Introduction - University of Agriculture Faisalabad · From the compressor, high-pressure gas is sent to the ... 2.2 Too-Low Condensor Pressure ... The high-pressure liquid refrigerant

1

Basic Refrigeration Cycle

Introduction

All air conditioning systems must have four basic elements:

Compressor

Condensor

Expansion Valve

Evaporator

From the compressor, high-pressure gas is sent to the condenser, where the heat is dissipated and

condensed to liquid. The high-pressure liquid flows on to the expansion valve, where it is metered

and its pressure is reduced. At the evaporator, the liquid absorbs heat from the air and evaporates to

gas. The cycle is then repeated, starting at the compressor.

Principles of Refrigeration

Liquids absorb heat when changed from liquid to gas

Gases give off heat when changed from gas to liquid.

For an air conditioning system to operate with economy, the refrigerant must be used repeatedly.

For this reason, all air conditioners use the same cycle of compression, condensation, expansion,

and evaporation in a closed circuit. The same refrigerant is used to move the heat from one area,

to cool this area, and to expel this heat in another area.

The refrigerant comes into the compressor as a low-pressure gas, it is compressed and

then moves out of the compressor as a high-pressure gas.

The gas then flows to the condenser. Here the gas condenses to a liquid, and gives off its

heat to the outside air.

The liquid then moves to the expansion valve under high pressure. This valve restricts the

flow of the fluid, and lowers its pressure as it leaves the expansion valve.

The low-pressure liquid then moves to the evaporator, where heat from the inside air is

absorbed and changes it from a liquid to a gas.

As a hot low-pressure gas, the refrigerant moves to the compressor where the entire cycle

is repeated.

Note that the four-part cycle is divided at the center into a high side and a low side This refers to

the pressures of the refrigerant in each side of the system

Performance Testing

http://www.swtc.edu/Ag_Power/air_conditioning/lecture/compressor.htm

http://www.swtc.edu/Ag_Power/air_conditioning/lecture/condensor.htm

http://www.swtc.edu/Ag_Power/air_conditioning/lecture/expansion_valve.htm

http://www.swtc.edu/Ag_Power/air_conditioning/lecture/evaporator.htm

2

Compound Gauge (Low Side)

The compound gauge derives its name from its function. It will register both pressure or

vacuum.

All air conditioning systems can, under certain conditions, drop from a pressure into a

vacuum on the low side. It is necessary that a gauge be used that will show either

pressure (psi and kPa) or inches of mercury vacuum (Hg.).

The vacuum side of the gauge must be calibrated to show 0 to 30 inches (0 to 762 mm)

Hg. The pressure side of the gauge must be calibrated to register accurately from 0

pressure to a minimum of 60 psi (414 kPa).

The maximum reading of the pressure should not exceed 160 psi (1103 kPa). Practically

all readings of the low side of the system will be less than 60 psi (414 kPa) with the

system in operation.

High Pressure Gauge (High Side)

The high pressure gauge is used to determine pressures in the high side of the system.

The gauge is calibrated to register accurately from zero pressure to a minimum of 300 psi

(2070 kPa).

A few systems operate under high head pressure during normal operation conditions.

This is why the high pressure gauge should have a reading of at least 600 psi (4140 kPa).

Gauge Manifold

The gauge manifold mounts the high and low side gauges and connects the gauges into

the high and low sides of the system by means of test hoses.

The gauges connect to the upper part of the manifold through holes drilled and tapped to

a 1/8-inch pipe thread.

Test hose connectors below the gauges on the lower side of the manifold direct the

refrigerant through the manifold to the gauges to obtain pressure readings.

A center test hose connector on the lower side of the manifold is connected to both

pressure gauges and the test hoses by a passage in the manifold.

Refrigerant flow into the high and low side is controlled by a shutoff hand valve at each

end of the manifold.

Components of refrigeration system 1. Compressor

The purpose of the compressor is to circulate the refrigerant in the system

under pressure, this concentrates the heat it contains.

At the compressor, the low pressure gas is changed to high pressure

gas.

This pressure buildup can only be accomplished by having a restriction in the high

pressure side of the system. This is a small valve located in the expansion valve.

The compressor has reed valves to control the entrance and exit of refrigerant gas during the

pumping operation. These must be firmly seated.

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An improperly seated intake reed valve can result in gas leaking back into the low side

during the compression stroke, raising the low side pressure and impairing the cooling

effect.

A badly seated discharge reed valve can allow condensing or head pressure to drop as it

leaks past the valve, lowering the efficiency of the compressor.

Two service valves are located near the compressor as an aid in servicing the system.

One services the high side, it is quickly identified by the smaller discharge hose routed to

the condenser.

One is used for the low side, the low side comes from the evaporator, and is larger than

the discharge hose

The compressor is normally belt-driven from the engine crankshaft. Most manufacturers use a

magnetic-type clutch which provides a means of stopping the pumping of the compressor when

refrigeration is not desired.

1.1 Compressor Relief Valve

Some compressors have a relief valve for regulating pressure. If the system discharge pressure

exceeds rated pressure, the valve will open automatically and stay open until the pressure drops.

The valve will then close automatically.

1.2 Compressor Noise Complaints

Many noise complaints can be traced to the compressor mount and drive.

If a unit is noisy at one speed and quiet at another, it is not compressor noise.

Many times this kind of noise can be eliminated or greatly reduced by changing the belt

adjustment.

Usually tightening mounts, adding idlers, or changing belt adjustment and length are

more successful in removing or reducing this type of noise, than replacing the

compressor.

Noises from the clutch are difficult to recognize because the clutch is so close to the

compressor. A loose bolt holding the clutch to the shaft will make a lot of noise.

The difference, between suction pressure and discharge pressure, also plays an important

part on sound level.

o A compressor with low suction pressure will be more noisy than one with a higher

pressure.

Consider whether the system is properly charged, whether the expansion valve is feeding

properly to use the evaporator efficiently, and whether enough air is being fed over the

evaporator coil.

2. Condensor

The purpose of the condenser is to receive the high-pressure gas

from the compressor and convert this gas to a liquid.

It does it by heat transfer, or the principle that heat will always

move from a warmer to a cooler substance.

Air passing over the condenser coils carries off the heat and the gas condenses.

The condenser often looks like an engine radiator.

Condensers used on R-12 and R-134a systems are not interchangeable. Refrigerant-134a has a

different molecular structure and requires a large capacity condenser.

As the compressor subjects the gas to increased pressure, the heat intensity of the refrigerant is

actually concentrated into a smaller area, thus raising the temperature of the refrigerant higher

than the ambient temperature of the air passing over the condenser coils. Clogged condenser fins

will result in poor condensing action and decreased efficiency.

A factor often overlooked is flooding of the condenser coils with refrigerant oil. Flooding results

from adding too much oil to the system. Oil flooding is indicated by poor condensing action,

causing increased head pressure and high pressure on the low side. This will always cause poor

cooling from the evaporator.

2.1 Too-High Condensor Pressure

Indicated By: Excessive head pressure on high side gauge.

Caused By: Restriction of refrigerant flow in high side of system or lack of air flow over

condenser coils.

2.2 Too-Low Condensor Pressure

Indicated By: Higher than normal pressure on low side gauge.

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Caused By: Failed compressor reed valve or piston. Heat exchange in the condenser will

be cut down, and the excessive heat will remain in the low side of the system.

3. Expansion Valve

The expansion valve removes pressure from the liquid refrigerant to allow expansion or change

of state from a liquid to a vapor in the evaporator.

The high-pressure liquid refrigerant entering the expansion valve is quite warm. This may be

verified by feeling the liquid line at its connection to the expansion valve. The liquid refrigerant

leaving the expansion valve is quite cold. The orifice within the valve does not remove heat, but

only reduces pressure. Heat molecules contained in the liquid refrigerant are thus allowed to

spread as the refrigerant moves out of the orifice. Under a greatly reduced pressure the liquid

refrigerant is at its coldest as it leaves the expansion valve and enters the evaporator.

Pressures at the inlet and outlet of the expansion valve will closely approximate gauge pressures

at the inlet and outlet of the compressor in most systems. The similarity of pressures is caused by

the closeness of the components to each other. The slight variation in pressure readings of a very

few pounds is due to resistance, causing a pressure drop in the lines and coils of the evaporator

and condenser.

Two types of valves are used on machine air conditioning systems:

Internally-equalized valve - most common

Externally-equalized valve special control

3.1 Internally-Equalized Expansion Valve

The refrigerant enters the inlet and screen as a high-pressure

liquid. The refrigerant flow is restricted by a metered orifice

through which it must pass.

As the refrigerant passes through this orifice, it changes from a

high-pressure liquid to a low-pressure liquid (or passes from the

high side to the low side of the system).

Let's review briefly what happens to the refrigerant as we change its pressure.

As a high-pressure liquid, the boiling point of the refrigerant has been raised in direct proportion

to its pressure. This has concentrated its heat content into a small area, raising the temperature of

the refrigerant higher than that of the air passing over the condenser. This heat will then transfer

from the warmer refrigerant to the cooler air, which condenses the refrigerant to a liquid.

The heat transferred into the air is called latent heat of condensation. Four pounds (1.8 kg) of

refrigerant flowing per minute through the orifice will result in 12,000 Btu (12.7 MJ) per hour

transferred, which is designated a one ton unit. Six pounds (2.7 kg) of flow per minute will result

in 18,000 Btu (19.0 MJ) per hour, or a one and one-half ton unit.

3.2 Let's look at each valve in detail.

The refrigerant flow through the metered orifice is extreamly important, anything restricting the

flow will affect the entire system.

If the area cooled by the evaporator suddenly gets colder, the heat transfer requirements

change. If the expansion valve continued to feed the same amount of refrigerant to the

evaporator, the fins and coils would get colder until they eventually freeze over with ice

and the air flow is stopped.

A thermal bulb has a small line filled with C02 is attached to the evaporator tailpipe. If

the temperature on the tail pipe raises, the gas will expand and cause pressure against the

diaphram. This expansion will then move the seat away from the orifice, allowing an

increased refrigerant flow. As the tail pipe temperature drops, the pressure in the thermal

bulb also drops, allowing the valve to restrict flow as required by the evaporator.

The pressure of the refrigerant entering the evaporator is fed back to the underside of the

diaphragm through the internal equalizing passage. Expansion of the gas in the thermal

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bulb must overcome the internal balancing pressure before the valve will open to increase

refrigerant flow.

A spring is installed against the valve and adjusted to a predetermined setting at the time

of manufacture. This is the superheat spring which prevents slugging of the evaporator

with excessive liquid.

Superheat is an increase in temperature of the gaseous refrigerant above the temperature

at which the refrigerant vaporizes. The expansion valve is designed so that the

temperature of the refrigerant at the evaporator outlet must have 8 to 12°F (4 to 7°C) of

superheat before more refrigerant is allowed to enter the evaporator.

The adjusted tension of this spring is the determining factor in the opening and closing of

the expansion valve. During opening or closing, the spring tension retards or assists valve

operation as required.

o Normally, this spring is never adjusted in the field. Tension is adjusted from four

to sixteen degrees as required for the unit on which it is to be installed. This

original setting is sufficient for the life of the valve, and special equipment is

required in most cases to accurately calibrate this adjustment

3.3. Externally-Equalized Expansion Valve

Operation of the externally-equalized valve is the same as the internal type except that

evaporator pressure is fed against the underside of the diaphragm from the tail pipe of the

evaporator by an equalizer line. This balances the temperature of the tail pipe through the

expansion valve thermal bulb against the evaporator pressure taken from the tail pipe.

4. Evaporator

The evaporator works the opposite of the condenser, here

refrigerant liquid is converted to gas, absorbing heat from the

air in the compartment.

When the liquid refrigerant reaches the evaporator its pressure has been reduced, dissipating its

heat content and making it much cooler than the fan air flowing around it. This causes the

refrigerant to absorb heat from the warm air and reach its low boiling point rapidly. The

refrigerant then vaporizes, absorbing the maximum amount of heat.

This heat is then carried by the refrigerant from the evaporator as a low-pressure gas through a

hose or line to the low side of the compressor, where the whole refrigeration cycle is repeated.

The evaporator removes heat from the area that is to be cooled. The desired temperature of

cooling of the area will determine if refrigeration or air conditioning is desired. For example,

food preservation generally requires low refrigeration temperatures, ranging from 40°F (4°C) to

below 0°F (-18°C).

A higher temperature is required for human comfort. A larger area is cooled, which requires that

large volumes of air be passed through the evaporator coil for heat exchange. A blower becomes

a necessary part of the evaporator in the air conditioning system. The blower fans must not only

draw heat-laden air into the evaporator, but must also force this air over the evaporator fins and

coils where it surrenders its heat to the refrigerant and then forces the cooled air out of the

evaporator into the space being cooled.

4.1 Fan Speeds

Fan speed is essential to the evaporation process in the system. Heat exchange, as we explained

under condenser operation, depends upon a temperature differential of the air and the refrigerant.

The greater the differential, the greater the amount of heat exchanged between the air and the

refrigerant. A high heat load, as is generally encountered when the system is turned on, will

allow rapid heat transfer between the air and the cooler refrigerant.

A blower fan turned on to its highest speed will deliver the most air across the fins and coils for

rapid evaporation.

For the coldest air temperature from the evaporator, operate the blower fan at the lowest speed

so the heat will be absorbed by the refrigerant from the air

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Problems of Flooded or Starved Evaporator Coils

Changing the state of the refrigerant in the evaporator coils is as important as the air flow over

the coils. Liquid refrigerant supplied to the coils by the expansion valve expands to a vapor as it

absorbs heat from the air. Some liquid refrigerant must be supplied throughout the total length of

the evaporator coils for full capacity.

A starved evaporator coil is a condition in which not enough refrigerant has been supplied

through the total coil length. Therefore, expansion of the refrigerant has not occurred through the

whole coil length, resulting in poor coil operation and too-low heat exchange.

A flooded evaporator is the opposite of the starved coil. Too much refrigerant is passed through

the evaporator coils, resulting in unexpanded liquid passing onto the suction line and into the

compressor.

Part Identification - Evaporator

http://www.swtc.edu/Ag_Power/air_conditioning/part_id/compressor.htm

Size Reduction

1

Prof. Dr. Muhammad Iqbal, Deptt. Farm

Machinery & Power, University of

Agriculture, Faisalabad

IntroductionSize reduction: Standardized small need based uniform size of product achieved

by mechanical means that includes:

1. Cutting 2. Crushing and grinding 3. Milling

Examples:

1. Cutting of fruits, cutting of vegetables {for earning}

2. Shredding sweet potatoes for drying.

3. Chopping green fodder for animals 4. Grinding lime stone for fertilizer

5. Grinding grain for livestock

6. Milling flour for bread making

Milling in trade terms is used relative to reduction of grain into meal or flour. Milling is an overall process including:

– Size reduction - Hulling Polishing

– Sorting

– Mixing and in some instances, certain chemical reactions

2




Objectives/Benefits• Increased surface area - grinding increases surface area of product

to cause faster chemical reaction in subsequent processes

• Improved palatability – to achieve feed size acceptable by animals.

• Allow mixing – Reduced smaller uniform size makes a material amenable to mixing e.g. ground alfalfa can be mixed with grain.

• Improved animal performance – Each animal and feed has some optimum feed particle size which optimizes digestive tract operation.

• Improved handling and processing qualities – Operations like pneumatic conveying, palleting, and oil extraction are improved by decreasing particle size.

3




Terminologies of Ground Materials• Particle – A particle is the state of subdivision of a material

where the shape depends on the process which formed it and /or the cohesive forces that hold it together; examples are-ears of corn, kernels of wheat, drops of water etc.

• Sieving – The process of classifying materials using a stack of sieves.

• Particle size – Particle is a representative dimension that describes the degree of comminuting of the particle e.g. sphere or any irregular shape. Particle size designation as a result of sieving is usually taken as a geometric mean of the size (GMD) of the smallest hole (d1) the particle passed through and the size the of the largest hole (d2) it would not pass through.

4

GMD =√ (d1 * d2)Prof. Dr. Muhammad Iqbal, Deptt. Farm



Particle statistics –

5

Table. Size distribution of 500 particles

Particle

size, di

micron

No of

particles, fi fi.di di-dm

fi(di-

dm) (di-dm)^2 fi(di-dm)^2 Cum p di^fi

2.8 4 11.20 0.01 1.149 -4.60 1.32 5.28 0.01 61.4656

3 15 45.00 0.02 0.949 -14.23 0.90 13.50 0.03 14348907

3.2 20 64.00 0.03 0.749 -14.98 0.56 11.21 0.06 1.27E+10

3.4 47 159.80 0.08 0.549 -25.79 0.30 14.16 0.14 9.54E+24

3.6 63 226.80 0.11 0.349 -21.97 0.12 7.66 0.26 1.11E+35

3.8 78 296.40 0.15 0.149 -11.61 0.02 1.73 0.41 1.67E+45

4 88 352.00 0.18 0.051 4.51 0.00 0.23 0.59 9.58E+52

4.2 69 289.80 0.15 0.251 17.33 0.06 4.35 0.73 1.01E+43

4.4 59 259.60 0.13 0.451 26.62 0.20 12.01 0.86 9.2E+37

4.6 35 161.00 0.08 0.651 22.79 0.42 14.84 0.94 1.57E+23

4.8 10 48.00 0.02 0.851 8.51 0.72 7.25 0.97 6492506

5 8 40.00 0.02 1.051 8.41 1.11 8.84 0.99 390625

5.2 4 20.80 0.01 1.251 5.00 1.57 6.26 1.00 731.1616

Sum 500 1974.4 1.00 0.00 7.31 107.33 9.58E+52

3.95 Avg SS 0.22

1.28

(GMD)

1974.4 SD 0.46Prof. Dr. Muhammad Iqbal, Deptt. Farm



6

Probabilty of particle size

0.00

0.05

0.10

0.15

0.20

2.5 3 3.5 4 4.5 5 5.5

Particle Size, micron

Pro

ba

bilit

y

Cumulative probability graph of particle size

0

0.2

0.4

0.6

0.8

1

2.5 3 3.5 4 4.5 5 5.5

Particle size, micron

Cu

mu

lati

ve

pro

ba

bili

ty,

cu

m p




Size characteristics

Size reduction machines performance includes:

– Machine capacity, (kg/hr)

– hp required per unit material reduced, (hp/kg)

– Size & shape of product before and after reduction

– The range in size and shape of resultant product

i.e. 6-Sigma quality term {for same mean ‘μ’ value}

7

6 σ1 < 6 σ2

More uniform less uniform

(Low range) (More range)




Six Sigma – what does it mean?• Six Sigma at many organizations simply means a

measure of quality that strives for near perfection. Six Sigma is a disciplined, data-driven approach and methodology for eliminating defects (driving toward six standard deviations between the mean and the nearest specification limit) in any process – from manufacturing to transactional and from product to service.

8




Effect of various parameters on size & shape size of individual grains after size reduction

The size & shape of individual grains after size reduction depends upon:

1. Physical characteristics of material (parent material)

a) Size; b) Shape; c) Moisture content

d) Shearing strength

2. Method of size reduction - The size and shape after reduction is never uniform. It is extremely improbable that even a small percentage of the grains would approximate any simple geometrical shape/figure.

Theoretically, an irregular figure is represented by:

1. An equivalent sphere

2. An equivalent cube or any other geometric figure. Then, surface area or volume is used as the basis of comparison.

9




Classes of reduced in size material

Dimension range – Particles accurately measured with minimum size of 1/8” = 0.125”

– Example: Diced fruit / vegetables, chopped forage

Sieve size – Particle size with minimum dimension range of 0.125” – 0.0029” approximately

– Example: ground feed, commercial fertilizer

Microscopic range – Particles with minimum range less than 0.0029”

10

Dimension range Sieve size Microscopic range

X ≥ 0.125̋ 1/8̋” 1/300”

(0.125 > X >0.0029)

X ≤ 0.0029 in




11

Tyler Standard Screen SievesMesh, No.

Openings to inch

Diameter of wire,

inch

Size of opening, inch

Actual Approximate

- 0.1480 1.0500 1

- 0.1350 0.7420 3/4

- 0.1050 0.5250 1/2

- 0.0920 0.3710 3/8

3 0.0700 0.2630 1/4

4 0.0650 0.1850 3/16

6 0.0360 0.1310 1/8

8 0.0320 0.0930 3/32

10 0.0350 0.0650 1/16

14 0.0250 0.0460 3/64

20 0.0172 0.0328 1/32

28 0.0125 0.0232 -

35 0.0122 0.0164 1/64

48 0.0092 0.0116 -

65 0.0072 0.0082 -100 0.0042 0.0058 -150 0.0026 0.0041 -200 0.0021 0.0029 -




• Tyler sieve series is based on 200-mesh screen (i.e.0.074 mm/ 0.0029 in opening)

• Progression of sieve openings in a stack of sieves is √2 : 1 i.e. next sieve in this series is 1.44 times larger than the previous one.

• Table- constitutes a normal set

• This opening size is based upon the 200 mesh sieve

• Each opening √2 = 1.414 times as larger the previous one e.g.:

12




Sieve No. Wire Diameter,

inch

Size of opening

=1.414 x previous size

65 0.0072 0.0082 x 1.414 = 0.00116

100 0.0042 0.0058 x 1.414 = 0.0082

150 0.0026 0.0041 x 1.414 = 0.0058

200 0.0021 0.0029 x 1.414 = 0.0041

0.0029

13




• The openings are square “□”, the size being the dimension of one side i.e. sieve no 200 mean each side of hole is 0.0029”

• Intermediate sieves with opening ratio of 4√2 = 1.189 are available, if added, would constitute a complete set i.e.

14

Mesh No. Wire Diameter,

inchOpening size, inch

100 0.0042 0.0041 * 1.414 or 0.0049*1.189 = 0.0058

0.0041*1.189 = 0.0049

150 0.0026 0.0029*1.414 or 0.0034481*1.189 = 0.0041

0.0029*1.189 = 0.0034481

200 0.0021 0.0029




Rule of Sieve development

Mesh 200 150 100

Sieve size,

inch

0.0029 * 4√2 = 0.03448

0.0029 * √2 = 0.0041

0.0029 * 2 = 0.0058

15




Procedure of Sieving Procedure:1-Take 250 g oven dry sample.

2-Grinding of sample is to be done. Operate

shaker for 10 minutes.

3-Weigh material on the finest sieve

containing material.

4. Operate the shaker for another 5-minutes

or epeat the procedure until weight in this

sieve does not change by more than 0.2 %

of total sample weight.

5-Weigh and record weights of material on

all the sieves.

16




Fineness Modulus and Uniformity Index

Fineness modulus and uniformity index indicate the uniformity of a grind feed or distribution of fines and coarse in the resultant product.

• FM = sum of weight fraction retained on each sieve / 100

• The ¾”,4,8,14,28,48, and 100 mesh sieves and pan are used in the set

17




18

GMD = 0.004e0.7005x ,

R2 = 0.99

0.0010

0.0100

0.1000

1.0000

0 1 2 3 4 5 6 7 8

Fineness Modulus

Geo

metric M

ean

D

iam

eter, in

ch

Effect of fineness modulus on geometric mean diameter of grain productFineness Modulus Graph

GMD=SQRT(Y1*Y2)

FM

Sieve

size GMD (in)

0 0.0000 0.0041

1 0.0058 0.0082

2 0.0116 0.0164

3 0.0232 0.0327

4 0.0460 0.0654

5 0.0930 0.1312

6 0.1850 0.2620

7 0.3710 0.5740




Mesh

Col-1

Opening

size, (IN)

Col-2

Percent of

material

retained

Col-3

Multiplier

Col-4

Multiply col-3 by col-4

Col-5=col-3 x col-4

3/8̋ 0.371 0.0 7 7 = 0.0

4 0.185 5.7 6 6 = 34.2

8 0.093 23.2 5 5 = 116.0

14 0.046 35.1 4 4 = 140.4

28 0.0232 18.4 3 3 = 55.2

48 0.0116 9.3 2 2 = 18.6

100 0.0058 5.8 1 1 = 5.8

PAN 2.5 0 0 = 0.0

Total 100 370.2

19

Note that if all the

material were fine

enough to pass

through including

No. 100 and retained on pan Then, FM=0

-------do----------- 48 and retained on No. 100 1





-------do----------- 3/8 and retained on No. 4 6

If all retained on 3/8” 7

FM = 370.2/100 = 3.7




Uniformity Index (UI)

• The FM gives guideline to find geometric mean diameter (GMD) but it does not indicate distribution of fines and coarse in any sample. For same FM and GMD, UI can be different. To find the relative properties of coarse, medium, and fine particles follow the following procedure.

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A

Screen

mesh

B

Percent

material

retained

C

Multiplier

B x C Sub

Total/10

3/8̋ 0 7 0 28.2/10 = 3

Coarse4 5.7 6 34.2

8 23.2 5 116.0

Sub

total=

28.2

14 35.1 4 140.4 53.2/10 = 5

Medium28 18.4 3 55.2

Sub

total=

53.2

48 9.3 2 18.6 17.6/10 = 2

Fine100 5.8 1 5.8

Pan 2.5 0 0.0

Sub

total=

17.6 370.2

Grand

total

100

21

A

Screen

mesh

B

Percent

material

retained

C

Multiplie

r

B x C Sub

Total/10

3/8̋ 0 7 00 13/10 =

1.3≈1

Coarse

4 4 6 24

8 9 5 25

Sub

total=

13

14 58 4 232 70/16 =

7≈7

Medium

28 12 3 36

Sub

total=

70

48 16 2 32 17/10 =

1.7≈2

Fine

100 1 1 1

Pan 0 0 0

Sub

total=

17

Grand

total

100 370

Sample-1 Sample-2

Sample No. FM UI GMD

1 370/10=3.7 3:5:2 0.0041 (2)FM = 0.0041 (2)3.7 = 0.053”

2 370/10=3.7 1:7:2 0.0041 (2)FM = 0.0041 (2)3.7 = 0.053”Prof. Dr. Muhammad Iqbal, Deptt. Farm



Value of ground feed

1. Advisable – Coarse grinding is normally advisable for more digestive processes.

2. Finely grain never advisable to feed farm animals except small chicks, and for them powdery feed never recommended.

3. Finally grain materials pass through the digestive tract too fast to be acted upon by various digestive processes.

22




Forage grinding • Coarsely chopped hay is advisable to minimize

the amount of material thrown out of the feed bunks by the animals.

• Fine chopping/grinding does not improve the forage and will probably lower its quality by exposure to oxidation. Also, the ability of the animal to digest the material will be decreased.

• Alfalfa crop for poultry be ground fine to provide necessary consistency for eating.

23




Fineness modulus for ground feeds

Material

Grind grains

Whole

grain

Coarse medium fine Very fine

Shelled

corn

6 4.8 3.6 2.4 1.8

Barley 5 4.1 3.2 2.3 1.5

Oats 4.5 3.7 2.9 2.1 1.4

Soybean 6 4.8 3.6 2.4 1.8

Wheat 5 4.1 3.2 2.3 1.5

Corn

fodder

- 5.5 4.2 2.9 -

Hay - 4.0 3.1 2.2 1.4 24




Energy Requirements Consider a small particle to be reduced into similar particles of smaller size.

The required energy is related to some function of the initial and reduced particle. Since the particles are assumed symmetrical a common dimension will be used

i.e. energy required to reduce (grind) a unit is proportional to a dimension of the reduced particle relative to a similar dimension of the original particle raised to some power ‘n’.

or

where,

E = energy per unit weight expended for size reduction

L = original particle dimension

C=constant

n = constant

integrate for a specific mass 25




1. For plastic deformation (Fibrous material)

Kick (1885) proposed that equivalent amounts of energy should result in equivalent geometrical changes in the size of pieces of a solid i.e. If one unit of energy produces 2 size L/2 lumps from size “L” then the same amount of energy should reduce these two lumps to 4 size L/4 lumps. Same amount of energy should reduce these 3 lumps to 8 size L/8 lumps and so on.

26

Size L L/2 L/4 L/8 L/16By same amount of

energyNo. 1 2 4 8 16




This is equivalent to assuming n=1

2. For crushing material

Rettinger (1867) postulated that: size reduction energy is proportional to new specific surface area produced, which in turn are proportional to the square of a common linear dimension. Therefore, n = 2

27




28

Where,

E = Energy required for crushing

C = Rettingers constant

L2 = Characteristic dimension of final product

L1 = Characteristic dimension of feed

3.General equation




Example-1:A 5-hp power is required to reduce a material from ¼ inch (0.25) size to 10 mesh (0.065 inch dia). How much power would be required if the reduction were to 20 mesh (0.0328 inch dia)

29

By Kick’s Law By Rettinger’s law

Put E=5 hp, L1=0.25, L2=0.065, find

C=8.65,

Then using C=8.65, L1=0.25 and new

L2=0.038, for above equation, find

E=7.55 hp

Put E=5 hp, L1=0.25, L2=0.065, find

C=0.438

Then using C=0.438, L1=0.25 and new

L2=0.038, for above equation, find

E=11.64 hp




Example-2:A material consisting of 20 mm particles is crushed to an average size of 5 mm and needs 18 kJ/kg fro this size reduction. Determine the energy required to crush the material from 25 mm to 3 mm considering other conditions are similar assuming:

30

By Kick’s Law By Rettinger’s law

Put E=18, L1=20, L2=5, find C=12.987

Then using C=12.987, L1=20 and new

L2=3, for above equation, find E=?

Put E=18, L1=20, L2=5, find C=120

Then using C=120, L1=20 and new

L2=3, for above equation, find E=?Prof. Dr. Muhammad Iqbal, Deptt. Farm



4.Bond, 1952 proposed that sine both kiks 8 Rittinger hypothesis did not seen correct for practical application, an exponent n=(1+2)/2=1.5 had been found more appropriate

31

Efficiency of size reductionWhere,

Af = surface area after grinding, m2

Ao = surface of before grinding, m2

E = grinding energy, Wh/kg

W = grain mass, kg

J = efficiency of size reduction m2/Wh




Size reduction procedures1. Cutting: (Separation or reduction process)

It is produced by – Pushing/forcing a thin, sharp knife through the material to be reduced.

Minimum deformation and rapture of reduced particle is resulted.

The newly produced are relatively undamaged e.g. cutting of fruits and vegetable and forge

Tool : Extremely sharp and thin knife

2. Crushing:

It is produced by applying a force in excess of the strength of the unit to be reduced.

The resulting particles are irregular in shape and size

Examples:

– Crushing limestone found other chemical fertilizers - Ground feed for livestock

– Flour and meal - Extract juice from sugarcane

3. Shearing: Combination of cutting and crushing

If shearing edge is thin and sharp performance approaches to cutting

Shearing used for reducing materials of a tough fibrous nature

Examples: cutting ensilage, threshing grain and straw

Tool: sharp knife and a bar

32




Requirements for size reduction1. Output is of uniform size: For the same geometric mean

diameter the standard deviation be low. If 6 ϭ1 < 6 ϭ2 then product-1 is more acceptable

2. Energy requirement is low/low cost involved: High energy use waste energy, dries up cost, may heat up the ground product

3. Machine be simple, safe, low cost and trouble free: Desirable features of machine are:

• Low wear rate on internal parts

• Tolerance to tramp iron (tolerance to pieces of iron in the feed to the grinder)

• Tolerance to empty running

• Tolerance to automatic startup

33




Size reduction devicesThree types of size reduction devices are:

1. Hammer Mill 2) The attrition / burr mill 3) Roller Mill

1. Hammer Mill: (A dynamic crushing mill)

• Most common size reduction devices

• Consists of a rotor with fixed or free swinging hammers attached confined in a housing.

• Material to be ground is fed into chamber where it is struck by Hammers. Hammers strike particles and break them into pieces.

• Particles escape from the grinding chambers only if they pass through the classifying screen partially surrounding the chamber

• Material is fed from the top into the grinding chamber

• The integral blower- thrower pulls in ground material passed through the screen and then blows and throws it through a tube to another location.

34




35




Advantages Disadvantages

1. Simplicity:

Simple design

Low purchase price

Trouble free operation

1. Fines – machine hammer

impact action produce,

relatively high proportion of

fines when grinding to certain

level of fineness.

2. Versatility: can grind variety

of materials brittle, fibrous

hay, mixtures also

2. Power requirement to achieve

a certain fineness is relatively

high

3. Foreign material tolerance –

Tolerant to tramp iron in

material being ground

3. High noise because of

interaction with air and

particle impacts.

4. Tolerant to empty operation.36

Hammer Mill: (A dynamic crushing mill)




2. Attrition MillMaterials introduced between a rotating disk and a second

stationary or counter rotating disk.

For grains, usually have plates mounted vertically. Size reduction is by combination of:

Crushing (compression)

Scrapping (attrition)

Cutting (shearing)

• Material fed from top into the center is propelled toward the outer edge of burrs by their rotation.

• In small mills – one burr turns and the other is held stationary

• In large mill – burrs are driven in opposite direction by different devices.

• Disk speed: 800 – 1200 RPM

• Disk outer dia: 4 – 60 inch 37




38




Attrition Mill (Advantages & Disadvantages


Relatively low cost

Relatively uniform product

1. Lack of versatility. Only

useful for brittle products and

not for mixtures/fibrous

material like hay.

2. In tolerance to foreign

objects, tramp iron in grain

cause sever damage.

3. Deterioration of performance

– as burrs wear, grinding

action deteriorates.

39




3. Roller Mill

• Crushes material between counter rotating cylinders

• Ground rolls specifications:

– Dia: 9.0 – 12 inch

– Length: 60 – 52 inch

– Speed: 350 – 1800 RPM

40




41




Roller Mill (Advantages & Disadvantages


1. Uniform product -

positive one pass crushing

action tend to produce

uniform size.

2. Low power requirement -

power is not wasted

generating fines.

1. In tolerance to foreign objects.

Recommendation: feed grain be

passed through screen and then

over a magnet to remove iron and

other foreign objects.

2. Startup – Start machine empty

since a few kernels at the pinch

point would require excessive

starting torque.

42




43

Surface area as affected by particle dia for same weight

1

2

3

4

5

6

3.141

3.959

4.530

4.991

5.377

5.699

GMD = 1.0006d-2.9992

R2 = 1

Surface area = 7.5d2 - 17.2d + 12.9

R2 = 0.95

0

1

2

3

4

5

6

7

0.500 0.600 0.700 0.800 0.900 1.000 1.100

Particle dia, mm

Surfa

ce a

rea

mm

2

0.000

1.000

2.000

3.000

4.000

5.000

6.000

Particle number

Sphere surface area calculationsCut D1 D2 D3 GMD A1 T.AREA

0 1 1 1 1.000 3.141 3.141

1 1 1 0.5 0.794 1.980 3.959

2 1 1 0.333 0.693 1.510 4.530

3 1 1 0.25 0.630 1.248 4.991

4 1 1 0.2 0.585 1.075 5.377

5 1 1 0.166 0.550 0.950 5.699




44

Rectangle surface area calculationsL W D A1 PARTICLES T.Area

1 1 1.00 6.00 1 6.00

1 1 0.50 4.00 2 8.00

1 1 0.33 3.33 3 10.00

1 1 0.25 3.00 4 12.00

1 1 0.20 2.80 5 14.00

1 1 0.17 2.66 6 15.98

Total Surface Area, mm2

y = 2x + 4

R2 = 1

0.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

16.00

18.00

0 1 2 3 4 5 6 7

Surface area, mm2

Particle, noProf. Dr. Muhammad Iqbal, Deptt. Farm



1

1. Belt conveyor

Belt Conveyors are used in many bulk material handling applications ranging from agricultural

application such as grain handling through to heavy duty applications, such as quarrying and mining.

They consist of 2 or more pulleys with a continuous loop of conveyor belt that rotates around them

Belt conveyor

Belt conveyor

Belt conveyor for small objects

Inclined Belt conveyor

Table top belt conveyor

2

2. Chain Conveyor

Roller chain conveyor

Scraper chain conveyor

Chain slat conveyor

Table chain conveyor

Bucket elevator and chain conveyor

Scraper conveyor

3

Apron chain conveyor for sacked material

Apron flat for sacked material

3. Bucket Elevator Bucket Elevator is widely used in grain & oil, food, feed, metallurgy, mine, plastic, chemical

engineering, building material and medicine industries, etc.

Silo

4

4. Screw conveyor 1. The common name of screw conveyor is screw

2. It is a conveying equipment widely used in grain & oil, feed, food, metallurgy, mine,

plastic, chemical engineering, building material, medicine industries,

3. It is suitable for the conveying of granular or powder material, including horizontal,

inclined and vertical conveying, etc.

Particles within the screw conveyor

Sweep auger conveyor in silo bin

Unloading auger

Grain silo system

5

1

HEAT TRANSFER THEORY

Heat transfer is an operation that occurs repeatedly in the food industry. Whether it is called

cooking, baking, drying, sterilizing or freezing, heat transfer is part of the processing of almost

every food. An understanding of the principles that govern heat transfer is essential to an

understanding of food processing.

Heat transfer is a dynamic process in which heat is transferred spontaneously from one body to

another cooler body. The rate of heat transfer depends upon the differences in temperature

between the bodies, the greater the difference in temperature, the greater the rate of heat transfer.

Temperature difference between the source of heat and the receiver of heat is therefore the

driving force in heat transfer. An increase in the temperature difference, increases the driving

force and therefore increases the rate of heat transfer. The heat passing from one body to another

travels through some medium which in general offers resistance to the heat flow. Both these

factors, the temperature difference and the resistance to heat flow, affect the rate of heat transfer.

As with other rate processes, these factors are connected by the general equation:

rate of transfer = driving force / resistance

rate of heat transfer = temperature difference/ heat flow resistance of medium

During processing, temperatures may change and therefore the rate of heat transfer will change.

This is called unsteady state heat transfer, in contrast to steady state heat transfer when the

temperatures do not change. Unsteady state heat transfer is more complex since an additional

variable, time, enters into the rate equations.

Heat can be transferred in three ways: by conduction, by radiation and by convection.

In conduction, the molecular energy is directly exchanged, from the hotter to the cooler regions,

the molecules with greater energy communicating some of this energy to neighbouring

molecules with less energy. An example of conduction is the heat transfer through the solid walls

of a refrigerated store.

Radiation is the transfer of heat energy by electromagnetic waves, which transfer heat from one

body to another, in the same way as electromagnetic light waves transfer light energy. An

example of radiant heat transfer is when a foodstuff is passed below a bank of electric resistance

heaters that are red-hot.

Convection is the transfer of heat by the movement of groups of molecules in a fluid. The groups

of molecules may be moved by either density changes or by forced motion of the fluid. An

example of convection heating is cooking in a jacketed pan: without a stirrer, density changes

cause heat transfer by natural convection; with a stirrer, the convection is forced.

In general, heat is transferred in solids by conduction, in fluids by conduction and convection.

Heat transfer by radiation occurs through open space, can often be neglected, and is most

significant when temperature differences are substantial. In practice, the three types of heat

transfer may occur together.

1. HEAT CONDUCTION

In the case of heat conduction, the equation, rate = driving force/resistance, can be applied

directly. The driving force is the temperature difference per unit length of heat-transfer path, also

known as the temperature gradient. Instead of resistance to heat flow, its reciprocal called

the conductance is used. This changes the form of the general equation to:

rate of heat transfer = driving force x conductance,

that is:

dQ/dt = kA dT/dx (5.1)

where

2

dQ/dt is the rate of heat transfer, the quantity of heat energy transferred per unit of

time,

A is the area of cross-section of the heat flow path,

dT/dx is the temperature gradient, that is the rate of change of temperature per unit

length of path, and

k is the thermal conductivity of the medium. Notice the distinction between thermal

conductance, which relates to the actual thickness of a given material (k/x)

and thermal conductivity, which relates only to unit thickness.

The units of k, the thermal conductivity, can be found from eqn. (5.1) by transposing the terms

k = dQ/dt x 1/A x 1/(dT/dx)

= J s-1

x m-2

x 1/(°C m-1

)

= J m-1

s-1

°C-1

Equation (5.1) is known as the Fourier equation for heat conduction.

Thermal Conductivity On the basis of eqn. (5.1) thermal conductivities of materials can be measured. Thermal

conductivity does change slightly with temperature, but in many applications it can be regarded

as a constant for a given material. In general, metals have a high thermal conductivity, in the

region 50-400 J m-1

s-1

°C-1

. Most foodstuffs contain a high proportion of water and as the

thermal conductivity of water is about 0.7 J m-1

s-1

°C-1

above 0°C, thermal conductivities of

foods are in the range 0.6 - 0.7 J m-1

s-1

°C-1

. Ice has a substantially higher thermal conductivity

than water, about 2.3 J m-1

s-1

°C-1

. The thermal conductivity of frozen foods is, therefore, higher

than foods at normal temperatures.

Most dense non-metallic materials have thermal conductivities of 0.5-2 J m-1

s-1

°C-1

. Insulating

materials, such as those used in walls of cold stores, approximate closely to the conductivity of

gases as they are made from non-metallic materials enclosing small bubbles of gas or air. The

conductivity of air is 0.024 J m-1

s-1

°C-1

at 0°C, and insulating materials such as foamed plastics,

cork and expanded rubber are in the range 0.03- 0.06 J m-1

s-1

°C-1

. Some of the new foamed

plastic insulating materials have thermal conductivities as low as 0.026 J m-1

s-1

°C-1

.

When using published tables of data, the units should be carefully checked. Mixed units,

convenient for particular applications, are sometimes used and they may need to be converted.

Conduction through a Slab

If a slab of material, as shown in Fig. 5.1, has two faces at different temperatures T1 and T2 heat

will flow from the face at the higher temperature T1 to the other face at the lower temperature T2.

Figure 5.1. Heat conduction through a slab The rate of heat transfer is given by Fourier's equation:

dQ/dt = kA T/x = kA dT/dx

Under steady temperature conditions dQ/dt = constant, which may be called q:

and so q = kA dT/dx

but dT/dx, the rate of change of temperature per unit length of path, is given by

(T1 -T2)/x where x is the thickness of the slab,

so q = kA(T1 - T2)/x

or q = kA T/x =k/x) A T (5.2)

3

This may be regarded as the basic equation for simple heat conduction. It can be used to

calculate the rate of heat transfer through a uniform wall if the temperature difference across it

and the thermal conductivity of the wall material are known.

EXAMPLE 5.1. Rate of heat transfer in cork A cork slab 10 cm thick has one face at -12°C and the other face at 21°C. If the mean thermal

conductivity of cork in this temperature range is 0.042 J m-1

s-1

°C-1

, what is the rate of heat

transfer through 1 m2 of wall?

T1 = 21°C T2 = -12°C T = 33°C

A = 1 m2 k = 0.042 J m

-1 s

-1 °C

-1 x = 0.1 m

q = 0.042 x 1 x 33

0.1

= 13.9 J s-1

Heat Conductances

In tables of properties of insulating materials, heat conductances are sometimes used instead of

thermal conductivities. The heat conductance is the quantity of heat that will pass in unit time,

through unit area of a specified thickness of material, under unit temperature difference, For a

thickness x of material with a thermal conductivity of k in J m-1

s-1

°C-1

, the conductance

is k/x = C and the units of conductance are

J m-2

s-1

°C-1

.

Heat conductance = C = k/x.

Heat Conductances in Series

Frequently in heat conduction, heat passes through several consecutive layers of different

materials. For example, in a cold store wall, heat might pass through brick, plaster, wood and

cork.

In this case, eqn. (5.2) can be applied to each layer. This is illustrated in Fig. 5.2.

A1 = A2 = A3 = ….. = A

q = AT1k1/x1 = AT2k2/x2 = AT3k3/x3 = ……..

So AT1 = q(x1/k1) and AT2 = q(x2/k2) and AT3 = q(x3/k3).…..

AT1 + AT2 + AT3 + … = q(x1/k1) + q(x2/k2) +q(x3/k3) + …

A(T1 + T2 + T3 + ..) = q(x1/k1 + x2/k2 +x3/k3 + …)

The sum of the temperature differences over each layer is equal to the difference in temperature

of the two outside surfaces of the complete system, i.e.

T1 + T2 + T3 + … = T

and since k1/x1 is equal to the conductance of the material in the first layer, C1, and k2/x2 is equal

to the conductance of the material in the second layer C2,

x1/k1 + x2/k2 + x3/k3 + ... = 1/C1 + 1/C2 + 1/C3 …= 1/U

where U = the overall conductance for the combined layers, in J m-2

s-1

°C-1

Therefore AT = q(1/U)

And so q = UAT (5.3)

This is of the same form as eqn (5.2) but extended to cover the composite slab. U is called the

overall heat-transfer coefficient, as it can also include combinations involving the other methods

of heat transfer – convection and radiation.

EXAMPLE 5.2. Heat transfer in cold store wall of brick, concrete and cork

A cold store has a wall comprising 11 cm of brick on the outside, then 7.5 cm of concrete and

then 10 cm of cork. The mean temperature within the store is maintained at -18°C and the mean

temperature of the outside surface of the wall is 18°C.

Calculate the rate of heat transfer through the wall. The appropriate thermal conductivities are

for brick, concrete and cork, respectively 0.69, 0.76 and 0.043 J m-1

s-1

°C-1

.

Determine also the temperature at the interfaces between the concrete and cork layers, and the

brick and concrete layers.

4

For brick x1/k1 = 0.11/0.69 = 0.16.

For concrete x2/k2 = 0.075/0.76 = 0.10.

Figure 5.2 Heat conductances in series

In the steady state, the same quantity of heat per unit time must pass through each layer.

q = A1T1k1/x1 = A2T2k2/x2 = A3T3k3/x3 = ……..

If the areas are the same,

For cork x3/k3 = 0.10/0.043 = 2.33

.

But 1/U = x1/k1 + x2/k2 + x3/k3

= 0.16 + 0.10 + 2.33

= 2.59

Therefore U = 0.38 J m-2

s-1

°C-1

T = 18 - (-18) = 36°C,

A = 1 m2

q = UAT

= 0.38 x 1 x 36

= 13.7 J s-1

Further, q = A3T3k3/x3

and for the cork wall A3 = 1 m2, x3/k3 = 2.33 and q = 13.7 J s

-1

Therefore 13.7 = 1 x T3 x 1/2.33 from rearranging eqn. (5.2)

T3 = 32°C.

But T3 is the difference between the temperature of the cork/concrete surface Tc and the

temperature of the cork surface inside the cold store.

Therefore Tc - (-18) = 32

where Tc is the temperature at the cork/concrete surface

and so Tc = 14°C.

If T1 is the difference between the temperature of the brick/concrete surface, Tb, and the

temperature of the external air.

Then 13.7 = 1 x T1 x 1/ 0.16 = 6.25 T1

Therefore 18 - Tb = T1 = 13.7/6.25 = 2.2

so Tb = 15.8 °C

Working it through shows approximate boundary temperatures: air/brick 18°C,brick/concrete

16°C, concrete/cork 14°C, cork/air -18°C

This shows that almost all of the temperature difference occurs across the insulation (cork): and

the actual intermediate temperatures can be significant especially if they lie below the

temperature at which the atmospheric air condenses, or freezes.

Heat Conductances in Parallel

5

Heat conductances in parallel have a sandwich construction at right angles to the direction of the

heat transfer, but with heat conductances in parallel, the material surfaces are parallel to the

direction of heat transfer and to each other. The heat is therefore passing through each material at

the same time, instead of through one material and then the next. This is illustrated in Fig. 5.3..

Figure 5.3 Heat conductances in parallel

An example is the insulated wall of a refrigerator or an oven, in which the walls are held together

by bolts. The bolts are in parallel with the direction of the heat transfer through the wall: they

carry most of the heat transferred and thus account for most of the losses.

EXAMPLE 5.3. Heat transfer in walls of a bakery oven The wall of a bakery oven is built of insulating brick 10 cm thick and thermal conductivity 0.22 J

m-1

s-1

°C-1

. Steel reinforcing members penetrate the brick, and their total area of cross-section

represents 1% of the inside wall area of the oven.

If the thermal conductivity of the steel is 45 J m-1

s-1

°C-1

calculate (a) the relative proportions of

the total heat transferred through the wall by the brick and by the steel and (b) the heat loss for

each m2 of oven wall if the inner side of the wall is at 230°C and the outer side is at 25°C.

Applying eqn. (5.1) q = ATk/x, we know that T is the same for the bricks and for the steel.

Also x, the thickness, is the same.

a. Consider the loss through an area of 1 m2 of wall (0.99 m

2 of brick, and 0.01 m

2 of steel)

For brick qb = AbT kb/x

= 0.99(230 - 25)0.22

0.10

= 446 J s-1

For steel qs = AsT ks/x

= 0.01(230 - 25)45

0.10

= 923 J s-1

Therefore qb /qs = 0.48

(b) Total heat loss

q = (qb + qs ) per m2 of wall

= 446 + 923

= 1369 J s-1

Therefore percentage of heat carried by steel

= (923/1369) x 100

= 67%

6

2. SURFACE HEAT TRANSFER

Newton found, experimentally, that the rate of cooling of the surface of a solid, immersed in a

colder fluid, was proportional to the difference between the temperature of the surface of the

solid and the temperature of the cooling fluid. This is known as Newton's Law of Cooling, and

it can be expressed by the equation, analogous to eqn. (5.2),

q = hsA(Ta– Ts) (5.4)

where hs is called the surface heat transfer coefficient, Ta is the temperature of the cooling fluid

and Ts is the temperature at the surface of the solid. The surface heat transfer coefficient can be

regarded as the conductance of a hypothetical surface film of the cooling medium of

thickness xf such that

hs = kf /xf

where kf is the thermal conductivity of the cooling medium.

Following on this reasoning, it may be seen that hs can be considered as arising from the

presence of another layer, this time at the surface, added to the case of the composite slab

considered previously. The heat passes through the surface, then through the various elements of

a composite slab and then it may pass through a further surface film. We can at once write the

important equation:

q = AT[(1/hs1) + x1/k1 + x2/k2 + .. + (1/hs2)] (5.5)

= UAT

where 1/U = (1/hs1) + x1/k1 + x2/k2 +.. + (1/hs2)

and hs1, hs2 are the surface coefficients on either side of the composite slab, x1, x2 ...... are the

thicknesses of the layers making up the slab, and k1, k2... are the conductivities of layers of

thickness x1, ..... . The coefficienths is also known as the convection heat transfer coefficient and

values for it will be discussed in detail under the heading of convection. It is useful at this point,

however, to appreciate the magnitude of hs under various common conditions and these are

shown in Table 5.1.

TABLE 5.1

APPROXIMATE RANGE OF SURFACE HEAT TRANSFER COEFFICIENTS

Fluid h (J m-2

s-1

°C-1

)

Boiling liquids 2400-24,000

Condensing liquids 1800-18,000

Still air 6

Moving air (3 m s-1

) 30

Liquids flowing through

pipes 1200-6000

EXAMPLE 5.4. Heat transfer in jacketed pan Sugar solution is being heated in a jacketed pan made from stainless steel, 1.6 mm thick. Heat is

supplied by condensing steam at 200 kPa gauge in the jacket. The surface transfer coefficients

are, for condensing steam and for the sugar solution, 12,000 and 3000 J m-2

s-1

°C-1

respectively,

and the thermal conductivity of stainless steel is 21 J m-1

s-1

°C-1

.

Calculate the quantity of steam being condensed per minute if the transfer surface is 1.4 m2 and

the temperature of the sugar solution is 83°C.

From steam tables, Appendix 8, the saturation temperature of steam at 200 kPa gauge(300 kPa

Absolute) = 134°C, and the latent heat = 2164 kJ kg-1

.

For stainless steel x/k = 0.0016/21 = 7.6 x 10-5

http://www.nzifst.org.nz/unitoperations/httrtheory1.htm#condslab

http://www.nzifst.org.nz/unitoperations/appendix8.htm

7

T = (condensing temperature of steam) - (temperature of sugar solution)

= 134 - 83 = 51°C.

From eqn. (5.5)

1/U = 1/12,000 + 7.6 x 10-5

+ 1/3000

U = 2032 J m-2

s-1

°C-1

and since A = 1.4 m2

q = UAT

= 2032 x 1.4 x 51

= 1.45 x 105 J s

-1

Therefore steam required

= 1.45 x 105 / (2.164 x 10

6) kg s

-1

= 0.067 kg s-1

= 4 kg min-1

3. UNSTEADY STATE HEAT TRANSFER

In food process engineering, heat transfer is very often in the unsteady state, in which

temperatures are changing and materials are warming or cooling. Unfortunately, study of heat

flow under these conditions is complicated. In fact, it is the subject for study in a substantial

branch of applied mathematics, involving finding solutions for the Fourier equation written in

terms of partial differentials in three dimensions. There are some cases that can be simplified and

handled by elementary methods, and also charts have been prepared which can be used to obtain

numerical solutions under some conditions of practical importance.

A simple case of unsteady state heat transfer arises from the heating or cooling of solid bodies

made from good thermal conductors, for example a long cylinder, such as a meat sausage or a

metal bar, being cooled in air. The rate at which heat is being transferred to the air from the

surface of the cylinder is given by eqn. (5.4)

q = dQ/dt = hsA(Ts - Ta)

where Ta is the air temperature and Ts is the surface temperature.

Now, the heat being lost from the surface must be transferred to the surface from the interior of

the cylinder by conduction. This heat transfer from the interior to the surface is difficult to

determine but as an approximation, we can consider that all the heat is being transferred from the

centre of the cylinder. In this instance, we evaluate the temperature drop required to produce the

same rate of heat flow from the centre to the surface as passes from the surface to the air. This

requires a greater temperature drop than the actual case in which much of the heat has in fact a

shorter path.

Assuming that all the heat flows from the centre of the cylinder to the outside, we can write the

conduction equation

Q = dQ/dt = (k/L)A( Tc– Ts )

where Tc is the temperature at the centre of the cylinder, k is the thermal conductivity of the

material of the cylinder and L is the radius of the cylinder.

Equating these rates:

hsA(Ts --Ta) = (k/L)A( Tc– Ts )

hs(Ts -- Ta) = (k/L)( Tc– Ts )

and so hsL/k = ( Tc– Ts )/ (Ts -- Ta)

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8

To take a practical case of a copper cylinder of 15 cm radius cooling in air kc = 380 J m-1

s-1

°C-

1, hs = 30 J m

-2 s

-1°C

-1 (from Table 5.1), L = 0.15 m,

(Tc– Ts)/ (Ts -- Ta) = (30 x 0.15)/380

= 0.012

In this case 99% of the temperature drop occurs between the air and the cylinder surface. By

comparison with the temperature drop between the surface of the cylinder and the air, the

temperature drop within the cylinder can be neglected. On the other hand, if the cylinder were

made of a poorer conductor as in the case of the sausage, or if it were very large in diameter, or if

the surface heat-transfer coefficient were very much larger, the internal temperature drops could

not be neglected.

This simple analysis shows the importance of the ratio:

heat transfer coefficient at the surface = hsL/k

heat conductance to the centre of the solid

This dimensionless ratio is called the Biot number (Bi) and it is important when considering

unsteady state heat flow. When (Bi) is small, and for practical purposes this may be taken as any

value less than about 0.2, the interior of the solid and its surface may be considered to be all at

one uniform temperature. In the case in which (Bi) is less than 0.2, a simple analysis can be used,

therefore, to predict the rate of cooling of a solid body.

Therefore for a cylinder of a good conductor, being cooled in air,

dQ = hsA(Ts -- Ta) dt

But this loss of heat cools the cylinder in accordance with the usual specific heat equation:

dQ = c .ρ .V.dT

where c is the specific heat of the material of the cylinder, ρ is the density of this material

and V is the volume of the cylinder.

Since the heat passing through the surface must equal the heat lost from the cylinder, these two

expressions for dQ can be equated:

c ρ VdT = hsA(Ts -- Ta) dt

Integrating between Ts = T1 and Ts = T2 , the initial and final temperatures of the cylinder during

the cooling period, t, we have:

- hsAt/c ρ V = loge (T2 - Ta)/(T1– Ta)

or (T2 - Ta)/(T1 – Ta) = exp( -hsAt/c ρ V ) (5.6)

For this case, the temperatures for any desired interval can be calculated, if the surface transfer

coefficient and the other physical factors are known. This gives a reasonable approximation so

long as (Bi) is less than about 0.2. Where (Bi) is greater than 0.2, the centre of the solid will cool

more slowly than this equation suggests. The equation is not restricted to cylinders, it applies to

solids of any shape so long as the restriction in (Bi), calculated for the smallest half-dimension, is

obeyed.

Charts have been prepared which give the temperature relationships for solids of simple shapes

under more general conditions of unsteady-state conduction. These charts have been calculated

from solutions of the conduction equation and they are plotted in terms of dimensionless groups

so that their application is more general. The form of the solution is:

9

ƒ{(T - T0)/( Ti - T0 )} = F{(kt/c ρ L2)(hsL/k)} (5.7)

where ƒ and F indicate functions of the terms following, Ti is the initial temperature of the

solid, T0 is the temperature of the cooling or heating medium, T is the temperature of the solid at

time t, (kt/c ρ L2) is called theFourier number (Fo) (this includes the factor k/c ρ the thermal

conductance divided by the volumetric heat capacity, which is called the thermal diffusivity) and

(hsL/k) is the Biot number.

A mathematical outcome that is very useful in these calculations connects results for two- and

three-dimensional situations with results from one-dimensional situations. This states that the

two- and three-dimensional values called F(x,y) and F(x,y,z) can be obtained from the individual

one-dimensional results if these are F(x), F(y) andF(z), by simple multiplication:

F(x,y) = F(x)F(y)

and

F(x,y,z) = F(x)F(y)F(z)

Using the above result, the solution for the cooling or heating of a brick is obtained from the

product of three slab solutions. The solution for a cylinder of finite length, such as a can, is

obtained from the product of the solution for an infinite cylinder, accounting for the sides of the

can, and the solution for a slab, accounting for the ends of the can.

Charts giving rates of unsteady-state heat transfer to the centre of a slab, a cylinder, or a sphere,

are given inFig. 5.4. On one axis is plotted the fractional unaccomplished temperature change,

(T - T0)/( Ti - T0 ). On the other axis is the Fourier number, which may be thought of in this

connection as a time coordinate. The various curves are for different values of the reciprocal of

the Biot number, k/hr for spheres and cylinders, k/hl for slabs.

More detailed charts, giving surface and mean temperatures in addition to centre temperatures,

may be found inMcAdams (1954), Fishenden and Saunders (1950) and Perry (1997).

Figure 5.4. Transient heat conduction

Temperatures at the centre of sphere,slab,and cylinder: adapted from Henderson and

Perry, Agricultural Process Engineering, 1955

EXAMPLE 5.5. Heat transfer in cooking sausages

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10

A process is under consideration in which large cylindrical meat sausages are to be processed in

an autoclave. The sausage may be taken as thermally equivalent to a cylinder 30 cm long and 10

cm in diameter. If the sausages are initially at a temperature of 21°C and the temperature in the

autoclave is maintained at 116°C, estimate the temperature of the sausage at its centre 2 h after it

has been placed in the autoclave.

Assume that the thermal conductivity of the sausage is 0.48 J m-1

s-1

°C-1

, that its specific gravity

is 1.07, and its specific heat is 3350 J kg-1

°C-1

. The surface heat-transfer coefficient in the

autoclave to the surface of the sausage is 1200 J m-2

s-1

°C-1

.

This problem can be solved by combining the unsteady-state solutions for a cylinder with those

for a slab, working from Fig. 5.3.

(a) For the cylinder, of radius r = 5 cm (instead of L in this case)

Bi = hsr/k = (1200 x 0.05)/0.48 = 125

(Often in these systems the length dimension used as parameter in the charts is the half-

thickness, or the radius, but this has to be checked on the graphs used.)

So 1/(Bi) = 8 x 10-3

After 2 hours t = 7200 s

Therefore Fo = kt/cr2 = (0.48 x 7200)/[3350 x 1.07 x 1000 x (0.05)

2] = 0.39

and so from Fig. 5.3 for the cylinder:

(T - T0)/( Ti - T0 ) = 0.175 = say, F(x).

(b) For the slab the half-thickness 30/2 cm = 0.15 m

and so Bi = hsL/k = (1200 x 0.15)/0.48 = 375

1/ Bi = 2.7 x 10-3

t = 7200 s as before and

kt/cL2 = (0.48 x 7200)/[3350 x 1.07 x 1000 x (0.15)

2]

= 4.3 x 10-2

and so from Fig. 5.3 for the slab:

(T - T0)/( Ti - T0 ) = 0.98 = say, F(y)

So overall (T - T0)/( Ti - T0 ) = F(x) F(y)

= 0.175 x 0.98

= 0.172

Therefore T2 - 116 = 0.172

21 - 116

Therefore T2 = 100°C

4. RADIATION HEAT TRANSFER

Radiation heat transfer is the transfer of heat energy by electromagnetic radiation. Radiation

operates independently of the medium through which it occurs and depends upon the relative

temperatures, geometric arrangements and surface structures of the materials that are emitting or

absorbing heat.

The calculation of radiant heat transfer rates, in detail, is beyond the scope of this book and for

most food processing operations a simplified treatment is sufficient to estimate radiant heat

effects. Radiation can be significant with small temperature differences as, for example, in freeze

drying and in cold stores, but it is generally more important where the temperature differences

are greater. Under these circumstances, it is often the most significant mode of heat transfer, for

example in bakers' ovens and in radiant dryers.

11

The basic formula for radiant-heat transfer is the Stefan-Boltzmann Law

q = A T 4 (5.8)

where T is the absolute temperature (measured from the absolute zero of temperature at -273°C,

and indicated in Bold type) in degrees Kelvin (K) in the SI system, and (sigma) is the Stefan-

Boltzmann constant = 5.73 x 10-8

J m-2

s-1

K-4

The absolute temperatures are calculated by the

formula K = (°C + 273).

This law gives the radiation emitted by a perfect radiator (a black body as this is called though it

could be a red-hot wire in actuality). A black body gives the maximum amount of emitted

radiation possible at its particular temperature. Real surfaces at a temperature T do not emit as

much energy as predicted by eqn. (5.8), but it has been found that many emit a constant fraction

of it. For these real bodies, including foods and equipment surfaces, that emit a constant fraction

of the radiation from a black body, the equation can be rewritten

q = A T 4 (5.9)

where (epsilon) is called the emissivity of the particular body and is a number between 0 and

1. Bodies obeying this equation are called grey bodies.

Emissivities vary with the temperature T and with the wavelength of the radiation emitted. For

many purposes, it is sufficient to assume that for:

*dull black surfaces (lamp-black or burnt toast, for example), emissivity is approximately 1;

*surfaces such as paper/painted metal/wood and including most foods, emissivities are about

0.9;

*rough un-polished metal surfaces, emissivities vary from 0.7 to 0.25;

*polished metal surfaces, emissivities are about or below 0.05.

These values apply at the low and moderate temperatures which are those encountered in food

processing.

Just as a black body emits radiation, it also absorbs it and according to the same law, eqn. (5.8).

Again grey bodies absorb a fraction of the quantity that a black body would absorb,

corresponding this time to their absorptivity (alpha). For grey bodies it can be shown

that = . The fraction of the incident radiation that is not absorbed is reflected, and thus, there

is a further term used, the reflectivity, which is equal to (1 – ).

Radiation between Two Bodies

The radiant energy transferred between two surfaces depends upon their temperatures, the

geometric arrangement, and their emissivities. For two parallel surfaces, facing each other and

neglecting edge effects, each must intercept the total energy emitted by the other, either

absorbing or reflecting it. In this case, the net heat transferred from the hotter to the cooler

surface is given by:

q = AC (T14- T2

4 ) (5.10)

where 1/C = 1/1 + 1/2 - 1, 1 is the emissivity of the surface at temperature T1 and 2 is the

emissivity of the surface at temperature T2.

Radiation to a Small Body from its Surroundings

In the case of a relatively small body in surroundings that are at a uniform temperature, the net

heat exchange is given by the equation

q = A(T14- T2

4 ) (5.11)

12

where is the emissivity of the body, T1 is the absolute temperature of the body and T2 is the

absolute temperature of the surroundings.

For many practical purposes in food process engineering, eqn. (5.11) covers the situation; for

example for a loaf in an oven receiving radiation from the walls around it, or a meat carcass

radiating heat to the walls of a freezing chamber.

In order to be able to compare the various forms of heat transfer, it is necessary to see whether an

equation can be written for radiant heat transfer similar to the general heat transfer eqn. (5.3).

This means that for radiant heat transfer:

q = hrA(T1 - T2) = hrA T (5.12)

where hr is the radiation heat-transfer coefficient, T1 is the temperature of the body and T2 is the

temperature of the surroundings. (The T would normally be the absolute temperature for the

radiation, but the absolute temperature difference is equal to the Celsius temperature difference,

because 273 is added and subtracted and so (T1 - T2) = (T1 - T2) =T

Equating eqn. (5.11) and eqn. (5.12)

q = hrA(T1 - T2) = A(T14- T2

4 )

Therefore hr = (T14- T2

4 )/ (T1 - T2)

= (T1 + T2 ) (T12 + T2

2)mmmmmm

If Tm = (T1 + T2)/2, we can write T1 + e = Tm and T2 - e = Tm

where

2e = T1 - T2

also

(T1 + T2) = 2 Tm

and then

(T12 + T2

2) = Tm

2 - 2eTm + e

2 + Tm

2 +2eTm +e

2

= 2Tm2 + 2e

2

= 2Tm2 + (T1 - T2)

2/2

Therefore hr = (2Tm)[2Tm2 + (T1 - T2)

2/2]

Now, if (T1 - T2) « T1 or T2, that is if the difference between the temperatures is small compared

with the numerical values of the absolute temperatures, we can write:

hr 4Tm3

and so

q = hrA T

= (x 5.73 x 10-8

x 4 xTm3 ) x A T

= 0.23 (Tm/100)3A T (5.13)

EXAMPLE 5.6.Radiation heat transfer to loaf of bread in an oven

Calculate the net heat transfer by radiation to a loaf of bread in an oven at a uniform temperature

of 177°C, if the emissivity of the surface of the loaf is 0.85, using eqn. (5.11). Compare this

result with that obtained by using eqn. (5.13). The total surface area and temperature of the loaf

are respectively 0.0645 m2 and 100°C.

q = A(T14- T2

4 )

13

= 0.0645 x 0.85 x 5.73 x 10-8

(4504- 373

4)

= 68.0 J s-1

.

By eqn. (5.13)

q = 0.23 (Tm/100)3A T

= 0.23 x 0.85(411/100)3 x 0.0645 x 77

= 67.4 J s-1

.

Notice that even with quite a large temperature difference, eqn. (5.13) gives a close

approximation to the result obtained using eqn. (5.11).

5. CONVECTION HEAT TRANSFER

Convection heat transfer is the transfer of energy by the mass movement of groups of molecules.

It is restricted to liquids and gases, as mass molecular movement does not occur at an

appreciable speed in solids. It cannot be mathematically predicted as easily as can transfer by

conduction or radiation and so its study is largely based on experimental results rather than on

theory.

The most satisfactory convection heat transfer formulae are relationships between dimensionless

groups of physical quantities. Furthermore, since the laws of molecular transport govern both

heat flow and viscosity, convection heat transfer and fluid friction are closely related to each

other.

Convection coefficients will be studied under two sections, firstly, natural convection in which

movements occur due to density differences on heating or cooling; and secondly, forced

convection, in which an external source of energy is applied to create movement. In many

practical cases, both mechanisms occur together.

Natural Convection

Heat transfer by natural convection occurs when a fluid is in contact with a surface hotter or

colder than itself. As the fluid is heated or cooled it changes its density. This difference in

density causes movement in the fluid that has been heated or cooled and causes the heat transfer

to continue.

There are many examples of natural convection in the food industry. Convection is significant

when hot surfaces, such as retorts which may be vertical or horizontal cylinders, are exposed

with or without insulation to colder ambient air. It occurs when food is placed inside a chiller or

freezer store in which circulation is not assisted by fans. Convection is important when material

is placed in ovens without fans and afterwards when the cooked material is removed to cool in

air.

It has been found that natural convection rates depend upon the physical constants of the fluid,

density , viscosity , thermal conductivity k, specific heat at constant pressure cp and

coefficient of thermal expansion (beta) which for gases = l/T by Charles' Law. Other factors

that also affect convection-heat transfer are, some linear dimension of the system, diameter D or

length L, a temperature difference term, T, and the gravitational acceleration g since it is

density differences acted upon by gravity that create circulation. Heat transfer rates are expressed

in terms of a convection heat transfer coefficient hc, which is part of the general surface

coefficient hs, in eqn. (5.5).

Experimentally, if has been shown that convection heat transfer can be described in terms of

these factors grouped in dimensionless numbers which are known by the names of eminent

workers in this field:

Nusselt number (Nu) = (hcD/k)

Prandtl number (Pr) = (cp/k)

Grashof number (Gr) = (D3

2g T/

2)

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14

and in some cases a length ratio (L/D).

If we assume that these ratios can be related by a simple power function we can then write the

most general equation for natural convection:

(Nu) = K(Pr)k(Gr)

m(L/D)

n (5.14)

Experimental work has evaluated K, k, m, n, under various conditions. For a discussion, see for

exampleMcAdams ( 1954) . Once K, k, m, n, are known for a particular case, together with the

appropriate physical characteristics of the fluid, the Nusselt number can be calculated. From the

Nusselt number we can find hc and so determine the rate of convection-heat transfer by applying

eqn. (5.5). In natural convection equations, the values of the physical constants of the fluid are

taken at the mean temperature between the surface and the bulk fluid. The Nusselt and Biot

numbers look similar: they differ in that for Nusselt, k and h both refer to the fluid, for Biot k is

in the solid and h is in the fluid.

Natural Convection Equations

These are related to a characteristic dimension of the body (food material for example) being

considered, and typically this is a length for rectangular bodies and a diameter for

spherical/cylindrical ones.

(1) Natural convection about vertical cylinders and planes, such as vertical retorts and oven

walls

(Nu) = 0.53(Pr.Gr)0.25

for 104 < (Pr.Gr) < 10

9 (5.15)

(Nu) = 0.12(Pr.Gr)0.33

for 109 < (Pr.Gr) < 10

12 (5.16)

For air these equations can be approximated respectively by:

hc = 1.3(T/L)0.25

(5.17)

hc = 1.8(T)0.25

(5.18)

Equations (5.17) and (5.18) are dimensional equations and are in standard units (T in °C

and L (or D) in metres and hc in J m-2

s-1

°C-1

). The characteristic dimension to be used in the

calculation of (Nu) and (Gr) in these equations is the height of the plane or cylinder.

(2) Natural convection about horizontal cylinders such as a steam pipe or sausages lying on a

rack

(Nu) = 0.54(Pr.Gr)0.25

for laminar flow in range 103 < (Pr.Gr) < 10

9. (5.19)

Simplified equations can be employed in the case of air, which is so often encountered in contact

with hotter or colder foods giving again:

For 104 < (Pr.Gr) < 10

9

hc = 1.3(T/D)0.25

(5.20)

and for 109< (Pr.Gr) < 10

12

hc = 1.8(T)0.33

(5.21)

(3) Natural convection from horizontal planes, such as slabs of cake cooling

The corresponding cylinder equations may be used, employing the length of the plane instead of

the diameter of the cylinder whenever D occurs in (Nu) and (Gr). In the case of horizontal

planes, cooled when facing upwards, or heated when facing downwards, which appear to be

working against natural convection circulation, it has been found that half of the value of hc in

eqns. (5.19) - (5.21) corresponds reasonably well with the experimental results.


15

Note carefully that the simplified equations are dimensional. Temperatures must be in °C and

lengths in m and then hc will be in J m-2

s-1

°C-1

. Values for , k and are measured at the film

temperature, which is midway between the surface temperature and the temperature of the bulk

liquid.

EXAMPLE 5.7. Heat loss from a cooking vessel

Calculate the rate of convection heat loss to ambient air from the side walls of a cooking vessel

in the form of a vertical cylinder 0.9 m in diameter and 1.2 m high. The outside of the vessel

insulation, facing ambient air, is found to be at 49°C and the air temperature is 17°C.

First it is necessary to establish the value of (Pr.Gr).

From the properties of air, at the mean film temperature, (49 + 17)/2, that is 33°C,

= 1.9 x 10-5

N s m-2

, cp = 1.0 kJ kg-1

°C-1

, k = 0.025 J m-1

s-1

° C-1

, = 1/308, =1.12 kg m-3

.

From the conditions of the problem, characteristic dimension = height = 1.2 m, T = 32°C.

Therefore (Pr.Gr) = (cp /k) (D3

2g T /

2)

= (L

2g T cp) / (k)

= [(1.2)3 x (1.12)

2 x 9.81 x 32 x 1.0 x 10

3 )/(308 x 1.9 x 10

-5 x 0.025)

= 5 x 109

Therefore eqn. (5.18) is applicable.

and so hc = 1.8T0.25

= 1.8(32)0.25

= 4.3 J m-2

s-1

°C-1

Total area of vessel wall = DL = x 0.9 x 1.2 = 3.4 m2

T = 32°C.

Therefore heat loss rate = hc A(T1 - T2)

= 4.3 x 3.4 x 32

= 468 J s-1

Forced Convection

When a fluid is forced past a solid body and heat is transferred between the fluid and the body,

this is called forced convection heat transfer. Examples in the food industry are in the forced-

convection ovens for baking bread, in blast and fluidized freezing, in ice-cream hardening rooms,

in agitated retorts, in meat chillers. In all of these, foodstuffs of various geometrical shapes are

heated or cooled by a surrounding fluid, which is moved relative to them by external means.

The fluid is constantly being replaced, and the rates of heat transfer are, therefore, higher than for

natural convection. Also, as might be expected, the higher the velocity of the fluid the higher the

rate of heat transfer. In the case of low velocities, where rates of natural convection heat transfer

are comparable to those of forced convection heat transfer, the Grashof number is still

significant. But in general the influence of natural circulation, depending as it does on

coefficients of thermal expansion and on the gravitational acceleration, is replaced by

dependence on circulation velocities and the Reynold’s number.

As with natural convection, the results are substantially based on experiment and are grouped to

deal with various commonly met situations such as fluids flowing in pipes, outside pipes, etc.

Forced convection Equations

(1) Heating and cooling inside tubes, generally fluid foods being pumped through pipes

In cases of moderate temperature differences and where tubes are reasonably long, for laminar

flow it is found that:

16

(Nu) = 4 (5.22)

and where turbulence is developed for (Re) > 2100 and (Pr) > 0.5

(Nu) = 0.023(Re)0.8

(Pr)0.4

(5.23)

For more viscous liquids, such as oils and syrups, the surface heat transfer will be affected,

depending upon whether the fluid is heating or being cooled. Under these cases, the viscosity

effect can be allowed, for (Re) > 10,000, by using the equation:

(Nu) = 0.027(/s)0.14

(Re)0.8

(Pr)0.33

(5.24)

In both cases, the fluid properties are those of the bulk fluid except for s, which is the viscosity

of the fluid at the temperature of the tube surface.

Since (Pr) varies little for gases, either between gases or with temperature, it can be taken as 0.75

and eqn. (5.23) simplifies for gases to:

(Nu) = 0.02(Re)0.8

. (5.25)

In this equation the viscosity ratio is assumed to have no effect and all quantities are evaluated at

the bulk gas temperature. For other factors constant, this becomes hc = k' v0.8

, as in equation

(5.28)

(2) Heating or cooling over plane surfaces

Many instances of foods approximate to plane surfaces, such as cartons of meat or ice cream or

slabs of cheese. For a plane surface, the problem of characterizing the flow arises, as it is no

longer obvious what length to choose for the Reynolds number. It has been found, however, that

experimental data correlate quite well if the length of the plate measured in the direction of the

flow is taken for D in the Reynolds number and the recommended equation is:

(Nu) = 0.036 (Re)0.8

(Pr)0.33

for (Re) > 2 x 104 (5.26)

For the flow of air over flat surfaces simplified equations are:

hc = 5.7 + 3.9v for v < 5 m s-1

(5.27)

hc = 7.4v0.8

for 5 < v < 30 m s-1

(5.28)

These again are dimensional equations and they apply only to smooth plates. Values for hc for

rough plates are slightly higher.

(3) Heating and cooling outside tubes

Typical examples in food processing are water chillers, chilling sausages, processing spaghetti.

Experimental data in this case have been correlated by the usual form of equation:

(Nu) = K (Re)n(Pr)

m (5.29)

The powers n and m vary with the Reynolds number. Values for D in (Re) are again a difficulty

and the diameter of the tube, over which the flow occurs, is used. It should be noted that in this

case the same values of (Re) cannot be used to denote streamline or turbulent conditions as for

fluids flowing inside pipes.

For gases and for liquids at high or moderate Reynolds numbers:

(Nu) = 0.26(Re)0.6

(Pr)0.3

(5.30)

whereas for liquids at low Reynolds numbers, 1 < (Re) < 200:

17

(Nu) = 0.86(Re)0.43

(Pr)0.3

(5.31)

As in eqn. (5.23), (Pr) for gases is nearly constant so that simplified equations can be written.

Fluid properties in these forced convection equations are evaluated at the mean film temperature,

which is the arithmetic mean temperature between the temperature of the tube walls and the

temperature of the bulk fluid.

EXAMPLE 5.8. Heat transfer in water flowing over a sausage

Water is flowing at 0.3 m s-1

across a 7.5 cm diameter sausage at 74°C. If the bulk water

temperature is 24°C, estimate the heat-transfer coefficient.

Mean film temperature = (74 + 24)/2 = 49°C.

Properties of water at 49°C are:

cp = 4.186 kJ kg-1

°C-1

, k = 0.64 J m-1

s-1

°C-1

, = 5.6 x 10-4

N s m-2

, = 1000

kg m-3

.

Therefore (Re) = (Dv/µ)

= (0.075 x 0.3 x 1000)/(5.6 x 10-4

)

= 4.02 x 104

(Re)0.6

= 580

(Pr) = (cp /k)

= (4186 x 5.6 x 10-4

)/0.64

= 3.66.

(Pr)0.3

= 1.48

(Nu) = (hcD/k)

= 0.26(Re)0.6

(Pr)0.3

Therefore hc = k/D x 0.26 x (Re)0.6

(Pr)0.3

= (0.64 x 0.26 x 580 x 1.48)/0.075

= 1904 J m-2

s-1

°C-1

EXAMPLE 5.9. Surface heat transfer to vegetable puree

Calculate the surface heat transfer coefficient to a vegetable puree, which is flowing at an

estimated

3 m min-1

over a flat plate 0.9 m long by 0.6 m wide. Steam is condensing on the other side of

the plate and maintaining the surface, which is in contact with the puree, at 104°C. Assume that

the properties of the vegetable puree are, density 1040 kg m-3

, specific heat 3980 J kg-1

°C-1

,

viscosity 0.002 N s m-2

, thermal conductivity 0.52 J m-1

s-1

°C-1

.

v = 3m min-1 = 3/60 ms-1

(Re) = (Lv/)

= (0.9 x 3 x 1040)/(2 x 10-3

x 60)

= 2.34 x 104

Therefore eqn. (5.26) is applicable and so:

(hcL/k) = 0.036(Re)0.8

(Pr)0.33

Pr = (cp/k)

= (3980 x 2 x 10-3

)/0.52

= 15.3

and so

(hcL/k) = 0.036(2.34 x 104)0.8

15.30.33

hc = (0.52 x 0.036) (3.13 x 103)(2.46)/0.9

= 160 J m-2

s-1

°C-1

18

EXAMPLE 5.10. Heat loss from a cooking vessel

What would be the rate of heat loss from the cooking vessel of Example 5.7, if a draught caused

the air to move past the cooking vessel at a speed of 61 m min-1

Assuming the vessel is equivalent to a flat plate then from eqn. (5.27)

v = 61/60 = 1.02 m s-1

, that is v < 1.02 m s-1

therefore hc = 5.7 + 3.9v

= 5.7 + (3.9 x 61)/60

= 9.7 J m-2

s-1

°C-1

So with A = 3.4m2, T = 32°C,

q = 9.7 x 3.4 x 32

= 1055 J s-1

19

OVERALL HEAT-TRANSFER COEFFICIENTS

It is most convenient to use overall heat transfer coefficients in heat transfer calculations as these

combine all of the constituent factors into one, and are based on the overall temperature drop. An

overall coefficient, U, combining conduction and surface coefficients, has already been

introduced in eqn. (5.5). Radiation coefficients, subject to the limitations discussed in the section

on radiation, can be incorporated also in the overall coefficient. The radiation coefficients should

be combined with the convection coefficient to give a total surface coefficient, as they are in

series, and so:

hs = (hr + hc) (5.32)

The overall coefficient U for a composite system, consisting of surface film, composite wall,

surface film, in series, can then be calculated as in eqn. (5.5) from:

1/U = 1/(hr + hc)1 + x1/k1 + x2/k2 + …+ 1/(hr + hc)2. (5.33)

EXAMPLE 5.11. Effect of air movement on heat transfer in a cold store In Example 5.2, the overall conductance of the materials in a cold-store wall was calculated.

Now on the outside of such a wall a wind of 6.7 m s-1

is blowing, and on the inside a cooling unit

moves air over the wall surface at about 0.61 m s-1

. The radiation coefficients can be taken as

6.25 and 1.7 J m-2

s-1

°C-1

on the outside and inside of the wall respectively. Calculate the overall

heat transfer coefficient for the wall.

Outside surface: v = 6.7 m s-1

.

And so from eqn. (5.28)

hc = 7.4v0.8

= 7.4(6.7)0.8

= 34 J m-2

s-1

°C-1

and hr = 6.25 J m-2

s-1

°C-1

Therefore hs1 = (34+6) = 40 J m-2

s-1

°C-1

Inside surface: v = 0.61 m s-1

.

From eqn. (5.27)

hc = 5.7 + 3.9v = 5.7 + (3.9 x 0.61)

= 8.1 J m-2

s-1

°C-1

and hr = 1.7 J m-2

s-1

°C-1

Therefore hs2 = (8.1 + 1.7) = 9.8 J m-2

s-1

°C-1

Now from Example 5.2 the overall conductance of the wall,

Uold = 0.38 J m-2

s-1

°C-1

and so

1/Unew = 1/hs1 + 1/Uold + 1/hs2

= 1/40 + 1/0.38 + 1/9.8

= 2.76.

Therefore Unew = 0.36 J m-2

s-1

°C-1

In eqn. (5.33) often one or two terms are much more important than other terms because of their

numerical values. In such a case, the important terms, those signifying the low thermal

conductances, are said to be thecontrolling terms. Thus, in Example 5.11 the introduction of

values for the surface coefficients made only a small difference to the overall U value for the

insulated wall. The reverse situation might be the case for other walls that were better heat

conductors.

EXAMPLE 5.12. Comparison of heat transfer in brick and aluminium walls Calculate the respective U values for a wall made from either (a) 10 cm of brick of thermal

conductivity 0.7 J m-1

s-1

°C-1

, or (b) 1.3mm of aluminium sheet, conductivity 208 J m-1

s-1

°C-1

.

Surface heat-transfer coefficients are on the one side 9.8 and on the other 40 J m-2

s-1

°C-1

.

http://www.nzifst.org.nz/unitoperations/httrtheory2.htm#ex5_2

20

(a) For brick

k = 0.7 J m-1

s-1

°C-1

x/k = 0.1/0.7 = 0.14

Therefore 1/U = 1/40 + 0.14 + 1/9.8

= 0.27

U = 3.7 J m-2

s-1

°C-1

(b) For aluminium

k = 208 J m-1

s-1

°C-1

x/k = 0.0013/208

= 6.2 x 10-6

1/U = 1/40 + 6.2 x 10-6

+ 1/9.8

= 0.13

U = 7.7 J m-2

s-1

°C-1

Comparing the calculations in Example 5.11 with those in Example 5.12, it can be seen that the

relative importance of the terms varies. In the first case, with the insulated wall, the thermal

conductivity of the insulation is so low that the neglect of the surface terms makes little

difference to the calculated U value. In the second case, with a wall whose conductance is of the

same order as the surface coefficients, all terms have to be considered to arrive at a reasonably

accurate U value. In the third case, with a wall of high conductivity, the wall conductance is

insignificant compared with the surface terms and it could be neglected without any appreciable

effect on U. The practical significance of this observation is that if the controlling terms are

known, then in any overall heat-transfer situation other factors may often be neglected without

introducing significant error. On the other hand, if all terms are of the same magnitude, there are

no controlling terms and all factors have to be taken into account.

Heat Transfer

Introduction• Heat transfer is employed in cooking, baking, drying,

sterilizing or freezing

Heat transfer is a dynamic process in which heat is transferred spontaneously from one body to another cooler body.

• The rate of heat transfer depends upon the differences in temperature between the bodies, the greater the difference in temperature, the greater the rate of heat transfer.

• rate of transfer = driving force / resistance

• rate of heat transfer = temperature difference/ heat flow

resistance of medium

Heat Transfer: conduction, radiation and convection.

• In conduction, the molecular energy is directly exchanged, from the hotter to the cooler regions, the molecules with greater energy communicating some of this energy to neighboring molecules with less energy.

• Radiation is the transfer of heat energy by electromagnetic waves, which transfer heat from one body to another.

• Convection is the transfer of heat by the movement of groups of molecules in a fluid. The groups of molecules may be moved by either density changes or by forced motion of the fluid.

HEAT CONDUCTION• rate = driving force/resistance, • The driving force is the temperature difference per unit

length of heat-transfer path, also known as the temperature gradient. Instead of resistance to heat flow, its reciprocal called the conductance is used. This changes the form of the general equation to:

• rate of heat transfer = driving force x conductance,that is:

dQ/dt = kA dT/dx (5.1)• Where,

– dQ/dt = is the rate of heat transfer, – A = is the area of cross-section of the heat flow path, – dT/dx = is the temperature gradient, that is the rate of

change of temperature per unit length of path, and– k = is the thermal conductivity of the medium.

• conductance, which relates to the actual thickness of a given material (k/x)

• and thermal conductivity, which relates only to unit thickness.

• The units of k, the thermal conductivity, can be found from eqn. (5.1) by transposing the terms

• k = dQ/dt x 1/A x 1/(dT/dx)

• = J s-1 x m-2 x 1/(°C m-1)= J m-1 s-1 °C-1

• Equation (5.1) is known as the Fourier equation for heat conduction.

Conduction through a Slab• The rate of heat transfer is given by• dQ/dt = kA T/x = kA dT/dx• Under steady temperature

conditions dQ/dt = constant, which may be called q:

• so q = kA dT/dx• but dT/dx, the rate of change of

temperature per unit length of path, is given by

• (T1 -T2)/x where x is the thickness of the slab,

• so q = kA(T1 - T2)/x• or

q = kA T/x =(k/x) A T (5.2)

EXAMPLE 5.1. Rate of heat transfer in cork

• A cork slab 10 cm thick has one face at -12°C and the other face at 21°C. If the mean thermal conductivity of cork in this temperature range is 0.042 J m-1 s-1 °C-1, what is the rate of heat transfer through 1 m2 of wall?– T1 = 21°C– T2 = -12°C– T = 33°C– A = 1 m2

– k = 0.042 J m-1 s-1 °C-1

– x = 0.1 m– q = 0.042 x 1 x 33/ 0.1 = 13.9 J s-1

Heat Conductances• Heat conductances are sometimes used instead

of thermal conductivities. • The heat conductance is the quantity of heat that

will pass in unit time, through unit area of a specified thickness of material, under unit temperature difference, For a thickness x of material with a thermal conductivity of –k = J m-1 s-1 °C-1, the conductance is k/x = C and the units of conductance areJ m-2 s-1 °C-1.

• Heat conductance = C = k/x.

Heat Conductances in Series• In a cold store wall, heat might pass through brick,

plaster, wood and cork.• In this case, eqn. (5.2) can be applied to each layer. This is

illustrated in Fig. 5.2.• A1 = A2 = A3 = ….. = A• q = AT1k1/x1 = AT2k2/x2 = AT3k3/x3 = ……..• So AT1 = q(x1/k1) and AT2 = q(x2/k2)

and AT3 = q(x3/k3).…..• AT1 + AT2 + AT3 + … = q(x1/k1) + q(x2/k2) +q(x3/k3) + …• A(T1 + T2 + T3 + ..) = q(x1/k1 + x2/k2 +x3/k3 + …)• The sum of the temperature differences over each layer is

equal to the difference in temperature of the two outside surfaces of the complete system, i.e.

• T1 + T2 + T3 + … = T

and since k1/x1 is equal to the conductance of the material in the first layer, C1, and k2/x2 is equal to the conductance of the material in the second layer C2,

• x1/k1 + x2/k2 + x3/k3 + ... = 1/C1 + 1/C2 + 1/C3 …= 1/U• where U = the overall conductance for the combined

layers, in J m-2 s-1 °C-1

• Therefore AT = q(1/U)• so q = UAT (5.3)

EXAMPLE 5.2 A cold store has a wall comprising 11 cm of brick on the outside, then 7.5 cm of concrete and then 10 cm of cork. The mean temperature within the store is maintained at -18°C and the mean temperature of the outside surface of the wall is 18°C.Calculate the rate of heat transfer through the wall. The appropriate thermal conductivities are for brick, concrete and cork, respectively 0.69, 0.76 and 0.043 J m-1 s-1 °C-1.Determine also the temperature at the interfaces between the concrete and cork layers, and the brick and concrete layers.

• For brick x1/k1 = 0.11/0.69 = 0.16.For concrete x2/k2 = 0.075/0.76 = 0.10.

• In the steady state, the same quantity of heat per unit time must pass through each layer.

• q = A1T1k1/x1 = A2T2k2/x2 = A3T3k3/x3 = ……..

If the areas are the same A1=A2=A3= ……………….• For cork x3/k3 = 0.10/0.043 = 2.33

.But 1/U = x1/k1 + x2/k2 + x3/k3

= 0.16 + 0.10 + 2.33= 2.59

• Therefore U = 0.38 J m-2 s-1°C-1

T = 18 - (-18) = 36°C,A = 1 m2

• q = UAT= 0.38 x 1 x 36

= 13.7 J s-1

• Further, q = A3T3k3/x3

• and for the cork wall A3 = 1 m2, x3/k3 = 2.33 and q = 13.7 J s-1

• Therefore 13.7 = 1 x T3 x 1/2.33 from rearranging eqn. (5.2)T3 = 32°C.

• But T3 is the difference between the temperature of the cork/concrete surface Tc (T3) and the temperature of the cork surface inside the cold store (T4)

• Therefore Tc - (-18) = 32 oC is ∆T3

• where Tc (T3) is the temperature at the cork/concrete surface

and so Tc = 14°C T3

• If T1 is the difference between the temperature of the brick/concrete surface, Tb (T2)and the temperature of the external air.

• Then 13.7 = 1 x T1 x 1/ 0.16 = 6.25 T1

• Therefore 18 - Tb = T1 = 13.7/6.25 = 2.2

• so Tb = 15.8 °C or T2

• Working it through shows approximate boundary temperatures: air/brick 18°C,brick/concrete 16°C, concrete/cork 14°C, cork/air -18°C

• This shows that almost all of the temperature difference occurs across the insulation (cork): and the actual intermediate temperatures can be significant especially if they lie below the temperature at which the atmospheric air condenses, or freezes.

Heat Conductances in Parallel• With heat conductances in

parallel, the material surfaces are parallel to the direction of heat transfer and to each other.

• The heat is therefore passing through each material at the same time, instead of through one material and then the next. This is illustrated in Figure 5.3.

EXAMPLE 5.3The wall of a bakery oven is built of insulating brick 10 cm thick and thermal conductivity 0.22 J m-1 s-1 °C-1. Steel reinforcing members penetrate the brick, and their total area of cross-section represents 1% of the inside wall area of the oven.If the thermal conductivity of the steel is 45 J m-1 s-1 °C-1

calculate (a) the relative proportions of the total heat transferred through the wall by the brick and by the steel and (b) the heat loss for each m2 of oven wall if the inner side of the wall is at 230°C and the outer side is at 25°C.– Applying eqn. (5.1) q = ATk/x, we know that T is the same

for the bricks and for the steel. Also x, the thickness, is the same.

(a) Consider the loss through an area of 1 m2 of wall (0.99 m2 of brick, and 0.01 m2 of steel)

• For brick qb = AbT kb/x = 0.99(230 - 25)0.22 / 0.10 = 446 J s-1

• For steel qs = AsT ks/x = 0.01(230 - 25)45 / 0.10 = 923 J s-1

• Therefore qb /qs = 0.48

(b) Total heat loss• q = (qb + qs ) per m2 of wall = 446 + 923 = 1369 J s-1

Therefore percentage of heat carried by steel= (923/1369) x 100 = 67%

SURFACE HEAT TRANSFER• Newton found, experimentally, that the rate of cooling of the

surface of a solid, immersed in a colder fluid, was proportional to the difference between the temperature of the surface of the solid and the temperature of the cooling fluid. This is known as Newton's Law of Cooling, and it can be expressed by the equation, analogous to eqn. (5.2),

q = hsA(Ta– Ts) (5.4)• where

– hs = surface heat transfer coefficient,– Ta = temperature of the cooling fluid – Ts = temperature at the surface of the solid.

The surface heat transfer coefficient can be regarded as the conductance of a hypothetical surface film of the cooling medium of thickness xf such that

hs = kf /xf

where kfthermal conductivity of the cooling medium.


• hs can be considered as arising from the presence of another layer, this time at the surface, added to the case of the composite slab considered previously. The heat passes through the surface, then through the various elements of a composite slab and then it may pass through a further surface film. We can at once write the important equation:

• q = AT[(1/hs1) + x1/k1 + x2/k2 +1/hs2)] = UAT• where 1/U = (1/hs1) + x1/k1 + x2/k2 +.. + (1/hs2)

and hs1, hs2 are the surface coefficients on either side of the composite slab, x1, x2 ...... are the thicknesses of the layers making up the slab, and k1, k2... are the conductivities of layers of thickness x1, ..... . The coefficienths is also known as the convection heat transfer coefficient and values for it will be discussed in detail under the heading of convection. It is useful at this point, however, to appreciate the magnitude of hs under various common conditions and these are shown in Table 5.1.

Table 5.1 Approximate Range of Surface Heat Transfer Coefficients

Fluid h (J m-2 s-1°C-1)

Boiling liquids 2400-24,000

Condensing liquids 1800-18,000

Still air 6

Moving air (3 m s-1) 30

Liquids flowing

through pipes1200-6000

Example 5.4. Heat Transfer In Jacketed Pan• Sugar solution is being heated in a jacketed pan

made from stainless steel, 1.6 mm thick. Heat is supplied by condensing steam at 200 kPa gauge in the jacket. The surface transfer coefficients are, for condensing steam and for the sugar solution, 12,000 and 3000 J m-2 s-1 °C-1 respectively, and the thermal conductivity of stainless steel is 21 J m-1 s-1 °C-1.Calculate the quantity of steam being condensed per minute if the transfer surface is 1.4 m2 and the temperature of the sugar solution is 83°C.

• From steam tables, the saturation temperature of steam at 200 kPa gauge(300 kPa Absolute) = 134°C, and the latent heat = 2164 kJ kg-1.

• For stainless steel x/k = 0.0016/21 = 7.6 x 10-5

• T = (condensing temperature of steam) - (temperature of sugar solution)= 134 - 83 = 51°C.

From eqn. (5.5)

• 1/U = [1/hsteam + x/k + 1/hsugar ]= [1/12,000 + 7.6 x 10-5 + 1/3000]U = 2032 J m-2 s-1 °C-1

• and since A = 1.4 m2

• q = UAT= 2032 x 1.4 x 51= 1.45 x 105 J s-1

• Therefore steam required

• = 1.45 x 105 / (2.164 x 106) kg s-1

= 0.067 kg s-1

= 4 kg min-1

UNSTEADY STATE HEAT TRANSFER

• In food process engineering, heat transfer is very often in the unsteady state, in which temperatures are changing and materials are warming or cooling.

• A simple case of unsteady state heat transfer arises from the heating or cooling of solid bodies made from good thermal conductors, for example a long cylinder, such as a metal bar, being cooled in air. The rate at which heat is being transferred to the air from the surface of the cylinder is given by eqn. (5.4)

• q = dQ/dt = hsA(Ts - Ta)

• where Ta is the air temperature and Ts is the surface temperature.


• Now, the heat being lost from the surface must be transferred to the surface from the interior of the cylinder by conduction. As an approximation, consider that all the heat is being transferred from the centre of the cylinder. In this instance, we evaluate the temperature drop required to produce the same rate of heat flow from the centre to the surface as passes from the surface to the air. This requires a greater temperature drop than the actual case in which much of the heat has in fact a shorter path.

• Assuming that all the heat flows from the centre of the cylinder to the outside, we can write the conduction equation

• dQ/dt = (k/L)A( Tc– Ts )

• where Tc is the temperature at the centre of the cylinder, k is the thermal conductivity of the material of the cylinder and L is the radius of the cylinder.

• Assuming that all the heat flows from the centre of the cylinder to the outside, we can write the conduction equation

• Q = dQ/dt = (k/L)A( Tc– Ts )

• where Tc is the temperature at the centre of the cylinder, k is the thermal conductivity of the material of the cylinder and L is the radius of the cylinder.

• Equating these rates:

• hsA(Ts --Ta) = (k/L)A( Tc– Ts )hs(Ts -- Ta) = (k/L)( Tc– Ts )

and so hsL/k = ( Tc– Ts )/ (Ts -- Ta)

• To take a practical case of a copper cylinder of 15 cm radius cooling in air kc = 380 J m-1 s-1 °C-1,

hs = 30 J m-2 s-1°C-1 (from Table 5.1), L = 0.15 m,

• (Tc– Ts)/ (Ts -- Ta) = (30 x 0.15)/380= 0.012

• In this case 99% of the temperature drop occurs between the air and the cylinder surface. By comparison with the temperature drop between the surface of the cylinder and the air, the temperature drop within the cylinder can be neglected. On the other hand, if the cylinder were made of a poorer conductor as in the case of the sausage, or if it were very large in diameter, or if the surface heat-transfer coefficient were very much larger, the internal temperature drops could not be neglected.

• This simple analysis shows the importance of the ratio:

• heat transfer coefficient at the surface = hsL/k

heat conductance to the centre of the solid

• This dimensionless ratio is called the Biot number (Bi) and it is important when considering unsteady state heat flow. When (Bi) is small, and for practical purposes this may be taken as any value less than about 0.2, the interior of the solid and its surface may be considered to be all at one uniform temperature. In the case in which (Bi) is less than 0.2, a simple analysis can be used, therefore, to predict the rate of cooling of a solid body.

• Therefore for a cylinder of a good conductor, being cooled in air,

• dQ = hsA(Ts - Ta) dt

But this loss of heat cools the cylinder in accordance with the usual specific heat equation:

• dQ = c .ρ .V.dT

• where c is the specific heat of the material of the cylinder, ρ is the density of this material and V is the volume of the cylinder.

• Since the heat passing through the surface must equal the heat lost from the cylinder, these two expressions for dQ can be equated:

• c .ρ.VdT = hsA(Ts -- Ta) dt

• Integrating between Ts = T1 and Ts = T2 , the initial and final temperatures of the cylinder during the cooling period, t, we have:

• - hsAt/c ρ V = loge (T2 - Ta)/(T1– Ta)

• or (T2 - Ta)/(T1 – Ta) = exp( -hsAt/c ρ V ) ----- 5.6

• For this case, the temperatures for any desired interval can be calculated, if the surface transfer coefficient and the other physical factors are known. This gives a reasonable approximation so long as (Bi) is less than about 0.2. Where (Bi) is greater than 0.2, the centre of the solid will cool more slowly than this equation suggests. The equation is not restricted to cylinders, it applies to solids of any shape so long as the restriction in (Bi), calculated for the smallest half-dimension, is obeyed.

• Charts have been prepared which give the temperature relationships for solids of simple shapes under more general conditions of unsteady-state conduction. These charts have been calculated from solutions of the conduction equation and they are plotted in terms of dimensionless groups so that their application is more general. The form of the solution is:

• ƒ{(T - T0)/( Ti - T0 )} = F{(kt/c ρ L2)(hsL/k)} (5.7)

• where ƒ and F indicate functions of the terms following, Ti is the initial temperature of the solid, T0 is the temperature of the cooling or heating medium, T is the temperature of the solid at time t, (kt/c ρ L2) is called the Fourier number (Fo) (this includes the factor k/c ρ the thermal conductance divided by the volumetric heat capacity, which is called the thermal diffusivity) and (hsL/k) is the Biot number.

• A mathematical outcome that is very useful in these calculations connects results for two- and three-dimensional situations with results from one-dimensional situations. This states that the two- and three-dimensional values called F(x,y) and F(x,y,z) can be obtained from the individual one-dimensional results if these are F(x), F(y) andF(z), by simple multiplication:

• F(x,y) = F(x)F(y)and

F(x,y,z) = F(x)F(y)F(z)

• Using the above result, the solution for the cooling or heating of a brick is obtained from the product of three slab solutions. The solution for a cylinder of finite length, such as a can, is obtained from the product of the solution for an infinite cylinder, accounting for the sides of the can, and the solution for a slab, accounting for the ends of the can.

• Charts giving rates of unsteady-state heat transfer to the centre of a slab, a cylinder, or a sphere, are given in Fig. 5.4. On one axis is plotted the fractional unaccomplished temperature change,(T - T0)/( Ti - T0 ). On the other axis is the Fourier number, which may be thought of in this connection as a time coordinate. The various curves are for different values of the reciprocal of the Biot number, k/hr for spheres and cylinders, k/hl for slabs.More detailed charts, giving surface and mean temperatures in addition to centre temperatures, may be found inMcAdams (1954), Fishenden and Saunders (1950) and Perry (1997)





FIGURE 5.4. TRANSIENT HEAT CONDUCTIONTEMPERATURES AT THE CENTRE OF SPHERE,SLAB,AND CYLINDER:

ADAPTED FROM HENDERSON AND PERRY, AGRICULTURAL PROCESS ENGINEERING, 1955

EXAMPLE 5.5. Heat transfer in cooking sausages

• A process is under consideration in which large cylindrical meat sausages are to be processed in an autoclave. The sausage may be taken as thermally equivalent to a cylinder 30 cm long and 10 cm in diameter. If the sausages are initially at a temperature of 21°C and the temperature in the autoclave is maintained at 116°C, estimate the temperature of the sausage at its centre 2 h after it has been placed in the autoclave.

• Assume that the thermal conductivity of the sausage is 0.48 J m-1 s-1 °C-1, that its specific gravity is 1.07, and its specific heat is 3350 J kg-1 °C-1. The surface heat-transfer coefficient in the autoclave to the surface of the sausage is 1200 J m-2 s-1 °C-1.

• This problem can be solved by combining the unsteady-state solutions for a cylinder with those for a slab, working from Fig. 5.3.

• (a) For the cylinder, of radius r = 5 cm (instead of L in this case)

• Bi = hsr/k = (1200 x 0.05)/0.48 = 125

• (Often in these systems the length dimension used as parameter in the charts is the half-thickness, or the radius, but this has to be checked on the graphs used.)

• So 1/(Bi) = 8 x 10-3

After 2 hours t = 7200 s

• Therefore Fo = kt/cr2 = (0.48 x 7200)/[3350 x 1.07 x 1000 x (0.05)2] = 0.39

• and so from Fig. 5.3 for the cylinder:(T - T0)/( Ti - T0 ) = 0.175 = say, F(x).

• (b) For the slab the half-thickness 30/2 cm = 0.15 mand so Bi = hsL/k = (1200 x 0.15)/0.48 = 375

1/ Bi = 2.7 x 10-3

t = 7200 s as before andkt/cρL2 = (0.48 x 7200)/[3350 x 1.07 x 1000 x (0.15)2]

= 4.3 x 10-2

• and so from Fig. 5.3 for the slab:(T - T0)/( Ti - T0 ) = 0.98 = say, F(y)

• So overall (T - T0)/( Ti - T0 ) = F(x) F(y)= 0.175 x 0.98= 0.172

• Therefore, (T2-116)/(21-116)=0.172

• Therefore T2 = 100°C

RADIATION HEAT TRANSFER

• Radiation heat transfer is the transfer of heat energy by electromagnetic radiation. Radiation operates independently of the medium through which it occurs and depends upon the relative temperatures, geometric arrangements and surface structures of the materials that are emitting or absorbing heat.

• The calculation of radiant heat transfer rates, in detail, is beyond the scope of this book and for most food processing operations a simplified treatment is sufficient to estimate radiant heat effects. Radiation can be significant with small temperature differences as, for example, in freeze drying and in cold stores, but it is generally more important where the temperature differences are greater. Under these circumstances, it is often the most significant mode of heat transfer, for example in bakers' ovens and in radiant dryers.

• The basic formula for radiant-heat transfer is the Stefan-Boltzmann Law

• q = A ϭT 4

• where T is the absolute temperature (measured from the absolute zero of temperature at -273°C, and indicated in Bold type) in degrees Kelvin (K) in the SI system, and ϭ (sigma) is the Stefan-Boltzmann constant = 5.73 x 10-8J m-2 s-1K-4 The absolute temperatures are calculated by the formula

K = (°C + 273).

• This law gives the radiation emitted by a perfect radiator (a black body as this is called though it could be a red-hot wire in actuality). A black body gives the maximum amount of emitted radiation possible at its particular temperature. Real surfaces at a temperature T do not emit as much energy as predicted by eqn. (5.8), but it has been found that many emit a constant fraction of it. For these real bodies, including foods and equipment surfaces, that emit a constant fraction of the radiation from a black body, the equation can be rewritten

• q = ϵA ϭT 4 (5.9)

• Where ϵ (epsilon) is called the emissivity of the particular body and is a number between 0 and 1. Bodies obeying this equation are called grey bodies

• Emissivities vary with the temperature T and with the wavelength of the radiation emitted. For many purposes, it is sufficient to assume that for:

*dull black surfaces (lamp-black or burnt toast, for example), emissivity is approximately 1;

*surfaces such as paper/painted metal/wood and including most foods, emissivities are about 0.9;

*rough un-polished metal surfaces, emissivities vary from 0.7 to 0.25;*polished metal surfaces, emissivities are about or below 0.05.

These values apply at the low and moderate temperatures which are those encountered in food processing.

• Just as a black body emits radiation, it also absorbs it and according to the same law, eqn. (5.8).Again grey bodies absorb a fraction of the quantity that a black body would absorb, corresponding this time to their absorptivity (alpha). For grey bodies it can be shown that = . The fraction of the incident radiation that is not absorbed is reflected, and thus, there is a further term used, the reflectivity, which is equal to (1 – ).

Pumps, Compressors, Fans,

Ejectors and Expanders

Pumps

• Moves Liquid, Creates Pressure– Vapor bubbles

• Causes Cavitations

• Erodes Impeller

– Solids Erode Impeller

• Pump Types– Centrifugal

– Positive Displacement

• Piston

• diaphragm

• Pump Power = Q*ΔP = brake (delivered) (horse) power from motor

Centrifugal Pumps

• Two Basic Requirements for Trouble-

Free Operation of Centrifugal Pumps

– no cavitation of the pump occurs throughout

the broad operating range

– a certain minimum continuous flow is always

maintained during operation

• Pump around loops

Reduced Flows

• Unfavorable conditions which may occur separately or simultaneously when the pump is operated at reduced flows

• Cases of heavy leakages from the casing, seal, and stuffing box

• Deflection and shearing of shafts

• Seizure of pump internals

• Close tolerances erosion

• Separation cavitation

• Product quality degradation

• Excessive hydraulic thrust

• Premature bearing failures

Centrifugal Pump

Electric Motor

Centrifugal Pump

Electric

Motor

Centrifugal Pump

• Converts

kinetic

energy to

pressure

energy

Impellers

Converts Kinetic Energy to

Pressure Energy

Different Types of Pump Head

• Total Static Head - Total head when the pump is not running

• Total Dynamic Head (Total System Head) - Total head when the pump is running

• Static Suction Head - Head on the suction side, with pump off, if the head is higher than the pump impeller

• Static Suction Lift - Head on the suction side, with pump off, if the head is lower than the pump impeller

• Static Discharge Head - Head on discharge side of pump with the pump off

• Dynamic Suction Head/Lift - Head on suction side of pump with pump on

• Dynamic Discharge Head - Head on discharge side of pump with pump on

Pump Head

• The head of a pump can be expressed in metric units as:

• head = (p2 - p1)/(ρg) + (v22- v1

2)/(2g) + (z2-z1)

• where

• h = total head developed (m)

• p2 = pressure at outlet (N/m2)

• p1 = pressure at inlet (N/m2)

• ρ = density of liquid (kg/m3)

• g = acceleration of gravity (9.81) m/s2

• v2 = velocity at the outlet (m/s)

http://www.engineeringtoolbox.com/pump-energy-equation-d_631.html

Pump Efficiency

• Centrifugal Pump

Pump Performance Curves

Resistance

Pump Design Scaling

• Pump Flow rate• Q2 = Q1 x [(D2xN2)/(D1xN1)]

• Pump Head• H2 = H1 x [(D2xN2)/(D1xN1)]2

• Pump Brake Horse Power • BHP2 = BHP1 x [(D2xN2)/(D1xN1)]3

– D = Impeller Diameter

– N = specific speed

Net Positive Suction Head-NPSH

• Pumps can not pump vapors!

• The satisfactory operation of a pump

requires that vaporization of the liquid

being pumped does not occur at any

condition of operation.

Net Positive Suction Head

Required, NPSHR

As the liquid passes from the pump suction to the eye of the impeller, the velocity

increases and the pressure decreases. There are also pressure losses due to

shock and turbulence as the liquid strikes the impeller. The centrifugal force of the

impeller vanes further increases the velocity and decreases the pressure of the

liquid. The NPSH required is the positive head (absolute pressure) required at the

pump suction to overcome these pressure drops in the pump and maintain the

liquid above its vapor pressure.

Net Positive Suction Head

Available, NPSHA

Net Positive Suction Head Available is a function of the system in which the

pump operates. It is the excess pressure of the liquid in feet absolute over its vapor

pressure as it arrives at the pump suction, to be sure that the pump selected does

not cavitate.

Head to Feed Pump Subcooling before Pump

To overcome suction head

Head

Designed

into

Installation

HX

Cool a few Degrees

To overcome suction head

Piston Pumps

Gear Pumps

Lobe Pumps

• food applications,

because they

handle solids

without damaging

the pump.

• Particle size

pumped can be

much larger in

these pumps than

in other PD types

Screw Pump

Centrifugal

Pump

Positive Displacement Pumps

• Piston Pumps

• Gear Pumps

– Lobe Pumps

• Diaphragm Pumps– The lower the speed of a PD

pump, the lower the NPSHR.

Pump Costs

• Cost based upon Size Factor

– Centrifugal Pump

• S=QH1/2

– Gear Pump

• S=Q

– Piston Pump

• S= Power (brake)

• Must cost Electric Motor also• S=Pc=PB/ηM

Compressors

• Types– Centrifugal

– Others• Piston

• Lobed

• Screw

– Methods of Calculation in Simulators• Polytropic, PVk-1/k= constant,

– Polytropic - This model takes into account both a rise in temperature in the gas as well as some loss of energy (heat) to the compressor's components. This assumes that heat may enter or leave the system, and that input shaft work can appear as both increased pressure (usually useful work) and increased temperature above adiabatic (usually losses due to cycle efficiency). Compression efficiency is then the ratio of temperature rise at theoretical 100 percent (adiabatic) vs. actual (polytropic). (k-1)/k = polytropic coefficient

• Isentropic, s(T1,P1)=s(T2,isentropic,P2)

• Theoretical Power– Powerisentropic= FlowRate*(h2,isentropic-h1)

• Efficiency ηs =Powerisentropic/Powerbrake

• ηs = (h2,isentropic-h1)/(h2-h1)

– Cost of Compressors• Size Factor is Compressor Power

s

k

k

P

PT

TT

1

1

1

21

12

http://en.wikipedia.org/wiki/Polytropic_process

Positive Displacement Compressor

Positive Displacement Compressor

http://www.city-compressors.co.uk/




Centrifugal Compressors

• Rotors

• Stators

• Jet

Engine

Design

Piston Compressor

Expander

• Reverse of Compressor

• Let flow produce shaft work

• Types– Centrifugal

– Positive Displacement• Piston

• Lobed

• Screw

– Methods of Calculation in Simulators• Polytropic, PVk-1/k= constant,

• Isentropic, s(T1,P1)=s(T2,isentropic,P2)

• Theoretical Power– Powerisentropic= f*(h2,isentropic-h1)

• Efficiency ηs=Powerbrake/Powerisentropic= (h2-h1) /(h2,isentropic-h1)

– Cost• Size factor = Power





Fans and Blowers

• Types

– Centrifugal (103-105 acfm, P=1-40 in H2O)• Backward Curved

• Straight radial

– Vane Axial

– Tube Axial

• Cost of Fans and Blowers

– Size factor = Volumetric Flow Rate

– Motor

Choice to Increase Pressure

• Heuristic 34– Use a Fan

• Atm to 1.47 psig

– Use a Blower• < 30 psig

– Compressor (or staged system)• > 30 psig

• Heuristic 34 - Number of Stages– Up to a Compression ratio 4 for each stage

• With intercooler between stages (ΔP=2 psi)

– Equal Hp for each stage (equal compression ratio)

Documents

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