heat transfer & thermodynamics

Heat Transfer

S. Y. B. Tech. Prod Engg.

Heat Transfer

ME0223 SEM-IV Applied Thermodynamics & Heat Engines

Applied Thermodynamics & Heat Engines

S.Y. B. Tech.

ME0223 SEM - IV

Production Engineering

Heat Transfer

S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines

Outline

• One – Dimensional Steady State Heat Transfer by conduction through plane wall,

Radial Heat Transfer by Conduction through hollow Cylinder / Sphere. Conduction

through Composite Plane and Cylindrical Wall.

• Heat flow by Convection. Free and Forced Convection. Nusselt, Reynolds and Prandtl

Numbers. Heat Transfer between two fluids separated by Composite Plane and

Cylindrical wall. Overall Heat Transfer Coefficient.

• Heat Exchangers, types of Heat Exchangers, Log Mean Temperature Difference.

• Radiation Heat Transfer, Absorptivity, Reflectivity and Transmissivity, Monochromatic

Emissive Power, Wein’s Law, Stefan-Boltzman’s Law and Kirchoff’s Law.

Heat Transfer


Heat Transfer

HEAT TRANSFER is a science that seeks to predict the energy transfer that may

take place between material bodies, as a result of temperature difference.

Heat Transfer RATE is the desired objective of an analysis that points out the

difference between Heat Transfer and Thermodynamics.

Thermodynamics is dealt with equilibrium, and does not predict how fast the

change will take place.

Heat Transfer supplements the First and Second Laws of Thermodynamics, with

additional rules to analyse the Energy Transfer RATES.

Heat Transfer


Conduction Heat Transfer

Fourier Law :

Heat Transfer (HT) Rate per unit cross – sectional (c/s) area is proportional

to the Temperature Gradient.

x

T

A

q

x

TAkq

Q = HT Rate,

∂T/∂T = Temperature Gradient in the direction of Heat Flow.

k = Constant of Proportionality, known as THERMAL CONDUCTIVITY, (W/mºC)

NOTE : Negative sign is to indicate that Heat flows from High – Temperature to

Low – Temperature region, i.e. to satisfy Second Law of Thermodynamics.

Heat Transfer



Heat Conduction through Plane Wall :

qx qx+dx

x dx

A

qgen = qi A dx

Generalised Case :

1. Temperature changes with time.

2. Internal Heat Sources.

Energy Balance gives;

Energy conducted in left face + Heat generated within element

= Change in Internal Energy + Energy conducted out right face.

Energy in left face =

Energy generated within element =

Change in Internal Energy =

Energy out right face =

x

TAkqx

dxAqi

dxT

Ac

dxx

Tk

xx

TkA

x

TAkq

dxxdxx

Heat Transfer



Heat Conduction through Plane Wall :

qx qx+dx

x dx

A

qgen = qi A dx

Combining the terms;

dxx

Tk

xx

TkAdx

TAcdxAq

x

TAk i

T

cqx

Tk

x i

This is One – Dimensional Heat Conduction Equation

T

cqz

Tk

zy

Tk

yx

Tk

x izyx

For Constant Thermal Conductivity, kx = ky = kz = k

T

k

q

z

T

y

T

x

T i 12

2

2

2

2

2

Where, α = ( k / ρc ) is called Thermal Diffusivity. (m2/sec)

(↑) α ; (↑) the heat will diffuse through the material.

Heat Transfer



Heat Conduction through Cylinder :

T

k

q

z

TT

rr

T

rr

T i 1112

2

2

2

22

2

Heat Conduction through Sphere :

T

k

qT

r

T

rrT

rri 1

sin

1sin

sin

112

2

2222

2

Special Cases :

1. Steady State One – Dimensional (No Heat Generation) :

02

2

dx

Td

2. Steady State One – Dimensional, Cylindrical co-ordinates (No Heat Generation) :

01

2

2

dr

dT

rdr

Td

3. Steady State One – Dimensional, with Heat Generation :

02

2

k

q

dx

Td i

Heat Transfer


Thermal Conductivity

GAS : Kinetic Energy of the molecules of gas is transmitted from High –

Temperature region to that of Low – Temperature through continuous random

motion, colliding with one another and exchanging Energy as well as

momentum.LIQUIDS : Kinetic Energy is transmitted from High – Temperature region to that

of Low – Temperature by the same mechanism. BUT the situation is

more complex; as the molecules are closely spaced and molecular

force fields exert strong influence on the Energy exchange.

SOLIDS : (a) Free Electrons : Good Conductors have large number of free

electrons, which transfer electric charge as well as Thermal Energy.

Hence, are known as electron gas. EXCEPTION : Diamond !

(b) Lattice Vibrations : Vibrational Energy in lattice structure of the

material.

NOTE : Mode (a) is predominant than Mode (b).

Heat Transfer


SOLIDS :


Heat Transfer


Multilayer Insulation

Alternate Layers of Metal and Non-Metal

Metal having higher Reflectivity.

Heat Transfer


LIQUIDS :


Heat Transfer


GASES :


Heat Transfer



Comparison :

Heat Transfer


Conduction through Plane Wall

qx qx+dx

x dx

A

qgen = qi A dx

For Const. k; Integration yields ;

)( 12 TTx

kAq

For k with some linear relationship, like k = k0(1+βT);

)(

2)( 2

12

212 TTTTx

kAq

x

TAkq

Fourier’s Law, Generalised Form

Heat Transfer


Conduction through Composite Wall

CC

BB

AA x

TTAk

x

TTAk

x

TTAkq

342312

Since Heat Flow through all sections must be SAME ;

Thus, solving the equations would result in,

Akx

Akx

Akx

TTq

C

C

B

B

A

A

41

ELECTRICAL ANALOGY :

1. HT Rate = Heat Flow

2. k, thickness of material & area = Thermal Resistance

3. ΔT = Thermal Potential Difference.

cesisThermal

DifferencePotentialThermalFlowHeat

tanRe

th

overall

R

Tq

A B C

1 2 3 4

q q

RA RB RC

q

T1 T4

Ak

x

A

AAk

x

B

BAk

x

C

CT2 T3

Heat Transfer


Conduction through Composite Wall

RA

RB

RE

T1T2 T3

qRC

RD

RF

RGT4 T5

A

B

C

D

E

F

G

q

1 2 3 4 5

Heat Transfer


Example 1An exterior wall of a house is approximated by a 4-in layer of common brick (k=0.7 W/m.ºC) followed by a 1.5-in layer of Gypsum plaster (k=0.48 W/m.ºC). What thickness of loosely packed Rockwool insulation (k=0.065 W/m.ºC) should be added to reduce the Heat loss through the wall by 80 % ?

Overall Heat Loss is given by;

thR

Tq

insulationwithR

insulationwithoutR

insulationwithoutq

insulationwithq

th

th2.0Because the Heat loss with the Rockwool insulation will be only 20 %, of that before insulation,

WCm

k

xR

WCmk

xR

p

b

/.079.048.0

0254.05.1

/.145.07.0

0254.04

2

2

For brick and Plaster, for unit area;

So that the Thermal Resistance without insulation is; WCmR /.224.0079.0145.0 2


Heat Transfer


WCmInsulationwithR /.122.12.0

224.0 2 Now;

Example 1…contd

This is the SUM of the previous value and the Resistance for the Rockwool.

rwR 224.0122.1

065.0898.0

x

k

xRrw

inmxrw 30.20584.0 …ANS


Heat Transfer


Conduction through Radial Systems

Cylinder with;

1. Inside Radius, ri.

2. Outside Radius, ro.

3. Length, L

4. Temperature Gradient, Ti-To

5. L >> r; → Heat Flow in Radial direction only.

Area for Heat Flow;Ar = 2πrL

Boundary Conditions :

T = Ti at r = ri

T = To at r = ro

Solution to the Equation is;

io

oi

rr

TTLkq

/ln

2

Fourier’s Law will be, r

TLrk

r

TAkq r

2

RA

q

Ti

kL

rrR ioth 2

/ln

To

ri

ro

q

r

dr

Heat Transfer


Conduction through Radial Systems

Thermal Resistance is,

kL

rrR ioth 2

/ln

For Composite Cylinder;

BBA krrkrrkrr

TTLq

//ln//ln//ln

2

342312

41

For Spheres;

oi

oi

rr

TTkq

114

RA RB RC

q

T1 T4

Lk

rr

A2/ln 12

Lk

rr

B2/ln 23

Lk

rr

C2/ln 34T2 T3

T1

T3

T2

T4

R1

R3

R2

R4

q

AB

C

Heat Transfer


Example 2

T1=600 ºC

T3=100 ºC

T2R1

R3

R2

q

Stainless SteelAsbestos

A thick-walled tube of stainless steel (18% Cr, 8% Ni, k=19 W/m.ºC) with 2 cm inner diameter (ID) and 4 cm outer diameter (OD) is covered with a 3 cm layer of asbestos insulation (k=0.2 W/m.ºC). If the inside wall temperature of the pipe is maintained at 600 ºC, calculate the heat loss per meter of length and the tube-insulation interface temperature.

Heat flow is given by;

mWkrrkrr

TT

L

q

AS

/6802.0/2/5ln19/1/2ln

)100600(2

//ln//ln

)(2

2312

31

mWkrr

TT

L

q

A

/6802//ln

)(

23

32

This Heat Flow is used to calculate the tube-insulation interface temperature as;

T2 = 595.8 ºC…ANS

Heat Transfer


Critical Thickness of Insulation

Consider a layer of Insulation around a circular pipe.

Inner Temperature of Insulator, fixed at Ti

Outer surface exposed to convective environment, T∞

From Thermal Network;

hrk

rrTTL

q

o

io

i

1/ln2

Expression to determine the outer radius of Insulation, ro

for maximum HT;

2

2

1/ln

112

0

hrkrr

krkrTTL

dr

dq

o

io

ooi

o

which gives;

h

kr 0

q

Ti

kL

rr io

2/ln

Lhro21

T∞

Ti

h, T∞

R1

R2

Heat Transfer


Example 3

Calculate the critical radius of asbestos (k=0.17 W/m.ºC) surrounding a pipe and exposed to room air at 20 ºC with h=3 W/m2

. ºC . Calculate the heat loss from a 200 ºC,

5 cm diameter pipe when covered with the critical radius of insulation and without insulation.

cmmh

kr 67.50567.0

0.3

17.00

Inside radius of the insulation is 5.0/2 = 2.5 cm.

mWL

q/7.105

)0.3)(0567.0(1

17.0)5.2/67.5ln(

)20200(2

Heat Transfer is calculated as;

Without insulation, the convection from the outer surface of the pipe is;

mWTTrhL

qi /8.84)20200)(025.0)(2)(0.3())(2( 0


Heat Transfer


Example 3…contd

Thus, the addition of (5.67-2.5) = 3.17 cm of insulation actually increases the

Heat Transfer by @ 25 %.

Alternatively, if fiberglass (k=0.04 W/m.ºC) is employed as the insulation

material, it would give;

cmmh

kro 33.10133.0

0.3

04.0

Now, the value of the Critical Radius is less than the outside radius of the

pipe (2.5 cm). So, addition of any fiberglass insulation would cause a decrease

in the Heat Transfer.


Heat Transfer


Thermal Contact Resistance

Two solid bars in contact.

Sides insulated to assure that Heat flows

in Axial direction only.

Thermal Conductivity may be different.

But Heat Flux through the materials

under Steady – State MUST be same.

Actual Temperature profile approx. as

shown.

The Temperature Drop at Plane 2, the

Contact Plane is said to be due to

Thermal Contact Resistance.

A B

ΔxA ΔxB

q q

T2A

T2B

T3

T1

x

T

1 2 3

Heat Transfer


Thermal Contact ResistanceEnergy Balance gives;

B

BB

C

BA

A

AA x

TTAk

Ah

TT

x

TTAkq

322221

/1

Akx

AhAkx

TTq

B

B

CA

A

1

31

No Real Surface is perfectly smooth.

Surface Roughness is exaggerated.

HT at joints can be contributed to :

1. Solid – Solid conduction at spots of contact.

2. Conduction through entrapped gases through

the void spaces.

Second factor is the major Resistance to Heat

Flow, as the conductivity of a gas is much lower

than that of a solid.

T2A

T2B

T3

T1

x

T

1 2 3

A

B

Heat Transfer


Thermal Contact Resistance

Designating;

1. Contact Area, Ac

2. Void Area, Av

3. Thickness of Void Space, Lg

4. Thermal Conductivity of the Fluid in Void space, kf

Ah

TT

L

TTAk

AkLAkL

TTq

C

BA

g

BAvf

CBgCAg

BA

/12/2/222222

Total C/s. Area of the bars is A.

Solving for hc;

f

v

BA

BAC

gC k

A

A

kk

kk

A

A

Lh

21

Most usually, AIR is the fluid in void spaces.

Hence, kf is very small compared to kA and kB.

A

B

Lg

T2A

T2B

Heat Transfer


Example 4Two 3.0 cm diameter 304 stainless steel bars, 10 cm long have ground surface and are exposed to air with a surface roughness of about 1 μm. If the surfaces are pressed together with a pressure of 50 atm and the two-bar combination is exposed to an overall temperature difference of 100 ºC, calculate the axial Heat Flow and Temperature Drop across the contact surface. Take 1/hc=5.28 X 10-4 m2.ºC/W.

The overall Heat Flow is subject to three resistances,

1. One Conducting Resistance for each bar

2. Contact Resistance.

WCXkA

xRth /679.8

)103()3.16(

)4)(1.0(220

For the bars;

Contact Resistance; WCX

X

AhR

CC /747.0

)103(

)4)(1028.5(122

4

Total Thermal Resistance; WCRth /105.8747.0)679.3)(2(

Heat Transfer


Example 4…contd

i.e. 6 % of the total resistance.

Temperature Drop across the contact is found by taking the ratio of the Contact

Resistance to the Total Resistance.

CTR

RT

th

CC

0544.6)100(338.12

)747.0(…ANS

Overall Heat Flow is; WR

Tq

th

338.12105.8

100

…ANS

Heat Transfer


Radiation Heat Transfer

Thermal Radiation is the electromagnetic radiation as result of its temperature.

There are many types of electromagnetic radiations, Thermal Radiation is one of them.

Regardless of the type, electromagnetic radiation is propagated at the speed of light,

3 X 108 m/sec. This speed is the product of wavelength and frequency of the radiation.

c = λ υc = Speed of lightλ = Wavelength υ = Frequency

NOTE : Unit of λ may be cm, A˚ or μm.

Physical Interpretation :

Heat Transfer


Thermal Radiation → 0.1 – 100 μm.

Visible Light Portion → 0.35 – 0.75 μm.


3 2 1 0 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10 -11 -12

log λ, m 1 μm 1 A˚

Thermal Radiation

Radio Waves Infrared Ultraviolet

X-Rays

γ-Rays

Visible

Heat Transfer



Propagation of Thermal radiation takes place in the form of discrete quanta.

Physical Interpretation :

h is Planck’s Constant, given by;

h = 6.625 X 10-34 J.sec

A very basic physical picture of the Radiation propagation →

Considering each quantum as a particle having Energy, Mass and momentum.

Each quantum having energy; E = hυ

E = mc2 = hυ

c2

m = hυ

& Momentum = c (hυ) = hυ

c2 c

Radiation Heat TransferPhysical Interpretation :

Heat Transfer


Stefan – Boltzmann Law

Applying the principles of Quantum-Statistical Thermodynamics;

When Energy Density is integrated over all wavelengths;

Total Energy emitted is proportional to the fourth power of absolute temperature.

Eb = σT4

Where, k is Boltzmann’s Constant; 1.38066 X 10-23 J/molecule.K

Energy Density per unit volume per unit wavelength;

1

8/

5

kThce

hcu

Equation is known as Stefan – Boltzmann Law.

Eb = Energy radiated per unit time per unit area by the ideal radiator, W/m2

σ = Stefan – Boltzmann Constant; 5.667 X 10-8 W/m2.K4

Heat Transfer


Stefan – Boltzmann Law

Subscript b in this equation → Radiation from a Blackbody.

Materials which obey this Law appear black to the eye, as they do not reflect any

radiation.

Thus, a Blackbody is a body which absorbs all the radiations incident upon it.

Eb is known as the Emissive Power of the Blackbody.

i.e. Energy radiated per unit time per unit area.

NOTE : “Blackness” of a surface to Thermal Radiation can be quite deceiving.

e.g. Lamp-black…..and Ice….!!!

Eb = σT4 qemitted = σ.A.T4

Heat Transfer


Radiation Properties

Radiant Energy incident on a surface;

(a) Part is Reflected,

(b) Part is Absorbed,

(c) Part is Transmitted.Incident Radiation Reflection

Transmission

Absorption

Reflectivity = ρ = Reflected Energy

Incident Energy

Absorptivity = α = Absorbed Energy

Incident EnergyAbsorptivity = α =

Absorbed Energy

Incident Energy

Incident EnergyTransmissivity = τ =

Transmitted EnergyIncident Energy

Transmissivity = τ = Transmitted Energy

ρ + α + τ = 1

NOTE : Most solids do not transmit Thermal Radiations. → ρ + α = 1

Heat Transfer



Types of Reflection :

1. Angle of Incidence = Angle of Reflection → Specular Reflection

2. Incident beam distributed uniformly in all directions → Diffuse Reflection

Source

Diffuse Reflection

θ1 θ2

Source

Specular Reflection

Offers the Mirror Image to the Observer

Heat Transfer



No REAL surface cab be perfectly Specular or Diffuse.

Ordinary mirror → Specular for visible region; but may not be in complete spectrum

of Thermal Radiation.

Rough surface exhibits the Diffused behaviour.

Polished surface exhibits the Specular behaviour.

Heat Transfer


Kirchhoff’s Law

SampleEA qiAα

Black Enclosure

Assume perfect Black enclosure.

Radiant flux incident is qi W/m2.

A Sample body placed inside Enclosure, in

Thermal Equilibrium with it.

Energy Absorbed by the Sample = Energy emitted.

i.e. E A = qi A α …(I)

Replacing the Sample Body with a Blackbody;

Eb A = qi A (1) …(II)

Dividing (I) by (II);

E / Eb = α

Heat Transfer


Ratio of Emissive Power of a body

to that of a Perfectly Black body, at

the same temperature is known as

Emissivity, ε of the body.

Kirchhoff’s Law

SampleEA qiAα

Black Enclosure

ε = E / Eb

Thus;

ε = α

This is known as Kirchhoff’s Identity.

Heat Transfer


Gray Body

Gray Body can be defined as the body whose Monochromatic Emissivity, ελ is

independent of wavelength.

Monochromatic Emissivity is defined as the ratio of Monochromatic Emissive

Power of the body to that of a Blackbody at the SAME wavelength and

temperature.ελ = Eλ / Ebλ

Total Emissivity of the body is related to the Monochromatic Emissivity as;

0

dEE bAnd

4

0

TdEE bb

Thus, 4

0

T

dE

E

Eb

B

Heat Transfer


Gray Body

When the GRAY BODY condition is applied,

ελ = ε

A functional relation for Ebλ was derived by Planck by introducing the concept of

QUANTUM in Electromagnetic Theory.

The derivation is by Statistical Thermodynamics.

By this theory, Ebλ is related to the Energy Density by;4

cuEb

1/

51

2

TCb e

CE

Where, λ = Wavelength, μm

T = Temperature, K

C1 = 3.743 X 108 ,W.μm/m2

C2 = 1.4387 X 104, μm.K

Heat Transfer


Wien’s Displacement Law

A plot of Ebλ as a function of T and λ.

Peak of the curve is shifted to

SHORTER Wavelengths at

HIGHER Temperatures.

Points of the curve are related by;

λmax T = 2897.6 μm.K

This is known as Wien’s

Displacement Law.

Heat Transfer


Wien’s Displacement Law

Shift in the maximum point of the Radiation Curve helps to explain the change in colour

of a body as it is heated.

Band of wavelengths visible to the eye lies between 0.3 – 0.7 μm.

Very small portion of the radiant energy spectrum at low temperatures is visible to eye.

As the body is heated, maximum intensity shifts to shorter wavelengths.

Accordingly, first visible sight of increase in temperature of a body is DARK RED

colour.

Further heating yields BRIGHT RED colour.

Then BRIGHT YELLOW.

And, finally WHITE…!!

Material also looks brighter at higher temperatures, as large portion of the total

radiation falls within the visible range.

Heat Transfer



General interest in amount of Energy radiated from a Blackbody in a certain

specified Wavelength range.

The fraction of Total Energy Radiated between 0 to λ is given by;

0

0

0

0

dE

dE

E

E

b

b

b

b

Diving both sides by T5; 1/51

5 2

TC

b

eT

C

T

E

This ratio is standardized in Graph as well as Tabular forms; with parameters as;

1. λT

2. Ebλ / T5

3. Eb 0-λT / σT4

Heat Transfer



If the Radiant Energy between Wavelengths λ1 and λ2 are desired;

0

0

0

00

12

21

b

b

b

bbb E

E

E

EEE

NOTE : Eb 0-∞ is the Total Radiation emitted over all Wavelengths = σT4

Heat Transfer


EXAMPLE :

Solar Radiation has a spectrum, approx. to that of a Blackbody at 5800 K.

Ordinary window glass transmits Radiation to about 2.5 μm.

From the Table for Radiation Function, λT = (2.5)(5800) = 14,500 μm.K

The fraction of then Solar Spectrum is about 0.97.

Thus, the regular glass transmits most of the Radiation incident upon it.

On the other hand, the room Radiation at 300 K has λT = (2.5)(300) = 750 μm.K

Radiation Fraction corresponding to this value is 0.001 per cent.

Thus, the ordinary glass essentially TRANSPARENT to Visible light, is almost OPAQUE

for Thermal Radiation at room temperature.


Heat Transfer


Heat Exchangers

Overall Heat Transfer Coefficient :

TA

TB

T1

T2q

Fluid A

Fluid B

h1

h2

RA RB RC

q

TA TB

Ah1

1

kA

xAh2

1T1 T2

)()()( 222111 BA TTAhTTx

kATTAhq

Heat Transfer is expressed as;

Applying the Thermal Resistance;

)/1()/()/1(

)(

21 AhkAxAh

TTq BA

Overall Heat Transfer by combined

Conduction + Convection is expressed in

terms of Overall Heat Transfer Coefficient ,

U, defined as;

q = U A ΔToverall

)/1()/()/1(

1

21 hkxhU

Heat Transfer


Heat Exchangers

Fluid A in

Fluid B in

RA RB RC

q

TA TB

iiAh

1

kL

rr io

2)/(ln

ooAh

1Ti To


Hollow Cylinder exposed to

Convective environment on its inner

and outer surfaces.

Area for Convection is NOT same for

both fluids.

→ ID and thickness of the inner tube.

Overall Heat Transfer would be;

oo

i

ii

BA

AhLkrr

Ah

TTq

12

)/(ln1)(

0

Heat Transfer


Heat Exchangers


Overall Heat Transfer Coefficient can be based on either INNER side or OUTER area

of the tube.

Based in INNER Area;

oo

iii

i

i

hAA

LkrrA

h

U1

2)/(ln1

1

0

Based in OUTER Area;

o

io

ii

oo

hLkrrA

hAA

U1

2)/(ln1

1

0

In general, for either Plane Wall or Cylinder,

thRUA

1

Heat Transfer


Example 5Water flows at 50 °C inside a 2.5 cm inside diameter tube such that hi = 3500 W/m2.°C. The tube has a wall thickness of 0.8 mm with thermal conductivity of 16 W/m.°C. The outside of the tube loses heat by free convection with ho = 7.6 W/m2.°C.Calculate the overall heat transfer coefficient and heat air at 20 °C.

3 Resistances in series.

L = 1.0 mtr, di = 0.025 mtr and do = 0.025 + (2)(0.008) mtr = 0.0266 mtr.

WCAh

Rii

i /00364.0)0.1)(025.0()3500(

11

WCLk

ddR iot /00062.0

)0.1)(16(2

)025.0/0266.0(ln

2

)/(ln

WCAh

Roo

o /575.1)0.1)(0266.0()6.7(

11

Heat Transfer


This clearly states that the controlling resistance for the Overall Heat Transfer

Coefficient is Outside Convection Resistance.

Hence, the Overall Heat Transfer Coefficient is based on Outside Tube Area.

Example 5….contd

TAUR

Tq o

th

overall

0

CmW

RAU

th

./577.7

575.100062.000364.0)0.1)(0266.0(

11

2

00

….ANS

Heat Transfer is obtained by;

WTAUq oo 19)2050)(0.1)(0266.0()577.7( ….ANS

Heat Transfer


Heat Exchangers

Fouling Factor :

After a period, the performance of the Heat Exchanger gets degraded as;

1. HT surface may become coated with various deposits.

2. HT surface may get corroded due to interaction between fluid and material.

This coating offers additional Resistance to the Heat Flow.

Performance Degradation effect is presented by introducing Fouling Factor or

Fouling Resistance, Rf.

Fouling Factor, Rf is defined as;cleandirty

f UUR

11

Heat Transfer


Types of Heat Exchangers

Shell-And-Tube Heat Exchanger :

Heat Transfer



Miniature / Compact Heat Exchanger :

Heat Transfer



Cross-Flow Heat Exchanger :

Heat Transfer


Log Mean Temperature Difference

1 2A

T

Hot Fluid

Cold Fluid

Th1

Tc1

Th2

Tc2

dqTh

Tc

dA

1 2A

T

Hot Fluid

Cold Fluid

Th1

Tc1

Th2

Tc2

dqTh

Tc

dA

TEMPERATURE PROFILES :

Parallel Flow Counter Flow

Heat Transfer


1 2A

T

Hot Fluid

Cold Fluid

Th1

Tc1

Th2

Tc2

dqTh

Tc

dA

dq = U dA (Th-Tc)


q = U A ΔTm

U = Overall Heat Transfer Coefficient.

A = Surface Area for Heat Transfer

consistent with definition of U.

ΔTm = Suitable Mean Temperature

Difference across Heat Exchanger.

As can be seen, the Temperature

Difference between the Hot and Cold

fluids vary between Inlet and Outlet.

Average Heat Transfer Area for the

above equation is required.

Heat Transfer


1 2A

T

Hot Fluid

Cold Fluid

Th1

Tc1

Th2

Tc2

dqTh

Tc

dA

dq = U dA (Th-Tc)


Heat transferred through elemental area dA;

ccchhh dTCmdTCmdq

)( ch TTdAUdq

hh

h

Cm

dqdT

And;

cc

c

Cm

dqdT

cchh

chch

CmCmdqTTddTdT

11)(

Solving for dq;

dACmCm

UTT

TTd

cchhch

ch

11)(

Heat Transfer



1 2A

T

Hot Fluid

Cold Fluid

Th1

Tc1

Th2

Tc2

dqTh

Tc

dA

dq = U dA (Th-Tc)

Equation can be integrated between

conditions 1 and 2 to yield;

cchhch

ch

CmCmAU

TT

TT 11

)(

)(ln

11

22

Again;

)( 21 hhhh

TT

qCm

)( 12 cccc

TT

qCm

&

This substitution gives;

)/()(ln

)()(

1122

1122

chch

chch

TTTT

TTTTAUq

)/()(ln

)()(

1122

1122

chch

chchm TTTT

TTTTT

OR;

Heat Transfer



This Temperature Difference, ΔTm , is known as Log Mean Temperature Difference.

TsbothofRationatural

TTLMTD HEofendotherHEofendone

log

Main Assumption :

1. Specific Heats (Cc and Ch) of fluids do not vary with Temperatures.

2. Convective HT Coefficients (h) are constant throughout the Heat Exchanger.

Serious concerns for validity due to : 1. Entrance Effects.

2. Fluid Viscosity.

3. Change in Th. Conductivity.

Heat Transfer



Heat Transfer


Example 6Water at the rate of 68 kg/min is heated from 35 to 75 °C by an oil having specific heat of 1.9 kJ/kg.°C. The fluids are used in a counterflow double-pipe heat exchanger, and the oil enters in the exchanger at 110 °C and leaves at 75 °C. The overall heat transfer coefficient is 320 W/m2.°C. Calculate the heat exchanger area.

Total Heat Transfer is calculated by Energy absorbed by water;

kWMJTCmq www 5.189min/37.11)3575)(18.4)(68(

2

Since all fluid temperatures are known, LMTD can be calculated.

C

TTTT

TTTTT

chch

chchm

44.37)75110/()3575(ln

)75110()3575(

)/()(ln

)()(

1122

1122

1A

T

Hot Fluid

Cold Fluid

110 °C

75 °C 75 °C

35 °C

dqTh

Tc

dAAnd; mTAUq Yields;

23

82.15)44.37)(320(

105.189m

X

TU

qA

m

….ANS

Heat Transfer


Example 7In stead of double-pipe heat exchanger, of Example 6, it is desired to use a shell-and-tube heat exchanger with water making one shell pass and oil making two tube passes. Calculate the heat exchanger area assuming other conditions same.

T1 = 35 °C T2 = 75 °C t1 = 110 °C t2 = 75 °C

467.011035

11075

11

12

tT

ttP

143.111075

7535

12

21

tt

TTR

mTFAUq yields;

23

53.19)44.37)(8.0)(320(

105.189m

X

TFU

qA

m

….ANS

Correction Factor from Chart = 0.8

Heat Transfer


Effectiveness-NTU Method

LMTD approach is suitable when both the inlet and outlet temperatures are known,

or can be easily computed.

However, when the temperatures are to be evaluated by an iterative method,

analysis becomes quite complicated as it involves the Logarithmic function.

In this case, the method of analysis is based on the Effectiveness of the Heat

Exchanger in transferring the given amount of Heat.

Effectiveness of the Heat Exchanger is defined as;

TransferHeatPossibleMaximum

TransferHeatActualessEffectiven

Actual Heat Transfer is calculated by;

1.Energy lost by HOT fluid.OR

2.Energy gained by COLD fluid.

Heat Transfer



For Parallel-Flow Heat Exchanger; )()( 1221 cccchhhh TTCpmTTCpmq

For Counter-Flow Heat Exchanger; )()( 2121 cccchhhh TTCpmTTCpmq

Maximum possible Heat Transfer Maximum possible Temperature Difference

Difference in INLET Temperatures of Hot and Cold fluids.

Maximum Temperature Difference Minimum value.)( Cm

Thus;Maximum Heat Transfer is given by; )()( minmax inletinlet ch TTCpmq

The fluid may be Hot or Cold, depending on their respective mass flow rates

and Specific Heats.

)( Cpm

Heat Transfer



For Parallel-Flow Heat Exchanger;11

21

11

21

)(

)(

ch

hh

chhh

hhhh

h TT

TT

TTCpm

TTCpm

11

12

11

12

)(

)(

ch

cc

chcc

cccc

c TT

TT

TTCpm

TTCpm

For Counter-Flow Heat Exchanger;21

21

21

21

)(

)(

ch

hh

chhh

hhhh

h TT

TT

TTCpm

TTCpm

21

21

21

21

)(

)(

ch

cc

chcc

cccc

c TT

TT

TTCpm

TTCpm

Heat Transfer



Effectiveness, ε, can be derived in a different way;

1 2A

T

Hot Fluid

Cold Fluid

Th1

Tc1

Th2

Tc2

dqTh

Tc

dA

dq = U dA (Th-Tc)

hh

cc

cc

cchhch

ch

Cpm

Cpm

Cpm

AU

CpmCpmAU

TT

TT

1

11

)(

)(ln

11

22For Parallel-Flow Heat Exchanger;

hh

cc

ccch

ch

Cpm

Cpm

Cpm

AU

TT

TT1exp

)(

)(

11

22OR

If Cold fluid is min fluid;)( Cpm

11

12

ch

ccc TT

TT

Heat Transfer



ccchhh dTCpmdTCpmdq

We know;

)()( 1221 cccchhhh TTCpmTTCpm

)( 2112 cc

hh

cc

hh TTCpm

CpmTT

This yields;)(

))(/(

)(

)(

11

2211

11

22

ch

ccchhcch

ch

ch

TT

TTTCpmCpmT

TT

TT

c

hh

cc

ch

cccchhccch

Cpm

Cpm

TT

TTTTCpmCpmTT

11)(

)())(/()(

11

212111

)/(1

)/(1/exp1

hhcc

hhcccc

c

CpmCpm

CpmCpmCpmUA

Heat Transfer



It can be shown that the SAME expression results if Hot fluid is min fluid;

EXCEPT that and are interchanged.

)( Cpm

)( cc Cpm

)( hh Cpm

In a General Form;

where, C = ; defined as CAPACITY RATE.)( Cm

Similar analysis for Counter-Flow Heat Exchanger yields;

The group of terms, (UA/Cmin ) is known as Number of Transfer Units (NTU).

This is so, since it is the indication of the size of the Heat Exchanger.

Heat Transfer


Effectiveness-NTU MethodHeat Exchanger Effectiveness Relations :

C

CN

1

)]1(exp[1

)]1(exp[1

)]1(exp[1

CNC

CN

1N

N

22.01)exp(exp1

NnwhereCn

NCn1

1

)exp(1)exp(1

1

NNC

C

N

)]}1(exp[1){/1( NeCC

)]}exp(1)[/1(exp{1 NCC

1

2/12

2/122/12

])1(exp[1

])1(exp[1)1(12

CN

CNXCC

Ne1

Heat Transfer


Effectiveness-NTU MethodHeat Exchanger NTU Relations :

C

CN

1

])1(1ln[

1

1ln

1

1

CC

N

1

N

)1(ln

11ln CC

N

)]1(ln1[ln1

CC

N

2/12

2/122/12

)1(1)/2(

)1(1)/2(ln)1(

CC

CCXCN

)1(ln N

Heat Transfer


Example 8A cross-flow heat exchanger is used to heat an oil in the tubes (C=1.9 kJ/kg.ºC) from 15 ºC to 85 ºC. Blowing across the outside of the tubes is steam which enters at 130 ºC and leaves at 110 ºC with a mass flow rate of 5.2 kg/sec. The overall heat transfer coefficient is 275 W/m2.ºC and C for steam is 1.86 kJ/kg.ºC. Calculate the surface area of the heat exchanger.

kWTCmq sss 193)110130)(86.1)(2.5(

Total Heat Transfer is calculated from Energy Balance of Steam;

1A

T

Hot Fluid

Cold Fluid

130 °C

85 °C 110 °C

15 °C

dqTh

Tc

dA

∆Tm is calculated by treating as a Counter-Flow Heat Transfer;

CTm

9.66

)15110()85130(

ln

)15110()85130(

Heat Transfer


Example 8….contd

t1 and t2 represent unmixed fluid (i.e. Oil) and T1 and T2 represent the mixed fluid (i.e. Steam). Hence;

T1 = 130 ºC; T2 = 110 ºC; t1 = 15 ºC; t2 = 85 ºC

From LMTD Correction Chart; F = 0..97

609.015130

1585

11

12

tT

ttP 286.0

1585

110130

12

21

tt

TTRAnd

Heat Transfer Area is;

2

3

82.10

)9.66)(97.0)(275(

10193

m

X

TFU

qA

m

….ANS

Heat Transfer


Example 9Calculate the heat exchanger performance in Example 8; if the oil flow rate is reduced to half while the steam flow rate is kept constant. Assume U remains same as 275 W/m2.ºC.

Calculating the Oil flow rate;

sec/45.1)1585)(9.1(

10193 3

kgX

mTCmq oooo

New Flow rate is half of this value. i.e. 0.725 kg/sec.

We assume the Inlet Temperatures remain same as 130 ºC for Steam and 15 ºC for Oil.

Hence, )130()15( ,, sessoeoo TCmTCmq

But, both the Exit Temperatures Te,o and Te,s are unknown.

The values of R and P can not be calculated without these temperatures. Hence, ∆Tm can not be calculated.

ITERATIVE procedure MUST be used to solve this example.

However, this example can be solved with Effectiveness-NTU Approach.

Heat Transfer


Example 10

Solve Example 9 by Effectiveness-NTU Method.

For Steam; CkWCpmC sss

/67.9)86.1)(2.5(

For Oil; CkWCpmC ooo

/38.1)9.1)(725.0(

Thus, the fluid having minimum is Oil.)( Cpm

It is observed that unmixed fluid (i.e. Oil) has Cmin and mixed fluid (i.e. Steam) has Cmax.

143.067.9/38.1/ maxmin CC

156.21380/)82.10)(275(/ min CAUNTU

Heat Transfer


Example 10….contd

Using the Effectiveness, we can calculate the Temperature Difference for Oil as;

CTTo 5.95)15130)(831.0()( max

Hence; from the Table; we get;

831.0)]1)(143.0(exp[1)143/0/1(

)]}1(exp[1){/1(156.2

e

eCC N

Thus, the Heat Transfer is;

kWTCpmq ooo 132)5.95)(38.1(

Thus,

Reduction in Oil flow rate by 50 % results in reduction in Heat Transfer by 32 %

only.

Heat Transfer


Example 11Hot oil at 100 ºC is used to heat air in a shell-and-tube heat exchanger the oil makes 6 tube passes and the air makes one shell-pass. 2 kg/sec of air are to be heated from 20 ºC to 80 ºC. The specific heat of the oil is 2100 kJ/kg.ºC, and its flow rate is 3.0 kg/sec. Calculate the area required for the heat exchanger for U = 200 W/m2.ºC.

Energy Balance is;

CT

T

TCpmTCpmq

e

oe

aaaoo

27.80

)2080)(1009)(0.2()100)(2100)(0.3(

0,

,

0

We have;

CWCpmCC

CWCpmCC

aac

ooh

/2018)1009)(0.2(

/6300)2100)(0.3(

min

max

And; 3203.0

6300

2018

max

C

CC main

Heat Transfer


Example 11….contd

Effectiveness is; 75.0)20100(

)2080(

max

T

Tc

99.1

)3203.01(3203.01)75.0/2(

)3203.01(3203.01)75.0/2(ln)3203.01(

)1(1)/2(

)1(1)/2(ln)1(

2/12

2/122/12

2/12

2/122/12

X

CC

CCXCN

From the NTU Table;

Thus;

2min

min

146.22)200(

)2018()1949.2( m

U

CNTUA

C

AUNTU

….ANS

Heat Transfer


Convection Heat Transfer

u

u∞

T∞

Tw

q

Consider a heated plate shown in Fig.

Temperature of the plate is Tw and that of

surrounding is T∞

Velocity profile is as shown in Fig.

Velocity reduces to Zero at the plate surface as

a result of Viscous Action.

Since no velocity at the plate surface, Heat is transferred by Conduction only.

Then , WHY Convection ?

ANS : Temperature Gradient depends on the rate at which fluid carries away the Heat.

h is known as the CONVECTIVE HEAT TRANSFER COEFFICIENT. (W/m2.K)

Overall Effect of Convection is given by Newton’s Law of Cooling.

q = h A ΔT = h A (Tw - T∞)

Heat Transfer


Convection Heat Transfer

u

u∞

T∞

Tw

q

Convective Heat Transfer has dependence

on Viscosity as well as Thermal properties of

the fluid.

1) Thermal Conductivity, k

2) Specific Heat, Cp

3) Density, ρ

Heated plate exposed to room air; without any external source of motion of fluid, the

movement of air will be due to the Density Gradient.

This is called Natural or Free Convection.

Heated plate exposed to air blown by a fan; i.e. with an external source of motion of fluid.

This is called Forced Convection.

Heat Transfer


Convection Energy Balance on Flow Channel

Te Ti

q

m

The same analogy can be used for evaluating the

Heat Loss / Gain resulting from a fluid flowing

inside a channel or tube, as shown in Fig.

Heated wall at temperature Tw loses heat to the

cooler fluid through the channel (i.e. pipe).

Temperature rise from inlet (Ti) to exit (Te).

)()( ,, avgfluidavgwie TTAhTTCpmq

Te, Ti and Tfluid are known as Bulk or Energy

Average Temperatures.

Heat Transfer


Convection Boundary Condition

We know that;

)( TTAhq wconv

With Electrical Analogy, as in case of Conduction;

Ah

TTq wconv 1

)(

The term is known as Convective Resistance;Ah1

Heat Transfer


Conduction – Convection System

Heat conducted through a body, frequently needs to be removed by Convection process.

e.g. Furnace walls, Motorcycle Engine, etc.

Finned Tube arrangement is the most common for such Heat Exchange applications.

Consider a One – Dimensional fin.

Surrounding fluid at T∞.

Base of fin at T0.

Energy Balance of element of fin with thickness dx ;

Energy in left face = Energy out right face +

Energy lost by Convection

Base

dx

L

qx Qx+x

A

t

dqconv =h P dx (T-T∞)

Heat Transfer



Convection Heat Transfer; )( TTAhq wconv

Where, Area of fin is surface area for Convection.

Let the C/s. area be A and perimeter be P.

Energy in left face; x

dTq k A

dx

Energy out right face;

2

2x dxx dx

dT dT d Tq k A k A dx

dx dx dx

Energy lost by Convection; ( )convq h Pdx T T

NOTE : Differential area of the fin is the product of Perimeter and the differential

length dx.

Heat Transfer



Combining the terms; we get; 2

20

d T hPT T

dx kA

Let, θ = (T - T∞)2

20

d hP

dx kA

Let m2 = hP/kA

Thus, the general solution of the equation becomes;

1 2mx mxC e C e

One boundary condition is;

θ = θ0 = (T - T∞) at x = 0.

Heat Transfer



Other boundary conditions are;

CASE 1 : Fin is very long.

Temperature at the fin end is that of surrounding.

CASE 2 : Fin has finite length.

Temperature loss due to Convection.

CASE 3 : Fin end is insulated.

dT/dx = 0 at x = L.


For CASE 1, Boundary Conditions are : θ = θ0 = (T - T∞) at x = 0.

θ = 0 at x = ∞.

And, the solution becomes;0 0

mxT Te

T T

Heat Transfer


For CASE 3, Boundary Conditions are : θ = θ0 = (T - T∞) at x = 0.

dθ/dx = 0 at x = L.


This yields;0 1 2

1 20 ( )mL mL

C C

m C e C e

Solving for C1 and C2, we get;2 2

0

cosh[ ( )]

1 1 cosh ( )

mx mx

mL mL

e e m L x

e e mL

where, the hyperbolic functions are defined as;

sinh2

x xe ex

cosh

2

x xe ex

tanh

x x

x x

e exe e

Heat Transfer



Solution for CASE 2 is;

0

cosh[ ( )] ( / )sinh[ ( )]

cosh ( ) ( / )sinh ( )

T T m L x h mk m L x

T T mL h mk mL

All the Heat loss by the fin MUST be conducted to the base of fin at x = 0.

Thus, the Heat loss is;

0x dx

x

dTq k A

dx

Alternate method of integrating Convection Heat Loss;

0 0

( )L L

q h P T T dx h P dx

Heat Transfer



Application of Conduction equation is easier than that for Convection.

(0)0 0[ ]mq k A m e hP k A For CASE 1 :

0 2 2

0

1 1

1 1

tanh ( )

mL mLq k A m

e e

h P k A mL

For CASE 3 :

0

sinh ( ) ( / ) cosh ( )

cosh ( ) ( / )sinh ( )

mL h mk mLq h P k A

mL h mk mL

For CASE 2 :

Heat Transfer


Viscous Flow

A) Flow over a Flat Plate :

At the leading edge of the plate, a region develops, where the influence of Viscous Forces

is felt.

These viscous forces are described in terms of Shear Stress, τ, between the fluid layers.

Laminar Sublayer

u∞

u

u∞

u

Laminar Transition Turbulent

x

y

dy

du

Heat Transfer


Viscous Flow

Region of flow, developed from the leading edge, in which the effects of Viscosity are

observed, is known as Boundary Layer.

Shear Stress is proportional to normal velocity gradient.

dy

du

The constant of proportionality, μ, is known as dynamic viscosity. (N-sec/m2)

The point for end of Boundary Layer is chosen as the y co-ordinate where the velocity

becomes 99 % of the free – stream value.

Initial development of Boundary Layer is Laminar.

After some critical distance from leading edge, small disturbances in flow get amplified.

This transition is continued till the flow becomes Turbulent.

Heat Transfer


Viscous Flow

Development of Flow Regimes


LaminarTransitionTurbulent

Heat Transfer


Viscous Flow

Heat Transfer


Viscous Flow

xuxu Re

Transition from Laminar to Turbulent takes place when;

5105 Xxuxu

where,

u∞ = Free – Stream Velocity (m/sec)

x = Distance from leading edge (m)

ν = μ / ρ = Kinematic Viscosity (m2/sec)

This particular group of terms is known as Reynold’s Number;

and denoted by (Re).

It is a dimensionless quantity.

Heat Transfer


Viscous Flow

Range for Reynlold’s No. (Re) transition from Laminar to Turbulent lies between

2 X 105 to 106; depending on;

1. Surface Roughness. 2. Turbulence Level.

NOTE : Generally, Transition ends at twice the Re where it starts.

Reynolds Number (Re) =

Ratio of Momentum Forces ( α ρu∞2 ) to Shear Stress ( α μu∞ / x ) .

Heat Transfer


Laminar profile is approximately Parabolic.

Turbulent profile has a initial part, close to plate, is very nearly Linear.

This is due to the Laminar Sublayer that adheres to the surface.

Portion outside this Sublayer is relatively Flat.

Laminar Sublayer

u∞

u

u∞

u


x

y

dy

du

Viscous Flow

Heat Transfer


Viscous Flow

Physical mechanism of Viscosity Momentum Transfer

Laminar flow Molecules move from one lamina to another,

carrying Momentum α Velocity

Net Momentum Transfer from High Velocity region to Low Velocity Region.

Force in direction of flow, i.e. Viscous shear Stress, τ

Turbulent flow has no distinct fluid layers.

Macroscopic chunks of fluid, transporting Energy and Momentum,

in stead of microscopic molecular motion.

Larger Viscous shear Stress, τ

Rate of Momentum Transfer α Rate of movement of molecules α T

Heat Transfer


Viscous Flow

B) Flow through a Pipe :

Starting Length

Boundary LayerUniform

Inlet Flow

Fully Developed Flow

Laminar Sublayer

Turbulent Core

(A)Laminar Flow

(B) Turbulent flow

(A)Laminar Flow

(B) Turbulent flow

Heat Transfer


Viscous Flow

Boundary Layer develops at the entrance of the pipe.

Eventually, the Boundary Layer fills entire tube. The flow is said to be fully developed.

For Laminar flow, Parabolic velocity profile is developed.

For Turbulent flow, a somewhat blunter profile is observed.

Velocity profiles can be mathematically expressed as :

For Laminar flow :

r

y

r

y

u

u

m

2

For Turbulent flow :7

1

r

y

u

u

m

Heat Transfer


Viscous Flow

Reynolds Number is used as criterion to check for Laminar or Turbulent flow.

Range of Reynolds Number for Transition :

2300Re

dumd

4000Re2000 d

Heat Transfer


Viscous Flow

Mean Velocity, G can be defined as;

muA

mG

Reynolds Number, Re can also be written as;

dG

d Re

where, m = Mass Flow Rate (kg / sec);

um = Mean Velocity (m / sec);

A = Cross-Sectional Area (m2).

Continuity Equation for One-dimensional flow in a tube ;

Aum m

Heat Transfer


Inviscid Flow

No Real fluid is inviscid.

Practically, we can assume the flow to be inviscid for certain conditions.

Flow at a sufficiently large distance from the flat plate, can be assumed to be inviscid.

Velocity Gradients, normal to the direction of flow are very small.

Viscous – Shear Forces are also very small.

Balance of Forces on an element of Incompressible fluid

= Change in Momentum of fluid element; yields Bernoulli’s Equation as;

where, ρ = Fluid Density, (kg/m3)

P = Pressure at particular point in flow, (Pa)

V = Velocity of flow at that point, (m/sec)

.2

1 2

Constg

VP

Heat Transfer


Inviscid Flow

Bernoulli’s Equation is considered as Energy Equation due to ;

1. The term ( V2 / 2g ) ≡ Kinetic Energy.

2. The term ( P / γ ) ≡ Potential Energy.

For a Compressible fluid, Energy Equation should take into account :

1. Changes in Internal Energy, h

2. Corresponding changes in Temperatures, T.

Heat Transfer


Inviscid Flow

where, u = Internal Energy, (Joule)

Q = Heat added to Control Volume, (Joule)

W = Net Work done in the Process, (Joule)

v = Specific Volume of fluid, (m3/kg)

h is the Enthalpy of the state and is defined as;

Pvuh

One – Dimensional Steady – Flow Energy Equation :

Wg

VhQ

g

Vh

22

22

2

21

1

Heat Transfer


Compressible Fluid Flow

Gas Constant for a gas is given as;

R = R / M

where, M = Molecular Weight of the gas.

R = Universal Gas Constant = 8314.5 J/kg.mol.K

For Air, the Ideal Gas Properties are :

Cv, air = 0.718 kJ/kg.K Rair = 287 J/kg.K

Cp, air = 1.005 kJ/kg.K γair = Cp / Cv = 1.4

Δh = Cp ΔT Δu = Cv ΔT

Equations of State of the fluid ;

P = ρ R T

Heat Transfer


Thermal Boundary Layer

On similar lines of Velocity Boundary Layers, there exist Thermal Boundary Layers also.

Flow regions where the fluid temperature changes from the free – stream value to the

value at the surface.

T∞

T

T∞

T


Thermal Boundary Layer thickness δT = Distance from surface in y – direction

where, ( T – Tw ) / ( T∞ - Tw ) = 0.99 or 99 %

Heat Transfer


Laminar Boundary Layer on Flat Plate

Exact Solution of Laminar

Flow Convection over a flat

plate needs differential

equations of Momentum and

Energy of the flow to obtain

the Temperature Gradient in

the fluid at the wall, and

hence, Convection Coefficient.

Assumptions :

1. Steady – State conditions,

2. Unit Depth,

3. Fluid Densiy, ρ

xu

Pdy

xx

uu dx

x

PP dx dy

x

yu

yy

uu dy

y

xudxy

2

2x xu u

dx dx dyy y

Heat Transfer


Similarly, net flow in the element in y – direction; dxdyy

u y


Total net flow in x – direction must be Zero. 0

dydxy

u

x

u yx

ρ, dx and dy can not be Zero. Hence; 0

y

u

x

u yx

Thus, net flow in the element in x – direction; dydxx

ux

dyux dydxx

uu xx

andMass Flow Rate in and out in x – direction;

Heat Transfer



Equation of Momentum can be derived from :

1.Newton’s Second Law of Motion. 2. Viscous Shear Stress in y - direction is negligible.

3. Newtonian fluid. 4. Absence of Pressure-Gradient in y – direction.

Flow across horizontal faces also contribute to the Momentum Balance in x – direction;

For bottom face, Momentum Flow entering is;y x

y x

u uu dy u dx dx

y x

For bottom face, Momentum Flow entering is; .x yu u dx

Rates of Momentum Flow in x – direction for left and right hand vertical faces are;

anddxdyux2 dxdydx

x

uu xx

2

Heat Transfer



Viscous Shear Force on bottom face is;

xu dxy

Viscous Shear Force on top face is;

x xu udy dx

y y y

Net Viscous Shear Force in x - direction is;

2

2xu dy dxy

Heat Transfer



Pressure Forces on left and right hand faces are;

PdyP

P dx dyx

and

Net Pressure Force in x - direction is;

.Pdx dy

x

Sum of the Net Forces = Momentum Flow out of the Control Volume in x - direction

Thus, in x – direction; and neglecting second – order differentials;

2

2

x x xx y

u u u Pu u

x y y x

Heat Transfer


Laminar Boundary Layer on Flat PlateOn the similar lines to that with Momentum Equation; the Energy Equation can also be

derived as;

Energy Balance =

Rate of Net Conduction in

+ Rate of Net Convection in

= 0

p xC u t dy xp x

u tC u dx t dx dy

x x

tk dy

x

2

2

t tk dy dx

x x

p yC u t dx

yp y

u tC u dy t dy dx

y y

tk dx

y

2

2

t tk dx dy

y y

Heat Transfer


i.e. 2 2

2 2

0

x xx

y yy

t tk dx dy

x y

u ut tCp u t dx dy

x x x x

u ut tCp u t dx dy

y y y y


OR2 2

2 2x y

t t t tu ux y x y

Heat Transfer



Conduction in x – direction is very small and can be neglected.

2

2

t

x

2 2

2 2x y

t t t tu ux y x y

Similarly, Pressure Gradient in x – direction is also small and can be neglected.P

x

2

2

x x xx y

u u u Pu u

x y y x

Heat Transfer



Similarities between Momentum Equation and energy Equation ;

2

2x y

t t tu ux y y

Pr / p

p

Ck Diffusion of Momentum

C k Diffusion of Energy

This dimensionless Number that relates Fluid Boundary Layer and

Thermal Boundary Layer is known as Prandtl No. and denoted by ( Pr ).

2 2

2 2

x x x xx y

u u u uu u

x y y y

ν is the Kinematic Viscosity or Momentum Diffusivity = μ / ρ��

Heat Transfer


Prandtl No. can vary from 4 X 10-3 – 0.2 and 1.0 - 4 X 104

Liquid Metals Viscous Oils


Gases have generally Pr = 0.7

u

T∞ δ δT

Pr < 1

δ = δT

Pr = 1

δδT

Pr > 1

Heat Transfer


Integral Momentum and Energy Equations

Consider a Control Volume that extends

from wall to just beyond the limit of

Boundary Layer in y – direction.

Thickness dx in x – direction.

Unit depth in z – direction.

Equation to relate Net Momentum Outflow

to Net Force acting in x – direction.

A

BC

D

dx

x

Unit Width

y = δ

v s

v s

y

Heat Transfer



Momentum Flow across face AB : 2

0

xu dy

Momentum Flow across face CD : 2 2

0 0

.x x

du dy u dy dx

dx

Fluid also enters the control Volume

across face BD with rate of :0

.x

du dy dx

dx

= Fluid leaving across face CD –

Fluid entering across face AB

Fluid entering across face BD has Velocity us in x – direction.

Momentum Flow into Control Volume in x – direction : 0

.s x

du u dy dxdx

Heat Transfer



Net Momentum Outflow in x – direction :

0 0

. .x s x

d du dy dx u u dy dx

dx dx

Pressure Force will act on face AB and CD.

Shear Force will act on face AC.

No Shear Force will act on face BD since it is at limit of Boundary Layer.

0xu

y

Net Force acting on Control Volume in x – direction : x x

x x w w

P PP P dx dx dx dx

x x

Heat Transfer



Pressure Gradient can be neglected as it is very small, compared to other terms.

Equality of the Net Momentum Outflow to the Net Force in x – direction :

0

x s x w

du u u dy

dx

This is know as the Integral Momentum Equation in Laminar Boundary Layer.

Heat Transfer



Integral Energy Equation can be derived in a similar way.

Control Volume similar to that of

Integral Momentum Equation is

considered, but extending beyond

the limits of both Velocity

Boundary Layer and Thermal

Boundary Layer.

Principle of Conservation of Energy

applied involves :

1.Enthalpy and Kinetic Energy of

Fluid entering and leaving.

2.Heat Transfer by Conduction.

Kinetic Energy can be neglected

as it is very small.

A

BC

D

x+dx

x

Unit Width

v s

v s

y t w t S

Velocity Boundary Layer, δ

Thermal Boundary Layer, δT

Heat Transfer



Enthalpy Flow Rate across face AB :

0

sy

p xC u t dy

Enthalpy Flow Rate across face CD :

0 0

.s sy y

p x p x

dC u t dy C u t dy dx

dx

Fluid also enters the control Volume

across face BD at the rate :0

.sy

x

du dy dx

dx

= Flow rate out face CD –

Flow rate in face AB

Enthalpy Flow will be :

0

.sy

p s x

dC t u dy dx

dx

Heat Transfer



Heat Transfer by Conduction is :0y

tk dx

y

Beyond the Thermal Boundary Layer, the temperature is constant at ts.

Integration needs to be carried out only up to y = δT.

Above equation changes to :

0 0

0T

s x

y

d tt t u dy

dx y

This is know as the Integral Energy Equation in Laminar Boundary Layer.

Conservation of Energy gives :

0 0 0

. . 0s sy y

p s x p x

y

d d tC t u dy dx C u t dy dx k dx

dx dx y

Heat Transfer


Laminar Forced Convection on Flat PlateIntegral Momentum Equation and Integral Energy Equation are applied to solve the

equation for Laminar Forced Convection.

Analysis assumes uniform Viscosity with Temperature.

Apply Integral Momentum Equation to derive for Velocity Boundary Layer Thickness.

STEP 1 :

Velocity profile assumed to be : ux = a + by + cy2 + dy3

Constants a, b, c and d are found by applying known Boundary Conditions :

ux = 0 at y = 0. a = 0.

ux = us at y = δ.

0xu

y

at y = δ

ux and uy = 02

20xu

y

at y = 0

This yields; b = 3

2

us

δc = 0 d =

- 1

2

us

δ3

Heat Transfer


Laminar Forced Convection on Flat Plate

This yields;

33 1

2 2x

s

u y y

u

Apply Integral Momentum Equation,

STEP 2 :

0

x s x w

du u u dy

dx

3 32

0

3 1 3 1. 1

2 2 2 2s

d y y y yu dy

dx

0

x

y

du

dy

Heat Transfer



Wall Shear Stress is found by considering Velocity Gradient at y = 0 and is = 3

2

us

δ

2 39 3

280 2s

s

udu

dx

2 3 280

2 39s

s

uu d dx

140

13 s

d dxu

Integration yields;

2 140

2 13 s

xC

u

δ = 0 at x = 0.C = 0

2 280

13 s

x

u

OR1/2

4.64

(Re)x

Heat Transfer


Laminar Forced Convection on Flat PlateTemperature Distribution in the Thermal Boundary Layer can be found out in similar manner.

Applying the known Boundary Conditions and solving for the constants d, e and f ;

33 1

2 2s T T

y y

Temperature profile assumed to be : θx = ( t – tw ) = dy + ey2 + fy3

STEP 1 :

Apply Integral Energy Equation,

STEP 2 :

0 0

0T

s x

y

du dy

dx y

Heat Transfer



0

3

2s

Tyy

From the Temperature Distribution Equation;

3 3

0

3 1 3 1 3.

2 2 2 2 2

T

sS S S S

T T T T T

d y y y yu u dy

dx

This yields;

Substituting as λ = δT / δ; gives;

33 3 3

20 280 2s

S S TT

du

dx

Heat Transfer


Neglecting as 3λ3 / 280 as very small term; and substituting for δ from; 1/2

4.64

(Re)x


3/430.93

1Pr

hT x

x

where, xh is the length of the start of the heated section.

If the plate is heating along its entire length, xh = 0

1/3

0.93

PrT

OR 1/3

1

(Pr)T xh

δT

δ

x

y

Heat Transfer



Heat Transfer at the wall is :0

3

2S

wTy

q k ky

…..from Temperature

Distribution Profile

This Heat Transfer rate is expressed as and is the Heat Transfer

Coefficient, h

3

2w

S T

q k

The group is a dimensionless number.h x

k

This is known as Nusselt Number, and is denoted by ( Nu )

h xNu

k

Heat Transfer


T

characterstic Linear Dimensionof the SystemNu

Equivalent conducting filmof thickness


θS

δT δT’

Tw

TS

Equivalent Conducting Film

Heat Transfer



Thus, .w S ST T

k kq h h

T

h x xNu

k

This gives Nu as ;

1/2 1/3

1/3

3 3 (Re) (Pr)

2 2(0.93) (4.64)w

xS T

q x x xNu

k x

This gives Local Nusselt Number at some point x from the leading edge of the plate.

The average value of the Convection Coefficient, h over the distance of 0 to x is given by;

0

1 x

h hdxx

This gives ; 3/12/1 (Pr))(Re332.0 xxNu

Thus, Average Nusselt No. is;3/12/1 (Pr))(Re664.0 xxNu

Heat Transfer


Example 12Air flows at 5 m/sec. along a flat plate maintained at 77 °C. Bulk air temperature is 27 °C. Determine at 0.1 mtr from the leading edge the velocity an temperature boundary layer thickness and local as well as average heat transfer coefficient.

Bulk Mean Temperature = 77 27

52 3252

C K

𝜌 =1.084 kg / m3

k = 28.1 X 10-6 kW / m.K

μ = 1.965 X 10-5 Pa.sec

Pr = 0.703

1/20.1

0.1

4.64

Rexx

1/20.1

0.1

4.64(0.1) 2.789

(Re )mm …..Ans (i)

0.10.1

Rexu x

5

(1.087)(5)(0.1)27,659.03

1.965 10X

Heat Transfer


Example 12….cntd

1/30.1 0.1

1

(Pr)T

1/30.1

1(2.789) 3.134

(0.703)T mm …..Ans (ii)

1/2 1/3

0.1 0.10.1

0.332 Re Prx x

h xNu

k

1/2 1/3

0.1 0.10.1

0.664 Re Prx x

h xNu

k

KmkWX

Xh

./108.13

1.0

101.28)703.0()03.659,27(332.0)(

23

63/12/1

1.0

…..Ans (iii)

…..Ans (iv)KmkWX ./106.27 23

Heat Transfer


Laminar Forced Convection in a TubeCase for :

1. Fully Developed Flow, and

2. Constant Heat Flux.

Velocity Profile of Parabolic shape.

To derive Energy Equation for flow in tube :

Consider a cylindrical element of flow.

Length, dx

Inner Radius, r

Outer Radius, r + dr

Energy Flow into and out of the element in :

1. Radial direction by Conduction, and 2. Axial direction by Convection.

rdr

dx

Q x

Q x+dx

Qr

Qr+dr

Heat Transfer


Laminar Forced Convection in a Tube

Conduction into the element : r

TdxrkQr

2

With change of radius dr, Conduction rate : drr

Tr

rdxkdr

r

Qr

2

This change in Conduction rate = Difference between Convection rates into and

out of the element in Axial Direction.

Temperature changes in Axial Direction.

Rate of Convection into the element : TCudrr p2

Rate of Convection out of the element :

dxx

TTCudrr p2

Hence, the difference is : dxx

TCudrr p

2

Heat Transfer



Summation of total Forces is zero.

x

T

k

C

r

Tr

rrup

1

This is the ENERGY EQUATION FOR LAMINAR FLOW IN TUBES.

Assumptions :

1.Constant Heat Flow, qw,

2.Constant Fluid properties,

tConsx

Ttan

Temperature of the fluid (at any radius) must

increase linearly in the direction of flow.

Heat Transfer



equation reduces to Total Differential Equation.tConsx

Ttan

Velocity, u α Velocity at the Axis of tube, ua and Radius, r.

2

1

wa r

r

u

u….assuming Parabolic Distribution.

….rw = wall radius.

Boundary Conditions :

00

ratr

Tww rratTT

wr

w rratr

Tkq

w

1. 2.

3. Heat Flux is related to Temperature Gradient.

Heat Transfer



Applying the Boundary Conditions :

00

ratr

T1. C1 =0

ww rratTT 2. 16

31 2

2w

aw

ru

x

TTC

Thus, Equation becomes :

www

wa Tr

r

r

rru

x

TT

16

3

16

1

4

1142

2

Substituting this value in the

above Equation would give;r

r

ru

x

T

r

Tr

r wa

2

11

Integrating for two times,

would give :212

42

ln164

1CrC

r

rru

x

TT

wa

Heat Transfer



Equation can be expressed in terms of Temperature Difference : θ = T - Tw

θa = Temperature Difference between axis ( r = 0 ) and wall.

16

31 2waa ru

x

T

Temperature Profile can be expressed non – dimensionally as :

42

3

1

3

41

wwa r

r

r

r

Heat Transfer at wall α Temperature Gradient at r = rw

w

a

wwa

r rrrdr

d

w

3

4

3

4

3

8

Heat Transfer



On the similar lines;

aw

a

rw h

r

k

rkq

w

3

4

wr

kh

3

4

In terms of Nud : 3

82

3

4

k

r

r

k

k

dhNu w

wd

This analysis is based on the Temperature Difference between the Axis and the Wall.

From practical point of view, analysis for Temperature Difference between the

Bulk and the Wall is important.

Bulk Temperature = Mean Temperature of the fluid.

Heat Transfer



Temperature Difference based on

Bulk Temperature is given as;

w

w

r

p

r

p

m

Cudrr

Cudrr

0

0

2

2

Solution of the Equation is : am 72

44

Heat Transfer

at the wall is : mm

ww

a

rw h

r

k

r

k

rkq

w

44

72

3

4

3

4

44

72

3

4

wr

kh

Heat Transfer



In terms of Nud :

44

722

3

4

k

r

r

k

k

dhNu w

wd

44

72

3

8dNu

36.4dNu

NOTE : The Nu is INDEPENDENT of Re as the Fully Developed flow,

Boundary Layer Thickness = Tube Radius.

Heat Transfer


Reynolds Analogy

Till now, the analysis of Forced Convection for Laminar Flow is carried out.

Turbulent flow demands for introduction of additional terms into Momentum

Equation and Energy Equation to take into consideration the presence of Turbulence.

Demands for Numerical Solution for Finite Difference Equations.

Approach for Turbulent Flow Convection α Similarities between Equations for :

1. Heat Transfer and

2. Shear Stress (OR Momentum Transfer)

Original Idea of such Analogy is put forth by Reynolds;

and hence named after him.

Heat Transfer


Reynolds Analogy

Equation for Shear Stress in Laminar Flow is :dy

du

dy

du

where, ν is the Kinematic Viscosity.

Similar equation for Shear Stress in Turbulent Flow is :dy

duT

The term, ε is known as Eddy Diffusivity.

ε α Shear Stress due to Random Turbulent Motion.

Turbulent Flow Presence of Viscous Shear Stress also.

Total Shear Stress is : dy

du

ε is not a Property of fluid, like μ

ε α Re and Turbulence Level.

ε generally >> ν

Heat Transfer


Shear Stress at Solid Surface

Reynolds Analogy implies that;

Heat Transfer at surface of Flat Plate / Tube = Shear Stress acting on the surface

Shear Stress Substituting in Equation of τ0y

u

y

Thus, for Laminar Flow on Flat Plate,

x from leading edge,

Reynolds Number (Rex),

Free – Stream Velocity us

1/2

0.647

Ref

x

C Cf is known as Skin Friction Coefficient,

212

wf

S

Cu

For Laminar Flow, the average value of Cd for length x = 2 . Cf

Heat Transfer



For Turbulent Flow; 1/50.0583 Ref xC

And, 2.58

0.455

log Red

x

C

These are Empirical Correlations α Laminar + Turbulent portion of Boundary Layer.

The RatioVelocity at limit of Boundary Layer

Free – Stream Velocity α (Rex)

0.1

2.12

Reb

S x

u

u

Heat Transfer



Corresponding relationships for Flow in Tubes :

Expressed in terms of Friction Factor, f:2

4.4.

12

wf

m

f Cu

um = Mean Velocity of Flow.

Laminar Flow :64

Redf

Turbulent Flow :

1/4

0.308

Redf

1/8

2.44

Reb

S d

u

u for Smooth Surfaces.

NOTE : Values for Rough Surfaces are much higher.

Heat Transfer


Heat Transfer across Boundary Layer

Laminar Flow : Heat Transfer across flow α Only by Conduction.

Fourier’s Law : P

dTq C

dy

Turbulent Flow : Energy will also be transmitted through Random Turbulent Motion.

P q

dTq C

dy

εq is known as Thermal Eddy Diffusivity.

Heat Transfer


Basis for Reynolds Analogy

Heat Transfer at surface of Flat Plate / Tube = Shear Stress acting on the surface

Laminar Flow, compare

P

dTq C

dy

du

dy and

Turbulent Flow, compare

P q

dTq C

dy

dudy

and

We know, ν / α = Prandtl Number, (Pr)

Similarly, ε / εq = Turbulent Prandtl Number.

NOTE : This is NOT a Property of the fluid.

Heat Transfer


Assumptions for Reynolds Analogy

1. ε = εq . An eddy of fluid with certain Temperature and Velocity is transferred

to a different state, then it assumes its new Temperature and Velocity

in equal time.

This assumption is found practically valid as;

ε / εq varies between 1.0 and 1.6

2. q and τ have same ratio at all values of y .

True when Temperature Profile and Velocity Profile are identical.

i.e. Pr = 1...….Laminar Flow

ε ≈ εq …..Turbulent Flow Since, ν and α << ε and εq

1q q

Heat Transfer


Simple Reynolds Analogy

Flow is assumed to be Full laminar OR Full Turbulent with Pr = 1.

Laminar Flow : Comparing the Equations, P

dTq C

dy

du

dy and

q k dT

du

This gives q / τ at any arbitrary plane

= qw / τw at wall…..according to Assumptions.

( )w S w

w S

q T Tk

u

Heat Transfer


Turbulent Flow : Comparing the Equations,


P q

dTq C

dy

dudy

and

P qCq dT

du

( )w S w

Pw S

q T TC

u

Heat Transfer



( )w S w

w S

q T Tk

u

( )w S w

Pw S

q T TC

u

and are clearly identical if Pr = 1

i.e. μ.Cp / k = 1

i.e. Cp = k / μRearranging the terms;

w w P

S S

q Ch

u

where, θs = ( Ts – Tw )

2f

P s

Ch C u 2

f

P s

Ch

C u

This gives Convection Coefficient h in terms of

Skin Friction Factor, Cf

P s

h

C uis known as Stanton Number and denoted by ( St ).

is dimensionless.

Re.

NuSt

Pr

Heat Transfer



Further re-arranging the terms;

2f

P s

Cx xh C uk k

2

f sC u xx

hk

as μ.Cp / k = 1

i.e. Cp = k / μ

Re2f

x x

CNu

1/2

0.647

Ref

x

C We know, 1/20.323 Rex xNu

With Integral Boundary Layer Equations for Laminar Flow on Flat Plate,

1/2 1/30.332 Re PrxNu

With Pr = 1, 1/20.332 RexNu

Heat Transfer



For the Flow in Tubes;

θs and us are replaced by corresponding Mean Values, θm and um.

2f mC u dd

hk

Re2f

d d

CNu

1/44. 0.308 Ref df C We know,

0.750.038 Red dNu

Heat Transfer


Prandtl – Taylor ModificationPrandtl and Taylor had modified the Reynolds Analogy to take into account the

variation of Prandtl Number.

Prandtl – Taylor Modification is valid for 0.5 < Pr < 250.

Reynolds Analogy modified by Prandtl and Taylor :

1

1 Pr 1

w SP

bw S

S

qC

uuu

Introducing the term Cf :

1

2 1 Pr 1

fwS P

bS w

S

Cqu C

u

u

Heat Transfer


Prandtl – Taylor Modification

Turbulent Flow on Flat Plate :

4/5

1/10

0.0292 Re Pr

1 2.12 Re Pr 1x

x

x

Nu

Turbulent Flow in Round Tubes :

3/4

1/8

0.0386 Re Pr

1 2.44 Re Pr 1x

x

x

Nu

Heat Transfer


Example 13Compare the heat transfer coefficients for water flowing at an average fluid temperature of 100 °C, and at a velocity of 0.232 m/sec. in a 2.54 cm bore pipe; using simple Reynolds Analogy and Prandtl – Taylor Modification.

At 100 °C, Pr = 1.74, k = 0.68 kW / m.K, ν = 0.0294 X 10-5 m2 / sec.

0.750.038 Red dNu

Simple Reynolds Analogy gives :

62.5dNu

.Nu kh

d

3(62.5) (0.68 10 )

0.02541.675 / .

X

kW m K

…..Ans (i)

Reynolds Number is : Reu d

000,20100294.0

)0254.0)(232.0(5

X

Heat Transfer


Example 13…cntd.

With Prandtl – Taylor Modification :

3/4

1/8

0.0386 Re Pr

1 2.44 Re Pr 1d

d

d

Nu

3/4

1/8

0.0386 20,000 1.7472.4

1 2.44 20,000 1.74 1dNu

.Nu kh

d

3(72.4) (0.68 10 )

0.02541.973 / .

X

kW m K

…..Ans (ii)

Heat Transfer


Dimensional Analysis

Convection Heat Transfer Analysis Difficult to approach analytically.

Easy to deal with Dimensional Analysis +

Experiments.

Dimensional Analysis Equations in terms of important physical quantities in

dimensionless groups.

Given Process α n different physical variables.

Q1, Q2, Q3, ….Qn.

Composed of k independent dimensional quantities.

(e.g. Length, Mass, Time, etc.)

Heat Transfer


Dimensional AnalysisBuckingham’s Pi Theorem :

Dimensionally Homogeneous Equation α ( n – k ) dimensional groups.

f1 ( Q1, Q2, Q3, …..Qn ) = 0

Then, f2 ( π1, π 2, π 3, ….. π n-k ) = 0

Each term, π composed of Q variables in form;

π = Q1a, Q2

b, Q3c, …..Qn

x ) = 0 and is dimensionless.

Thus, a set of π terms includes all independent dimensionless groups.

No π term can be formed by combining other π terms.

Set of Equations for a, b, c, …..x by equating the sum of components of each

independent dimensions to Zero.

k Equations for n unknowns.

Heat Transfer


Dimensional Analysis….Example

Consider the Differential Equation for Momentum and Energy Transfer for Forced

Convection in Laminar Flow.

2

2

x x xx y

u u u Pu u

x y y x

2 2

2 2x y

t t t tu ux y x y

Dependent Variable : Convection Coefficient, h

Independent : 1. Velocity, u

Variables 2. Linear Dimension, l

3. Thermal Conductivity, k

4. Viscosity, μ

5. Specific Heat, Cp

6. Density, ρ

Independent : 1. Mass, M

Dimensional 2. Length, L

Quantities 3. Time, T

4. Temperature, θ

5. Heat, H…..Assumed.

6. H / θ in case of ( h, k & Cp)

Heat Transfer



Thus; 7 Physical Variables.

4 Independent Quantities. ( n – k ) = 3 π terms obtained.

4 Variables, which involve all 4 dimensions and DO NOT form any dimensionless group

within, are : u, l, k and μ.

hklu dcba 11111

Pdcba Cklu 2222

2

33333

dcba klu

The term π1 can be written as :

TL

H

TL

M

TL

HL

T

Ldc

ba

2

11

1

1

Heat Transfer


Dimensional Analysis….ExampleFollowing Equations for a1, b1, c1 and d1 can be obtained :

L : a1 + b1 – c1 – d1 – 2 = 0

T : – a1 – c1 – d1 – 1 = 0

H / θ : c1 + 1 = 0

M : d1 = 0 This implies : a1 = 0

b1 = 1

c1 = (-1)

d1 = 0

π1 term is = Nuk

lh

Heat Transfer



Thus, the result is :

φ2 ( Nu, Pr, Re ) = 0

Nu = φ ( Pr, Re )

This agrees with Reynolds Analogy, i.e.

Nu = f ( Pr, Re )

Similarly, π2 term is = Prk

CP

And, π3 term is = Re

lu

Heat Transfer



Scale Model Testing is the valuable practical application of use of such Dimensionless

Analysis.

With such models, the performance of the projected design can be estimated.

Pre – requisites : 1. Model must be geometrically similar to the full – scale design.

2. Re, Pr must be reproduced correctly.

This helps to predict : 1. Flow Patterns.

2. Thermal Boundary Layer.

3. Fluid Boundary Layer.

4. Nusselt Number ( Nu ).

Heat Transfer


Empirical Relations for Forced Convection

A. Laminar Flow in Tubes :

Average Nusselt Number at distance x from the entry is given by :

0.141/3

1/3 1/31.86 Re Prd d

w

dNu

x

All physical properties are to be evaluated at Arithmetic Mean Bulk Temperature, θm

except μw at Wall Temperature.

Equation is valid for Heating as well as Cooling, in the range,

1/3

1/3 1/3100 Re Pr 10,000d

d

x

Heat Transfer



B. Turbulent Flow in Tubes :

For 1. Fluids with Pr = 1

2. Moderate Temperature Difference between fluid and wall

( 5 °C for Liquids and 55 °C for Gases )

0.80.023 Re Pr

n

d dNu


n = 0.4…..Heating

= 0.3…...Cooling

Re 10,000d Equation is valid for :

Equation is for Fully Developed Flow, i.e. ( x / d ) >> 60.

Heat Transfer



B. Turbulent Flow in Tubes :

For larger Temperature Difference and wide range of Prandtl Numbers,

0.14

0.8 1/30.027 Re Prd d

w

Nu

Equation is valid for :

0.7 < Pr < 16,700


except μw at Wall Temperature.

Heat Transfer



C. Turbulent Flow along Flat Plate :

1/3 0.80.036 Pr Re 18,700d dNu


Equation is based on :

1. Laminar Flow, i.e.

2. Turbulent Flow after Transition at Re = 40,000.

3. 10 > Pr > 0.6

1/2 1/30.664 Re Prx xNu

Heat Transfer



D. Heat Transfer to Liquid Metals :

Liquid Metals very low Prandtl NumberFor 1. Turbulent Flow

2. Smooth Pipes / Tubes

a ) Uniform Wall Heat Flux : 0.40.625 Re Prd dNu

b ) Constant Wall Temperature : 0.85.0 0.025 Re Prd dNu


Equation is for : 1. ( x / d ) >> 60.

2. 102 < (Red Pr) < 104

Heat Transfer


Natural Convection

Energy Exchange between a body and an essentially stagnant fluid surrounding it.

Fluid Motion is due entirely to Buoyancy Forces caused by Density Variation of the fluid.

Natural Convection Object dissipating its Energy to the surrounding.

1. Intentional : Cooling of any Machine.

Heating of house or room

2. Unintentional : Loss through Steam Pipe.

Dissipation of warmth to the cold air

outside the window or room

Heat Transfer


Fluid Flow due to Natural Convection has both Laminar and Turbulent regimes.

Natural Convection

Boundary Layer produced has ZERO Velocity at both, 1. Solid Surface and

2. At the Outer Limit.

u = 0 u = 0Tw

Bulk Fluid Temperature

Velocity Distribution

Direction of induced Motion

Heat Transfer


Laminar Flow over Flat PlateSolution for Boundary Layer Momentum Equation and Energy Equation is possible

with introduction of a term known as Body Force.

This is then followed by Dimensional Analysis.

Body Force : ρs = Density of Cold Undisturbed Fluid.

ρ = Density of warmer fluid.

θ = Temperature Difference between the two fluid regimes.

( ).S g Buoyancy Force =

𝜌s is related to 𝜌 by : (1 )S β = Coefficient of Cubical Expansion of fluid.

1 . . . .g g Buoyancy Force =

Independent Variables for calculation of h

Addition of Buoyancy Force ( β, g, θ )

along with fluid properties ( , 𝜌 Cp, μ, k, and linear dimension l )

Heat Transfer



8 Physical Variables

5 Dimensionless Quantities 3 π terms are expected.

H and θ are not combined; as the Temperature Difference is now an important

Physical Variable.

5 Physical Variables selected common to all π terms are : ( 𝜌, μ, k, θ and l )

h , C� p and ( βg ) each appear in separate π terms.

1 1 1 1 11

a b c d ek l h

2 2 2 2 22

a b c d ePk l C

3 3 3 3 33

a b c d ek l g

Heat Transfer



Solving for a, b, c, d, and e as Constants, and substituting;

1

hlNu

k 2

PC Prk

2 3 3

3 2 2

g l g l

The Dimensionless Relationship obtained is :

( ,Pr, ) 0Nu Gr ( ,Pr)Nu GrOR

( )Buoyancy Force

Grashof Number GrShear Force

Buoyancy Force in Natural Convection ≡ Momentum Force in Forced Convection.

This π3 is known as Grashof Number and denoted by ( Gr ).

Heat Transfer



By experimental studies, it is found that,

( ,Pr)Nu Gr

is corrected to :

( ,Pr)bNu a Gr

where a and b are Constants.

This product, ( Gr . Pr ) is known as Rayleigh Number,

and denoted by ( Ra ).

Transition from Laminar to Turbulent Flow takes place in

the range of :

107 < ( Gr . Pr ) < 109

Heat Transfer


Formulae for Natural Convection

0.250.525( .Pr)d dNu Grwhen 104 < ( Grd . Pr ) < 109 (Laminar Flow)

0.330.129( .Pr)d dNu Grwhen 109 < ( Grd . Pr ) < 1012 (Turbulent Flow)

Below ( Grd . Pr ) = 104 ; No such relationship exists and Nu reduces to 0.4

With such low values of ( Grd . Pr ) the Boundary Layer Thickness becomes

appreciable as compared to the diameter.

In case of thin wires, Heat Transfer occurs in the limit by Conduction through the

stagnant film.

All physical properties are to be evaluated at Average of Surface and Bulk Fluid

Temperature; which is the Mean Film Temperature.

A. Horizontal Cylinder :

d

Heat Transfer



B. Vertical Surfaces :

l

l

Characteristic Linear Dimension is the Length or Height of the surface, l

Boundary Layer results from the vertical motion of the fluid.

Length of Boundary Layer is important than its Width.



0.250.59( .Pr)d lNu Grwhen 104 < ( Grd . Pr ) < 109 (Laminar Flow)

0.330.129( .Pr)d dNu Grwhen 109 < ( Grd . Pr ) < 1012 (Turbulent Flow)

Heat Transfer



C. Horizontal Flat Surfaces :

Fluid Flow is most restricted in case of horizontal surfaces.

Also, Heat Transfer Coefficient varies depending whether

the horizontal surface is above or below the fluid.

Square / Rectangular Surface up to l = 2 ft ( Mean Length of side )

For Cold fluid above Hot surface

OR Hot fluid below Cold surface

0.250.54( .Pr)d lNu Grwhen 105 < ( Grd . Pr ) < 108 (Laminar Flow)

0.330.14( .Pr)d dNu GrWhen ( Grd . Pr ) > 108 (Turbulent Flow)

Heat Transfer



C. Horizontal Flat Surfaces :

For Cold fluid below Hot surface

OR Hot fluid above Cold surface

Convective motion is surely restricted surface itself prevents

vertical motion.

Therefore, only Laminar Flow is possible with,

0.250.25( .Pr)d dNu GrWhen ( Grd . Pr ) > 105



Heat Transfer


D. Approximate Formulae for Air :


Convective Convection mainly deals with Air as a fluid medium;

Air properties do not vary greatly over limited temperature range.It is possible to derive simplified formulae for Air as :

13

132

1

tan

tan

bb

bb

b

Pb

lXtCons

lCg

ktConsh

It could be found out that

b = 0.25…..Laminar Flow

b = 0.33….Turbulent Flow

Index for l = ( -0.25 )….Laminar Flow

= 0 …..Turbulent Flow

25.0

l

Ch

….Laminar Flow33.0Ch ….Turbulent FlowAND

Heat Transfer


25.0

00131.0

d

h

33.0)(00124.0 h25.0

00141.0

l

h

33.0)(00131.0 h

25.0

00131.0

l

h

33.0)(00152.0 h25.0

00058.0

l

h

h is given in terms of ( kW / m2.K ) θ is given in terms of ( °C )

l, the linear dimension is given in terms of (m )

D. Approximate Formulae for Air :


Heat Transfer


Thank You !

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