Combustion basics.docx

7/29/2019 Combustion basics.docx

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Combustion basics: The Fire Triangle / Fire

Tetrahedron

Fire Triangle

For many years, it has been widely accepted that the combination of the following three (3)

factors contribute to the initiation of a fire:

- fuel

- a source of ignition/energy/heat- oxygen

The combination of all these three (3) factors has therefore led to the widely known Triangle ofCombustion. Consequently, when trying to put out a fire, we strive to remove at least one of

these three components

Picture 1Fire triangle

Fire Tetrahedron
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Further research over the years revealed a fourth necessary component of fire, the chemical chain

reaction. The Fire Triangle was consequently transformed to the Fire Tetrahedron. Put it simply,a Tetrahedron is a solid pyramid with four plane faces (from the Greek words tesseris edres),each one representative of the four necessary elements.

In short, a fire begins by an external ingition source which is usually in the form of a flame orspark. With its turn, the external ignition source heats the fuel under the presence of oxygen. As

both fuel and oxygen are heated, molecular activity increases. If properly heated, a self-sustaining chemical reaction is developed . The consequent chemical reaction will then escalate

at a point where the external ignition source is no longer necessary for the propagation of the

fire.

Once ignition has occurred, it will continue until:

- all the available fuel has been consumed or

- the fuel and/or oxygen is removed or- the temperature is reduced by cooling or

- the number of excited molecules is reduced and the chain reaction is broken

Picture 2Fire tetrahedron

Although simplistic in nature, presented diagrams are in fact a good example of how to actually

extinguish a fire in the real world. For example, we can create a barrier using foam in order to

reduce/eliminate the fueling of fire and therefore deprive the fire from one of its necessaryelements (oxygen). By applying water, we can also reduce the temperature below the ignition

temperature. Finally, by using a Halon extinguisher, we can create an inert gas barrier which will

interfere with the chemical chain reaction.

Related Standards and Codes

There are various international standards referring to fire protection. Most widely used standardsnowadays are National Fire Protection Association (NFPA) standards. However, reference can

also be made to other US standards (like ANSI standards or Code of Federal Regulation (CFR)

standards) or European standards (ISO and DIN standards). Leading commercial insurance
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companies have also developed standards of their own (for example Factory Mutual) and their

requirements may sometimes be stricter than those of international standards.

Understanding power factor

A motor always consumes active power, which is converted into mechanical action. Activepower is expressed in Watt and is usually represented by the letter P. In some textbooks, activepower can also be referred to as true or real power.

Reactive power is also required for the magnetisation of the motor, but this does not produceany action. Reactive power is expressed in VAR (Volt Ampere Reactive) and is usually

represented by the letter Q. Reactive power is power stored in and discharged by electrical

equipments like inductive motors, transformers and solenoids.

If both active and reactive power are combined together, they yield the apparent powerS. In

other words, the apparent power is the vector sum of the active and the reactive power. Apparentpower is expressed in VA. (please refer to the graph below)
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Figure 1Active, Reactive and Apparent power vectors

Definition of power factor

The ratio between the active power and the reactive power is known as the power factor, and is

often designated as cos . A typical value for power factor ranges between 0.7-0.9: the lowervalue is typically valid for small motors, whereas the bigger value for larger motors.

In general, power factor (PF) can be expressed with the following formula:

PF = P / S ..(1)

where,

P = Magnitude of active (true or real) powerS = Magnitude of apparent power

The reactive power required by inductive loads increases the amounts of apparent power in thedistribution system. Increasing of the value of reactive and apparent power will cause the power

factor to drop.


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Importance of power factor

The value of power factor is important for electrical systems because:

An overall power factor value less than 1 demonstrates that the electricity supplier has toprovide more generating capacity than actually required

Voltage waveform distortion and overheating in the cables causes current waveform distortion,thus contributing to reduced power factor.

Low power factor is very inefficient and some public power utilities may charge extra when thepower factor value drops below 0.95.

Leading and Lagging Power Factors

Power factors are usually referred to as leading or lagging so as to indicate the sign of thephase angle. Inductive loads like transformers and motors consumes reactive power with current

waveform lagging the voltage. Capacitive loads like capacitor banks typically produce reactive

power with current phase leading the voltage.

Squirrel cage induction motors

Perhaps the most well-known type of electric motor in use nowadays is the three-phase squirrelcage induction motor.

It typically consists of two (2) main components: the stator and the rotor. As their name

implies, the stator part is stationary, whereas the rotor part is rotating. The stator produces therotating magnetic field compared to the rotor, which transforms this type of energy into

movement, i.e. mechanical energy. Please refer to Figure 1 (D-end stands for motor driven end,
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i.e the equipment driven by the motor, e.g. a pump, is located at that side, whereas N-end stands

for motor non-driven end)

Figure 1Electric motor structure

How an electric motor works

To explain how an electric motor operates, one must first note that the stator is connected to the

three-phase power supply. The current which is applied at the stator windings produces a rotating

magnetic force field, which by its turn creates current also at the rotor of the motor. As a result, amagnetic field is also created at the rotor. The interaction between these two magnetic fields (i.e.
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from the stator and from the rotor) yields the turning torque, which is the cause that makes the

rotor shaft rotate.

Motor rotation speed and motor slip

Due to various losses inherent to the motors nature, the speed of the motor is alwaysapproximately 1 to 3 % lower compared to the magnetic field synchronous speed. Thisdifference is usually referred to as motor slip (s).

Slip is given by the following formula:

s = (n1n) / n1. (1), where:

n1=synchronous speed

n= asynchronous speed

As a result, this type of motors is usually known as asynchronous motors.

However, it is to be noted that permanent magnet motors do not produce any slip at all

(permanent magnet motors will not be discussed at this article)

The synchronous speed (n), expressed in revolutions per minute (or rpm) is given by the

following formula:

n = (120 * f)/ p. (2), where:

f = motor supply frequency (in Hz) and

p = number of motor poles (even number)

Electric motor efficiency

As expected, the energy conversion at a motor is not loss-free. Several losses occur, being amongothers the result of resistance losses, ventilation losses, friction losses etc.

Consequently, the motor efficiency (n) is expressed from the following formula:

n = P2 / P1. (3),where

P2 = motor shaft power (in W)P1 = applied electric power (in W)

P2 is mentioned at the motors name plate.


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Insulation class

Insulation class refers to the insulation material of the motors windings. Several insulationclasses, namely B, F and H, exist, in accordance with IEC (International Electrotechnical

Commission) standards. A letter corresponding to the temperature, which is the upper limit for

the insulation application area, characterizes each insulation class.

Insulation class B corresponds to a maximum winding temperature of 130C (ambient

temperature 40C + temperature increase 80C + temperature margin 10C)

Insulation class F corresponds to a maximum winding temperature of 155C (ambient

temperature 40C + temperature increase 105C + temperature margin 10C)

Insulation class H corresponds to a maximum winding temperature of 180C (ambienttemperature 40C + temperature increase 125C + temperature margin 15C)

Protection class

A motors protection class is stated with the letters IP (IP stands for Ingress Protection)followed by two digits, the first of which indicates the degree of protection against contact and

penetration of solid objects, whereas the second states the motors degree of protection against

water. This classification is in accordance with IEC 60034-5 standard. For more details about IPprotection designation, please refer to Enggyclopedia article about theIngress Protection

Enclosures.

Cooling method

Cooling methods is accordance with IEC 60034-6 can tell how a motor is cooled. The coolingmethod is designated with the letters IC (IC stands for International Cooling) followed by a

series of digits, representing the cooling type (e.g. self-ventilated, forced cooling etc) and the

cooling mode of operation (e.g. surface cooling, liquid cooling etc). Please refer to Figure 2,highlighting the most common methods of motor cooling.
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Figure 2Common methods of electric motor cooling
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Steam generation fundamentals

The process of boiling water to make steam is a familiar phenomenon. Thermodynamicallyspeaking, the heat energy used results in a change of phase from liquid to gaseous state, i.e. from

water to steam. A steam generating system has to provide a continuous and uninterrupted heat

source for this conversion.

The simplest steam generating equipment used to carry out this conversion is a kettle type boiler

for heating a specific quantity of water. As a result of the applied heat, the water temperature

increases. Eventually, for a given presssure, theboiling (saturation) temperatureis reached andbubbles begin to form. As heat continues to be applied, the temperature remains unchanged and

steam begins to be created, escaping the water surface. At this point all the heat energy coming

into the system is used for vaporization of water and not for increase in temperature. If the steam

is continuously removed out of the vessel, the temperature will remain unaltered until all thewater has evaporated. In order to guarantee the continuous conversion of water to steam, all we

have to do is provide a regulated supply of water to equal the amount of steam being produced

and removed out of the vessel.

For a given pressure, steam heated above the saturation temperature is called superheated

steam, whereas water cooled below the saturation temperature is called subcooled water. If the

boiling water is a closed system, then after converting all the water to steam, new heat energycoming into the system is used for increasing steam temperature i.e. superheating the saturated

steam.

Production of high pressure steam

Technical and economic considerations indicate that the most efficient and econominc way of

producing high pressure (HP) steam is to heat relatively small diameter tubes containing a

continuous flow of water.

Two inherently different boiling systems are used in order to accomplish this task:

- The first system uses a steam drum or a fixed steam-water separation point and- The second system that does not use a steam drum, referred to as once-through steam

generator (OTSG).

Steam Drum

Perhaps the most widely used system is that of the steam drum. In this system, the drum serves

as the point of separation of steam from water. Subcooled water enters the tube to which heat is
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applied. As the water flows across the tube, it is heated to the boiling point. Consequently,

bubbles are formed and wet steam is generated. In most boilers, a steam-water mixture leaves thetube and enters the steam drum. The remaining water is then mixed with the makeup

(replacement) water, returned back to the heated tube and the process is repeated.

Once Through Steam Generator (OTSG)

Without the use of a steam drum, subcooled water also enters the heated tube and gets converted

to steam along the flow path (lenght of tube). The point where water turns into steam depends on

the water flow rate (boiler load) and heat input rate. With close control of both flow rate and heat

input rate, we can make sure that all of the water is evaporated withing the tube and only steamleaves the tube. Therefore, there is no need for having a steam drum.

Figure-1 presents a schematic view of steam generators with steam drum and without drum(OTSG).
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Figure1 - Schematic for steam generator with and without steam drum (OTSG)

At very high pressures, a point is reached where water no longer exhibits boiling behaviour.

Above this pressure (approximately 221 bar or 3200 psi), the water temperature keeps increasing

with added heat. Boilers designed to operate above this critical pressure are referred to assupercritical boilers. Drums are no longer required at supercritical boilers: these boilers operate

effectively n the once-through principle.


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Sample ProblemPump power calculations

Problem Statement

Estimate the shaft power and motor power requirement to pump 200,000 kg/hr of water available

at 250C and atmospheric pressure from a storage tank. The rated differential head requirement is

30 m.

Assume the mechanical efficiency of the pump to be 70%.

Assume the motor efficiency to be 90%.

Solution

The motor power required to run the pump as specified in the sample problem statement, iscalculated by first determining the theoretical power requirement and then dividing this

requirement by pump efficiency and motor efficiency.

Step1

The first step is to determine the important physical properties of water at given conditions. Theonly important physical property for solving this problem is the mass density of water.

Using EnggCyclopediasLiquid Density Calculator, waterdensityat 250C =994.72 kg/m

3

Using water density, the mass flow rate is converted to volumetric flow rate.

Volumetric flow = 200,000 / 994.72 = 201.06 m3/hr

Also the differential pressure is determined using differential head as,

P = gh = 994.72 9.81 30/105 = 2.93 bar

Step2

The next step is to determine the theoretical power requirement which is essentially the productof volumetric flow (Q) and differential pressure (P).

Power requirement = Q P = 201.06/3600 m3/s 2.93 10

5N/m

2

Theoretical power requirement = 16350 Watt = 16.35 kW
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Step3

Pump shaft power requirement = Theoretical power requirement / pump efficiency.

For a pump that has been already purchased or has been ordered for manufacturing, the

efficiency can be determined using the pump performance curves provided by pumpmanufacturer. Here the problem statement has specified pump efficiency to be 70%.

Hence, pump shaft power requirement = 16.35 kW / 0.7 = 23.36 kW

Similarly, motor power requirement = Pump shaft power requirement / motor efficiency

Similar to pump efficiency, electric motor efficiency for motors already purchased or ordered,

can be provided by the manufacturer of motor. However for purpose of this sample problem the

efficiency is to be taken as 90% as per problem statement.

Motor power requirement = 23.36 / 0.9 = 25.95 kW = 25.95 1.3596 HP = 35.28 HP

Electric motors are available for following standard Horsepower ratings.

1 1.5 2 3 5 7.5 10 15 20 25 30 40 50

60 75 100 125 150 200 250 300 350 400 450 500 600

700 800 900 1000 1250 1500 1750 2000 2250 2500 3000 3500 4000

Hence in order to have satisfy the minimum power requirement, the motor to be purchased has to

have power rating of 40 HP or higher.

Sample ProblemPump sizing calculations

Sample Problem Statement -
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Estimate the pump differential pressure, shaft power and motor power requirement to pump

200,000 kg/hr of water. The water stream is available from a storage tank which operates atatmospheric pressure and 250C.

Minimum liquid level in the storage tank above pump suction nozzle is kept as 3m. Suction line

is 6 in size and 10m long.

The discharge from pump is to be sent to another vessel with a top connection for water inlet.

The maximum height for the 6 discharge line above the pump discharge nozzle is 12m. Thedischarge vessel operates at a pressure of 3 barg. There is no control valve in the discharge line.

Discharge line to be assumed 100m long considering all the fittings and valves.

Assume pump efficiency to be close to 70% and motor efficiency to be close to 90%.

Solution -

This sample problem is solved in following 3 basic steps. Various calculators fromEnggCyclopedia are used for solving this sample problem.

Step 1.

First step of solving this sample problem requires determination of the important physical

properties of given fluid (water) at given temperature and pressure conditions.

Using EnggCyclopediasLiquid Density Calculator, water density at 250C =994.72 kg/m3

Using EnggCyclopediasLiquid Viscosity Calculator, water viscosity at 250C =0.90 cP
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Using EnggCyclopediasVapor Pressure Calculator, water vapor pressure at 250C =0.032 bara

Step 2.

Second step to solve this sample problem is to calculate various pressure drop values in the

suction side. Line pressure drop is to be calculated usingEnggCyclopedias pipe pressure drop

calculator. For help regarding the use of this calculator, refer tosolved example for line sizing.The line pressure drop for discharge line is also to be calculated in the same way. In the present

case, the pressure drop for 6 suction and discharge lines is around 5 bar/km. For 10m suction

line the pressure drop becomes 0.05 bar and for 100m discharge line, it becomes 0.5 bar.
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For suction line strainer, pressure drop can be calculated usingEnggCyclopedias Strainer

pressure drop calculator. For this example, the strainer pressure drop is around 0.09 bar.

Step 3.

All these inputs should now be entered inEnggCyclopedias Pumps sizing calculatoras the final

step of problem solving. In the present example only the difference between heights of liquid

level and pump suction/discharge nozzles is considered instead of considering the absolute

heights. Hence the pump suction and discharge nozzles are considered as reference levels onsuction and discharge sides respectively. Hence, the heights of these nozzles are considered as

zero.

The following image presents the calculation performed along with all the inputs and outputs.
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observed from the solved sample problem, the required differential pressure is 4.52 bar. Pump

shaft power and motor power requirements are 36.1 and 40.1 kW respectively. NPSHa andsuction pressure values are also available for checking the suction line adequacy. A high NPSHa

value (11.5m) indicates that the pump suction line and height of the liquid on suction side are

adequate to run the pump.

Liquid density

This is a sample of theliquid density calculator. To access the working calculator, please sign

up for free membership trial.

The application presented here can be used to calculate liquid phase densities of various organic

and inorganic substances at a specified temperature. For any sunstance the temperature must be

within an applicable range displayed upon the selection of liquid. The densities given here are in

kg/m3 and temperatures are in Kelvin (K).

An easy way of browsing this list is to know the type of substance you are looking for. In adescending order this list presents Alkanes, Alkenes, Alkynes, Cycloalkanes, Aromatics,

Alcohols, Ethers, Aldehydes, Ketones, Organic Acids, Esters, Amines, Amides, Nitriles, OrganicSulphides, Haloalkanes, Haloaromatics and a small group of inorganic compounds at the end.

Select Liquid from the list:

CO2 Carbon dioxide

CS2 Carbon disulphide

HF Hydrogen fluoride

HCl Hydrogen Chloride

HBr Hydrogen bromide

HCN Hydrogen cyanide

H2S Hydrogen sulphide

SO2 Sulphur dioxide

SO3

Sulphur trioxideH2O Water

Minimum applicable temperature = K

Maximum applicable temperature = K

Calculate the liquid density at temperature = K

Liquid density = kg/m3
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Also refer toEnggCyclopedias gas density calculator.

Explosive Levels of GasesUpper and Lower

Limits

Fire Triangle

The widely known Fire Triangle shows the three necessary components required to supportcombustion: Fuel, Oxygen and Ignition source. All three elements must be present in order to

support combustion. What the triangle does not show is that both fuel and oxygen have to be

mixed in the relevant proportion in order to burn.

Picture 1Fire Triangle

A combustible mixture can only be produced within a limited band of gas/air concentration. This

band is unique for each gas and vapor and is characterised by an upper level, characterised as the

Upper Explosive Level (UEL) and a lower level, known as Lower Explosive Level (LEL). Thesevalues are also sometimes referred to as the lower and upper flammability limits (LFL and UFLrespectively).

At concentration levels below the LEL, there is not enough fuel (gas) to initiate an explosion and

the gas/air mixture is poor. At concentration levels above UEL, there is not enough oxygenand the gas/air mixture is rich.
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The flammable range falls between the limits of the LEL and UEL for every specific gas or

mixture of gases. An increase in pressure, temperature or oxygen content will usually broadenthe flammability range.

Typical flammability levels of some gases

FuelLower Explosive Limit (LEL) (%

volume in air)Lower Explosive Limit (LEL) (%

volume in air)

Methane 5.0 15.0

Butane 1.6 8.4

Propane 2.1 9.6

Ethanol 3.3 19.0

Gasoline (100

Octane)1.4 7.8

IsopropylAlcochol 2.0 12.7

Ethyl Ether 1.9 36.0

Xylene 0.9 7.0

Toluene 1.0 7.1

Hydrogen 4.0 75.0

Acetylene 2.5 85.0

It is self-evident that for safety reasons, use of any gas detection systems will have to be set so as

to detect levels from zero percent of gas concentration up to the LEL. By the time this

concentration is reached, shut-down procedures or facility evacuation should have been put inplace. However, this will typically take place at a concentration usually less than 50% of the LEL

so that adequate safety margin and time is provided.

Special attention needs to be taken at enclosed or not properly ventilated spaces/enclosures. At

such places, a concentration higher than the UEL is possible to occur.Therefore, extreme

vigilence and attention is required at times of inspection of such spaces, since merely the openingof a door or a hatch will force air to the inspected enclosure which could dilute the gases to a

hazardous, combustible mixture.


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Safety Integrity Levels (SIL)

Probability of Failure on Demand (PFD)

SIL deals with the safety of any product-equipment consisting of an E/E/PES(Electrical/Electronic/Programmable Electronic System) system. SIL is a quantifiable measure of

the E/E/PES of a product, testing if the product is able to carry out its intended safety function-

operation when called to do so. This is where the term Probability of Failure on Demand (PFD)

comes in. There are four (4) levels of SIL rating (please refer to the table below). The higher theSIL rating the smaller the PFD of the equipment and therefore the safer the equipment. A

number of methods for determining SIL requirements (e.g. risk graph, hazardous event severity

matrix etc) is provided at standards IEC 61508 and IEC 61511.

Safety Integrity Levels from standard IEC 61508-1

Safety

Integrity Level

(SIL)

Demand Mode of Operation (average

probability of failure to perform design

function on demandPFD)

Continuous / High demand Mode of

Operation (probability of dangerous

failure per hour)

4 10-

to 10-4

10-

to 10-

3 10-4 to 10- 10- to 10-

2 10- to 10- 10- to 10-

1 10-

to 10-1

10-

to 10-

Equipment designed/produced before the introduction of SIL rating system may receive, when

tested, a low or no SIL rating. In order to overcome this problem, there are various techniquesthat can be recommended, always in consultation with the Owner and the Manufacturer of the

equipment. Such techniques include among others the following: frequent proof testing or

combining systems with different types of technologies in order to eliminate common failuremodes.
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