transfer augmentation for fluid flowing
through pipes using CFD .using CFD for modeling the heat and
fluid flow is an efficient tool for predicting equipment performance
.CFD offers a convenient means to study the detailed flows and
heat exchange process which takes place inside the tube .friction
factor and Nusselt number for air flowing through the specified
pipe (diameter 0.06 m ,length 0.8m)were obtained first for smooth
pipe and second for roughened pipe.in this seminar the factor that
affect the enhancement techniques using roughened pipes are
studied. these factor are the ratio of(pitch/pipe diameter),Reynolds
numbers
ABSTRACT


 The results showed that there is an increase in heat
transfer coefficient is related to the decreasing of
ratio of pitch/pipe length ,increasing of Reynolds
number .the performance of roughened pipe is
evaluated depending on the calculation of thermo-
hydraulic performance(THP)and its found that the
thermo-hydraulic performance increase as Reynolds
number increase and (pitch/pipe daimeter)decrease
ABSTRACT


 High performance heat transfer system is great
importance in many industrial applications. The
performance of conventional heat exchangers can be
substantially improved by a number of heat transfer
enhancement techniques.. The process industry is
continuously working to incorporate enhancement in
heat transfer. Enhancement techniques can be
classified as active methods, which require external
power and Passive methods, which require no direct
application of external power.
INTRODUCTION


 The enhanced surfaces are routinely used to improve
thermal and hydraulic performance of heat
exchangers. Experimental investigation of heat
transfer of circular tube(smooth and roughened
pipe) have been studied under uniform heat flux
conditions. Air is used as working fluid.
INTRODUCTION

 To make the equipment compact
 To achieve a high heat transfer rate using minimize pumping
power
 Minimize the cost of energy and material
 A need for miniaturization of a heat exchanger in specific
applications space, OTEC
 Working fluids of low thermal conductivity (gases and oils
)and desalination plants
 Increase efficiency of process and system
 Design optimum heat exchanger size
 Transfer required amount of heat with high effectiveness
 Reduce the volume and weight
 For given temperature difference improved heat transfer
 Effective utilization of energy – minimum operating cost
Why need heat transfer enhancement


 Active method:-external power input for the
enhancement of heat transfer
 Passive method:-surface or geometrical
modification to the flow channel by incorporating
inserts or additional devices
 Compound method:-when any two or more
techniques employed simultaneously
Heat transfer enhancement
techniques


 Treated surface are heat transfer surface that have affine –
scale alteration to their finish or coating .the alteration
could be continuous or discontinuous, where the
roughness is much smaller than what affects single-phase
heat transfer, and they are used primarily for boiling and
condensing duties..
 Rough surface are generally surface modification that
promote turbulence in the flow field, primarily in the
single phase flows , and do not increase the heat transfer
surface area. Their geometric features range from random
sand-grain roughness to discrete three-dimensional,
PASSIVE TECHNIQUES


 Extended surfaces, more commonly referred to as finned
surfaces, provide an effective heat transfer surface area
enlargement. Plain fins have been used routinely in many
heat exchangers. The newer developments, however,
have led to modified finned surfaces that also tend to
improve the heat transfer coefficients by disturbing the
flow field in addition to increasing the surface area
 Displaced enhancement device are inserts that are used
primarily in confined forced convection, and they
improve energy transport indirectly at the heat exchange
surface by “displacing” the fluid from the heated or
cooled surface of the duct with bulk fluid from the core
flow
PASSIVE TECHNIQUES


 Swirl flow devices produce and superimpose swirl or
secondary recirculation on the axial flow in a channel.
They include helical strip or cored screw-type tube
inserts, twisted ducts, and various forms of altered
(tangential to axial direction) flow arrangements, and they
can be used for single-phase as well as two-phase flows
 Coiled tubes are what the name suggests, and they lead to
relatively more compact heat exchangers. The tube
curvature due to coiling produces secondary flows, which
promote higher heat transfer coefficients in single-phase
flows as well as in most regions of boiling.
PASSIVE TECHNIQUES


 Surface tension devices consist of wicking or grooved
surfaces, which direct and improve the flow of liquid to
boiling surfaces and from condensing surfaces
 Additives for liquids include the addition of solid
particles, soluble trace additives, and gas bubbles in
single-phase flows, and trace additives, which usually
depress the surface tension of the liquid, for boiling
systems
 Additives for gases include liquid droplets or solid
particles, which are introduced in single-phase gas flows
in either a dilute phase (gas–solid suspensions) or dense
phase (fluidized beds).
PASSIVE TECHNIQUES


 Mechanical aids are those that stir the fluid by
mechanical mean or by rotating the surface. The
more prominent examples include rotating tube heat
exchangers and scraped-surface heat and mass
exchangers
 Surface vibration has been applied primarily, at
either low or high frequency, in single phase flows to
obtain higher convective heat transfer coefficients
Active techniques:-


 Fluid vibration or fluid pulsation, with vibrations ranging
from 1.0 Hz to ultrasound, used primarily in single-phase
flows, is considered to be perhaps the most practical type
of vibration enhancement technique
 Electrostatic fields which could be in the form of electric
or magnetic fields, or a combination of the two, from dc
or ac sources, can be applied in heat exchange systems
involving dielectric fluids. Depending on the application,
they can promote greater bulk fluid mixing and induce
forced convection (corona “wind”) or electromagnetic
pumping to enhance heat transfer
Active techniques


 Injection, used only in single-phase flow, pertains to the
method of injecting the same or a different fluid into the
main bulk fluid either through a porous heat transfer
interface or upstream of the heat transfer section
 Suction involves either vapor removal through a porous
heated surface in nucleate or film boiling, or fluid
withdrawal through a porous heated surface in single-
phase flow
 Jet impingement involves the direction of heating or
cooling fluid perpendicularly or obliquely to the heat
transfer surface. Single or multiple jets (in clusters or
staged axially along the flow channel) may be used in
both single-phase and boiling applications
Active techniques


 1.These techniques generally use simple surface or
geometrical modifications to the flow channel by
incorporation inserts or additional devices
 2. It does not need any external power input
 3.Insert manufacturing process is simple and these
techniques can be easily employed in an existing
heat exchanger
Why passive techniques


 4.Passive insert configuration can be selected
according to the heat exchanger working condition
 5.It can be used in design of compact heat
exchangers
 6.It is not only applicable in heat exchanger but also
in solar air heater and cooling of electronic
components(heat sink)
Why passive techniques

 Use of secondary heat transfer surface
 Disruption the laminar sub layer in the turbulent
boundary layer
 Disruption of the unenhanced fluid velocity
 Introduction secondary flows
 Promoting boundary –layer separation
 Enhancing effective thermal conductivity of the
fluid under static conditions
 Enhancing effective thermal conductivity of the
fluid under dynamic
 Delaying the boundary layer development
Mechanisms of augmentation of heat transfer

 Thermal dispersion
 Increasing the order of the fluid molecules
 Redistribution of the flow
 Modification of radiative property of the convective
medium
 Increasing the difference between the surface and
fluid temperature
 Increasing fluid flow rate passively
 Increasing the thermal conductivity of the solid
phase using special nanotechnology fabrications
Mechanisms of augmentation of heat
transfer


 Resistance to heat transfer should be minimized
 Contingencies should be anticipated via safety margins;
for example, allowance for fouling during operation
 The equipment should be sturdy
 Cost and material requirements should be low.
 Corrosion should be avoided
 Pumping cost should be kept low
 Space required should be kept low.
 Required weight should be kept low
DESIGN CONSIDERATIONS


 Power plant
 Air conditioning
 Refrigeration
 Process industry
 Solar water heater
 Shell and tube heat exchanger
 Nuclear reactor
application


 Design the smooth pipe and roughened pipe on
Gambit with given technical properties,
 R= 30mm L1=150 L2= 500mm L3=150mm a= 5mm
P= 30,60 ,120,180 and solve the problem at Fluent.
Smooth pipe


 L1: Adiabatic wall (Aluminium)
 L2: Constant heat flux(1500 W/m2) (Aluminium)
 L3: Adiabatic wall (Aluminium)
 Velocity inlet: For Velocity 0.5, 1.0 1.5, 2.0, 2.5 m/s
values analysis should be done seperately.
 Pressure outlet: Gauge Pressure 0 kPa
 Fluid: Air (ρ=1.225 kg/m3, Cp=1007 j/kgK,
k=0.0242 W/mK, μ=1.7894e-5 kg/ms)
Boundry Conditions

 Re=ρDU/μ
 Nu=h*D/k h=q/((Ty-Tb))
 q: Heat flux (W/m2)
 Ty: Area-weighted average wall temperature
 Tb: Area-weighted average test region temperature
(T1+T2)/2
 f=∆P/(1/2) ρU^2 ( L/D)
 ΔP: Pressure drop of test region(P1-P2)
 L: Length(L2)
 D: Diameter of pipe (2R)
 U: Mean velocity (V1+V2)/2
 THP=(〖Nu〗s⁄〖Nu〗r )/(fs⁄fr )^□(1/3)
 THP : Thermo-Hydraulic Performance
 s index: (smooth) r index: (roughened)
Expressions:

 FROM FLUENT
 u=(u1+u2)/2=(0.50000525+0.50000036)/2=0.500002805
 Re=(1.225*0.500002805*0.06)/2=2053.77
 Nu=(h*D)/K
 Tb=(T1+T2)/2=(301.17181+451.3706)/2=376.27
 h=q/(Ty-Tb )=1500/(632.99-376.27)=5.8429
 Nu=(5.8429*0.06)/0.0242=14.4866
 f=∆P/(□(1/2)*ρ*(0.500002805)^2*□(L/D))
 ∆P=P1-P2=0.280411-0.203235=0.077176
 f=0.077176/(□(1/2)*1.225*(0.500002805)^2*0.06)=0.604
SMOOTH PIPE: FOR VELOCITY 0.5

 FROM FLUENT
 u=(u1+u2)/2=(1.0000066+1.0000004)/2=1.0000035
 Re=(ρ*u*D)/μ=(1.225*1.0000035*0.06)/(1.7894e^(-5)
)=4107.5
 Tb=(T1+T2)/2=(300.15268+347.32898)/2=323.74
 h=q/(Ty-Tb )=1500/(374.38297-323.74)=29.619
 Nu=(h*D)/K=(29.619*0.06)/0.0242=73.435
 ∆P=P1-P2=1.31646-0.78725529=0.52920471
 f=∆P/(□(1/2)
ρ(u)^2*□(L/D))=0.52920471/(□(1/2)*1.225*(1.000003
5)^2*□(0.5/0.06))=0.103
-SMOOTH PIPE:- FOR VELOCITY 1.0

U=0.5 U=1.0 U=1.5 U=2.0 U=2.5
Re 2053 4107.5 6161.2 8215 10268.8
Nu 14.486 73.435 97.457 120.05 140.38
f 0.604 0.103 0.08 0.065 0.056
SMOOTH PIPE

 FROM FLUENT
 u=(u1+u2)/2=(0.5335362+0.50558513)/2=0.519560665
 Re=(ρ*u*D)/μ=(1.225*0.519560665*0.06)/(1.7894e^(-5)
)=2134.106
 Tb=(T1+T2)/2=(303.85016+405.08405)/2=354.467105
 h=q/(Ty-Tb )=1500/(370.85272-354.467105)=91.5437
 Nu=(h*D)/K=(91.5437*0.06)/0.0242=226.9678
 ∆P=P1-P2=11.311311-0.52985072=10.78146028
 f=∆P/(□(1/2) ρ〖*(u)〗
^2*□(L/D))=10.78146028/(□(1/2)*1.225*(0.519560665)^2*
□(0.5/0.06))=8.125949
ROUGHENED PIPE (PITCH/PIPE
LENGTH=0.5=30)FOR VELOCITY=0.5

 u=(u1+u2)/2=(1.0416229+1.0158457)/2=1.0287343
 Re=(ρ*u*D)/μ=(1.225*1.0287343*0.06)/(1.7894e^(-5)
)=4225.5488
 Tb=(T1+T2)/2=(301.69254+353.12662)/2=327.40958
 h=q/(Ty-Tb )=1500/(344.46912-327.40958)=87.92734
 Nu=(h*D)/K=(87.92734*0.06)/0.0242=218
 ∆P=P1-P2=19.574051-1.1792716=18.3947794
 f=∆p/(□(1/2)*ρ*(u)^2*□(L/D))=18.3947794/(□(1/2)*1.225*(
1.0287343)^2*□(0.5/0.06))=3.405
-ROUGHENED PIPE (PITCH/PIPE
LENGTH=0.5=30)FOR VELOCITY=1.0


U=0.5 U=1.0 U=1.5 U=2.0 U=2.5
Re 2134.10 4225.548 6339.8 8523.2 10697
Nu 226.967 218 199.076 194.055 199.88
f 8.12594 3.405 2.16318 1.61839 1.2454
diameter=0.5=30)

 FROM FLUENT:- P1=7.2058 P2=0.4881 u1=0.50002795
u2=0.50007921 Ty=360.847 T1=303.6272 T2=392.6641
 u=(u1+u2)/2=(0.50002795+0.50007921)/2=0.50005358
 Re=(ρ*u*D)/μ=(1.225*0.50005358*0.06)/(1.7894e^(-5) )=2053
 Tb=(T1+T2)/2=(303.6272+392.6641)/2=348.14565
 h=q/(Ty-Tb )=1500/(360.847-348.14565)=118.09
 Nu=(h*D)/K=(118.09*0.06)/0.0242=292.80
 ∆P=P1-P2=7.2058-0.4881=6.7177
 f=∆P/(□(1/2)*ρ*(u)^2*□(L/D))=6.7177/(□(1/2)*1.225*(0.5000
5358)^2*□(0.5/0.06))=5.2633
LENGTH=1=60) FOR VELOCITY=0.5

 FROM FLUENT:-P1=14.4194 P2=1.321674 u1=1.0000315
u2=1.0001024Ty=336.01 T1=300.73727 T2=346.61212
 u=(u1+u2)/2=(1.0000315+1.0001024)/2=1.00006695
 Re=(ρ*u*D)/μ=(1.225*1.00006695*0.06)/(1.7894e^(-5)
)=4107.79
 Tb=(T1+T2)/2=(300.73727+346.61212)/2=323.6746
 h=q/(Ty-Tb )=1500/(336.01-323.6746)=121.6
 Nu=(h*D)/K=(121.6*0.06)/0.0242=301.49
 ∆P=P1-P2=14.4194-1.321674=13.097726
 f=∆P/(□(1/2)*ρ*(u)^2*□(L/D))=13.097726/(□(1/2)*1.225*(1.
00006695)^2*□(0.5/0.06))=2.565


U=0.5 U=1.0 U=1.5 U=2.0 U=2.5
Re 2053 4107.79 6162.21 8217.84 10273
Nu 292.8 301.49 281.721 285.6 301.227
f 5.263 2.565 1.675 1.43243 1.3199
diameter=1=60)

 FROM FLUENT:-P1=5.0953159 P2=0.3721281 u1=0.53281337
u2=0.50563174 Ty=358.05173 T1=303.96246 T2=388.91751
 u=(u1+u2)/2=(0.53281337+0.50563174)/2=0.519222555
 Re=(ρ*u*D)/π=(1.225*0.519222555*0.06)/(1.7894e^(-5) )=2132.718
 T_B=(T1+T2)/2=(303.96246+388.91751)/2=346.439985
 h=q/(Ty-Tb )=1500/(358.05173-346.439985)=129.17955
 Nu=(h*D)/K=(129.17955*0.06)/0.0242=320.2798
 ∆P=P1-P2=5.0953159-0.35721281=4.73810
 f=∆P/(□(1/2)
ρ*(u)^2*□(L/D))=4.73810/(□(1/2)*1.225*(0.519222555)^2*□(0.5/0.
06))=3.4432

 FROM FLUENT:-P1=10.686235 P2=1.1565763 u1=1.0408984
u2=1.0172341 Ty=333.39319 T1=301.64767 T2=344.94949
 u=(u1+u2)/2=(1.0408984+1.0172341)/2=1.02906625
 Re=(ρ*u*D)/μ=(1.225*1.02906625*0.06)/(1.7894e^(-5)
)=4226.91
 Tb=(T1+T2)/2=(301.64767+344.94949)/2=323.29858
 h=q/(Ty-Tb )=1500/(333.39319-323.29858)=148.594
 Nu=(h*D)/K=(148.594*0.06)/0.0242=368.594
 ∆P=P1-P2=10.686235-1.1565763=9.5296587
 f=∆P/(□(1/2)*ρ*(u)^2*□(L/D))=9.5296587/(□(1/2)*1.225*(1.0
2906625)^2*□(0.5/0.06))=1.763


U=0.5 U=1.0 U=1.5 U=2.0 U=2.5
Re 2132.71 4226.9 6370.3 8572 10770
Nu 320.279 368.41 366.22 368.96 376.68
f 3.4432 1.763 1.2396 1.0306 0.9215
diameter=2=120)

 FROMFLUENT:-P1=3.43750 P2=0.45387 u1=0.50002003
u2=0.50008273 Ty=356.07758 T1=304.05145 T=385.88266
 u=(u1+u2)/2=(0.50002003+0.50008273)/2=0.50005138
 Re=(ρ*u*D)/μ=(1.225*0.50005138*0.06)/(1.7894e^(-5)
)=2053.97
 Tb=(T1+T2)/2=(304.05145+385.88266)/2=344.967055
 h=q/(Ty-Tb )=1500/(356.7758-344.967055)=131.8
 Nu=(h*D)/K=(131.8*0.06)/0.0242=326.787
 ∆P=P1-P2=3.4375029-0.45387191=2.98363099
 f=∆P/(□(1/2)*ρ*(u)^2*□(L/D))=2.98363099/(□(1/2)*1.225*(0.
50005138)^2*□(0.5/0.06))=2.3377

 FROM FLUENT:- P1=7.174204 P2=1.201369 u1=1.0000595
u2=1.0001125 Ty=331.98926 T1=300.78568 T2=343.16086
 u=(u1+u2)/2=(1.0000595+1.0001125)/2=1.000086
 Re=(ρ*u*D)/μ=(1.225*1.000086*0.06)/(1.7894e^(-5)
)=4107.875
 Tb=(T1+T2)/2=(300.78568+343.16086)/2=321.97327
 h=q/(Ty-Tb )=1500/(331.98926-321.97327)=149.760
 Nu=(h*D)/K=(149,76*0.06)/0.0242=371.307
 ∆P=P1-P2=7.1742043-1.2013698=5.9728345
 f=∆P/(□(1/2)*ρ*(u)^2*□(L/D))=5.9728345/(□(1/2)*1.225*(1.
000086)^2*□(0.5/0.06))=1.16998


U=0.5 U=1.0 U=1.5 U=2.0 U=2.5
Re 2053.97 4107.875 6161.5 8215.22 10269
Nu 326.787 371.307 373.319 377.27 385.8
f 2.3377 1.16998 0.85634 0.71938 0.6283
diameter=3=180)

 For v=0.5 (roughened pipe P/D=0.5=30)
 THP=(14.4860⁄226.9678)/(0.604⁄8.125949)^□(1/3) =0.85
 For v=1.0 (roughened pipe P/D=0.5=30)
 THP=(73.435⁄218)/(0.103⁄3.405)^□(1/3) =1.08
 THP=(97.4577⁄199.076)/(0.08⁄2.16318)^□(1/3) =1.4
 THP=(120.059⁄194.055)/(0.065⁄1.61839)^□(1/3) =1.8
 THP=(140.38⁄199.88)/(0.056⁄1.2454)^□(1/3) =1.975
THP) FOR ROUGHENED PIPE P/D=0.5=30(
𝑇𝐻𝑃 =
𝑁𝑢 𝑠
𝑁𝑢 𝑟
𝑓 𝑠
𝑓 𝑟
1
3

U=0.5 U=1.0 U=1.5 U=2.0 U=2.5
𝑷
𝑫 = 𝟎. 𝟓 = 𝟑𝟎 0.85 1.08 1.4 1.8 1.975 THP
𝑷
𝑫 = 𝟏 = 𝟔𝟎 0.1 0.6 0.95 1.17 1.33 THP
𝑷
𝑫 = 𝟐 = 𝟏𝟐𝟎 0.08 0.5 0.66 0.81 0.94 THP
𝑷
𝑫 = 𝟑 = 𝟏𝟖𝟎 0.07 0.44 0.576 0.7 0.81 THP
Thermo-Hydraulic Performance


Contours of velocity magnitude(0.5
m/s)


Contours of velocity magnitude(2.5 m/s)


Contours of total temperature(0.5 m/s)


Contours of total temperature(2.5 m/s)


 Heat transfer be better when we use roughened pipe
 Increase in heat transfer companion with increase in
friction factor because of roughened surface
 The results show that better heat transfer when the
pitch/diameter is lower( (P/D)=0.5=30) and
velocity=2.5
CONCLUSIONS


Kind of passive techniques


 Augmentation of heat transfer, single
phase(www.thermopedia/content/574)
 Saunders, S.A.E “Heat Exchangers’’ John Wiley &
Sons, 2002.
reference

heat transfer enhancement in pipe flow

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