Download - MEC 3609 H T Conduction
MEC 3609: HEAT TRANSFER
Conduction
Convection
Radiation
What is Heat Transfer?
• Energy in Transit is called Heat.
• Heat Transfer is a process by which energy transfer takes place.
• A knowledge of heat transfer is necessary in order to evaluate cost, the feasibility, and the size of the equipment to transfer a specified amount of heat in a given time.
• For example, power generation, heating and cooling, and many industrial and domestic applications.
Conduction, Convection and
Radiation
Heat Transfer Conduction Heat Transfer:
Fourier's law of conduction, one dimensional heat conduction through composite wall, tubes and spheres.
Derivation of general transient conduction equation with a heat source. Steady state 1D conduction with and without energy generation; overall heat transfer coefficient, critical and economic thickness of insulation.
Extended Surfaces: derivation of fin equation for simpler cases, fin efficiency and effectiveness.
Unsteady heat conduction, lumped -system analysis, numerical method.
Convection Heat Transfer
Forced and free convection;
Laminar and turbulent flow;
Mass, Momentum and Energy Equation;
External and internal flows;
Heat Exchangers;
Boiling and condensation.
Radiation Heat Transfer
Laws of blackbody and gray body radiation; semi-
transparent and opaque material.
Intensity, emissive power, emittance, absorptance,
reflectance, transmittance; Shape factor.
Radiation exchange between blackbody and gray
surfaces; radiation shield.
Thermal Power Plant
Steam Turbine
Heat Exchangers
Basic Modes of Heat Transfer
• Three distinct modes of heat transfer: conduction,
convection and radiation.
• Heat transfer in a solid or fluid at rest takes place by
conduction.
• Heat transfer in a fluid in motion takes place by
convection.
• Both conduction and convection require a medium,
whereas, radiation heat transfer requires no medium.
Conduction Heat Transfer
• Conduction heat transfer
Qx A dT/dx
• Heat transfer in x direction is given by
Qx = -kA dT/dx
(Fourier Law of Conduction)
k, constant of proportionality, called thermal
conductivity, W/m.K
T
x
Conduction Heat Transfer
T
x
T
x
dT/dx >0
Q<0 dT/dx <0
Q>0
Sign Convention for the direction of heat flow
Thermal Conductivities
Materials k at 300K (W/m.K)
• Copper 399.0
• Aluminium 237.0
• Carbon steel, 1%C 43.0
• Glass 0.81
• Plastics 0.2 - 0.3
• Water 0.6
• Air 0.026
• Styrofoam 0.030
Convective heat transfer between a surface
and a fluid can be calculated by
Qc = hc A T, Newton Law of Cooling
Qc = rate of heat transfer, W
hc = average convective heat
transfer coefficient, W/m2.K
A = heat transfer area, m2
T = Temp diff between surface
(Ts) and the fluid (Tf), K
Convective Heat Transfer
Fluid W/m2K
• Air, free convection 6 - 30
• Superheated steam or
air, forced convection 30 - 300
• Oil, forced convection 60 - 1,800
• Water, forced convection 300 - 6,000
• Water, boiling 3,000 - 60,000
• Steam, condensing 6,000 - 120,000
Convective heat transfer
coefficient
Radiation Heat Transfer • The amount of radiation leaving a surface
depends on the absolute temperature (K)
and the nature of the surface.
• A perfect radiator (blackbody) emits
radiation from the surface at the rate
Qr = A1T14, W
A1 = surface area, m2
T1= surface temperature, K
= Stefan-Boltzmann constant
5.67x10-8,W/m2K4
Basic Equations of Heat
Transfer
• Conduction: Fourier Law of Conduction
Qx = -kA dT/dx, W
• Convection: Newton Law of Cooling
Qc = hc A T, W
• Radiation: Stefan-Boltzmann Equation
Qr = A1T14, W
Example 1: In a manufacturing plant, the walls and ceiling of an oven are made of 200 mm thick fire-clay brick having a thermal conductivity of 1.5 W/m.K. During steady-state operation, measurements reveal an inner surface temperature of 1200oC and an outer surface temperature of 200oC. The internal dimensions of the oven are as follows: Length = 4m, Width=3m and the Height =3 m. What is the rate of heat input required to maintain steady-state temperature inside the oven?
Solution of Example 1:
4 m
3 m 3 m
1200oC 200oC
200 mm
Tm = 0.5(1200+200)
= 700oC
km = 1.5 2(4x3+3x3)+4x3 = Internal HT area = 54 m2
Four walls Ceiling
Heat Flux, Q/A = -k dT/dx = - k (To - Ti)/t
= 1.5(W/m.K)[(1200 - 200)(K)]/0.20( m)
= 7500 W/m2
Rate of heat input required = 7500 x 54 = 405 kW
Example 2: An insulated pipe supplying steam from a boiler runs through a room where the air and walls are at 30oC. The outer diameter of the pipe is 100 mm and its surface temperature is 250oC. The natural convection heat transfer coefficient from the surface to the air is 20 W/m2K. Find the rate of heat transfer from the surface due to convection and radiation per unit length of pipe. For radiation heat loss, the outer surface of the pipe may be treated as blackbody surface.
Solution for Example 2:
Tœ=30oC
250oC Do = 100 mm
h=20 W/m2K
Heat loss due to convection, Q
= hA (To - Tœ),(W/m2K)(m2)(K);
= h( DoL) (To - Tœ)
Q/L = 1382.3 W/m = Qcon
Heat loss due to radiation,Q
= A (To 4 - Tœ
4 ), Temp, T in K
= ( DoL) (To 4 - Tœ
4 )
Q/L = (5.67x10-8,W/m2K4) (0.1m)( 5234 -3034)(K4)
= 1182.58, W/m = Qrad
Qtotal = Qcon + Qrad =1382.3 + 1182.58 = 2564.88, W/m
Example 3 • The temperature in a house located at latitude 40o N is
maintained at 23oC with a temperature controller. The temperature of the inner surfaces of walls, floors and the ceiling of the house are found to be at an average temperature of 12oC in winter and 27oC in summer. A person with an external body surface area of 1.2 m2 and temperature of 32oC remains in standing position for fifteen minutes inside the room, where the temperature is 23oC. Find the rate of radiation exchange between the person and surrounding surfaces.
Solution Example 3
The emittance of person (external surfaces of the skin) =0.95
The rate of radiation heat exchange is given by
)(44
surrss TTAQ
For summer,
))(300305)(2.1)(./1067.5)(95.0( 4442428 KmKmWxQ
= 35.78 W
For winter,
))(285305)(2.1)(./1067.5)(95.0( 4442428 KmKmWxQ
=132.9W
Note: T is the Absolute Temperature in K
Although the thermostat setting is same, one feels chilly in winter
and warm in summer.
Basic Equations of Heat
Transfer
• Conduction: Fourier Law of Conduction
Qx = -kA dT/dx, W
•Convection: Newton Law of Cooling
Qc = hc A T, W
•Radiation: Stefan-Boltzmann Equation
Qr = A1T14, W
Exercise-1A
• 1. In a cold climate, a house is heated either using electricity or gas or coal to maintain
the desired temperature. The roof of such a house is 5 m long and 7 m wide, and 0.20 m thick, and is made of concrete having a thermal conductivity of 0.8 W/m.K. On a winter night , the temperatures of the inner and outer surfaces of the roof are measured as 16oC and 2oC, respectively, for a period of 8 hours. Determine (i) the rate of heat loss through the roof and (ii) the cost of heat loss to the home owner if the cost of electricity is $0.17 per kWh.
• 2. An electrical heater, which consists of a rod 300 mm long and 10 mm in diameter, is
placed in room at 12oC in steady state operation. Heat is generated in the rod as a result
of resistance heating and the surface temperature is 140oC under steady state operation. The voltage drop and the current through the rod are measured and found to be 50 V and 2 A, respectively. Considering negligible heat losses by radiation, estimate the convective heat transfer coefficient between the outer surface of the rod and the air in the room
• 3. A blackbody at 25oC is exposed to solar radiation and the temperature increased to 95o C. Estimate the increase in radiation heat transfer.
Exercise-1A
• 1. In a cold climate, a house is heated either using electricity or gas or coal to maintain the desired temperature. The roof of such a house is 5 m long and 7 m wide, and 0.20 m thick, and is made of concrete having a thermal conductivity of 0.8 W/m.K. On a winter night , the temperatures of the inner and outer surfaces of the roof are measured as 16oC and 2oC, respectively, for a period of 8 hours. Determine (i) the rate of heat loss through the roof and (ii) the cost of heat loss to the home owner if the cost of electricity is $0.17 per kWh.
Exercise-1A
Brief outline of solution
• Question 1
• Heat loss from the roof
• q = kA[(T1 –T2)/t], where t = thickness of
in kW the roof
Amount of heat lost during 8 hours
Q = q x no of hours, kWh
Cost = (amount of energy in kWh) x(unit cost of
energy)
Exercise-1A
• Question 2
• An electrical heater, which consists of a rod 300 mm long and 10 mm in diameter, is placed in room at 12oC in steady state operation. Heat is generated in the rod as a result of resistance heating and the surface temperature is 140oC under steady state operation. The voltage drop and the current through the rod are measured and found to be 50 V and 2 A, respectively. Considering negligible heat losses by radiation, estimate the convective heat transfer coefficient between the outer surface of the rod and the air in the room.
•
Exercise-1A
Brief outline of solution
• Question 2
• Neglect radiation heat loss
• Under steady state operation, heat loss from the surface by convection equals energy generated within the rod due to resistance heating.
• q = energy generated = VI = (voltage drop, V) x
• (current, A), W
• = heat lost =hAs(Ts – Ta), W
• h = q /[As(Ts – Ta)] = ,W/m2 K
Exercise-1A
• Question 3
• A blackbody at 25oC is exposed to solar
radiation and the temperature increased to
95o C. Estimate the increase in radiation
heat transfer.
Exercise-1A
Brief outline of solution
• Question 3
• Calculate emissive power at both temperatures. Increase in radiant heat transfer is equal to the difference in emissive power.
• E1 = σ T14 , W/m2 E2 = σ T2
4 , W/m2
• Increase in radiant heat transfer = E2 –E1
SECTION 2 CONDUCTION EQUATIONS
Conduction equation is a mathematical tool.
Describe energy distribution within the body
of the material.
Derived by performing an energy
balance on an elemental volume.
Steady state problem - the temperature
of the body of the material is independent
of time.
Transient or unsteady problem - the
temperature of the material is a function
of time.
CONDUCTION EQUATIONS (contd)
• T = f(x,t) ...one dimensional,
transient problem
T = f(x) One dimensional,
steady state
• T = f(x,y,z,t) three dimensional,
transient problem
• T = f(r,,t) Cylindrical co-
ordinate, 2D, transient
• T = f(r,) 2D, steady
x
θ
r
1-D CONDUCTION EQUATIONS
y
x
z
dx
dy dz
x x
q q+dq
Figure 1.1: Control volume in rectangular co-ordinates
T(x,t)
Qx >>Qy, Qz
CONDUCTION EQUATIONS
Plane Wall
Considering temperature variation in the control
volume in x-direction only, T = f(x)
Rate of energy conducted into control volume
= Rate of energy conducted out of control volume
x
y
At x, heat conducted into the element
= Qx
=-kA dT/dx
At x+dx, heat conducted out of the element
= Qx+dx = -kA dT/dx
+[d(-kAdT/dx)/dx]dx
x = 0
T=To
x=L
T=TL
L
For steady state conduction,
Qx – Qx+dx =0
x x+dx
21
2
2
tan
0
0
CxCT
solution
tconsk
dx
Tdk
dx
dTk
dx
d
becomes
d2T/dx2=0
Integrate
twice
STEADY STATE CONDUCTIO0N
1-D system without heat generation
Rectangular Coordinates
Boundary Conditions:
1. T(0) = T0
2. T(L) = TL
Fluid Flow
at T , h
x
x=0 x=L
T0 TL
T
Heat Flux
W/m2K
T = C1x + C2
x = 0, T=To, C2 = To;
x = L, T=TL,
TL = C1 L + To C1 =( TL –To)/L
Hence, T = x ( TL –To)/L + To
(T – To)/(TL – To) = x/L
STEADY STATE CONDUCTIO0N
1-D system without heat generation
Rectangular Coordinates
Solution:
T(x) – T0
TL - T0
T0 TL
Fluid Flow
at T , h
x
x=0 x=L
T0 TL
T
Heat Flux
W/m2K
= x/L (2.3)
Q = -kA dT/dx
=kA(T0 - TL)/L
= (T0 - TL)/(L/ kA)
Rt.cond = L/kA (2.6)
Rt,cond
When convection at x=0 and x=L is
taken into account
1/hiA L/kA 1/hoA
Ti To TL T
Qc=hiA(Ti-T0)
= (Ti-T0)/(1/hiA)
STEADY STATE CONDUCTIO0N
1-D system without heat generation
Rectangular Coordinates-Composite walls
T1 T2 T3 T4 T
Fluid Flow
at T , h
x
T
Heat Flux
W/m2K
1 2 3 4
L1
L2 L3
q = Ttotal/ Rt
Rt = R1 + R2 + R3 + R4
R1 = L1/k1A1
R2= L2/k2A2
R3= L3/k3A3
R4= 1/hA
Q = hA(T4 - T)
R4= 1/hA
R1 = L1/k1A1 R2= L2/k2A2 R3= L3/k3A3 R4= 1/hA
q
Taking convection into account on
the left surface
• Qx=
Ti-T1 =Qx(1/hiA)
T1-T2= Qx(L1/k1A)
T2-T3 = Qx(L2/k2A)
T3-T4=Qx(L3/k3A)
T4-To =Qx(1/hoA)
Ti –To = Qx(1/hiA+L1/k1A+L2/k2A+L3/k3A+1/hoA) =∆T
Qx = UA ∆T where UA= 1/ Rtotal (2.13)
1 2 3
4
Ti To
Example 2.1B
• A double glazed window (height:1m and
width:1.5m consists of two 4mm-thick layers of glass (k=0.78 W/mK) separated by a 10mm thick stagnant air space (k=0.026 W/mK). Determine the steady rate of heat transfer through this double-glazed window. The temperature inside the room is maintained at 22oC while the ambient (outdoor) is 32oC. The convective heat transfer coefficient of the inner and outer surfaces of the window are hi=12 W/m2K and h0=48 W/m2K, respectively, which include the effect of radiation.
Solution Example 2.1B
Solution:
W
WK
WK2
mK2
mW
WK
WK2
mmKW
m
WK2
mmKW
m
WK2
mK2
mW
15.303317.0
2232
R
TQ
3317.0RR
0139.05.148
1
Ah
1R
0034.0RR
256.05.1026.0
001.0
AkR
0034.05.178.0
004.0
AkR
055.05.112
1
Ah
1R
totoal
total
o
0
13
2
22
1
11
i
i
l
l
l1 l3 l2
22oC 32
oC
Ri R1 R2 R3 Ro
Exercise 2A • 1. In an aluminum pan placed on a heater, heat is transferred
steadily to the boiling water. Find the outer surface temperature of the bottom of the pan and the boiling heat transfer coefficient under the following conditions:
• The inner surface temperature of the bottom of the pan: 108oC.
• Rate of heat transfer to the bottom of the pan : 600 J/s
• Thermal conductivity of the aluminum pan material : 237 W/m.K
• The diameter of the pan : 250 mm
• Temperature of water inside the pan : 95 oC
• The thickness of the pan material : 5 mm
•
• Ans: 942 W/m2K;108 oC
Exercise 2A-contd
• 2. The temperature inside a house is maintained at 20 oC with a gas heater, where the outside temperature is 10 oC on a winter night. Heat is lost through roof, 300 m2 in area and thickness 150 mm, to the surroundings air at 10 oC. The roof may be considered black with an emittance value of 1. The thermal conductivity of the roof material is 2 W/m.K. The roof exchange radiation with the sky at 260K. The convective heat transfer coefficients for the inner and outer surfaces of the roof are 5 and 12 W/m2K, respectively.
• Determine – the rate of heat loss from the roof;
– for a period of 14 hours, find the amount of money lost, if the energy cost is 2 cents/MJ.
• Ans:37.44 kW; $38.00
Steady state conduction
Cylindrical coordinate
Fig 2.4
Heat Trasnsfer from a
hollow cylinder
Steady state heat conduction
Cylindrical coordinate system
• For steady heat transfer in radial direction
• qr - qr+dr = 0
• qr – [qr+dr] = 0 (2.16)
• qr – [qr+(dqr/dr)dr]=0
• Fourier law of conduction gives
• qr = - kA (2.17)
•
•
•
dx
dT
where A = 2 r.L
L = length of cylinder
Substitution of equation (2.17) into equation (2.16) results in the following equation:
-(2 rL)kdr
dT + (2 rL)k
dr
dT + (2 kL)
d dTr dr
dr dr
=0
(2 kL)d dT
r drdr dr
=0
d
drrdT
dr
0
d
drrdT
dr
0 (2.18)
Integrating equation (2.18) twice gives
T= C1Ln r + C2
Boundary Conditions:
(1) T(ri) =Ti
(2) T(ro ) =To 21
1
1
0
CLnrCT
drr
CdT
Cdr
dTr
Integrate
dr
dTr
dr
d
21 CLnrCT
Boundary Conditions (BC)
1 ii TrT
2 oTrT
BC 1 gives 21 CLnrCT ii (1)
BC 2 gives 21 CLnrCT oo (2)
Subtract (2) from (1)
o
i
oioir
rLnCLnrLnrCTT 11 )(
Hence, )/(1
oi
oi
rrLn
TTC
Insert C1 in equation 1, giving
2
2
)/(
)/(
CLnrrrLn
TTT
CLnrrrLn
TTT
i
oi
oii
i
oi
oii
Hence, the solution for the local temperature, T(r) is
T(r T T T
Ln r ri
Ln r ri i o
i o
)
where,
CT T
Ln r r
i o
i o
1
and
C T T TLn r
Ln r ri i o
i
i o
2
Hence, heat flow through the wall
Q k A rd T
d rk rL
d T
d r
kLT T
L n r r
T T
L n r r
k L
T T
RR
L n r r
k L
r
i o
o i
i o
o i
i o
tt
o i
( ) ( )
;
2
2
2
2
Rr h L1
1 1
1
2
R
Ln r r
k L2
2 1
12
R
Ln r r
k L3
3 2
22
Lk
rrR
3
34
42
)/ln(
LhrR
o4
52
1
Rt = R1 + R2 + R3 + R4 + R5
Example 2.2B Steam at a temperature of 300oC flows through a
cast iron pipe (k=75 W/mK) whose inner and
outer diameters are 50mm and 55mm
respectively. The pipe is covered with 25mm-
thick glass wool insulation (k=0.05 W/mK).
Heat is lost to the surroundings at 30oC by
convection and radiation, with a combined heat
transfer coefficient of 25 W/m2K. The heat
transfer coefficient at the inner wall of the pipe
is 65 W/m2K. Find the rate of heat loss from the
steam per unit length of the pipe. Find the
temperature drop across the wall of the pipe and
the insulation. Also find (UA) for the pipe.
Example 2.2B: Solution
Ti = Fluid temperature inside the tube T1 = Temperature of the inner wall of the tube
T2 = Temperature of the outer wall of the tube
T3 = Temperature of the outer surface of
T3
To
T2 r2
r1
Ti T1 T2 T3 T0
R1
=1/h1A1
R2 R3 R4
T3
Ti
T1
T2
To
insulation
To = Surrounding fluid temperature
1i
1Ah
1R ;
Lk2
rrLnR
1
12
2
;
Lk2
rrLnR
2
23
3
; oo
4Ah
1R
Considering L=1m , A1 = 2r1 =
2(0.025) = 0.1571 m2
A0 = A3 = 2r3 = 2(0.0525) = 0.3299 m
2
WK
2mmKW
098.01571.065
1R1
WKmKW
0002.0752
255.27Ln
R 2
WKmKW
058.205.02
5.275.52Ln
R 3
WK1212.03299.0x25
1R 4
WK
RRtotal
2774.2
1212.0058.2002.0098.0
0
Q = steady state heat loss
= (300 – 30) / 2.2774 = 118.55W
Temperature drop across pipe wall (pw)
2pw QRT
= (118.55W)(0.0002 K/W)
= 0.02 oC
3insulation QRT
= (118.5W)(2.058 K/W)
= 243.97 oC
T1 T2
Q
R2
total
oi
R
TTTUAQ
, where oi TTT
Hence, totalR
1UA
mKW439.02774.2
1
Basic Equations of Heat
Transfer
• Conduction: Fourier Law of Conduction
Qx = -kA dT/dx, W
• Convection: Newton Law of Cooling
Qc = hc A T, W
• Radiation: Stefan-Boltzmann Equation
Qr = A1T14, W
Steady state conduction
Spherical coordinate
Steady state conduction
Spherical coordinate
dr
dTkAqr
drdr
dTkA
dr
d
dr
dTkAq drr
For steady state, qr – q r+dr = 0
For A=4πr2
04 2
dr
dr
dTkr
dr
d
For constant thermal conductivity
02
dr
dTr
dr
d
1-D system without heat generation
Spherical Coordinates
ro
ri
1-D system without heat generation
Spherical Coordinates
21)( C
r
CrT
Boundary Conditions:
1) T(ri) = Ti; 2
i
1
rC
CT i
2) T(ro) = To To = - 2
o
1
rC
C
Hence,
)/1()/1(
1
io
oi
rr
TTC
and
)/1()/1()/1(2
io
oi
iirr
TTrTC
]1[r
r
rr
r
TT
TT i
oi
o
oi
i
STEADY STATE CONDUCTIO0N
1-D system without heat generation
Summary of Results
Confuguration Heat flow rate q Thermal Resistance
Plane wall q kA
L T
L
kA
Hollow Cylinder
q
kL
Ln r r T
o i
2
Ln r r
kL
o i
2
Hollow sphere q kr r
r r T
i o
i o
4
r r
kr r
o i
i o
4
Class Test: upto this point
(Just before this slide)
Date of Class test:
Thursday 6 Sept 2007 at 5pm
Class Test: 20%
Lab: 20%
Final Exam:60%
1-D system without heat generation
Overall Heat Transfer Coefficient
q = UA (T)total (21)
Like Newton's law of cooling
q = hA (T)
both U and h have the same dimensions, W/m2K.
STEADY STATE CONDUCTIO0N
1-D system without heat generation
Overall Heat Transfer Coefficient For composite walls:
)/()/()/(
1
,
)/()/()/(
1
332211
321
333222111
kLkLkLU
AAAFor
AkLAkLAkLUA
STEADY STATE CONDUCTIO0N
1-D system without heat generation
Overall Heat Transfer Coefficient For composite cylinders, Figure 2.5
q UA T)T T
R t
(1
;where UAR t
1
(2.23)
and
LrhLk
rrLn
Lk
rrLn
Lk
rrLn
LrhR
oii
t
33
34
2
23
1
12
2
1
2
)/(
222
1
Ti
Steady state heat conduction
critical thickness of insulation
LrhkL
rrLn
TTq
o
io
i
2
1
2
q q LT T
Ln r r
k h r
i
o i
o
2
1
2
(2.24)
1/2hπro Ln(ro/ri)/2πk
Critical thickness of insulation
Resistance
Condition for optimum heat flow
The condition of optimum heat flow can be obtained by
differentiating equation (2.24) with respect to ro and setting
the resultant equation to zero.
This condition is h ro/k = 1.0, where h ro/k is known as
Biot number, Bi
Hence the optimum heat transfer occurs from the cylinder
when the Biot Number is 1 and the critical radius,
rcrit = k/h (2.25)
At Bi = 1, heat transfer rate reaches a maximum;
resistance to heat transfer is minimum.
At Bi < 1, addition of insulation increases heat
transfer rate. Electric cables are designed
for maximum heat dissipation, hence the
insulation value should be around the
critical value.
At Bi > 1, addition of insulation decreases heat
transfer rate. To reduce heat losses, the
insulation thickness should be much
greater than the critical thickness of
insulation.
Example 2.2C An electric wire, diameter d=3mm and length L=5m, is
tightly wrapped with a 2mm-thick plastic cover (thermal
conductivity k=0.15 W/mK). Measurements indicate that
a current of 10A passes through the wire causing a
voltage drop of 8V. The wire is exposed to an
environment at 32oC with a convective heat transfer
coefficient h=12 W/m2K. Determine the temperature at
the interface of the wire and the plastic cover in steady
operation. Also, evaluate whether doubling the thickness
of the plastic cover will increase or decrease this
interface temperature.
Example 2.2C - solution
Under steady operating condition, rate of heat transfer is equal to the heat generated within the wire.
Q = VI = (8V)(10A)
= 80 W
A2 = (2r2)L = 2(0.0035m)(5m)
= 0.1099 m2
WK2
mmKW758.0
)1099.0)(12(
1
hA
1R
2
con
mmKW 515.02
1535Ln
kL2
rrLnR 12
plastic
= 0.1798 WK
WKplasticcontotal RRRR 9378.01798.0758.0
r1
k
T1
T2
r2
T1 T2 Ta
Rplastic Rcon
Q
totalo1
total
o1 QRTTR
TTQ
WKW 9378.08032
= 107.02 oC
The critical thickness of insulation:
K
2mW
mKW
12
15.0
h
krcr
= 0.0125 m
= 12.5 mm
1-D steady state heat conduction
with energy generation
Considering an element in a wall with a uniformly
distributed heat source
in x-direction dxdx
dTkA
dx
d
dx
dTkA
dx
dTkAq
xnet )(
energy generated within the control volume = qg A dx
Energy balance gives, k 02
2
gqdx
Td
STEADY STATE CONDUCTIO0N 1-D system with heat generation
]
2 0Tq
k
g
Poisson’s Equation T1
qg
X= L
X= 0
x
insulated
Boundary Conditions
d2T/dx2 +qg/k = 0
1. T(0) = T1
2. q x=L= - k dT/dx at x=L
= 0 T(x) = - [qg/2k]x2 + C1x +C2
2
2
1
2
1 2
0
2
g
g
g
qd T
dx k
qdTx C
dx k
qT x C x C
k
1
1 2 1
2
1
(1) (0)
(2)( ) 0
;
2
x L
g
g g
T T
dT
dx
q LC C T
k
q q LT x T
k k
STEADY STATE CONDUCTIO0N
1-D system with heat generation
L
x
kT
xLq
T
TxT g
21
)(
11
1
The temperature distribution is parabolic
with x
and the maximum value occurs at the
insulated
surface, x = L.
k
LqTTLT
g
2)(
2
1max
Conduction in cylinders with
energy generation
1-D steady state conduction equation in cylindrical coordinate is given by
0 gdrrr QQQ
02
])(2)(2[)(2
grLdrq
drdr
dTr
dr
dkL
dr
dTrkL
dr
dTrkL
After rearrangements
0)(1
k
q
dr
dTr
dr
d
r
g
STEADY STATE CONDUCTIO0N 1-D system with heat generation
Cylindrical Coordinates
Current flowing through a wire
qg = I2R/V
d (rdT/dr)/rdr + qg/k = 0
T(r) = C1 Ln r - qgr2/4k +C2
ro
Boundary conditions:
1. T(ro) = To
2. dT/drr=o = 0
STEADY STATE CONDUCTIO0N 1-D system with heat generation
Cylindrical Coordinates
2
22
14
)(
oo
og
o
o
r
r
kT
rq
T
TrT
Maximum temperature occurs at the centre
T(o) = Tmax = To + qgro2/4k
STEADY STATE CONDUCTIO0N 1-D system with heat generation
Cylindrical Coordinates
Example 5:
A 2-kW resistance water heater is used to boil
water in a kettle. The cylindrical heating element
has a diameter of 5 mm and length 0.6 m, where
thermal conductivity, k= 15 W/m.K. If the outer
surface temperature is 110oC, estimate the
temperature at the centre of the element.
STEADY STATE CONDUCTIO0N 1-D system with heat generation
Cylindrical Coordinates
Solution:
qg = Qgen/Vele =2000(W)/π (0.0025)2(m2) 0.6 (m)
= 0.1697x109 W/m3
T(r=0) = To + qgro2/4k
=110+0. 1697x109 (W/m3) (0.0025)2(m2)
/4x15 (W/m.K)
= 127.7oC
STEADY STATE CONDUCTIO0N 1-D system without heat generation
Heat Transfer from Fins
• Heat transfer from a surface to air or gas is
rather slow due to lower heat transfer
coefficient.
• One of the ways of increasing heat transfer
is to provide additional surface area,
perhaps in the form of fins, which may be
of different designs and geometries.
STEADY STATE CONDUCTIO0N 1-D system without heat generation
Heat Transfer from Fins
Fins in Electronic Cooling
STEADY STATE CONDUCTIO0N 1-D system without heat generation
Heat Transfer from Fins
x x
-kA dT/dx -kA dT/dx
+d(-kAdT/dx) x/dx
hAs[T(x ])-T
qx qc qx+ x
Energy balance across
the element:
,)(
TxThAdxdx
dTkA
dx
d
dx
dTkA
dx
dTkA
qqq
s
xx
cxxx
or
d
dxkA
dT
dxh P T x T
( ) = 0 (2.34)
where A = Pdx P= Perimeter
As
STEADY STATE CONDUCTIO0N 1-D system without heat generation
Heat Transfer from Fins
02
2
2
mdx
dEquation representing
temperature distribution in
a straight fin of constant
x-section Where = T - T
m2 = h P/ kA
Solution, = C1 e-mx + C2 e mx
= C1 cosh mx + C2 sinh mx
STEADY STATE CONDUCTIO0N 1-D system without heat generation
Heat Transfer from Fins
Two boundary conditions are required:
1. x=0, = To - T
2. Other boundary condition depends
on the physical situation
Case 1: Very long fin, T x= T
Case 2: Negligible heat losses from the tip of the fin,
dT/dx] x=L= 0
Case 3: Fin of finite length, heat losses from the tip by
convection, -k dT/dx] x=L= h [ TL - T]
x
Different Boundary Conditions
STEADY STATE CONDUCTIO0N 1-D system without heat generation
Heat Transfer from Fins
Case 1. Long Fin
x , = To - T 0
x=0, = o
Temperature distribution,
= C1 e-mx + C2 e mx
C1 = o and C2 = 0
/ o = (T- T)/( To - T) = e -mx
Heat Transfer
dxhPQLx
x
0 = -kA d/dx| x=0
x
Q = kA om = o[hpkA] 1/2
STEADY STATE CONDUCTIO0N 1-D system without heat generation
Heat Transfer from Fins
Case 2. Fins with insulated tip
x
= C1 e-mx + C2 e mx
C1= o /(1+e -2mL)
C2= o /(1+e 2mL)
/ o = [e-mx /(1+e -2mL)] + [e mx /(1+e 2mL)]
Q = -kA d/dx| x=0
= kA om[1/(1+e 2mL) - 1/(1+e -2mL)]
= o [hpkA] 1/2 tanh (mL)
dT/dx] x=L= 0
x=0, T = To
= o
STEADY STATE CONDUCTIO0N 1-D system without heat generation
Heat Transfer from Fins
Case 3. Fins with convection at the tip
x
-kdT/dx]x=L
= hL(T x=L - T)
Temp Distribution:
/ o = (T- T)/( To - T) = A/B
where,
A = cosh m(L-x) +(h/mk) sinh m(L-x)
B = cosh mL + (h/mk) sinh mL
Heat Transfer:
Q = (hpkA)1/2 ( To - T) C/D
where,
C = sinh mL + (h/mk) cosh mL
D = cosh mL + (h/mk) sinh mL
STEADY STATE CONDUCTIO0N 1-D system without heat generation
Heat Transfer from Fins
Fin Efficiency:
A ratio of actual heat transfer to heat which would
be transferred if the entire fin were at the base
temperature
f = o (hpkA)1/2/(hpL o )
=(1/L)(kA/hp)1/2
=1/mL, for Long Fins
STEADY STATE CONDUCTIO0N 1-D system without heat generation
Heat Transfer from Fins
Effectiveness of Fins:
= ( Q with fin)/ (Q without fin)
= f Af h o /(hAb o )
= f Af h /(hAb)
= f Af/Ab = [kP/hA]0.5
where, Af = PL
Ab = A
STEADY STATE CONDUCTIO0N 1-D system without heat generation
Heat Transfer from Fins
Temperature Profile and Heat Transfer Equation
Straight Fins with constant area
Infinite fin (L) = 0
θ/ θb = e-mx θ = T - T
θb= Tb - T
qf = M, where M= (hpkAc)1/2 θb
Finite fin – no heat loss from the tip,
dθ/dx = 0 at x=L
θ/ θb = A/B, A=cosh m(L-x); B=cosh mL
qf = M tanh mL m2= hp/kAc
Temperature Profile and Heat Transfer Equation
Straight Fins with constant area
Finite fin – convective heat loss from the tip of
the fin
h(L) = -k d/dx x=0
θ/ θb = C/D, C= coshm(L-x)+(h/mk)sinh m(L-x)
D= coshmL+(h/mk)sinmL
qf = M /, = sinhmL+(h/mk)coshmL
= coshmL+(h/mk)sinmL
STEADY STATE CONDUCTIO0N 1-D system without heat generation
Heat Transfer from Fins
Example 2.3A:
A copper pin fin 2.5 mm in diameter protrudes from a
wall at 100oC into an air at 28oC. The heat transfer is
mainly by natural convection with a heat transfer
coefficient of 12 W/m2.K.
Calculate heat losses from the fin assuming-
i) the fin is infinitely long;
ii) the fin is 30 mm long and the tip of the
fin is insulated;
iii) the fin 30 mm long having convective
losses, with h = 12 W/m2.K
Solution 2.3 A
The fin temperature is a function of length and varies
between 100oC and 28
oC. We evaluate thermal
conductivity at an average temperature of 642
28100
oC,
k=396.0 W/mK.
PhkA = (0.0025m) x (12 W/mK)
x (396 W/mK) x (/4)(0.0025m)2
= 1.834x10-4
(W2/K
2)
phkA = 0.0135 W/K
i) Considering the fin infinitely long, L
PhkAQ 0 = To-T = 72
= 72 x 0.0135 W
= 0.9745 W
2.3A- (ii)
•
i) The tip of the fin is insulated, 0dx
d
Lx
mLtanhPhkAQ 0
= 0.9745 x 0.2058
= 0.2005 W
kA
hPm
963.6
)0025.0(4
x396
)0025.0(x12
2
mL = 6.963 x 0.03
= 0.2088
2.3A-(iii)
i) Convective heat transfer from the tip,
mLsinh
mkhmLcosh
mLcoshmk
hmLsinhPhkAQ 0
W205.0
02.1
2144.0x9745.0
21.0x00435.002.1
02.1x00435.021.0x9745.0
sinh mL = 0.21
cosh mL = 1.02
396x963.6
12
mk
h
= 0.00435
Example 2.3B
An array of 10 aluminum alloy fins, each 3 mm wide, 0.4 mm thick, and 40 mm long, is used to cool a transistor. When the base is at 67oC and the ambient is at 27oC, how much power do they dissipate if the combined convection and radiation heat transfer coefficient is estimated to be 8 W/m2K? The alloy has a conductivity of 175 W/m.K. The heat transfer from the tip of the fin is negligible. Also, find the efficiency and effectiveness of the fin.
Solution Outline For one fin:
A= (0.003)(0.0004)=1.2x10-6m2
P=2(0.003+0.0004)=6.8x10-3m
m2= hP/kA =(8.0W/m2.K)(6.8x10-3m) /(175W/m.K)(1.2x10-6m2)
=259 m -2
m = 16.1 m -1; mL= (16.1 m-1)(0.040 m)
= 0.644
PL=(6.8x10-3m)(0.040 m)= 2.72x10-4m2
1. Finite fin – negligible tip losses:
Qmax = h(PL)(To – Tα)
= (8 W/m2K)(2.72x10-4m2)(340-300)K
= 0.087 W for 10 fins, QT=0.87 W
η = (1/mL)tanh mL
= (1/0.644)tanh (0.644)
= 0.881
Q = ηQmax= 0.0764W per fin
QT= 10x Q=0.764 W
TRANSIENT HAET CONDUCTIO0N
General Heat Transfer Equation:
2T + qg/k = (1/)T/ t
Newton’s Law,
q/As =h (Ts - T) = T/(1/h)
Fourier Law,
q/A = -k dT/dx = T/(Lc/k)
Bi = Biot Number
= h Lc/k = (Lc/k)/(1/h) < 0.1
TRANSIENT HAET CONDUCTIO0N
Lumped Capacitance Method
q = hAs (T - T) = Vc dT/dt
dT/(T - T) = -(hAs/ Vc) dt
Integration between, t=0, T=Ti
and t=t, T =T
(T - T)/(Ti - T) =e - t
where, = (hAs/ Vc)
T
=density
c=sp ht
V=volume
T
TRANSIENT HAET CONDUCTIO0N
Lumped Capacitance Method
T
t
T
Heat transfer over a period of time t’
Qc = dtTThA
t
s )(
'
0
Qc = Vc (Ti - T)
[1 - exp(-hAs/ Vc)t’]
Qmax = Vc (Ti - T)
Qc/Qmax =1- exp(-hAs/Vc)t’
• Steel ball bearings are required to be subjected to heat treatment to obtain the desired surface characteristics. The balls are heated to a temperature of 650oC and then quenched in a pool of oil that has a temperature 55oC. The ball bearings have a diameter of 40 mm. The convective heat transfer coefficient between the ball bearings and oil is 300 W/m2.K. Determine
• a) the length of time that the bearings must remain in the oil before their temperature drops to 200oC,
• b) total amount of heat removed from each bearing during this time interval, and
• c) instantaneous heat transfer rate from the bearings when they are first placed in the oil and when they reach 200oC.
• The properties of steel ball bearings are as follows:
• k = 50W/m.K; α = k/ρcp =1.3x10-5 m2/s
• Outline of Solution:
• hAst/ρVcp =(hLc/k) (αt/Lc2) = (Bi) (Fo)
• Where, (Bi) = (hLc/k = h(ro/3)/k =0.04
• (Fo) = (αt/Lc2) = αt/(ro/3)2 = 0.293t
• a) (200-55)/(650-55) = e –(0.04)(0.293t)
• t = 120.5 s
• Find t from above eqn
• b) Q =hAs(Ti - T)[1- e -(Bi) (Fo) ] t/ (Bi) (Fo)
• = 5.79x104 J
• t = 0, Q = 897 W
• t = 120.5 s, Q = 218 W
TRANSIENT HAET CONDUCTIO0N
Lumped Capacitance Method
Example:
The hot plate of a cooker has a surface area of 0.05 m2 and
is made of steel (density:7,820 kg/m3) having a total weight
of 1.4 kg. The convection heat transfer coefficient is 17
W/m2K between the plate and its surroundings at 27oC.
How long, after being switched on, would the plate take to
attain a temperature of 117oC? The plate heater is rated at
500Watts and initially at the temperature of the
surroundings. The thermal conductivity of the plate is 17.3
W/m.K.
Solution
sg hAVq
dt
dVc
where = T - T
badt
d
where c
qa
g
Vc
hAb s
Separating the variable and integrating
t
oodtb
ba
bd
bta
baLn
sec1261
a
baLn
bt
where,
775.0c
qa
g
310318.1 xVc
hAb s
Lumped Capacitance Method
Example 2
• The temperature of a stream of natural gas flowing through a pipe at 100oC is to be measured by a thermocouple whose junction can be approximated as a 1-mm diameter sphere. The properties of the junction are as follows: k=35 W/mK, ρ=8500 kg/m3 and Cp = 320 J/kg.K. The convection heat transfer coefficient between the junction and the gas, h=210 W/m2K. The thermocouple is initially at 28oC. Determine the time constant of the thermocouple. Also, find the time taken to read 99% of the initial gas temperature difference.
Example 2- solution outline
3
4
2
1
/ 61.67 10
0.1
0.01
0.462
10
c
s
c
i
s
bt
i
V DL x m
A D
hLBi check
k
T T
T T
hAs b
Vc
T Te t s
T T