Basic Information for Engineering

8/8/2019 Basic Information for Engineering

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Engineer/ Misfer Mohammed Al-Dossary(2007)

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Basic i n fo r m a t i on f o r

Eng ineer in g & Design o f

Tech n ica l App l icat ion

Prepar ed by :

Eng inee r / M i sfe r Mohamm ed Al -Dossary


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1 . Mech an ics


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Mass and WeightWhat is weight and what is mass? An explanation of the difference between weight andmass

Mass and weight are two common, and often misused and misunderstood terms in mechanics and fluid

mechanics.

The fundamental relation between the mass and the weight is defined from the Newton's Second Law and can beexpressed in SI-units as

F = m a (1)

Where

F = force (N)

m = mass (kg)

a = acceleration (m/s2)

Mass

Massis a measure of the amount of material in an object, being directly related to the number and type of atomspresent in the object. Mass does not change with a body's position, movement or alteration of its shape unlessmaterial is added or removed.

The mass is a fundamental property of an object, a numerical measure of its inertia and a fundamental measureof the amount of matter in the object.

Weight

Weightis the gravitational force acting on a body mass. Transforming Newton's Second Law regarding the weightas a force due to gravity gives

w = m g (2)

Where

w = weight (N)

m = mass (kg)

g = acceleration of gravity (m/s2)

The handling of mass and weight depends on the systems of units that are used. The most common systems ofunits are the

International System - SI

British Gravitational System - BG

English Engineering System - EE


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The International System - SIIn the SI system the mass unit is the kg and since the weight is a force - the weight unit is the Newton (N).Equation (2) for a body with 1 kg mass can then be expressed as:

w = 1 (kg) 9.807 (m/s2) = 9.807 (N) (2b)

Where

9.807 (m/s2) = standard gravity close to earth in the SI system

As a result:

a 9.807 N force acting on a body with 1 kg mass will give the body an accelerationof 9.807 m/s

2

A body with mass of 1 kg weights 9.807 N

The British Gravitational System - BG

The British Gravitational System of units is used by engineers in the English-speaking world with the samerelation to the foot - pound - second system as the meter kilogram - force second system (SI) has to the meter -kilogram - second system. For engineers who deal with forces, instead of masses, it's convenient to use a systemthat has as its base unit's length, time, and force, instead of length, time and mass.

The three base units are the foot, the second, and the pound-force.

In the BG system the mass unit is the slug and is defined from the Newton's Second Law (1). The unit of mass,the slug, is derived from the pound-force by defining it as the mass that will accelerate at 1 foot per second persecond when a 1 pound-force acts upon it:

1 lb = (1 slug)(1 ft/s2)

In other words, 1 lb force acting on 1 slug mass will give the mass an acceleration of 1 ft/s

2

.

The weight of the mass from equation (2) in BG units can be expressed as:

w (lb) = m (slugs) g (ft/s2)

With a standard gravity 32.174 ft/s2

- the mass of 1 slug weights 32.174 lb.

The English Engineering System - EE

In the English Engineering system of units the primary dimensions are force, mass, length, time, and temperature.The units for force and mass are defined independently:

the basic unit of mass is pound mass (lbm)

the unit of force is the pound (lb) alternatively pound force (lbf).

In the EE system 1 lb of force will give a mass of 1 lbm a standard acceleration of 32.174 ft/s2.

Since the EE system operates with these units of force and mass, the Newton's Second Law can be modified to

F = m a / gc (3)

Where


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gc = a proportionality constantor transformed to weight

w = m g / gc (4)

The proportionality constant gc makes it possible to define suitable units for force and mass. We can transform (4)

to

1 lb = (1 lbm)(32.174 ft/s2) / gc

or

gc = (1 lbm)(32.174 ft/s2)/(1 lb)

Since 1 lb force gives a mass of 1 lbm an acceleration of 32.174 ft/s2

and a mass of 1 slug an acceleration of 1ft/s

2, then

1 slug = 32,174 lbm

Example - Weight versus Mass

A car's mass is 1,644 kg. The weight can be calculated:

w = (1,644 kg)(9.807 m/s2) = 16.122,7 N = 16.1 kN

In other words - there is a force of 16.1 kN between the car and the earth.

Density, Specific Weight and Specific Gravity

An introduction and definition of density, specific weight and specific gravity - formulaswith examples

Density

Density is defined as an object mass per unit volume. Mass is a property.

The density can be expressed as

= m / V = 1 / vg (1)

Where

= density (kg/m3)

m = mass (kg)

V = volume (m3)

vg = specific volume (m3/kg)


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The SI units for density are kg/m3. The imperial (BG) units are lb/ft

3(slugs/ft

3). While people often use pounds per

cubic foot as a measure of density in the U.S., pounds are really measures of force, not mass. Slugs are thecorrect measure of mass. You can multiply slugs by 32.2 for a rough value in pounds.

The higher the density, the tighter the particles are packed inside the substance. Density is a physicalproperty constant at a given temperature and density can help to identify a substance.

Example - Use the Density to Identify the Material:

An unknown liquid substance has a mass of 18.5 g and occupies a volume of 23.4 ml. (milliliter).

The density can be calculated as

= [18.5 (g) / 1000 (g/kg)] / [23.4 (ml) / 1000 (ml/l) 1000 (l/m3) ]

= 18.5 10-3

(kg) / 23.4 10-6

(m3)

= 790 kg/m3

If we look up densities of some common substances, we can find that ethyl alcohol, or ethanol, has a density of790 kg/m

3. Our unknown liquid may likely be ethyl alcohol!

Example - Use Density to Calculate the Mass of a Volume

The density of titanium is 4507 kg/m3

. Calculate the mass of 0.17 m3

titanium!

m = 0.17 (m3) 4507 (kg/m

3)

= 766.2 kg

Specific Weight

Specific Weight is defined as weight per unit volume. Weight is a force.

Specific Weight can be expressed as

= g (2)

Where

= specific weight (kN/m3)

g = acceleration of gravity (m/s2)

The SI-units of specific weight are kN/m3. The imperial units are lb/ft

3. The local acceleration g is under normal

conditions 9.807 m/s2

in SI-units and 32.174 ft/s2

in imperial units.Example - Specific Weight Water

Specific weight for water at 60oF is 62.4 lb/ft

3in imperial units and 9.80 kN/m

3in SI-units.


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Example - Specific Weight Some other MaterialsSpecific Weight -

Product

Imperial Units(lb/ft3)

SI Units(kN/m3)

Ethyl Alcohol 49.3 7.74

Gasoline 42.5 6.67

Glycerin 78.6 12.4

Mercury 847 133

SAE 20 Oil 57 8.95

Seawater 64 10.1

Water 62.4 9.80

Specific Gravity

The Specific Gravity - SG - is a dimensionless unit defined as the ratio of density of the material to the density ofwater at a specified temperature. Specific Gravity can be expressed as

SG = = / H2O (3)

Where

SG = specific gravity

= density of fluid or substance (kg/m3)

H2O = density of water (kg/m3)

It is common to use the density of water at 4oC (39

oF) as reference - at this point the density of water is at the

highest.

Since Specific Weight is dimensionless it has the same value in the metric SI system as in the imperial Englishsystem (BG). At the reference point the Specific Gravity has same numerically value as density.

Example - Specific Gravity

If the density of iron is 7850 kg/m3, 7.85 grams per cubic centimeter (cm

3), 7.85

kilograms per liter, or 7.85 metric tons per cubic meter - the specific gravity of iron is:

SG = 7850 kg/m3/ 1000 kg/m

3

= 7.85


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(The density of water is 1000 kg/m3)

Newton's Second Law

F = m g (1)

Where

F = force

m = mass

g = acceleration of gravity

The force caused by gravity - g - is called weight. Note!Mass - m - is a property.

The acceleration of gravity can be observed by measuring the change velocity of a free falling object:

g = dv / dt (2)

Where

dv = change in velocity

dt = change in time

If we drop an object we will observe that it will accelerate to a speed of 9,81 m/s or 32,174 ft/s in one second.

Acceleration of Gravity in SI Units

g = 9.81 m/s2

Acceleration of Gravity in Imperial Units

g = 32.174 ft/s2

Velocity and Distance Traveled of a Free Falling Object

The velocity of a free falling object can be expressed as:

v = g t (3)

Where

v = velocity

The distance traveled by a free falling object can be expressed as:

s = 1/2 g t2

(4)

Where

s = distance traveled by the object


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The velocity and distance traveled by a free falling object:Velocity Distance

Time (s)

m/s ft/s m ft

1 9.8 32.2 4.9 16.1

2 19.6 64.3 19.6 64.3

3 29.4 96.5 44.1 144.8

4 39.2 128.7 78.5 257.4

5 49.1 160.9 122.6 402.2

6 58.9 193.0 176.6 579.1

7 68.7 225.2 240.3 788.3

8 78.5 257.4 313.9 1,029.6

9 88.3 289.6 397.3 1,303.0

10 98.1 321.7 490.5 1,608.7

Note! The velocity is achieved without any aero-dynamical resistance (vacuum). Obviously the air resistance willbe significant for increasing velocities.

Belt

Belt driven fans are common in heating, ventilation and cooling systems.

Belt Length

The length of the belt can be calculated as

lb = ((df+ dm) 1.5708 ) + (2 lfm) (1)

Where

lb = length of belt (mm, inches)


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df= pulley diameter fan (mm, inches)dm = pulley diameter motor (mm, inches)

lfm = center to center distance of fan and motor pulleys (mm, inches)

Belt Velocity

The velocity at which a belt travels may be expressed as

v = dm nm / 12 (2)

Where

v = velocity of belt (ft/min)

nm = velocity motor (rpm)

Belt Gearing

The relationship between the speed of the motor, the fan and their disc diameters can be expressed as

df nf= dm nm (3)


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Electrical FormulasThe most common used electrical formulas - Ohms Law and their combinations

Common electrical units used in formulas and equations are:

Volts - The units of electrical potential or motive force. The force is required to sendone ampere of current through one ohm of resistance.

Ohms - The units of resistance. One ohm is the resistance offered to the passage ofone ampere when impelled by one volt.

Amperes - The units of current. One ampere is the current which one volt can sendthrough a resistance of one ohm.

Watts - The unit of electrical energy or power. One watt is the product of oneampere and one volt. One ampere of current flowing under the force of one voltgives one watt of energy.

Volt Amperes - The product of the volts and amperes as shown by a voltmeter andammeter. In direct current systems, volt ampere is the same as watts or the energydelivered. In alternating current systems, the volts and amperes may or may not be100% synchronous. When synchronous, the volt amperes equal the watts on awattmeter. When not synchronous, volt amperes exceed watts. More about reactivepower.

Kilovolt Ampere - One kilovolt ampere - KVA - is equal to 1,000 volt amperes.

Power Factor- is the ratio of watts to volt amperes.

Electric Power Formulas

W = E I (1a)

W = R I2

(1b)

W = E2/ R (1c)

where

W = power (Watts)

E = voltage (Volts)

I = current (Amperes)

R = resistance (Ohms)

Electric Current Formulas

I = E / R (2a)

I = W / E (2b)

I = (W / R)1/2

(2c)

Electric Resistance Formulas

R = E / I (3a)

R = E2/ W (3b)


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R = W / I2

(3c)

Electrical Potential Formulas - Ohms Law

Ohms law can be expressed as:

E = R I (4a)

E = W / I (4b)

E = (W R)1/2

(4c)

Example - Ohm's law

A 12 volt battery supplies power to a resistance of 18 ohms.

I = (12 Volts) / (18 ohms)

= 0.67 Ampere

Electrical Motor Formulas

Electrical Motor Efficiency

= 746 Php / Winput (5)

Where

= efficiency

Php = output horsepower (hp)

Winput = input electrical power (Watts)

Or alternatively

= 746 Php / (1.732 E I PF) (5b)

Electrical Motor - Power

W3-phase = (E I PF 1.732) / 1,000 (6)

Where

W3-phase = electrical power 3-phase motor (kW)

PF = power factor electrical motor

Electrical Motor - Amps

I3-phase = (746 Php) / (1.732 E PF) (7)

Where


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I3-phase = electrical current 3-phase motor (Amps)PF = power factor electrical motor

Electrical Motor Efficiency

Calculating electric motor efficiencies

Electrical motor efficiency is the ratio between the shaft output power - and the electrical input power.

Electrical Motor Efficiency when Shaft Output is measured in Watt

If the power output is measured in Watt (W), efficiency can be expressed as:

m = Pout / Pin (1)

Where

m = motor efficiency

Pout = shaft power out (Watt, W)

Pin = electric power in to the motor (Watt, W)

Electrical Motor Efficiency when Shaft Output is measured in Horsepower

If the power out is measured in horsepower (hp), efficiency can be expressed as:

m = Pout 746 / Pin (2)

where

Pout = shaft power out (horsepower, hp)

Pin = electric power in to the motor (Watt, W)

Primary and Secondary Resistance Losses

The electrical power lost in the primary rotor and secondary stator winding resistance are also called the copperlosses. The copper loss vary with the load in proportion to the current squared and can be expressed as

Pcl = R I2

(3)

where

Pcl = stator winding - copper loss (W)

R = resistance ()

I = current (Amp)

Iron Losses

These losses are the result of magnetic energy dissipated when when the motors magnetic field is applied to thestator core.


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Stray LossesStray losses are the losses that remain after primary copper and secondary losses, iron losses and mechanicallosses. The largest contribution to the stray losses is harmonic energies generated when the motor operatesunder load. These energies are dissipated as currents in the copper windings, harmonic flux components in theiron parts, leakage in the laminate core.

Mechanical Losses

Mechanical losses include friction in the motor bearings and the fan for air cooling.

NEMA Design B Electrical Motors

Electrical motors constructed according NEMA Design B must meet the efficiencies below:

Power(hp)

Minimum Nominal Efficiency1)

1 - 4 78.8

5 - 9 84.0

10 - 19 85.5

20 - 49 88.5

50 - 99 90.2

100 - 124 91.7

> 125 92.4

1)NEMA Design B, Single Speed 1200, 1800, 3600 RPM. Open Drip Proof (ODP) or Totally Enclosed Fan Cooled

(TEFC) motors 1 hp and larger that operate more than 500 hours per year.


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Electrical Motors and Heat LossHeat gain from electrical motors to the surroundings

As a rule of thumb the heat loss from an electrical motor during operation can be estimated as

Size of Motor(kW)

Heat Loss(Watts/kW)

0 - 2 250

3 - 15 150

15 - 150 100

150 - 80

1 kW = 1.34 hp

1 hp = 0.746 k W

Electrical Units

Definition of common electrical units as Ampere, Volt, Ohm, Siemens and more.

Ampere - (A)

The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, ofnegligible circular cross section, and placed 1 meter apart in vacuum, would produce between these conductors aforce equal to 2 x 10

-7Newton per meter of length.

Coulomb - (C)

The standard unit of quantity in electrical measurements. It is the quantity of electricity conveyed in one second bythe current produced by an electro-motive force of one volt acting in a circuit having a resistance of one ohm, orthe quantity transferred by one ampere in one second.

Farad - (F)

The farad is the standard unit of capacitance. Reduced to base SI units, one farad is the equivalent of one secondto the fourth power ampere squared per kilogram per meter squared (s

4A

2kg

-1m

-2).

When the voltage across a 1 F capacitor changes at a rate of one volt per second (1 V/s), a current flow of 1 Aresults. A capacitance of 1 F produces 1 V of potential difference for an electric charge of one coulomb (1 C).

In common electrical and electronic circuits, units of microfarads (F), where 1 F = 10-6

F, and picofarads (pF),where 1 pF = 10

-12F, are used.


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Ohm - ()The derived SI unit of electrical resistance; the resistance between two points on a conductor when a constantpotential difference of 1 volt between them produces a current of 1 ampere.

Henry - (H)

The Henry is the unit of inductance. Reduced to base SI units, one Henry is the equivalent of one kilogram metersquared per second squared per ampere squared (kg m

2s

-2A

-2).

Inductance

An inductor is a passive electronic component that stores energy in the form of a magnetic field.

The standard unit of inductance is the Henry, abbreviated H. This is a large unit. More common units are themicro-Henry, abbreviated H (1 H =10

-6H) and the mille- Henry, abbreviated mH (1 mH =10

-3H). Occasionally,

the nano-henry (nH) is used (1 nH = 10-9

H).

Joule - (J)

The unit of energy work or quantity of heat done when a force of one Newton is applied over a displacement ofone meter. One joule is the equivalent of one watt of power radiated or dissipated for one second.

In imperial units, the British thermal unit (Btu) is used to express energy. One Btu is equivalent to approximately1,055 joules.

Siemens - (S)

The unit of electrical conductance S = A / V

Watt

The watt is used to specify the rate at which electrical energy is dissipated, or the rate at which electromagneticenergy is radiated, absorbed, or dissipated.

The unit of power W or Joule/second

Weber - Wb

The unit of magnetic flux.

The flux that, when linking a circuit of one turn, produces in it an electromotive force (Emf) of 1 volt as it isreduced to zero at a uniform rate in one second.

1 Weber is equivalent to 108

Maxwells

Tesla - T

The unit of magnetic flux density. The Tesla is equal to 1 Weber per square meter of circuit area.

Volt

The Volt (V) is the Standard International (SI) unit of electric potential or electromotive force. A potential of onevolt appears across a resistance of one ohm when a current of one ampere flows through that resistance.Reduced to SI base units, 1 V = 1 kg times m

2times s

-3times A

-1(kilogram meter squared per second cubed per

ampere).


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Frequency and Speed of Electrical Motors

The velocity of 2, 4, 6 and 8 poles electrical motors at 50 Hz and 60 Hz

The velocity at which an induction motor will operate depends on the input power frequency and the number of electricalmagnetic poles in the motor.

Speed (RPM)

Frequency 50 Hz Frequency 60 Hz

Poles

Synchronous Full Load Synchronous Full Load

2 3,000 2,850 3,600 3,450

4 1,500 1,425 1,800 1,725

6 1,000 950 1,200 1,150

8 750 700 900 850

KVA

Real and reactive power

Total electrical power consumption depends on real power - electrical energy consumption, and reactive power -imaginary power consumption, and can be expressed as

W = Wapplied + Wreactive (1)

Where

W = total power (Kilovolt Amps)

Wapplied = true applied power (Kilowatts)

Wreactive = apparent reactive power (Kilovolt Amps reactive)

Reactive Power

Reactive Power - Wreactive - is measured in volt-amperes reactive (VAR) and is the power stored in and dischargedby the inductive motors, transformers or solenoids.


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Reactive power required by inductive loads increases the amount of apparent power - measured in kilovolt amps(kVA) - in the distribution system. Increasing the reactive and apparent power causes the power factor - PF - todecrease.

Power and Horsepower in Electrical Motors

Electrical power is rated in Horsepower or Watts.

Electrical power is in general rated in Watts or Horsepower. A horsepower is a unit of power equal to 746 watts or33,000 lb.ft per minute (or 550 lb.ft per second).

A watt is a unit of measure equal to the power produced by a current of 1 amp across the potential difference of 1volt. A watt is 1/746 of 1 horsepower.

Even if the watt is the base unit of electrical power, its common to rate motor power is in either horsepower orwatts.

Power in Watts

Electric power of a motor can be expressed as:

Pw = m U I (1)

where

Pw = power (W, watt)

m = motor efficiency

U = voltage (V)

I = current (A, amps)

Power in Horsepower

Horse power of a motor can be expressed as:

Php = Pw / 746 (2)

Or

Php = m U I / 746 (2b)

Where

Php = horsepower (hp)

Example - The Horsepower of an Electrical Motor

The horse power of a 230 V electrical motor with 85% efficiency pulling 10 amps can be calculated as:

Php = 0.85 (230 V) (10 amps) / 746

= 2, 62 hp


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Power Factor for a Three-Phase Electrical Motor

A Power Factor definition and the Power Factor for a three-phase electrical motor

The power factor of an AC electric power system is defined as the ratio of the active (true or real) powerto theapparent power.

Active (Real or True) Poweris measured in watts (W) and is the power drawn by the electricalresistance of a system that does useful work.

Apparent Poweris measured in volt-amperes (VA) and is the voltage on an AC system multiplied by allthe current that flows in it. It is the vector sum of the true and the reactive power.

The third component of the AC power flow, the

Reactive Power, is measured in volt-amperes reactive (VAR). Reactive Power is the power stored inand discharged by the inductive motors, transformers or solenoids.

The reactive power required by an inductive load will increase the amount of apparent power - measured inkilovolt amps (kVA) - in the distribution system. Increasing the reactive and apparent power will cause the powerfactor - PF - to decrease.

Power Factor

It is common to define the Power Factor - PF - as the cosine of the phase angle between voltage and current - orthe "cos". The power factor defined by IEEE and IEC is the ratio between the applied true power - and theapparent power, and can in general be expressed as:

PF = Wactive / Wapparent (1)

where

PF = power factor

Wactive = active (true or real) power (Watt)

Wapparent = apparent power (VA, volts amps)

A low power factor is the result of inductive loads such as transformers and electric motors. Unlike resistive loadscreating heat by consuming kilowatts, inductive loads require a current flow to create magnetic fields to producethe desired work.

Power factor is an important measurement in electrical AC systems because

an overall power factor less than 1 indicates that the electricity supplier need to provide more generatingcapacity than actually required

the current waveform distortion that contributes to reduced power factor is caused by voltage waveformdistortion and overheating in the neutral cables of three-phase systems

International standards such as IEC 61000-3-2 have been established to control current waveform distortion byintroducing limits for the amplitude of current harmonics.


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Example - Power FactorA industrial plant draws 200 A at 400 V and the supply transformer and backup UPS is rated 200 A 400 V = 80kVA.

If the power factor - PF - of the loads is only 0.7 - only 80 kVA 0.7 = 56 kVA of real power is consumed by thesystem. If the power factor was close to 1, the supply system with transformers, cables, switchgear and UPS

could have been done considerably smaller.

A low power factor is expensive and inefficient and some utility companies may charge additional fees when thepower factor is less than 0.95. A low power factor will reduce the electrical system's distribution capacity byincreasing the current flow and causing voltage drops.

Power Factor for a Three-Phase Motor

The total power required by an inductive device as a motor or similar consists of

Active (true or real) power (measured in kilowatts, kW)

Reactive power - the nonworking power caused by the magnetizing current, required to operate thedevice (measured in kilovars, kVAR)

The power factor for a three-phase electric motor can be expressed as:

PF = Wapplied / [(3)1/2

U I] (2)

where

PF = power factor

Wapplied = power applied (W, watts)

U = voltage (V)

I = current (A, amps)


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3 . Com bu st ion


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Chimney SizingCalculating chimney draft and chimney area

The capacity of a chimney depends on the potential chimney draft - the pressure difference created by the insideand outside air temperature difference - and the chimney area and height.

The pressure difference will force air and fuel gas from the boiler or fireplace through the chimney and out of thebuilding.

Chimney Draft

The difference pressure draft in a chimney can be expressed as:

dpch = h (o - i) g (1)

Where

dpch = pressure draft in the chimney (Pa)

h = height of chimney (m)

o = density of outside air (kg/m3)

i = density of inside air (kg/m3)

g = 9.81 - gravitation (m/s2)

(1) can alternatively be expressed as:

dpch = 0.0465 dt h (2)

Where

dt = temperature difference between inside and outside air (K,oC)

Chimney Area

The design velocity of air and flue gases in a small furnace should be below 2 m/s. The design velocity of a largefurnace should be below 10 m/s.

The chimney area can be calculated as

A = Q / v (3)

Where

A = cross-sectional area of chimney (m2)

Q = volume of flue gases at chimney temperature (m3/s)

v = velocity (m/s)

It is a general rule to provide 1100 mm2

chimney area per 1 kW boiler rating.


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Combustion Efficiency and Excess Air

Optimizing a boilers efficiency is important to minimize fuel consumption and unwanted

excess to the environment

To ensure complete combustion of the fuel used, combustion chambers are supplied with excess air. Excess airincrease the amount of oxygen and the probability of combustion of all fuel.

when fuel and oxygen in the air are in perfectly balance - the combustion is said to be stoichiometric

The combustion efficiency will increase with increased excess air, until the heat loss in the excess air is largerthan the heat provided by more efficient combustion.

Typical excess air to achieve highest efficiency for different fuels is

5 - 10% for natural gas

5 - 20% for fuel oil

15 - 60% for coal

Carbon dioxide - CO2 - is a product of the combustion and the content in the flue gas is an important indication ofthe combustion efficiency.


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An optimal content of carbon dioxide - CO2 - after combustion is approximately 10% for natural gas, andapproximately 13% for light oils.

Normal combustion efficiencies for natural gas at different amounts of excess air and flue gas temperatures areindicated below:

Combustion Efficiency (%)

Excess % Flue Gas Temperature (oF)

Air Oxygen 200 300 400 500 600

9.5 2.0 85.4 83.1 80.8 78.4 76.0

15 3.0 85.2 82.8 80.4 77.9 75.4

28.1 5.0 84.7 82.1 79.5 76.7 74.0

44.9 7.0 84.1 81.2 78.2 75.2 72.1

81.6 10.0 82.8 79.3 75.6 71.9 68.2

Flue Gas Loss Combustion Oil

The relationship between temperature difference flue gas and supply air, CO2 concentration in the flue gas, andthe efficiency loss in the flue gas combustion oil, is expressed in the diagram below.


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Combustion Testing

Combusting testing of fuel oil and gas burners

To secure safe and efficient operation of gas or fuel oil burners it is common to test for

carbon dioxide

smoke

excess air (O2/CO2)

stack temperature

draft

NOx

Carbon Dioxide - CO2

When the CO2 content of the flue gas is low, less than 8 percent, heat is lost up the chimney and the operation isinefficient.

Low carbon dioxide content may be caused by

to small burner nozzle

air leakage into the furnace or boiler

under firing in the combustion chamber


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When the CO2 content of the flue gas is too high it is common with excess smoke in the flue gas. High carbondioxide content may be caused by insufficient draft

over fired burner

Smoke (only fuel oil burners)

Smoke in flue gas indicates poor burner performance. The amount of smoke can be measured with a smoketester where smoke particles set on a filter paper are interpreted according a Bacharach scale.

Smoky combustion can be caused by

soot formation on the heating surfaces

insufficient draft, incorrectly adjusted draft regulator, improper fan delivery

poor fuel supply, malfunctioning fuel pump

defective firebox

oil-burner nozzle defect, wrong size

excessive air leaks in boiler or furnace

wrong fuel-air ratio

Stack Temperature

The "net stack temperature" is the difference between the flue gas inside the chimney and the room temperatureoutside the burner. Net stack temperatures above 700

oF are in general to high. Typical values are between 330 -

500oF.

High stack temperatures may be caused by

undersized furnace

defective combustion chamber

incorrectly sized combustion chamber

excessive draft

over fired burner

draft regulator improperly adjusted

soot formation on the heating surfaces


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Energy Content some Common SourcesCommon energy sources and their energy content

The energy content in some common energy sources can be found in the table below:

Energy Source Unit Energy Content (BTU)

Electricity 1 Kilowatt-hour 3,412

Coal 1 Short Ton 20,681,000

Crude Oil 1 Barrel - 42 gallons 5,800,000

Diesel Fuel 1 Gallon 139,000

Gasoline 1 Gallon 124,000

Natural Gas 1 Cubic Foot 1,026

Heating Oil 1 Gallon 139,000

Propane 1 Gallon 91,000

Residual Fuel Oil1)

1 Barrel - 42 gallons 6,287,000

1)Residual Fuel Oil - The liquid or semi-liquid, high-boiling fraction of residue from the distillation of crude oil.

BTU - British Thermal Unit

The unit of heat in the imperial system - the BTU - is

the amount of heat required to raise the temperature of one pound of water through 1 degF (58.5oF -

59.5oF) at sea level (30 inches of mercury).

1 Btu (British thermal unit) = 1055.06 J = 107.6 kpm = 2.931 10-4

kWh = 0.252 kcal = 778.16 ft.lbf =1.05510

10ergs = 252 cal = 0.293 watt- hours

An item using one kilowatt-hour of electricity will generate 3412 BTU.


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Flash Point - FuelsSome common fuels and their flash points

The flash point of a chemical is the lowest temperature where it will evaporate enough fluid to form a combustibleconcentration of gas. The flash point is an indication of how easy a chemical may burn. Materials with higher flash

points are less flammable or hazardous than chemicals with lower flash points.

T(oC) = 5/9[T(oF) - 32]

Fuel Gases - Heating Values

Heating values of more or less common fuel gases as acetylene, blast furnace gas,ethane, biogas and many more

1 Btu/ft3

= 8.9 kcal/m3

= 3.73x104

J/m3

1 Btu/lb = 2,326.1 J/kg = 0.55556 kcal/kg

Heating Value

The gross (high) and net (low) heating values

The heating value is

the amount of heat produced by combustion of a unit quantity of a fuel

We differentiate in gross and net heating values:

Gross (or high, upper) Heating Value

The gross or high heating value is the amount of heat produced by the complete combustion of a unit quantity offuel.

The gross heating value is obtained when

all products of the combustion are cooled down to the temperature before thecombustion

the water vapor formed during combustion is condensed

Net (or lower) Heating Value

The net or lower heating value is obtained by

subtracting the latent heat of vaporization of the water vapor formed by thecombustion

from the gross or higher heating value.


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Common UnitsCommon units used for heating value are

1 Btu/lb = 2,326.1 J/kg = 0.55556 kcal/kg

1 J/kg = 0.43 Btu/lb = 2.39x10-4

kcal/kg

1 kcal/kg = 1.80 Btu/lb = 4,187 J/kg

Natural Gas Consumption

Natural gas consumption in some common equipment as boiling pans, ovens, cookers,kettles and more

Approximate natural gas demand for common appliances can be found in the table below.

Natural Gas ConsumptionHeat

Dissipated

Type of Consumer

(ft3/h)

(m3/sx 10

-6)

(liter/s) (Btu/hour) (kW)

Small broiler 30000 9

Large broiler 61000 18

Boiler and roaster, combined 68000 20

Wash Boiler 30 -50

230 -400

0.23 -0.40

27000 -51000

8 -15

Furnace 120000

Storage water heater, 50 gallon (115 - 190liter)

50000

1 kW = 3,412 Btu/h

Documents

Basic Information for Engineering