DOE Fundamentals Handbook, Thermodynamics, Heat Transfer and Fluid Flow
Thermodynamics Chapter 3- Heat Transfer
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Transcript of Thermodynamics Chapter 3- Heat Transfer
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Conduction 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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Conduction 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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Conduction 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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
SOLIDS :
Thermal Conductivity
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Multilayer Insulation
Alternate Layers of Metal and Non-Metal
Metal having higher Reflectivity.
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
LIQUIDS :
Thermal Conductivity
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
GASES :
Thermal Conductivity
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Thermal Conductivity
Comparison :
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.
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
ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Heat Transfer
S. Y. B. Tech. Prod Engg.
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
ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.
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
ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Heat Transfer
S. Y. B. Tech. Prod Engg.
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.
ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.
Thermal Radiation → 0.1 – 100 μm.
Visible Light Portion → 0.35 – 0.75 μm.
Radiation Heat Transfer
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Radiation 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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Radiation Properties
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Radiation Properties
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Radiation 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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Radiation 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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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.
Radiation Heat Transfer
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Heat Exchangers
Fluid A in
Fluid B in
RA RB RC
q
TA TB
iiAh
1
kL
rr io
2)/(ln
ooAh
1Ti To
Overall Heat Transfer Coefficient :
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Heat Exchangers
Overall Heat Transfer Coefficient :
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Types of Heat Exchangers
Shell-And-Tube Heat Exchanger :
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Types of Heat Exchangers
Shell-And-Tube Heat Exchanger :
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Types of Heat Exchangers
Shell-And-Tube Heat Exchanger :
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Types of Heat Exchangers
Shell-And-Tube Heat Exchanger :
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Types of Heat Exchangers
Miniature / Compact Heat Exchanger :
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Types of Heat Exchangers
Cross-Flow Heat Exchanger :
Heat Transfer
S. Y. B. Tech. Prod Engg.
Types of Heat Exchangers
Cross-Flow Heat Exchanger :
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
1 2A
T
Hot Fluid
Cold Fluid
Th1
Tc1
Th2
Tc2
dqTh
Tc
dA
dq = U dA (Th-Tc)
Log Mean Temperature Difference
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
1 2A
T
Hot Fluid
Cold Fluid
Th1
Tc1
Th2
Tc2
dqTh
Tc
dA
dq = U dA (Th-Tc)
Log Mean Temperature Difference
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Log Mean Temperature Difference
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Log Mean Temperature Difference
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Log Mean Temperature Difference
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Log Mean Temperature Difference
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Log Mean Temperature Difference
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Log Mean Temperature Difference
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Effectiveness-NTU Method
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Effectiveness-NTU Method
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Effectiveness-NTU Method
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Effectiveness-NTU Method
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Effectiveness-NTU Method
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.
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
S. Y. B. Tech. Prod Engg.
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
S. Y. B. Tech. Prod Engg.
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Conduction – Convection System
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
S. Y. B. Tech. Prod Engg.
Conduction – Convection System
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
S. Y. B. Tech. Prod Engg.
Conduction – Convection System
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.
ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
For CASE 3, Boundary Conditions are : θ = θ0 = (T - T∞) at x = 0.
dθ/dx = 0 at x = L.
Conduction – Convection System
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Conduction – Convection System
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Conduction – Convection System
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Viscous Flow
Development of Flow Regimes
Laminar Transition Turbulent
LaminarTransitionTurbulent
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Viscous Flow
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Viscous Flow
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Viscous Flow
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
Laminar Transition Turbulent
x
y
dy
du
Viscous Flow
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
Laminar Transition Turbulent
Thermal Boundary Layer thickness δT = Distance from surface in y – direction
where, ( T – Tw ) / ( T∞ - Tw ) = 0.99 or 99 %
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Similarly, net flow in the element in y – direction; dxdyy
u y
Laminar Boundary Layer on Flat Plate
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Laminar Boundary Layer on Flat Plate
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Laminar Boundary Layer on Flat Plate
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Laminar Boundary Layer on Flat Plate
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
Laminar Boundary Layer on Flat Plate
OR2 2
2 2x y
t t t tu ux y x y
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Laminar Boundary Layer on Flat Plate
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Laminar Boundary Layer on Flat Plate
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Prandtl No. can vary from 4 X 10-3 – 0.2 and 1.0 - 4 X 104
Liquid Metals Viscous Oils
Laminar Boundary Layer on Flat Plate
Gases have generally Pr = 0.7
u
T∞ δ δT
Pr < 1
δ = δT
Pr = 1
δδT
Pr > 1
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Integral Momentum and Energy Equations
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Integral Momentum and Energy Equations
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Integral Momentum and Energy Equations
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Integral Momentum and Energy Equations
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Integral Momentum and Energy Equations
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Integral Momentum and Energy Equations
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Laminar Forced Convection on Flat Plate
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Laminar Forced Convection on Flat Plate
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Neglecting as 3λ3 / 280 as very small term; and substituting for δ from; 1/2
4.64
(Re)x
Laminar Forced Convection on Flat Plate
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Laminar Forced Convection on Flat Plate
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
T
characterstic Linear Dimensionof the SystemNu
Equivalent conducting filmof thickness
Laminar Forced Convection on Flat Plate
θS
δT δT’
Tw
TS
Equivalent Conducting Film
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Laminar Forced Convection on Flat Plate
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Laminar Forced Convection in a Tube
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Laminar Forced Convection in a Tube
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Laminar Forced Convection in a Tube
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Laminar Forced Convection in a Tube
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Laminar Forced Convection in a Tube
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Laminar Forced Convection in a Tube
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Laminar Forced Convection in a Tube
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.
Shear Stress at Solid Surface
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
S. Y. B. Tech. Prod Engg.
Shear Stress at Solid Surface
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
S. Y. B. Tech. Prod Engg.
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Turbulent Flow : Comparing the Equations,
Simple Reynolds Analogy
P q
dTq C
dy
dudy
and
P qCq dT
du
( )w S w
Pw S
q T TC
u
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Simple Reynolds Analogy
( )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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Simple Reynolds Analogy
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Simple Reynolds Analogy
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Dimensional Analysis….Example
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Dimensional Analysis….Example
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Dimensional Analysis
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Empirical Relations for Forced Convection
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
All physical properties are to be evaluated at Arithmetic Mean Bulk Temperature, θm
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Empirical Relations for Forced Convection
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
All physical properties are to be evaluated at Arithmetic Mean Bulk Temperature, θm
except μw at Wall Temperature.
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Empirical Relations for Forced Convection
C. Turbulent Flow along Flat Plate :
1/3 0.80.036 Pr Re 18,700d dNu
All physical properties are to be evaluated at Arithmetic Mean Bulk Temperature, θm
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Empirical Relations for Forced Convection
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
All physical properties are to be evaluated at Arithmetic Mean Bulk Temperature, θm
Equation is for : 1. ( x / d ) >> 60.
2. 102 < (Red Pr) < 104
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Dimensional Analysis
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Dimensional Analysis
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Dimensional Analysis
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Formulae for Natural Convection
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.
All physical properties are to be evaluated at Average of Surface and Bulk Fluid
Temperature; which is the Mean Film Temperature.
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Formulae for Natural Convection
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
Formulae for Natural Convection
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
All physical properties are to be evaluated at Average of Surface and Bulk Fluid
Temperature; which is the Mean Film Temperature.
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
D. Approximate Formulae for Air :
Formulae for Natural Convection
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
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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 :
Formulae for Natural Convection
Heat Transfer
S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines
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