R. Shanthini 17 May 2010 1 Content of Lectures 1 to 6: Heat transfer: Source of heat Heat transfer Steam and electricity as heating media Determination

R. Shanthini 17 May 2010

1

Content of Lectures 1 to 6:

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

PM3125: Lectures 4 to 6


2

Heat Transferis the means by which energy moves from

a hotter object to

a colder object


3

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)


4

Latent heat Conduction

Convection

Radiation

Mechanisms of Heat Transfer


5

Conduction

HOT(lots of vibration)

COLD(not much vibration)

Heat travels along the rod


6

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


7

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?


8

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


9

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( )


10

T

T T1

x

10

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


11

T

T

x

11

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


12

12

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

=


13

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


14

Example 1 continued

(1000 – 40) AQ.

+ (0.03/0.07) (0.21/1.04)

=

= 1522.6 W/m2Q.

A


15

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


16

L

Conduction through hollow-cylinder

Ti – ToQ.

[ln(ro/ri)] / 2πkL

=

ri

ro

Ti

To


17

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


18

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



19

Example 3 continued

2 π L ( 600 – 100)Q.

[ln(2/1)] / 19

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

Q.

L= 680 W/m


20








21

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.


22

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

Convection


23

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


24

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


25

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 :


26

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


27

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.


28

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


29

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.


30

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)


31

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.


32

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


33

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?


34

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


35

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


36








37

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.


38

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


39

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?


40

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


41

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


42

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 )


43

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

=


44

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


45

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.


46

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


47

Mathematical Problems on Heat ExchangerTc,in

Tc,out

Th,in

Th,out

. .Q = mc cc (Tc,out – Tc,in) = mh ch (Th,in – Th,out)

.


48

Mathematical Problems on Heat Exchanger

Tc,in

Th,out

Th,in

Tc,out

Tc,in

Tc,out

Th,in

Th,out

Parallel-flow heat exchanger

high heat transfer

low heat transfer


49


Tc,in

Th,out

Th,in

Tc,out

a b

Q = U A ΔT .

where ΔT = ΔTa - ΔTb

ln(ΔTa / ΔTb)is the log mean temperature difference (LMTD)

ΔTa ΔTb

Parallel-flow heat exchanger


50

Mathematical Problems on Heat ExchangerTc,out

Tc,in

Th,in

Th,out


.

Counter-flow heat exchanger


51


Tc,in

Tc,out

Th,in

Th,out

Th,in

Th,out


Tc,in

Tc,out


52


a b

Q = U A ΔT .

where ΔT = ΔTa - ΔTb

ln(ΔTa / ΔTb)is the log mean temperature difference (LMTD)

ΔTa

ΔTb


Tc,in

Tc,out

Th,in

Th,out


53

An exhaust pipe, 75 mm outside diameter, is cooled by surrounding it by an annular space containing water.

The hot gases enters the exhaust pipe at 350oC, gas flow rate being 200 kg/h, mean specific heat capacity at constant pressure 1.13 kJ/kg K, and comes out at 100oC.

Water enters from the mains at 25oC, flow rate 1400 kg/h, mean specific heat capacity 4.19 kJ/kg K.

The heat transfer coefficient for gases and water may be taken as 0.3 and 1.5 kW/m2 K and pipe thickness may be taken as negligible.

Calculate the required pipe length for (i) parallel flow, and for (ii) counter flow.

Example in heat Exchanger Design


54

Solution:



.

(1400 kg/hr) (4.19 kJ/kg K) (Tc,out – 25)oC

= (200 kg/hr) (1.13 kJ/kg K) (350 – 100)oC

The temperature of water at the outlet = Tc,out = 34.63oC.


55

Solution continued:

(i) Parallel flow:


ΔTa = 350 – 25 = 325oC

ΔTb = 100 – 34.63 = 65.37oC

ΔT = ΔTa - ΔTb

ln(ΔTa / ΔTb)

325 – 65.37

ln(325 / 65.37)= = 162oC

Q = U A ΔT .

= (UA) 162oC

What is UA?


56

Solution continued:

1/U = 1/hwater + 1/hgases

= 1/1.5 + 1/0.3 = 4 (kW/m2 K)-1

Therefore, U = 0.25 kW/m2 K

A = π (outer diameter) (L) = π (0.075 m) (L m)


Q .

= (UA) 162oC

What is Q?

= (0.25) π (0.075) L (162) kW

.


57

Solution continued:


Q .

= (UA) 162oC = (0.25) π (0.075) L (162) kW


.

= (200 kg/h) (1.13 kJ/kg K) (350 – 100)oC = 15.69 kW

Substituting the above in

we getL = 1.64 m


58

Solution continued:

(ii) Counter flow:


ΔTa = 350 – 34.63 = 315.37oC

ΔTb = 100 – 25 = 75oC

ΔT = ΔTa - ΔTb

ln(ΔTa / ΔTb)

315.37 – 75

ln(315.37 / 75)= = 167.35oC

Q = U A ΔT .

= (UA) 167.35oC

Q = 15.69 kW; U = 0.25 kW/m2 K ;.

A = π (0.075) L m2

Therefore, L = 1.59 m


59

Other Heat Exchanger Types

Cross-flow heat exchanger with both fluids unmixed

The direction of fluids are perpendicular to each other.

The required surface area for this heat exchanger is usually calculated by using tables.

It is between the required surface area for counter-flow and parallel-flow heat exchangers.


60


One shell pass and two tube passes

Th,in

Th,out

Tc,in

Tc,out

The required surface area for this heat exchanger is calculated using tables.


61


Two shell passes and two tube passes

Th,inTc,in

The required surface area for this heat exchanger is calculated using tables.

Th,out

Tc,out


62

Batch Sterilization (method of heating):

Steamheating

Electricalheating

Direct steam sparging


63

For batch heating with constant rate heat flow:

M c (T - T0) q t =

M - mass of the medium

T0 - initial temperature of the medium

T - final temperature of the medium

c - specific heat of the medium

q - rate of heat transfer from the electrical coil to the medium

t - duration of electrical heating

Electricalheating

Total heat lost by the coil to the medium

= heat gained by the medium

.

.


64

Direct steam sparging = (M + ms t) c (T – T0)

M - initial mass of the raw medium

T0 - initial temperature of the raw medium

ms - steam mass flow rate

t - duration of steam sparging

H - enthalpy of steam relative to the enthalpy at the initial temperature of the raw medium (T0)

T - final temperature of the mixture

c - specific heat of medium and water

For batch heating by direct steam sparging:

.

(ms t) (H + cT0).

+ M c T0 = (M + mst) c T.

ms t H. .


65

For batch heating with isothermal heat source:

M - mass of the medium

T0 - initial temperature of the medium

TH - temperature of heat source (steam)

T - final temperature of the medium

c - specific heat of the medium

t - duration of steam heating

U - overall heat transfer coefficient

A - heat transfer area

Steamheating

U A t = M c ln

Could you prove the above?

T0 - TH ( ) T - TH


66

For batch heating with isothermal heat source:

Steamheating

T = TH + (T0 - TH) exp - U A tc M ( )

U A t = M c lnT0 - TH ( ) T - TH


67

Example of batch heating by direct steam sparging:

A fermentor containing 40 m3 medium at 25oC is going to be sterilized by direct injection of saturated steam. The steam at 350 kPa absolute pressure is injected with a flow rate of 5000 kg/hr, which will be stopped when the medium temperature reaches 122oC. Determine the time taken to heat the medium.

Additional data required:

Enthalpy of saturated steam at 350 kPa = ??

Enthalpy of water at 25oC = ??

The heat capacity of the medium 4.187 kJ/kg.K

The density of the medium are 4.187 kJ/kg.K and 1000 kg/m3, respectively.)


68

Example of batch heating by direct steam sparging:

A fermentor containing 40 m3 medium at 25oC is going to be sterilized by direct injection of saturated steam. The steam at 350 kPa absolute pressure is injected with a flow rate of 5000 kg/hr, which will be stopped when the medium temperature reaches 122oC. Determine the time taken to heat the medium.

Additional data: The enthalpy of saturated steam at 350 kPa and water at 25oC are 2732 and 105 kJ/kg, respectively. The heat capacity and density of the medium are 4.187 kJ/kg.K and 1000 kg/m3, respectively.

Solution:

= (M + ms t) c (T – T0)ms t H. .

Use the equation below:


69

(5000 kg/hr) (th) (2732-105) kJ/kg

= [(40 m3)(1000 kg/m3) + (5000 kg/hr)(th)](4.187 kJ/kg.K)(122-25)K

(5000 th) (2627) kJ = [40000 + 5000 t](4.187)(97)kJ

Taking the heating time (th) to be in hr, we get

(5000 th) [2627 – 4.187 x 97] = 40000 x 4.187 x 97

th = 1.463 hr

= (M + ms t) c (T – T0)ms t H. .

Therefore, the time taken to heat the medium is 1.463 hours.


70

Example of batch heating with isothermal heat source:

A fermentor containing 40 m3 medium at 25oC is going to be sterilized by an isothermal heat source, which is saturated steam at 350 kPa absolute pressure. Heating will be stopped when the medium temperature reaches 122oC. Determine the time taken to heat the medium.

Additional data: The saturated temperature of steam at 350 kPa is 138.9oC. The heat capacity and density of the medium are 4.187 kJ/kg.K and 1000 kg/m3, respectively.

Solution:

Use the equation below:



71


Taking the heating time (th) to be in hr, we get

(2500 kJ/hr.m2.K) (40 m2) (tc)

= (40 m3) (1000 kg/m3) (4.187 kJ/kg.K) ln[(25-138.9)/(122-138.9)]

(2500 kJ/K) (40) (th) = (40) (1000) (4.187 kJ/K) ln[113.9/16.9]

(2500 kJ/K) (40) (th) = (40) (1000) (4.187 kJ/K) (1.908)

th = 3.1955 hr

Therefore, the time taken to heat the medium is 3.1955 hours.


72

Explain why heating with isothermal heat source takes twice the time taken by heating with steam sparging, even though we used the same steam.


73

A steel pipeline (inside diameter = 52.50 mm; outside diameter = 60.32 mm) contains saturated steam at 121.1oC. The line is insulated with 25.4 mm of asbestos. Assume that the inside surface temperature of the metal wall is at 121.1oC and the outer surface of the insulation is at 26.7oC. Taking the average value of ksteel as 45 W/m.K and that of

kasbestos as 0.182 W/m.K, calculate the following:

(a) Heat loss for 30.5 m of pipe length. [10 marks]

(b) Mass (in kg) of steam condensed per hour in the pipe due to the heat loss. [10 marks]

Additional data given on the next slide:

Question from PM3125 / Jan 2010 past paper


74

Additional Data:

i) Heat transfer rate through the pipe wall is given by,

asbestos

23

steel

12

21

)/ln()/ln(

)(2

k

rr

k

rr

TTLQ

where L is the length of pipe, T1 and T2 are the respective temperatures at

the inner and outer surfaces of the insulated pipe, r1 and r2 are the

respective inner and outer radius of the steel pipe, and r3 is the outer

radius of the insulated pipe. ii) Latent heat of vapourization of steam could be taken as 2200 kJ/kg.

Question from PM3125 / Jan 2010 past paper


75

Group Assignment will be uploaded at

http://www.rshanthini.com/PM3125.htm

(keep track of the site)


76

End of slides for the heat transfer lecture


77

Additional material not used in the lectures.


78

Critical Radius of Insulation

Ti – ToQ.

[ln(ro/ri)] /2πkPL

=

ri

ro

Ti

Tor Pipe

+ [ln(r/ro)] /2πkIL + 1/hairA

Insulation

Pipe resistance could be neglected

A = 2 π r L


79

Critical Radius of InsulationTi – ToQ

.=

[ln(r/ro)] /2πkIL + 1/(hair 2πrL)

2π L ( Ti – To) =

[ln(r/ro)] /kI + 1/(hair r)Convective resistance

Insulationresistance

Increasing r decreases convective resistance and increases heat transfer.

Increasing r increases insulation resistance and decreases heat transfer.


80

Critical Radius of Insulation

= 0 at the critical radius of insulation,

which leads to rcr = kI / hair

If the outer radius of the pipe (ro) < rcr and if insulation is added to the pipe, heat losses will first

increase and go through a maximum at the insulation radius of rcr and then decrease.

dQ.

/dr

If the outer radius of the pipe (ro) > rcr and if insulation is added to the pipe, heat losses will

continue to decrease.

Documents

R. Shanthini 17 May 2010 1 Content of Lectures 1 to 6: Heat transfer: Source of heat Heat transfer Steam and electricity as heating media Determination