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

QQ

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