Prof. R. Shanthini 5 & 12 March 2012 1 PM3125: Lectures 6 to 9 Content of Lectures 6 to 12: Heat transfer: - Source of heat - Heat transfer - Steam and

Prof. R. Shanthini 5 & 12 March 2012

1

PM3125: Lectures 6 to 9

Content of Lectures 6 to 12:

Heat transfer: - Source of heat - Heat transfer - Steam and electricity as heating media- Determination of requirement of amount of steam/electrical energy - Steam pressure- Mathematical problems on heat transfer


2

What is Heat?


3

What is Heat?

Heat is energy in transit.


4

Units of Heat• The SI unit is the joule (J),

which is equal to Newton-metre (Nm).

• Historically, heat was measured in terms of the ability to raise the temperature of water.

• The calorie (cal): amount of heat needed to raise the temperature of 1 gramme of water by 1 C0 (from 14.50C to 15.50C)

• In industry, the British thermal unit (Btu) is still used: amount of heat needed to raise the temperature of 1 lb of water by 1 F0 (from 630F to 640F)


5

Conversion between different

units of heat:

1 J = 0.2388 cal = 0.239x10-3 kcal = 60.189 Btu

1 cal = 4.186 J = 3.969 x 10-3 Btu


6

Sensible Heat

• What is 'sensible heat‘?

Sensible heat is associated with a temperature change


7

Specific Heat Capacity

• To raise the temperature by 1 K, different substances need different amount of energy because substances have different molecular configurations and bonding (eg: copper, water, wood)

• The amount of energy needed to raise the temperature of 1 kg of a substance by 1 K is known as the specific heat capacity

• Specific heat capacity is denoted by c


8

Calculation of Sensible Heat

where ΔT is the temperature change in the substance

Q = m c dT ∫ Q is the heat lost or gained by a substance

m is the mass of substance

c is the specific heat of substance which changes with temperature

T is the temperature

When temperature changes causes negligible changes in c,

Q = m c dT ∫ = m c ∆T


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When temperature changes causes significant changes in c,

Q = m c ∆T cannot be used.

Q = ∆H = m ∆h Instead, we use the following equation:

where ΔH is the enthalpy change in the substance

and ∆h is the specific enthalpy change in the substance.

To apply the above equation, the system should remain at constant pressure and the associated

volume change must be negligibly small.


10


Q = m c ΔT (since c is taken as a constant)

= (300 g) (0.896 J/g oC)(70 - 25)oC

= 12,096 J

= 13.1 kJ

Calculate the amount of heat required to raise the temperature of 300 g Al from 25oC to 70oC.

Data: c = 0.896 J/g oC for Al


11

Exchange of Heat

Heat lost by iron = Heat gained by water

(m c ΔT)iron = (m c ΔT)water

(100 g) (0.452 J/g oC)(80 - tf)oC

= (53.5 g) (4.186 J/g oC)(tf - 25)oC

80 - tf = 4.955 (tf -25)

tf = 34.2oC

Calculate the final temperature (tf), when 100 g iron at 80oC is tossed into 53.5g of water at 25oC.

Data: c = 0.452 J/g oC for iron and 4.186 J/g oC for water


12

Latent Heat

• What is ‘latent heat‘?

Latent heat is associated with phase change of matter


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Phases of Matter


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Phase Change• Heat required for phase changes:

» Melting: solid liquid

» Vaporization: liquid vapour

» Sublimation: solid vapour

• Heat released by phase changes:» Condensation: vapour liquid

» Fusion: liquid solid

» Deposition: vapour solid


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Phase Diagram: Water


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Phase Diagram: Water

Saturated steam

Superheated steam

Saturated liquidCompressed liquid


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Phase Diagram: WaterExplain why water is at liquid

state at atm pressure


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Phase Diagram: Carbon DioxideExplain why CO2 is at gas state

at atm pressure

Explain why CO2 cannot be made a

liquid at atm pressure


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Latent Heat

Latent heat is the amount of heat added per unit mass of substance during a phase change

Latent heat of fusion is the amount of heat added to melt a unit mass of ice OR it is the amount of heat removed to freeze a unit mass of water.

Latent heat of vapourization is the amount of heat added to vaporize a unit mass of water OR it is the amount of heat removed to condense a unit mass of steam.


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Water: Specific Heat Capacities and Latent Heats

Specific heat of ice ≈ 2.06 J/g K (assumed constant)

Heat of fusion for ice/water ≈ 334 J/g (assumed constant)

Specific heat of water ≈ 4.18 J/g K (assumed constant)

Latent heat of vaporization cannot be assumed a constant since it changes significantly with the pressure,

and could be found from the Steam Table

How to evaluate the sensible heat gained (or lost) by superheated steam?


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Water: Specific Heat Capacities and Latent Heats

How to evaluate the sensible heat gained (or lost) by superheated steam?

Q = m c ∆T cannot be used since changes in c with changing

temperature is NOT negligible.

Q = ∆H = m ∆h Instead, we use the following equation:

provided the system is at constant pressure and the associated volume change is negligible.

Enthalpies could be referred from the Steam Table


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Properties of SteamLearnt to refer to Steam Table to find properties of steam such as saturated (or boiling point) temperature and latent heat of vapourization at give pressures, and enthalpies of superheated steam at various pressures and temperatures.

Reference:

Chapter 6 of “Thermodynamics for Beginners with worked examples” by R. Shanthini

(published by Science Education Unit, Faculty of Science, University of Peradeniya)

(also uploaded at http://www.rshanthini.com/PM3125.htm)


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Warming curve for waterWhat is the amount of heat required to change 2 kg of ice

at -20oC to steam at 150oC at 2 bar pressure?

-20oC ice


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-20oC ice

0oC melting point of ice


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-20oC ice


120.2oC boiling point of water at 2 bar

Boiling point of water at 1 atm pressure is 100oC. Boiling point of water at 2 bar is 120.2oC. [Refer the Steam Table.]


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-20oC ice


120.2oC boiling point of water at 2 bar

150oC superheated steam

Specific heat

Specific heat

Specific heat

Latent heat

Latent heat


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Specific heat required to raise the temperature of ice from -20oCto 0oC= (2 kg) (2.06 kJ/kg oC) [0 - (-20)]oC = 82.4 kJ

Latent heat required to turn ice into water at 0oC= (2 kg) (334 kJ/kg) = 668 kJ

Specific heat required to raise the temperature of water from 0oC to 120.2oC

= (2 kg) (4.18 kJ/kg oC) [120.2 - 0)]oC = 1004.9 kJ


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Latent heat required to turn water into steam at 120.2oC and at 2 bar= (2 kg) (2202 kJ/kg) = 4404 kJ

[Latent heat of vapourization at 2 bar is 2202 kJ/kg as could be referred to from the Steam Table]

Specific heat required to raise the temperature of steam from 120.2oC to 150oC

= (2 kg) (2770 – 2707) kJ/kg = 126 kJ

[Enthalpy at 120.2oC and 2 bar is the saturated steam enthalpy of 2707 kJ/kg and the enthalpy at 150oC and 2 bar is 2770 kJ/kg as could be referred to from the Steam Table]


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Total amount of heat required

= 82.4 kJ + 668 kJ + 1004.9 kJ + 4404 kJ + 126 kJ

= 6285.3 kJ


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Application: Heat ExchangerIt is an industrial equipment in which heat is transferred from a hot fluid (a liquid or a gas) to a cold fluid (another liquid or gas) without the two fluids having to mix together or come into direct contact.

Hot fluid at TH,in Hot fluid

at TH,out

Cold fluid at TC,out

Cold fluid at TC,in

Heat lost by the hot fluid = Heat gained by the cold fluid


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Application: Heat Exchanger


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Heat Exchanger

mhot chot (TH,in – TH,out) = mcold ccold (TC,out – TC,in). .


mass flow rate of hot fluid

Specific heat of hot fluid

mass flow rate of cold fluid

Specific heat of cold fluid

Temperature decrease in the hot fluid

Temperature increase in the cold fluid


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Heat Exchanger

mhot chot (TH,in – TH,out) = mcold ccold (TC,out – TC,in). .


The above is true only under the following conditions:

(1) Heat exchanger is well insulated so that no heat is lost to the environment

(2) There are no phase changes occurring within the heat exchanger.


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Heat Exchanger


+ Heat lost to the environment

If the heat exchanger is NOT well insulated, then


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High pressure liquid water at 10 MPa (100 bar) and 30oC enters a series of heating tubes. Superheated steam at 1.5 MPa (15 bar) and 200oC is sprayed over the tubes and allowed to condense. The condensed steam turns into saturated water which leaves the heat exchanger. The high pressure water is to be heated up to 170oC. What is the mass of steam required per unit mass of incoming liquid water? The heat exchanger is assumed to be well insulated (adiabatic).

Worked Example 1 in Heat Exchanger


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Solution to Worked Example 1 in Heat Exchanger


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High pressure (100 bar) water enters at 30oC and leaves at 198.3oC.Boiling point of water at 100 bar is 311.0oC. Therefore, no phase changes in the high pressure water that is getting heated up in the heater.

Heat gained by high pressure water

= ccold (TC,out – TC,in)

= (4.18 kJ/kg oC) x (170-30)oC

= 585.2 kJ/kg

[You could calculate the above by taking the difference in enthalpies at the 2 given states from tables available.]

Solution to Worked Example 1 in Heat Exchanger contd.


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Superheated steam at 1.5 MPa (15 bar) and 200oC is sprayed over the tubes and allowed to condense. The condensed steam turns into saturated water which leaves the heat exchanger.

Heat lost by steam

= heat lost by superheated steam to become saturated

steam

+ latent heat of steam lost for saturated steam to turn into

saturated water

= Enthalpy of superheated steam at 15 bar and 200oC

– Enthalpy of saturated steam at 15 bar

+ Latent heat of vapourization at 15 bar

= (2796 kJ/kg – 2792 kJ/kg) + 1947 kJ/kg = 1951 kJ/kg



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Since there is no heat loss from the heater,

Heat lost by steam = Heat gained by high pressure water

Mass flow rate of steam x 1951 kJ/kg

= Mass flow rate of water x 585.2 kJ/kg

Mass flow rate of steam / Mass flow rate of water

= 585.2 / 1951

= 0.30 kg stream / kg of water



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Give the design of a heat exchanger which has the most effective heat

transfer properties.

Assignment

Learning objectives:1) To be able to appreciate heat transfer applications in pharmaceutical

industry

2) To become familiar with the working principles of various heat exchangers

3) To get a mental picture of different heat exchangers so that solving heat transfer problems in class becomes more interesting


41

Steam enters a heat exchanger at 10 bar and 200oC and leaves it as saturated water at the same pressure. Feed-water enters the heat exchanger at 25 bar and 80oC and leaves at the same pressure and at a temperature 20oC less than the exit temperature of the steam. Determine the ratio of the mass flow rate of the steam to that of the feed-water, neglecting heat losses from the heat exchanger.

If the feed-water leaving the heat exchanger is fed directly to a boiler to be converted to steam at 25 bar and 300oC, find the heat required by the boiler per kg of feed-water.

Worked Example 2 in Heat Exchanger


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- Steam enters at 10 bar and 200oC and leaves it as saturated water at the same pressure. - Saturation temperature of water at 10 bar is 179.9oC. - Feed-water enters the heat exchanger at 25 bar and 80oC and leaves at the same pressure and at a temperature 20oC less than the exit temperature of the steam, which is 179.9oC.- Boiling point of water at 25 bar is (221.8+226.0)/2 = 223.9oC. - Therefore, no phase changes in the feed-water that is being heated.

Heat lost by steam = Heat gained by feed-water (with no heat losses)

Mass flow rate of steam x [2829 – 2778 + 2015] kJ/kg

= Mass flow rate of feed-water x [4.18 x (179.9-20-80) ] kJ/kg

Mass flow of steam / Mass flow of feed-water = 333.98 / 2066 = 0.1617 kg stream / kg of water

Solution to Worked Example 2 in Heat Exchanger


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If the feed-water leaving the heat exchanger is fed directly to a boiler to be converted to steam at 25 bar and 300oC, find the heat required by the boiler per kg of feed-water.

- Temperature of feed-water leaving the heat exchanger is 159.9oC- Boiling point of water at 25 bar is (221.8+226.0)/2 = 223.9oC - The feed-water is converted to superheated steam at 300oC

Heat required by the boiler per kg of feed-water

= {4.18 x (223.9-159.9) + (1850+1831)/2 + [(3138+3117)/2 – (2802+2803)/2]} kJ/kg

= {267.52 + 1840.5 + [3127.5 – 2802.5]} kJ/kg

= 2433 kJ/kg of feed-water



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Heat Transfer

is the means by which

energy moves from

a hotter object to

a colder object


45

Mechanisms of Heat TransferConduction

is the flow of heat by direct contact between a warmer and a cooler body.

Convectionis the flow of heat carried by moving gas or liquid.

(warm air rises, gives up heat, cools, then falls)

Radiationis the flow of heat without need of an intervening medium.

(by infrared radiation, or light)


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Latent heat Conduction

Convection

Radiation

Mechanisms of Heat Transfer


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Conduction

HOT(lots of vibration)

COLD(not much vibration)

Heat travels along the rod


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Conduction is the process whereby heat is transferred directly through a material, any bulk motion of the

material playing no role in the transfer.

Those materials that conduct heat well are called thermal conductors, while those that conduct heat poorly

are known as thermal insulators.

Most metals are excellent thermal conductors, while wood, glass, and most plastics are common thermal

insulators.

The free electrons in metals are responsible for the excellent thermal conductivity of metals.

Conduction


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Conduction: Fourier’s Law

Q = heat transferredk = thermal conductivityA = cross sectional area T = temperature difference between two endsL = lengtht = duration of heat transfer

Cross-sectional area A

L

ΔTQ = Lk A t( )What is the unit of k?


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Substance ThermalConductivityk [W/m.K]

Substance ThermalConductivityk [W/m.K]

Syrofoam 0.010 Glass 0.80

Air 0.026 Concrete 1.1

Wool 0.040 Iron 79

Wood 0.15 Aluminum 240

Body fat 0.20 Silver 420

Water 0.60 Diamond 2450

Thermal Conductivities


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T

T T1

x

Conduction through Single Wall

k A (T1 – T2)

Δx

Q.

Q.

= Δx

Q.

Use Fourier’s Law:

ΔTQ = Lk A t( )


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T

T T1

x

52

Conduction through Single Wall

Δx

Q.

T1 – T2=

Δx/(kA)

Q.

Thermal resistance (in K/W) (opposing heat flow)

k A (T1 – T2)Q.

= Δx


53

T

T

x

53

Conduction through Composite Wall

ΔxA

Q.

T1 – T2=

(Δx/kA)A

Q.

Q.

A B C

T

T

ΔxB ΔxC

kA kB kC

= T2 – T3

(Δx/kA)B

= T3 – T4

(Δx/kA)C


54

54

Conduction through Composite Wall

+ (Δx/kA)B (Δx/kA)A + (Δx/kA)C[ ]Q

.

= T1 – T2 + T2 – T3 + T3 – T4

T1 – T4

T1 – T2=

(Δx/kA)A

Q.

= T2 – T3

= T3 – T4

(Δx/kA)C (Δx/kA)B

Q.

+ (Δx/kA)B (Δx/kA)A + (Δx/kA)C

=


55

An industrial furnace wall is constructed of 21 cm thick fireclay brick having k = 1.04 W/m.K. This is covered on the outer surface with 3 cm layer of insulating material having k = 0.07 W/m.K. The innermost surface is at 1000oC and the outermost surface is at 40oC. Calculate the steady state heat transfer per area.

Example 1

Tin – ToutQ.

+ (Δx/kA)insulation (Δx/kA)fireclay

=

Solution: We start with the equation


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Example 1 continued

(1000 – 40) AQ.

+ (0.03/0.07) (0.21/1.04)

=

= 1522.6 W/m2Q.

A


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We want to reduce the heat loss in Example 1 to 960 W/m2. What should be the insulation thickness?

Example 2

Tin – ToutQ.

+ (Δx/kA)insulation (Δx/kA)fireclay

=


(1000 – 40)

+ (Δx)insulation /0.07) (0.21/1.04)

=Q.

A= 960 W/m2

(Δx)insulation = 5.6 cm


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L

Conduction through hollow-cylinder

Ti – ToQ.

[ln(ro/ri)] / 2πkL

=

ri

ro

Ti

To


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Conduction through the composite wall in a hollow-cylinder

Ti – ToQ.

[ln(r2/r1)] / 2πkAL

=

r1

r2

Ti

To

r3

Material A

+ [ln(r3/r2)] / 2πkBL

Material B


60

A thick walled tube of stainless steel ( k = 19 W/m.K) with 2-cm inner diameter and 4-cm outer diameter is covered with a 3-cm layer of asbestos insulation (k = 0.2 W/m.K). If the inside-wall temperature of the pipe is maintained at 600oC and the outside of the insulation at 100oC, calculate the heat loss per meter of length.

Example 3

Ti – ToQ.

[ln(r2/r1)] / 2πkAL

= + [ln(r3/r2)] / 2πkBL



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Example 3 continued

2 π L ( 600 – 100)Q.

[ln(2/1)] / 19

= + [ln(5/2)] / 0.2

Q.

L= 680 W/m


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Convection Convection is the process in which heat is carried from place to place by the bulk movement of a fluid (gas or liquid).

Convection currents are set up when a pan of water is heated.


64

It explains why breezes come from the ocean in the day and from the land at night

Convection


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Convection: Newton’s Law of Cooling

Q.

conv. = h A (Tsurface – Tfluid)

Area exposed

Heat transfer coefficient (in W/m2.K)

Heated surface at Tsurface

Flowing fluid at Tfluid


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Q.

conv.Tsurface – Tfluid

Heated surface at Tsurface

Flowing fluid at Tfluid

=1/(hA)

Convective heat resistance (in K/W)

Convection: Newton’s Law of Cooling


67

The convection heat transfer coefficient between a surface at 50oC and ambient air at 30oC is 20 W/m2.K. Calculate the heat flux leaving the surface by convection.

Example 4

Solution:

Heated surface at Tsurface = 50oC

Flowing fluid at Tfluid = 30oC

h = 20 W/m2.K

Q.

conv.= h A (Tsurface – Tfluid)

= (20 W/m2.K) x A x (50-30)oC

.Q

conv.

A= 20 x 20 = 400 W/m2

Heat flux leaving the surface:

Use Newton’s Law of cooling :


68

Air at 300°C flows over a flat plate of dimensions 0.50 m by 0.25 m. If the convection heat transfer coefficient is 250 W/m2.K, determine the heat transfer rate from the air to one side of the plate when the plate is maintained at 40°C.

Example 5

Solution:

h = 250 W/m2.K

A = 0.50x0.25 m2

Q.

conv.= h A (Tsurface – Tfluid)

= 250 W/m2.K x 0.125 m2

x (40 - 300)oC

= - 8125 W/m2

Heat is transferred from the air to the plate.

Use Newton’s Law of cooling :

Heated surface at Tsurface = 40oC

Flowing fluid at Tfluid = 300oC


69

Forced Convection

In forced convection over external surface:Tfluid = the free stream temperature (T∞), or a

temperature far removed from the surface

In forced convection through a tube or channel:Tfluid = the bulk temperature

In forced convection, a fluid is forced by external forces such as fans.


70

Free ConvectionIn free convection, a fluid is circulated due to buoyancy effects, in which less dense fluid near the heated surface rises and thereby setting up convection.

In free (or partially forced) convection over external surface:

Tfluid = (Tsurface + Tfree stream) / 2

In free or forced convection through a tube or channel:

Tfluid = (Tinlet + Toutlet) / 2


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Change of Phase Convection

Change-of-phase convection is observed with boiling or condensation. It is a very complicated mechanism and therefore will not be covered in this course.


72

Overall Heat Transfer through a Plane Wall

T

T

xΔx

Q.

Q.

Fluid Aat TA > T1

Fluid Bat TB < T2

Q. TA – T1

=1/(hAA)

T2– TB=

1/(hBA)

T1 – T2=

Δx/(kA)


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Overall Heat Transfer through a Plane Wall

Q. TA – T1

=1/(hAA)

T2– TB=

1/(hBA)

T1 – T2=

Δx/(kA)

Q. TA – TB

=1/(hAA) + 1/(hBA) + Δx/(kA)

(TA – TB)= U A

1/U = 1/hA + 1/hB + Δx/k

where U is the overall heat transfer coefficient given by

Q.


74

Overall heat transfer through hollow-cylinder

ri

ro

Ti

To

L(TA – TB)= U A

1/UA = 1/(hAAi) + 1/(hBAo)

Q.

Fluid A is inside the pipeFluid B is outside the pipe TA > TB

+ ln(ro/ri) / 2πkL

where


75

Steam at 120oC flows in an insulated pipe. The pipe is mild steel (k = 45 W/m K) and has an inside radius of 5 cm and an outside radius of 5.5 cm. The pipe is covered with a 2.5 cm layer of 85% magnesia (k = 0.07 W/m K). The inside heat transfer coefficient (hi) is 85 W/m2 K, and the outside coefficient (ho) is 12.5 W/m2 K. Determine the heat transfer rate from the steam per m of pipe length, if the surrounding air is at 35oC.

Example 6

(TA – TB)= U A

Solution: Start with

Q.

(120 – 35)= U A

What is UA?


76

Example 6 continued

1/UA = 1/(85Ain)

+ 1/(12.5Aout)

+ ln(5.5/5) / 2π(45)L

+ ln(8/5.5) / 2π(0.07)L

1/UA = 1/(hAAi) + 1/(hBAo)+ ln(ro/ri) / 2πkL + …

Ain = 2π(0.05)L and Aout = 2π(0.08)L

1/UA = (0.235 + 0.0021 +5.35 + 1) / 2πL


77

Example 6 continued

Q.

(120 – 35)= U A

= 81 L

Q.

= 81 W/m

UA = 2πL / (0.235 + 0.0021 +5.35 + 1)

(120 – 35) / (0.235 + 0.0021 +5.35 + 1)= 2πL

steam

steel

insulation

air

/ L


78








79

Radiation Radiation is the process in which energy is transferred by means of electromagnetic waves of wavelength band between 0.1 and 100 micrometers solely as a result of the temperature of a surface.

Heat transfer by radiation can take place through vacuum. This is because electromagnetic waves can propagate through empty space.


80

The Stefan–Boltzmann Law of Radiation

ε = emissivity, which takes a value between 0 (for an ideal reflector) and 1 (for a black body).

σ = 5.668 x 10-8 W/m2.K4 is the Stefan-Boltzmann constant

A = surface area of the radiator

T = temperature of the radiator in Kelvin.

Q t

= ε σ A T4


81

Ratio of the surface area of a cub to its

volume is much larger than for its mother.

Why is the mother shielding

her cub?


82

What is the Sun’s surface temperature?

The sun provides about 1000 W/m2 at the Earth's surface.

Assume the Sun's emissivity ε = 1Distance from Sun to Earth = R = 1.5 x 1011 m

Radius of the Sun = r = 6.9 x 108 m


83

T4 =

(4 π 1.52 x 1022 m2)(1000 W/m2)

= 2.83 x 1026 W

(4 π 6.92 x 1016 m2)

= 5.98 x 1018 m2

(1) (5.67 x 10-8 W/m2.K4) (5.98 x 1018 m2)

2.83 x 1026 W

T = 5375 K

What is the Sun’s surface temperature?

Q t

= ε σ A T4

εσ


84

If object at temperature T is surrounded by an environment at temperature T0, the net radioactive heat flow is:

Temperature of the radiating surface

Temperature of the environment

Q t

= ε σ A (T4 - To4 )


85

What is the rate at which radiation is emitted by a surface of area 0.5 m2, emissivity 0.8, and temperature 150°C?

Example 7

Solution:

0.5 m2

(0.8) (5.67 x 10-8 W/m2.K4) (0.5 m2) (423 K)4

= 726 W

0.8

Q t

= ε σ A T4

5.67 x 10-8 W/m2.K4

Q t

[(273+150) K]4

=


86

If the surface of Example 7 is placed in a large, evacuated chamber whose walls are maintained at 25°C, what is the net rate at which radiation is exchanged between the surface and the chamber walls?

Example 8

Solution:

(0.8) x (5.67 x 10-8 W/m2.K4) x (0.5 m2)

x [(423 K)4 -(298 K)4 ]

= 547 W

Q t

[(273+150) K]4

=

Q t

= ε σ A (T4 - To4 )

[(273+25) K]4


87

Example 8 continued

Note that 547 W of heat loss from the surface occurs at the instant the surface is placed in the chamber. That is, when the surface is at 150oC and the chamber wall is at 25oC.

With increasing time, the surface would cool due to the heat loss. Therefore its temperature, as well as the heat loss, would decrease with increasing time.

Steady-state conditions would eventually be achieved when the temperature of the surface reached that of the surroundings.


88

Under steady state operation, a 50 W incandescent light bulb has a surface temperature of 135°C when the room air is at a temperature of 25°C. If the bulb may be approximated as a 60 mm diameter sphere with a diffuse, gray surface of emissivity 0.8, what is the radiant heat transfer from the bulb surface to its surroundings?

Example 9

Solution:

(0.8) x (5.67 x 10-8 J/s.m2.K4) x [π x (0.06) m2]

x [(408 K)4 -(298 K)4 ]

= 10.2 W (about 20% of the power is dissipated by radiation)

Q t

[(273+135) K]4

=

Q t

= ε σ A (T4 - To4 )

[(273+25) K]4

Documents

Prof. R. Shanthini 5 & 12 March 2012 1 PM3125: Lectures 6 to 9 Content of Lectures 6 to 12: Heat transfer: - Source of heat - Heat transfer - Steam and