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PM3125: Lectures 4 to 6. 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. Heat Transfer.

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PM3125: Lectures 4 to 6


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pm3125 lectures 4 to 6
PM3125: Lectures 4 to 6
  • 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
heat transfer
Heat Transfer

is the means by which energy moves from

a hotter object to

a colder object

mechanisms of heat transfer
Mechanisms of Heat Transfer

Conduction

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

Convection

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

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

Radiation

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

(by infrared radiation, or light)

slide4

Mechanisms of Heat Transfer

Latent heat

Conduction

Convection

Radiation

slide5

Conduction

HOT

(lots of vibration)

COLD

(not much vibration)

Heat travels along the rod

slide6

Conduction

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.

slide7

Conduction: Fourier’s Law

( )

ΔT

k A t

Q

=

L

Cross-sectional area A

L

Q = heat transferred

k = thermal conductivity

A = cross sectional area

DT = temperature difference

between two ends

L = length

t = duration of heat transfer

What is the unit of k?

slide9

Conduction through Single Wall

.

.

.

Q

Q

Q

( )

ΔT

k A t

Q

=

L

Use Fourier’s Law:

T1

k A (T1 – T2)

T2 T1

=

x

Δx

Δx

slide10

Conduction through Single Wall

.

.

.

Q

Q

Q

k A (T1 – T2)

T1

=

Δx

T1 – T2

=

Δx/(kA)

T2 T1

x

Δx

Thermal resistance (in k/W)

(opposing heat flow)

10

slide11

Conduction through Composite Wall

.

.

.

Q

Q

Q

B

C

A

T1

T2

T3

T4

kA

kB

kC

x

ΔxA

ΔxB

ΔxC

T1 – T2

T3 – T4

T2 – T3

=

=

=

(Δx/kA)A

(Δx/kA)C

(Δx/kA)B

11

slide12

Conduction through Composite Wall

T1 – T2

T3 – T4

T2 – T3

=

=

=

(Δx/kA)A

(Δx/kA)C

(Δx/kA)B

[

]

+ (Δx/kA)B

+ (Δx/kA)B

+ (Δx/kA)C

+ (Δx/kA)C

(Δx/kA)A

(Δx/kA)A

= T1 – T2 + T2 – T3 + T3 – T4

.

.

.

Q

Q

Q

T1 – T4

=

12

slide13

.

Q

Example 1

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.

Solution: We start with the equation

Tin – Tout

=

+ (Δx/kA)insulation

(Δx/kA)fireclay

slide14

.

.

Q

Q

Example 1 continued

(1000– 40) A

=

+ (0.03/0.07)

(0.21/1.04)

= 1522.6

W/m2

A

slide15

.

.

Q

Q

Example 2

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

Solution: We start with the equation

Tin – Tout

=

+ (Δx/kA)insulation

(Δx/kA)fireclay

(1000– 40)

W/m2

= 960

=

A

+ (Δx)insulation/0.07)

(0.21/1.04)

(Δx)insulation

cm

= 5.6

slide16

Conduction through hollow-cylinder

.

Q

ro

Ti

ri

To

L

Ti – To

=

[ln(ro/ri)] / 2πkL

slide17

.

Q

Conduction through the composite wall in a hollow-cylinder

r3

r2

To

Material A

Ti

r1

Material B

Ti – To

=

+ [ln(r3/r2)] / 2πkBL

[ln(r2/r1)] / 2πkAL

slide18

.

Q

Example 3

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.

Solution: We start with the equation

Ti – To

=

+ [ln(r3/r2)] / 2πkBL

[ln(r2/r1)] / 2πkAL

slide19

.

.

Q

Q

Example 3 continued

2 π L (600 – 100)

=

+ [ln(5/2)] / 0.2

[ln(2/1)] / 19

= 680 W/m

L

mechanisms of heat transfer1
Mechanisms of Heat Transfer

Conduction

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

Convection

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

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

Radiation

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

(by infrared radiation, or light)

convection
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.

slide22

Convection

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

slide23

Convection: Newton’s Law of Cooling

= h A (Tsurface – Tfluid)

conv.

.

Q

Flowing fluid at Tfluid

Heated surface at Tsurface

Area exposed

Heat transfer coefficient (in W/m2.K)

slide24

Convection: Newton’s Law of Cooling

conv.

.

Q

Flowing fluid at Tfluid

Heated surface at Tsurface

Tsurface – Tfluid

=

1/(hA)

Convective heat resistance (in k/W)

slide25

.

Q

= h A (Tsurface – Tfluid)

conv.

.

Q

conv.

A

Example 4

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.

Solution:

Use Newton’s Law of cooling :

Flowing fluid at Tfluid = 30oC

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

Heated surface at Tsurface = 50oC

Heat flux leaving the surface:

= 20 x 20

= 400 W/m2

h = 20 W/m2.K

slide26

.

Q

= h A (Tsurface – Tfluid)

conv.

Example 5

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.

Solution:

Use Newton’s Law of cooling :

Flowing fluid at Tfluid = 300oC

Heated surface at Tsurface = 40oC

= 250 W/m2.K x 0.125 m2

x (40 - 300)oC

= - 8125 W/m2

h = 250 W/m2.K

A = 0.50x0.25 m2

Heat is transferred from the air to the plate.

slide27

Forced Convection

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

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

slide28

Free Convection

In 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

slide29

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.

slide30

Overall Heat Transfer through a Plane Wall

.

.

.

Q

Q

Q

T1 – T2

T2– TB

TA – T1

=

=

=

1/(hBA)

1/(hAA)

Δx/(kA)

Fluid A

at TA > T1

T1

T2

Fluid B

at TB < T2

x

Δx

slide31

Overall Heat Transfer through a Plane Wall

.

.

Q

Q

T1 – T2

T2– TB

TA – T1

=

=

=

1/(hBA)

1/(hAA)

Δx/(kA)

.

TA – TB

Q

=

1/(hAA)

+ Δx/(kA)

+ 1/(hBA)

(TA – TB)

= U A

where U is the overall heat transfer coefficient given by

1/U = 1/hA

+ Δx/k

+ 1/hB

slide32

Overall heat transfer through hollow-cylinder

ro

Ti

ri

To

.

Q

Fluid A is inside the pipe

Fluid B is outside the pipe

TA > TB

L

(TA – TB)

= U A

where

1/UA = 1/(hAAi)

+ ln(ro/ri) / 2πkL

+ 1/(hBAo)

slide33

.

Q

Example 6

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.

Solution: Start with

(120 – 35)

(TA – TB)

= U A

= U A

What is UA?

slide34

Example 6 continued

1/UA = 1/(hAAi)

+ ln(ro/ri) / 2πkL + …

+ 1/(hBAo)

+ ln(5.5/5) / 2π(45)L

1/UA = 1/(85Ain)

+ 1/(12.5Aout)

+ ln(8/5.5) / 2π(0.07)L

Ain = 2π(0.05)L and

Aout = 2π(0.08)L

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

slide35

.

.

Q

Q

Example 6 continued

UA = 2πL / (0.235 + 0.0021 +5.35 + 1)

(120 – 35)

= U A

air

steel

=2πL

(120 – 35) / (0.235 + 0.0021 +5.35 + 1)

insulation

steam

= 81 L

/ L

= 81 W/m

mechanisms of heat transfer2
Mechanisms of Heat Transfer

Conduction

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

Convection

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

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

Radiation

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

(by infrared radiation, or light)

radiation
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.

the stefan boltzmann law of radiation

Q

= εσ A T4

t

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.

why is the mother shielding her cub
Why is the mother shielding her cub?

Ratio of the surface area of a cub to its volume is much larger than for its mother.

what is the sun s surface temperature
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    

what is the sun s surface temperature1

Q

= εσ A T4

t

What is the Sun’s surface temperature?

(4 π 6.92 x 1016 m2)

= 5.98 x 1018 m2

(4 π 1.52 x 1022 m2)(1000 W/m2)

= 2.83 x 1026 W

2.83 x 1026 W

T4 =

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

ε

σ

T = 5375 K

slide42

Q

= εσ A (T4 - To4 )

t

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

slide43

Q

= εσ A T4

t

Example 7

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

Solution:

[(273+150) K]4

0.5 m2

0.8

5.67 x 10-8 W/m2.K4

Q

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

=

t

= 726 W

slide44

Q

= εσ A (T4 - To4 )

t

Example 8

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?

Solution:

[(273+25) K]4

[(273+150) K]4

Q

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

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

=

t

= 547 W

slide45

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.

slide46

Q

= εσ A (T4 - To4 )

t

Example 9

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?

Solution:

[(273+25) K]4

[(273+135) K]4

Q

(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 ]

=

t

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

slide47

Mathematical Problems on Heat Exchanger

Q = mc cc (Tc,out – Tc,in)

= mh ch (Th,in – Th,out)

Tc,in

Th,out

Th,in

Tc,out

.

.

.

slide48

Mathematical Problems on Heat Exchanger

Th,in

Th,out

Tc,out

Tc,in

Tc,in

Parallel-flow heat exchanger

Th,out

Th,in

Tc,out

high heat transfer

low heat transfer

slide49

Mathematical Problems on Heat Exchanger

Th,in

Th,out

Tc,out

Tc,in

Parallel-flow heat exchanger

ΔTa

ΔTb

a

b

.

Q = U A ΔT

ΔTa - ΔTb

is the log mean temperature difference (LMTD)

where ΔT =

ln(ΔTa / ΔTb)

slide50

Mathematical Problems on Heat Exchanger

Q = mc cc (Tc,out – Tc,in)

= mh ch (Th,in – Th,out)

Tc,out

Counter-flow heat exchanger

Th,out

Th,in

Tc,in

.

.

.

slide51

Mathematical Problems on Heat Exchanger

Tc,out

Counter-flow heat exchanger

Th,out

Th,in

Th,in

Tc,in

Th,out

Tc,out

Tc,in

slide52

Mathematical Problems on Heat Exchanger

Th,in

Th,out

Tc,out

Tc,in

Counter-flow heat exchanger

ΔTa

ΔTb

a

b

.

Q = U A ΔT

ΔTa - ΔTb

is the log mean temperature difference (LMTD)

where ΔT =

ln(ΔTa / ΔTb)

slide53

Example in heat Exchanger Design

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.

slide54

Q = mc cc (Tc,out – Tc,in)

= mh ch (Th,in – Th,out)

Example in heat Exchanger Design

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.

slide55

Example in heat Exchanger Design

  • Solution continued:
  • Parallel flow:

ΔTa = 350 – 25 = 325oC

ΔTb = 100 – 34.63 = 65.37oC

ΔTa - ΔTb

325 – 65.37

ΔT =

= 162oC

=

ln(ΔTa / ΔTb)

ln(325 / 65.37)

.

= (UA) 162oC

Q = U A ΔT

What is UA?

slide56

Example in heat Exchanger Design

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)

.

= (0.25) π (0.075) L (162) kW

= (UA) 162oC

Q

.

What is Q?

slide57

Q = mc cc (Tc,out – Tc,in)

= mh ch (Th,in – Th,out)

Example in heat Exchanger Design

Solution continued:

.

.

.

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

= 15.69 kW

Substituting the above in

.

= (0.25) π (0.075) L (162) kW

= (UA) 162oC

Q

we get

L = 1.64 m

slide58

Example in heat Exchanger Design

Solution continued:

(ii) Counter flow:

ΔTa = 350 – 34.63 = 315.37oC

ΔTb = 100 – 25 = 75oC

ΔTa - ΔTb

315.37 – 75

ΔT =

= 167.35oC

=

ln(ΔTa / ΔTb)

ln(315.37 / 75)

.

= (UA) 167.35oC

Q = U A ΔT

.

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

A = π (0.075) L m2

Therefore, L = 1.59 m

slide59

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.

slide60

Other Heat Exchanger Types

One shell pass and

two tube passes

Th,in

Tc,in

Tc,out

Th,out

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

slide61

Other Heat Exchanger Types

Two shell passes and

two tube passes

Th,in

Tc,in

Tc,out

Th,out

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

slide62

Batch Sterilization (method of heating):

Electrical

heating

Steam

heating

Direct steam

sparging

slide63

For batch heating with constant rate heat flow:

Total heat lost by the coil to the medium

= heat gained by the medium

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

.

Electrical

heating

.

q t =

M c (T - T0)

slide64

For batch heating by direct steam sparging:

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

.

.

.

+ M c T0

= (M + mst) c T

(ms t) (H + cT0)

Direct steam

sparging

.

.

= (M + ms t) c (T – T0)

ms t H

slide65

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

( )

T0 - TH

Steam

heating

U A t = M c ln

T- TH

Could you prove the above?

slide66

For batch heating with isothermal heat source:

( )

T0 - TH

U A t = M c ln

T- TH

( )

U A t

T = TH + (T0 - TH) exp -

c M

Steam

heating

slide67

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.)

slide68

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:

Use the equation below:

.

.

= (M + ms t) c (T – T0)

ms t H

slide69

.

.

= (M + ms t) c (T – T0)

ms t H

(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

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

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

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

th = 1.463 hr

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

slide70

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:

( )

T0 - TH

U A t = M c ln

T- TH

slide71

( )

T0 - TH

U A t = M c ln

T- TH

(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)]

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

(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.

slide72

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

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Question from PM3125 / Jan 2010 past paper

  • 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:
slide74

Question from PM3125 / Jan 2010 past paper

Additional Data:

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

where L is the length of pipe, T1 and T2are the respective temperatures at the inner and outer surfaces of the insulated pipe, r1and 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.

slide75

Group Assignment will be uploaded at

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

(keep track of the site)

slide78

.

Q

Critical Radius of Insulation

To

r

Pipe

Insulation

ro

Ti

ri

Ti – To

=

[ln(ro/ri)] /2πkPL

+ [ln(r/ro)] /2πkIL+ 1/hairA

Pipe resistance could be neglected

A = 2 π r L

slide79

.

Q

Critical Radius of Insulation

Ti – To

=

[ln(r/ro)] /2πkIL+ 1/(hair 2πrL)

2π L ( Ti – To)

=

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

Convective resistance

Insulation

resistance

Increasing r increases insulation resistance and decreases heat transfer.

Increasing r decreases convective resistance and increases heat transfer.

slide80

.

Q

Critical Radius of Insulation

d

/dr

= 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.

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