1. ENGR. YURI G. MELLIZA
Heat Transfer
It is that area of mechanical engineering that deals
with the different principles and mechanisms involved
in transferring heat from one point to another.

2. Modes of Heat Transfer
1. 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.

3. Conduction
Metal rod
fire
t1 t2
Hotter body Colder body

4. Convection
1•
2 • surface
t1
t2
Fluid

5. Radiation
Hot body
Cold body

6. Conduction
From Fourier' s Law
t1 L
1 − kA(t2 − t1 ) kA(t1 − t2 )
A
Q= =
k
2t L L
A ( t1 − t 2 ) ( t 1 − t 2 )
2
Q Q= =
L L
k kA
Where:
L – thickness, meters
A – surface area, m2
k – thermal conductivity, mW°C or mWK
− -
Q – conductive heat flow, Watts

7. Thermal Circuit Diagram
R
1 2
Q
−(t2 − t1 ) − Δt (t1 − t2 )
Q= = =
L R R
kA
°C K
R −resistance, or
W W
− ∆t − temperatur e potential, °C or K
L
R=
kA

8. Conduction through a Composite Plane Wall
L1 L2 L3
1
A 2
3
k1 k2 k3 4
Q
( t1 −t 4 ) ( t1 −t 4 )
Q= =
L1 L L R1 +R2 +R 3
+ 2 + 3
k1 A k2 A k 3 A
(t1 −t 4 ) A ( t1 −t 4 )
Q= =
1 L 1 L 2 L 3  L 1 L 2 L 3 
k +k +k  k +k +k 
A 1 2 3  1 2 3
Q (t1 −t 4 )
=
A L 1 L 2 L 3 
k +k +k 
 1 2 3

9. Thermal Circuit Diagram
R1 R2 R3
4
1 3
2
Q
L1
R1 =
k1A
L2
R2 =
k2A
L3
R3 =
k 3A

10. 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 650°C. 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 L3 R1 R2 R3
1
A
2 4
3 1 2 3
Q
4
Q
Given:
L1 =0.20 m ; L2 = 0.10 m ; L3 = 0.20 m
k1 = 1.46 ; k2 = 0.26 ; k3 = 0.69
t1 = 650°C
Q/A = 56 W/m2

11. At 1 to 2 At 1 to 3
Q ( t1 − t 2 ) Q ( t1 − t 3 )
= =
A L1 A L1 + L 2
k1 k1 k2
Q  L 1  Q  L 1 L 2 
t2 = t1 −    t3 = t1 −   + 
 
A  k 1 
    A  k1 k 2 
  0.20    0.20 0.10 
t2 = 650 − 56  = 641.7°C t3 = 650 − 56 +  = 620.2°C
  1.36    1.36 0.26 
At 1 to 4
Q ( t1 − t 4 )
=
A L1 + L2 + L 3
k1 k 2 k 3
Q  L L L 
t 4 = t1 −   1 + 2 + 3 
 
 A  k 1 k 2 k 3 
  0.20 0.10 0.20 
t 4 = 650 − 56 + +  = 604°C
  1.36 0.26 0.69 

12. Convection
A
Fluid
If t1 > t2
Q = hA ( t1 − t2 ) Watts
1•
t1 If t1 < t2 (heat flows in opposite direction)
2•
t2 Q = hA ( t2 − t1 ) Watts
h
( t 2 − t1 )
Q Q= Watts
1
hA
Where:
Q – convective heat flow, Watts
A – surface area in contact with the fluid, m2
h – convective coefficient, W/m2-°C or W/m2-K
t1, t2 – temperature, °C

13. Conduction from Fluid to Fluid separated
by a composite plane wall
L1 L2 L3
1
A
2
3 o• ho,
i• to
hi ti
k1 k2 k3 4
Q
( ti − t o ) (ti − to )
Q= =
1 L L L 1 R i + R1 + R 2 + R 3 + R o
+ 1 + 2 + 3 +
hi A k 1 A k 2 A k 3 A ho A
( ti − t o ) A ( ti − t o )
Q= =
1  1 L1 L 2 L 3 1   1 L1 L 2 L 3 1 
h + k + k + k + h  h + k + k + k + h 
A i 1 2 3 o  i 1 2 3 o
Q ( ti − t o )
=
A  1 L1 L 2 L 3 1 
h + k + k + k + h 
 i 1 2 3 o

14. Thermal Circuit Diagram
Ri R1 R2 R3 Ro
1 2 3 4
i o
Q
1 L1
Ri = R1 =
hi A k1A
1 L2
Ro = R2 =
ho A k2A
L3
R3 =
k3A

15. Overall Coefficient of Heat Transfer
(ti − t o )
Q=
1 L1 L2 L3 1
+ + + +
hi A k 1 A k 2 A k 3 A ho A
A (ti − t o )
Q=
 1 L1 L2 L 3 1 
h + k + k + k + h 
 i 1 2 3 o
Q = UA (−∆t)
1
U=
 1 L1 L2 L 3 1 
h + k + k + k + h 
 i 1 2 3 o
Where:
U – overall coefficient of heat transfer, W/m2-°C or W/m2-K

16. CONDUCTION THROUGH CYLINDRICAL COORDINATES
Q
( t1 − t 2 )
Q=
ln r2
1 2 r1 Where:
2πkL r1 – inside radius, m
− (∆t) r2 – outside radius, m
Q=
R L – length of pipe, m
t1 − (∆t) = (t1 − t2 ) k – thermal conductivity of
r1 material, W/m-°C
k ln r2
r1
r2 t2 R=
2πkL

17. For composite cylindrical pipes (Insulated pipe)
( t1 − t 3 ) 2πL (t1 − t 3 )
Q= =
Q
r r
ln r2 ln 3 ln r2 ln 3
r1 r2 r1 r2
+ +
1 2 3 2πk 1L 2πk 2L k1 k2
− (∆t)
Q=
R1 + R 2
k1
t1 − (∆t) = (t1 − t3 )
r1
r2 r3
ln ln
k2 r2 t2
r1 r2
R1 = ; R2 =
r3 t3 2πk 1L 2πk 2L

18. Heat Flow from fluid to fluid separated by a composite cylindrical
wall
(ti − to )
Q=
Q r r
ln 2 ln 3
1 r1 r2 1
o + + +
hiA i 2π k1L 2π k 2L ho A o
i ti 1 2 3 to
− ( ∆ t)
hi ho Q=
R i + R1 + R 2 + R o
− ( ∆ t) = ( t i − t o )
t1 r3
k1 r1 ln r2 ln
1 r1 r2 1
Ri = ; R1 = ; R2 = ; Ro =
r2 t2 hi A i 2π k1L 2π k 2L ho A o
k2
t3
A i = 2π r1L ; A o = 2π r3L
r3

19. Overall Coefficient of Heat Transfer
Q = Uo A o (−∆t)
Q =Ui A i (−∆t)
1
Uo A o =Ui A i =
r2 r3
1 ln r1 ln r2 1
+ + +
hi A i 2πk 1L 2πk 2L ho A o
where :
Uo -coefficie nt of heat transfer based on outside area
Ui - coefficien t of heat transfer based on inside area

20. Radiation
From 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 = δAT4 Watts →1
where:
δ = 5.67x 10-8 W/m2-°K4
δ -Stefan-Boltzmann constant
A- surface area,m2
T - absolute temperature, -°K

21. Body # 1
T1
A1
Body # 2
The net radiant heat transfer T2
between two bodies or surfaces is;
→2
Q = δA1(T14 - T24)
Real bodies, surfaces, are not perfect radiators and
absorbers,
but emit, for the same surface temperature, a fraction of the
blackbody radiation. This fraction is called the “emittance” ∈

22. Emittance(∈)
Actual surface radiation at T
∈=
Black surface radiation at T
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 = δ ∈ 1 A1(T14 - T24) 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 = δ F 1-2 A1(T14 - T24) Watts

23. 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,
T -T
1 2'
Q=
R
T -T
1 2'
R=
Q
Where T2’ is an arbitrary reference temperature.

24. Heat Exchangers
Types of Heat Exchangers
1. 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 Contact
m1, h1
m2, h2
m3, h3

25. Applying First Law for an OPEN SYSTEM
Mass Balance
m1 +m2 = m3 → Eq. 1
Energy balance, ∆KE and ∆PE negligible
m1h1 +m2h2 = m3h3 → Eq. 2
m1h1 +m2h2 = m1h3 + m 2 h 3
m1 (h1 − h 3 ) = m 2 ( h 3 − h 2 ) →Eq. 3

26. Shell and Tube Type
th1
mh
1 mc
tc2
B
A t
c1
2
mh mc
th2

27. By energy balance
Heat rejected by the hot fluid = Heat absorbed by the cold fluid
Qh = Qc
Q h = m h C ph (t h1 − t h 2 ) →Eq .1
Q c = m c C pc (t c 2 − t c1 ) →Eq .2
Where:
mc – mass flow rate of cold fluid, kg/sec
mh – mass flow rate of hot fluid, kg/sec
h – enthalpy, kj/kg
t – temperature,°C
Cpc – specific heat of the cold fluid, KJ/kg-°C
Cph – specific heat of the hot fluid, KJ/kg-°C
Q – heat transfer, KW
h, c – refers to hot and cold, respectively
1, 2 – refers to entering and leaving conditions of hot fluid
A, B – refers to entering and leaving conditions of cold fluid

28. Heat Transfer in terms of OVERALL COEFFICIENT
Of HEAT TRANSFER U
UA (LMTD )
Q= KW
1000
where :
W
U - overall coefficient of heat transfer, 2 or
m - °C
W
m2 - K
A - total heat transfer surface area, m2
LMTD −log mean temperatur e difference

29. Log Mean Temperature Difference (LMTD)
θ2 − θ1
LMTD =
θ2
ln
θ1
Where:
θ1 – small terminal temperature difference, °C
θ2 – large terminal temperature difference, °C

30. Terminal Temperature Difference
1. For Counter Flow 2. For Parallel Flow
T th1 T th1
Ho
tF
θ2 Ho
tF
luid
th2
luid
tc2 th2 θ2 θ1
Co
ld θ1 Flu
id tc2
Flu ld
id Co
tc1
tc1
A A
θ1 = t h 2 − t c1 θ1 = t h 2 − t c 2
θ 2 = t h1 − t c 2 θ 2 = t h1 − t c 2

31. 3. For Constant temperature HEATING 4. For Constant temperature COOLING
T T th1
t Hot Fluid t
θ1 Ho
tF
θ2 luid
θ2 tc2 th2
id
Flu
Co
l d θ1
t Cold Fluid t
tc1
A A
θ1 = t − t c 2 θ1 = t h 2 − t
θ2 = t − t c1 θ2 = t h1 − t

32. Correction Factor
FUA(LMTD )
Q= KW
1000
where :
W
U - overall coefficient of heat transfer, 2 or
m - °C
W
m2 - K
A - total heat transfer surface area, m2
LMTD −log mean temperatur e difference
F - Correction Factor

33. Exhaust gases flowing through a tubular heat exchanger at the rate of 0.3 kg/sec are
cooled from 400 to 120°C by water initially at 10°C. 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)
Given
θ2 - θ1 390 − 53.3
mc = 0.4 kg/sec ; Cpc = 4.19 KJ/kg - C LMTD = = = 169.2°C
θ2 390
ln ln
mh= 0.3 kg/sec ; Cph = 1.13 KJ/kg - C θ1 53.3
th1 = 400°C ; th2 = 120°C For Counter Flow
tc1 = 10°C θ2 = th 1 - tc2 = 400 − 66.7 = 333.3°C
Qh = Qc θ1 = th2 − tc1 = 120 −10 = 110°C
0.3(1.13)(400 −120) = 0.4(4.19)(t c2 −10) 333.3 −110
LMTD = = 201.43°C
333.3
tc2 = 66.7°C ln
110
For Parallel Flow Q = UA(LMTD)
θ2 = th1 - t c1 = 400 - 10 = 390°C Q
A=
θ1 = th2 - tc2 = 120 - 66.7 = 53.3°C U(LMTD)

34. Q = 0.3(1.13)(400 −120) = 94.92 KW
Q = 94920 Watts
94 ,920
A= = 4.01 m2
140(169.2)
94 ,920
A= = 3.4 m2
140(201.43)