HEAT TRANSFER

Content

• Modes of heat transfer?

• Fourier Law of heat conduction

• Convective heat coefficient

• Radiant heat coefficient

• Overall heat transfer coefficient

• Hands-on example

Temperature

• A measure of energy due to level of heat

– Freezing point of water is 0 ˚ C

– Boiling point of water is 100 ˚ C

Common Temperature Scales

What is Heat?

Heat is the total internal kinetic energy due to

molecular motion in an object

Quantity of heat is BTU or Kilo Joule (kJ)

• One BTU is the amount of heat required to raise 1

lb of water by 1 ˚ F

• One calorie is required to raise 1 g of water by 1 ˚

C

1 cal = 4.187 J

• 1 BTU= 1.055 kJ= 1055 J

Heat Vs Temperature

• Heat energy depends on mass. Temperature is

independent of mass.

– 2 litres of boiling water has more heat energy than

1 litre of boiling water

• Temperature is not energy, but a measure of it

• Heat is energy

Heat is Energy

When heat (ie energy) goes into a substance,

one of two things can happen:

1. Temperature goes up

2. Change of state

Temperature Goes Up

• Heat that causes a rise in temperature e.g.

heating water before boiling

• The heat energy is used to increase the kinetic

energy of the molecules in the substance

• This is also known as the sensible heat

Change Of State

• Heat that brings about a change in potential

energy of the molecules (temperature remains

constant). Also called the latent heat.

Specific Heat

• It is the heat required to the temperature of 1

kg (lb) a substance by 1 ˚ K (F)

• Example:

water’s specific heat is 1 btu/ lb F (4.2 kJ/kg K)

air’s specific heat is 0.24 btu/ lb F (1.0 kJ/kg K)

Sizing Heating Capacity

T heat x specific x mass required heat of Quantity A =

Example:

What is the heat required to raise air

temperature from 15 ˚C to 25 ˚C at a

flow rate of 2000 l/s?

Heat Transfer

• If there is a temperature difference in a

system, heat will always move from higher to

lower temperatures

What is actually flowing?

Heat Transfer Modes

There are 3 modes

of heat transfer.

1. Conduction

2. Convection

3. Radiation

Conduction

• Heat transfer through a solid medium via

direct contact

• Expressed by Fourier’s Law

Fourier’s Law

Q

X

T2 T1

dx

dT

k q ÷ = "

k = thermal conductivity (W/ m K)

T = temperature (K)

q” = heat flux (W/m

2

)

Heat flow rate = q” x area (W)

Fourier’s law at steady state

kA L

T T

q

k L

T T

q

L

T T

k q

dx

dT

k q

in out

in out

in out

/

flow heat of Area x " Q

rate fer Heat trans

/

"

State) (Steady "

Law) (Fourier "

÷

÷ =

=

÷

÷ =

÷

÷ =

÷ =

T1

T2

q

R=L/k

Unit thermal resistance

Example 1

• Temperature of 35 C and 22 C are maintained

on opposite sides of a steel floor of 6mm

thick. Compute the heat flux through the

floor.

• Thermal conductivity for steel = 50 W/m K

Thermal Conductivity, k (W/m K)

Liquids

Water: 0.556

Ammonia: 0.54

Gases

Air : 0.024

Water vapor: 0.021

Common Metals

Copper: 385

Aluminum: 221

Steel: 50

Non-metals

Common brick: 0.6

Mineral wool: 0.04

Ceiling board: 0.06

Quiz

• Suppose a human could live for 2 h unclothed

in air at 45 ˚F. How long could she live in water

at 45 ˚F?

Electrical- Thermal Analogy

ce sis

ce sis

tan Re

difference e Temperatur

q flux, Heat

Thermal

tan Re

Potential Voltage

I Current,

Law) s (Ohm' Electrical

=

=

T1

T2

q

R=L/kA

Composite Wall

Using the resistance concept,

T1

T2

R1 R2

Q

2

2

2

1

1

1

2 1

1 2

"

k

x

R

k

x

R

R R

T T

q

=

=

+

÷

=

Example 2

A wall of a Switchgear room consists the

following:

Q

Q

q

2

k

2

35 C 22 C

6mm 25mm 100mm

TNF panel

k = 0.02 W/m K

Firebatt

k = 0.04 W/m K

Steel plate

k = 50 W/m K

Q

Determine Q, if the wall is 3m x 4m ?

Convection

• Energy transfer by fluid

motion

• Two kinds of convection

– Forced convection: Fluid is

forced

– Natural or free convection:

fluid is induced by

temperature difference

where:

h

c

is convection coefficient (W/m

2

C),

T

s

is surface temperature (°C),

T

a

is surrounding air temperature (°C)

Rc= unit convective resistance.

Convective Heat Transfer

air flow

T

a

T

s

y

q

c

c

C

a s

a s

c

h

R

h

T T

q

T T h q

1

1

(

"

) ( "

cooling of Law s Newton'

)

=

÷

=

÷ =

Magnitude of Convection Coefficients

Arrangement h, W/m2 K Btu/(h.ft2.F)

Air, free (indoor) 10-30 1-5

Air, forced

(outdoor)

30-300 5-50

Oil, forced 60-1800 10-300

Water, forced 300-6000 50-1000

Steam, condensing 6000-120000 1000-20000

Example 3

The same as Example 2. Consider convection of

the exposed surfaces, calculate Q.

Q

Q

q

2

k

2

35 C 22 C

6mm 25mm 100mm

TNF panel

k = 0.02 W/m K

Firebatt

k = 0.04 W/m K

Steel plate

k = 50 W/m K

Q

Radiation

• Energy emitted by object that is at any

temperature above absolute zero

• Energy is in the form electromagnetic waves

• No medium needed and travel at speed of

light

Hot Body

Radiator

radiation olar

: Example

S

Radiation

• Important mode of heat at high temperatures,

e.g. combustion furnace

• At room temperature it may just be

measurable.

• Intensity depends on body temperature and

surface characteristics

• Solar radiation is the radiation emitted by the

sun due to nuclear fusion reaction

• Solar Constant: The amount of solar energy

arriving at the top of the atmosphere

perpendicular to the sun’s rays.

• = 1375 W m

-2

Solar Radiation

Solar Radiation Spectrum

99% of solar radiation is between 0.3 to 3 µm.

Wien’s Law

m µ ì

T

2900

m=

Wien’s Law

The Black Body

E = AoT

4

• E =The amount of energy (W )

emitted by an object

• o = Stefan-Boltzmann constant =

5.67 x 10

-8

W m

-2

K

-4

• T = Temperature (K)

• A= area (m2)

The Grey Body

metals polished for 0.07 - 0.02

materials common for 0.9 - 0.8

ty emissitivi

where ) ( E E

body, actual an For

4

b

=

=

=

= =

c

co c T A

Net Radiant Heat

• If a hot object is radiating to a cold

surrounding, the radiation loss is

) ( q

4 4

c h T T A ÷ =co

Quiz

How much energy does human body radiate?

• Body temperature is 37 C

• Body area is 1.5 m2

• ε= 0.7

Radiant Heat Transfer

• Unit thermal resistance for radiation is written

as

r

c

r

h

1

R

T) ( h q"

=

A =

Radiation coefficient is a function of

temperature, radiation properties and

geometrical arrangement of the enclosure

and the body in question.

Combined convection and radiant

Coefficient

• The heat transfer is combination of convection

and radiation

r c h h

1

R

, resistance thermal Combined

) )( ( "

"

+

=

A + =

+ =

T h h q

q q q

r c

r c

Combined Surface Coefficients

Air velocity Emissivity, ε=0.9

3.5 m/s h = 22.7 W/m2 K

7 m/s h = 35 W/m2 K

Still air h = 8.5 W/ m2 K

• Some practical values of surface coefficients:

(source: ASHRAE Fundamentals 1989)

k2

Combined modes

T

k1

Outside

Inside

T

hot

T

cold

T

1

T

3

R2=L1/k1 + L2/K2

R3=1/h

hot

R1=1/h

cold

T

2

T

1

T

cold

T

hot

Resistance in parallel, R= R1 + R2 +R3

T

3

Compute

2 2 1 1

2

1

2 2 1 1

2

2

1

1

3 2 1

/ / / 1

"

/ 1

"

/ / / 1 / 1

"

1

/

1

k L k L h

T T

q

h

T T

q

k L k L h h

T T

q

h k

L

k

L

h

R

R R R R

cold

cold

cold

cold

cold hot

cold hot

hot cold

+ +

÷

=

÷

=

+ + +

÷

=

+ + + =

+ + =

T

hot

T

cold

T

1

T

2

R2=L1/k1 + L2/k2

R3=1/h

hot

R1=1/h

cold

Overall Heat Transfer Coefficient

• Heat transfer processes includes conduction,

convection and radiation simultaneously

• The total conduction heat transfer for a wall or

roof is expressed as

Q = A x U x ∆T where

U is the overall heat transfer coefficient (or U-

value)

R

U

R R R R

1

....... 3 2 1

=

+ + = +

Example

• Find the overall heat

transfer coefficient of

a flat roof having the

construction shown in

the figure.

Solution

T

2

T

1

R3

R1

R2

R4

R5

R6

Solution

Resistance Construction Unit resistance (m2 K/ W)

R1 Outside air

R2 steel

R3 Mineral wool

R4 Air space

R5 Ceiling board

R6 Inside air

Total R

Solution

K W/m 40 . 0

48 . 2

1

R

1

U

t coefficien fer heat trans Overall

2

= = =

Heat Transfer Loop

in a DX System

Heat Exchanger Coil

Heat is exchanged between

2 fluids.

Q= UA ∆T

For cross flow,

Q= UA (LMTD)

Heat Exchanger- Mean Temperature

Difference

LTD

GTD

Ln

LTD - GTD

x Area x U Q

LMTD x Area x U Q Transfer, Heat

=

=

Heat transfer optimization

• We have the following relations for heat transfer:

– Conduction: Q = k A ∆T /d

– Convection: Q = A h

c

∆T

– Radiation: Q = A h

r

∆T

• As a result, when equipment designers want to improve

heat transfer rates, they focus on:

– Increasing the area A, e.g. by using profiled tubes and ribbed

surfaces.

– Increasing AT (which is not always controllable).

– For conduction, increasing k /d.

– Increase h

c

by not relying on natural convection, but

introducing forced convection.

– Increase h

r

, by using “black” surfaces.