heat transfer
DESCRIPTION
TRANSCRIPT
It is that area of mechanical engineering that dealswith the different principles and mechanisms involvedin transferring heat from one point to another.
Heat Transfer
ENGR. YURI G. MELLIZA
Modes of Heat Transfer1. Conduction: Is the transfer of heat from one point to another point within
a body or from one body to another body when they are physical contact with each other.
2. Convection: Is the transfer of heat from one point to another within a fluid.a. Natural or Free convection – motion of the fluid is due to the
difference in density because of a difference in temperature.b. Force Convection – motion of fluid is accomplished by
mechanical means, such as a fan or a blower.3. Radiation: It is the flow of heat from one body to another body separated
by a distance due to electromagnetic waves.
fire
Metal rod
t1
Hotter body
t2
Colder body
Conduction
Fluid
Convection
2
t2
t1surface
1
Radiation
Hot body
Cold body
Conduction
L1
2k
A
Q
t1
t2
kAL
tt
kL
ttAQ
LttkA
LttkA
Q
Law sFourier' From
2121
2112
)()(
)()(
Where: L – thickness, meters A – surface area, m2
k – thermal conductivity, Q – conductive heat flow, Watts
K-mW
or Cm
W
Thermal Circuit Diagram
1 2R
Q
kAL
R
K or C potential, etemperaturtWK
orWC
,resistanceR
Rtt
RΔt
kAL
)t(tQ 2112
)(
Conduction through a Composite Plane Wall
L1L2 L3
k1 k2 k3
1A 2
3
4
Q
3
3
2
2
1
1
41
3
3
2
2
1
1
41
3
3
2
2
1
1
41
321
41
3
3
2
2
1
1
41
kL
kL
kL
ttAQ
kL
kL
kL
ttA
kL
kL
kL
A1
ttQ
RRRtt
AkL
AkL
AkL
ttQ
)(
)()(
)()(
Thermal Circuit Diagram
12
R1
Q
43
R2 R3
AkL
R
AkL
R
AkL
R
3
33
2
22
1
11
A furnace is constructed with 20 cm of firebrick, k = 1.36 W/m-K, 10 cm of insulating brick, k = 0.26 W/m-K, and 20 cm of building brick, k = 0.69 W/m-K. The inside surface temperature is 650C. The heat loss from the furnace wall is 56 W/m2. Determine a. the interface temperature and the outside wall temperature, C b. the total resistance Rt, for 1 m2
L1 L2 L31
2
3
4
Q
A
1 2
R1
Q
43
R2 R3
Given: L1 =0.20 m ; L2 = 0.10 m ; L3 = 0.20 m k1 = 1.46 ; k2 = 0.26 ; k3 = 0.69 t1 = 650C Q/A = 56 W/m2
At 1 to 2
C7641361200
56650t
kL
AQ
tt
kL
ttAQ
2
1
112
1
1
21
..
.
)(
C2620260100
361200
56650t
kL
kL
AQ
tt
kL
kL
ttAQ
3
2
2
1
113
2
2
1
1
31
..
.
.
.
)(
At 1 to 3
At 1 to 4
C604690200
260100
361200
56650t
kL
kL
kL
AQ
tt
kL
kL
kL
ttAQ
4
3
3
2
2
1
114
3
3
2
2
1
1
41
.
.
.
.
.
.
)(
Convection
FluidA
1
2
Q
t2
t1
h
Watts
hA1
ttQ
Watts tthAQdirection) opposite in flows (heat ttIf
Watts tthAQt t If
12
12
21
21
21
)(
)(
)(
Where:Q – convective heat flow, WattsA – surface area in contact with the fluid, m2
h – convective coefficient, W/m2-C or W/m2-Kt1, t2 – temperature, C
Conduction from Fluid to Fluid separated by a composite plane wall
L1L2 L3
k1 k2 k3
1A 2
3
4
Q
i hi ti
o ho, to
o3
3
2
2
1
1
i
oi
o3
3
2
2
1
1
i
oi
o3
3
2
2
1
1
i
oi
o321i
oi
o3
3
2
2
1
1
i
oi
h1
kL
kL
kL
h1
ttAQ
h1
kL
kL
kL
h1
ttA
h1
kL
kL
kL
h1
A1
ttQ
RRRRRtt
Ah1
AkL
AkL
AkL
Ah1
ttQ
)(
)()(
)()(
Thermal Circuit Diagram
1 2R1
Q
43R2 R3
i o
Ri Ro
AkL
R
AkL
R
AkL
R
3
33
2
22
1
11
Ah1
R
Ah1
R
oo
ii
Overall Coefficient of Heat Transfer
o3
3
2
2
1
1
i
o3
3
2
2
1
1
i
oi
o3
3
2
2
1
1
i
oi
h1
kL
kL
kL
h1
1U
tUAQ
h1
kL
kL
kL
h1
ttAQ
Ah1
AkL
AkL
AkL
Ah1
ttQ
)(
)(
)(
Where:U – overall coefficient of heat transfer, W/m2-C or W/m2-K
CONDUCTION THROUGH CYLINDRICAL COORDINATES
kL2r
r
R
tttR
tQ
kL2r
rtt
Q
1
2
21
1
2
21
ln
)()(
)(
ln
)(
Where: r1 – inside radius, m r2 – outside radius, m L – length of pipe, m k – thermal conductivity of material, W/m-Cr1
r2
1 2
t1
t2
Q
k
For composite cylindrical pipes (Insulated pipe)
r1
r2
1 2
t1
t2
Q
k1
3
r3t3
k2
Lk2r
r
R ; Lk2r
r
R
tttR Rt
Q
kr
r
kr
rttL2
Lk2r
r
Lk2r
rtt
Q
2
2
3
21
1
2
1
31
21
2
2
3
1
1
2
31
2
2
3
1
1
2
31
lnln
)()(
)(
lnln
)(
lnln
)(
Heat Flow from fluid to fluid separated by a composite cylindrical wall
r1
r2
1 2
t1
t2
Q
k1
3
r3t3
k2
ihi
ti
o
ho
to
Lr2A ; Lr2A
Ah1
R ; Lk2r
r
R ; Lk2r
r
R ; Ah1
R
tttRR RR
tQ
Ah1
Lk2r
r
Lk2r
r
Ah1
ttQ
3o1i
ooo
2
2
3
21
1
2
1ii
i
oi
o21i
oo2
2
3
1
1
2
ii
oi
lnln
)()(
)(
lnln
)(
Overall Coefficient of Heat Transfer
area inside on based transfer heatof tcoefficien-Uarea outside on based transfer heatof nt-coefficieU
:whereAh1
Lk2r
rln
Lk2r
rln
Ah1
1AUAU
)t(AUQ)t(AUQ
i
o
oo2
2
3
1
1
2
ii
iioo
ii
oo
RadiationFrom Stefan - Boltzmann Law:The radiant heat flow Q for a blackbody is proportional to the surface area A, times the absolute surface temperature to the 4th power. A blackbody, or black surface is one that absorb all the radiation incident upon it.
Q = AT4 Wattswhere:
= 5.67x 10-8 W/m2-°K4
-Stefan-Boltzmann constantA- surface area,m2 T - absolute temperature, -°K
1
Body # 2
T2
Body # 1T1
A1
The net radiant heat transfer between two bodies or surfaces is;
Q = A1(T14 - T2
4)
2
Real bodies, surfaces, are not perfect radiators and absorbers,but emit, for the same surface temperature, a fraction of theblackbody radiation. This fraction is called the “emittance”
Tat radiation surface BlackTat radiation surface Actual
Emittance()
Actual bodies or surfaces are called “Gray Bodies” or “Gray surfaces”. Thus, the net rate of heat transfer between gray surface at temperature T1 to a surrounding black surface at temperature T2 is;
Q = 1 A1(T14 - T2
4) Watts
The enclosure being total and a black surface may be modified by the modulus F1-2, which accounts for the relative geometries of the surfaces(not all radiation leaving 1 reaches 2) and the surface emmitances, thus the equation becomes,
Q = F1-2 A1(T14 - T2
4) Watts
Radiant heat transfer frequently occurs with other modes of heat transfer, and the use of a radiative resistance R is very helpful. Let us now define Q to also be,
Where T2’ is an arbitrary reference temperature.
Q
TT=R
R
TT=Q
2'-
1
2'-
1
Heat Exchangers
Types of Heat Exchangers1. Direct Contact Type: The same fluid at
different states are mixed.2. Shell and Tube Type: One fluid flows inside
the tubes and the other fluid on the outside.
Direct Contactm1, h1
m2, h2
m3, h3
3 Eq. hmhmhm
2 Eq. hmhmhmnegligible PE and KE balance, Energy
1 Eq. mm mBalance Mass
SYSTEM OPEN an for Law First Applying
312211
332211
321
)hh(m)hh(m
hm
232311
32
Shell and Tube Type
mc
mc
mh
mh
1
2A
B
tc1
tc2
th1
th2
By energy balanceHeat rejected by the hot fluid = Heat absorbed by the cold fluid
2EqttCmQ
1EqttCmQ
12
21
ccpccc
hhphhh
ch
.)(
.)(
Where:mc – mass flow rate of cold fluid, kg/secmh – mass flow rate of hot fluid, kg/sech – enthalpy, kj/kgt – temperature,CCpc – specific heat of the cold fluid, KJ/kg-CCph – specific heat of the hot fluid, KJ/kg-CQ – heat transfer, KWh, c – refers to hot and cold, respectively1, 2 – refers to entering and leaving conditions of hot fluidA, B – refers to entering and leaving conditions of cold fluid
Heat Transfer in terms of OVERALL COEFFICIENTOf HEAT TRANSFER U
difference etemperatur mean log m area, surface transfer heat total - A
K-m
W
or C-m
W transfer, heatof tcoefficien overall - U
:where
KW
2
2
2
LMTD
1000
)LMTD(UAQ
Log Mean Temperature Difference (LMTD)
1
2
12
lnLMTD
Where:1 – small terminal temperature difference, C2 – large terminal temperature difference, C
Terminal Temperature Difference
1. For Counter Flow
T
A
Hot Fluid
Cold Fluid
2
1
th1
th2tc2
tc1
2c1h2
1c2h1
tt
tt
T
A
Hot Fluid
Cold Fluid
21
th1
th2
tc2
tc1
2. For Parallel Flow
2c1h2
2c2h1
tt
tt
3. For Constant temperature HEATING
T
A
Hot Fluid
Cold Fluid21
t
tc2
tc1
t
1c2
2c1
tt
tt
4. For Constant temperature COOLING
T
A
Hot Fluid
Cold Fluid
2
1
t
th2
th1
t
tt
tt
1h2
2h1
Correction Factor
Factor Correction-Fdifference etemperatur mean log
m area, surface transfer heat total - A
K-m
W
or C-m
W transfer, heatof tcoefficien overall - U
:where
KW )(
2
2
2
LMTD
1000
LMTDFUAQ
Exhaust gases flowing through a tubular heat exchanger at the rate of 0.3 kg/sec are cooled from 400 to 120C by water initially at 10C. The specific heat capacities of exhaust gases and water may be taken as 1.13 and 4.19 KJ/kg-K respectively, and the overall heat transfer coefficient from gases to water is 140 W/m2-K. Calculate the surface area required when the cooling water flow is 0.4 kg/sec; a. for parallel flow (4.01 m2) b. for counter flow (3.37 m2)
C3.5366.7-120t-tC39010-400t-t
Flow Parallel ForC7.66t
)10t)(19.4(4.0)120400)(13.1(3.0QcQh
C10tC120t ; C400t
C-KJ/kg 1.13C ; kg/sec 0.3m
C-KJ/kg 4.19C ; kg/sec 4.0m
Given
c2h21
c1h12
2c
2c
c1
h2h1
phh
pcc
)LMTD(UQ
A
)LMTD(UAQ
C43.201
1103.333
ln
1103.333LMTD
C11010120ttC3.3337.66400t-1t
Flow Counter For
C2.169
3.53390
ln
3.53390
ln
-LMTD
1c2h1
c2h2
1
2
12
2
2
m 4.3)43.201(140
920,94A
m 01.4)2.169(140
920,94A
Watts 94920QKW 92.94)120400)(13.1(3.0Q