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Fuels and combustion

The consumption of energy by chemical and biochemical process industries are

quite large

The process energy requirements are met by the combustion of fossil fuels

The fossil fuels consist chiefly of carbon and hydrogen, with small amounts of

oxygen, nitrogen ,sulphur ,mineral matter and moisture.

The composition of fuels vary widely depending upon the type and source of the

fuels.

During combustion ,the chemical energy stored in the fuels is released as heat

energy.

The combustion process is the rapid chemical reaction of oxygen with materials

and is used generally for the purpose of liberating heat energy.

Majority of fossil fuels are being used in transportation, industries heating and

generation of electricity.

Crude petroleum is refined into gasoline; diesel and jet fuel that power the

worlds transportation system.

Today, fossil fuels are considered to be non-renewable for the reason that their

consumption rate is far in excess of the rate of their formation.

Coal

About 250 to 350 million years ago coal was formed on earth in hot, damped

regions.

Almost 27350 billion metric tones of known coal deposits occur on our planet.

Out of which about 56% are located in Russia, 28% in USA and Canada.

India has about 5% of worlds coal reserve and that too not of very good quality

in term of heat capacity.

West Bengal, Jharkhand, Orissa, Andhra Pradesh, Madya Pradesh and

Maharastra are the major coal producing states of India.

Mainly, there are three types of coal:

Anthracite or hard coal ( 90% carbon content)

Bituminous or soft coal (85% carbon content)


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Lignite or brown coal (70% carbon content)

The present annual extraction rate of coal is about 3000 million metric tones, at

this rate coal reserves may lasts for about 200 hundred years and if its use is

increased by 2% per year then it will last for another 65 years.

Oil:

Crude oil, also called petroleum, is a thick liquid found in underground rock

formations.

The petroleum industry extracts crude oil out of the ground and then refines it

into products such as gasoline.

Crude oil contains a complex mixture of compounds made of carbon chains with

hydrogen molecules attached to each link in the chain.

Extracted crude oil also contains small amounts of sulfur, oxygen, and nitrogen

compounds mixed with the hydrocarbons.

The principle of oil refining is to remove crude oils impurities, that is, anything

that is not a hydrocarbon.

The following nations hold the largest oil reserves, in order:

Saudi Arabia, Canada, Iran, Iraq, United Arab Emirates, Kuwait, Venezuela,

Russia, Libya, and Nigeria.

Petroleum:

Convenience of petroleum or mineral oil and its greater energy content as

compared to coal on weight basis has made it the lifeline of global economy.

Petroleum is cleaner fuel when compared to wood or coal as it burns completely

and leaves no residue.

Petroleum is unevenly distributed like any other mineral. There are 13 countriesin the world having 67% of the petroleum reserves which together form the

OPEC (Organization of Petroleum Exporting Countries)

Six regions in the world are rich in petroleum USA, Mexico, Russia and West

Asian countries. Saudi Arabia has one fourth of the world oil reserves.


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The oil bearing potential of India is estimated to be above one million square

kilometers is about one third of the total geographic area.

Northern plains in the Ganga-Brahmaputra valley, the coastal strips together with

their off-shore continental shelf (Bombay High), the plains of Gujarat, the Thar

Desert and the area around Andaman and Nicobar Islands are the places where

oil reserves exist..

Advantages and disadvantages of conventional oil:

Advantage Disadvantage

Ample supply for 40-50years Need to find substitute within

Low cost (with huge substitutes) Air pollution when burnt,

Releases carbon dioxide

Easily transported within and between countries ,

Efficient distribution system

Moderate water pollution

Low land use

Technology is well developed

Natural gas:


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Natural gas mainly consists of Methane (CH4) along with other inflammable

gases like Ethane and propane.

Natural gas is least polluting due to its low Sulphur content and hence is cleanest

source of energy.

It is used both for domestic and industrial purposes.

Natural gas is used as a fuel in thermal plants for generating electricity, as a

source of hydrogen gas in fertilizer industry and as a source of carbon in tyre

industry

The total natural gas reserves of the world is about 600 000 billion m3, out of this

Russia has 34%, Middle East 18%, North America 17%, Africa and Europe 9%

each and Asia 6%.

Annual production of natural gas is about 1250 billion m3 and hence it is

expected to last for about 50-100 years.

In India gas reserves are found in Tripura, Jaisalmer, off shore areas of Bombay

and Krishna-Godavari Delta.

Advantages and disadvantages of Natural gas:

Advantage Disadvantage

Ample supply for 125 years Non renewable resource

High net energy yield Methane ( a green house gas)


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Low cost (with huge substitutes) Air pollution when burnt,

Releases carbon dioxide

Less air pollution than other fossil fuels Shipped across ocean as highly

Easily transported by pipelines Requires pipelines

Low land use

Proximate and Ultimate Analysis of Coal

The proximate analysis of coal is an empirical procedure for analysis of coal in order to

list the composition in terms of the following items, all expressed as percentages by

weight of coal The different items are Moisture, Volatile matter, Ash and Fixed carbon .

Moisture: The loss of weight when coal is heated in an oven at 105 degree C.

Volatile matter: The loss in weight when coal sample is heated in a covered crucible

for about seven minutes at 950 centigrade minus the weight of the moisture.

Ash:

The weight of residue obtained when the sample of coal is subjected to complete

combustion in a muffle furnace at 700-750 degree centigrade.

Fixed carbon: Calculated as 100-the sum of moisture, Volatile matter and ash already

obtained.

Or FC= 100-(M+VCM +A)FC -Fixed carbon M-Moisture

VCM-Volatile combustible matter A-Ash

Ultimate Analysis of Coal

The ultimate analysis of coal gives the composition as the percentage by weight of the

various elements such as C, H, N, O, S, etc. and the ash. The ultimate analysis


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provides the necessary data required for material balance in the combustion of coal.

The analysis is carried out by elaborate and standard chemical techniques. The carbon

dioxide, water and sulphur dioxide formed by burning coal in a sufficient supply of

oxygen determine respectively the amounts of carbon, hydrogen and sulphur in coal.

Carbon dioxide is absorbed in KOH solution to find the quantity of Carbon dioxide

present. Water is found out by absorbing water in concentrated sulphuric acid or any

suitable adsorbent like anhydrous calcium chloride or silica gel. Sulphur dioxide is

absorbed in lead chromate.

The nitrogen in coal is estimated by Kjeldahl method.

The percentage of oxygen in the coal is found by subtracting the percentages of all

other constituents from 100.

Combustion Reaction

Type I Problems:

The compositions and quantities of the various streams entering the process are known.

It is required to calculate the quantity and composition of streams leaving the process

for a specified degree of completion of the operation.

Air with known excess

Solid fuel, liquid fuel or gaseous fuel Flue gases,

Exhaust gas

Chimney gas,

Product Gas,

combustion gases

Residue, cinder, ash

Type II problems:


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The composition of exit stream is known. The composition and quantities of entering

streams are not known, partially known or completely known. It is required to calculate

the composition and quantities of all the entering streams and degree of completion of

reaction are to be determined.

Steps to solve type I problems:

1. Select a suitable basis

2. Write the possible reactions

3. Calculate oxygen required for the reaction

4. Calculate the oxygen required from air and call it the theoretical oxygen

5. Calculate the oxygen supplied from the known excess used

6. Calculate nitrogen entering along with oxygen

7. Identify the gases leaving from reactions and input data

8. Calculate the quantity of all the components leaving

9. Calculate the % compositions and find the other quantities asked for

Note: While calculating Oxygen required always assume complete conversion

1. A coal has the ultimate composition C-67.34%, H2-4.67%, O2-8.47%, N2-1.25%, S-

4.77% and the rest ash. Find the theoretical air fuel ratio. If 20% excess air is used find

the composition of the flue gases leaving.

20% excess air

C-67.34%

H2-4.67% Flue gases

O2-8.47%

N2-1.25%

S-4.77% ash

Ash-13.50%

Basis: 100 kg coal

Carbon entering= 67.34 kg= 67.34/12= 5.612 kg moles


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H2 entering = 4.67 kg = 4.67/2 = 2.335 kg moles

O2 entering = 8.47 kg = 8.47/32 = 0.265 kg moles

N2 entering = 1.25 kg = 1.25/28 = 0.045 kg moles

S entering = 4.77 kg = 4.77/32 = 0.150 kg moles

Reactions: C+O2 CO2

H2+1/2 O2 H2O

S+O2 SO2

O2 required= 5.612x1/1+2.335x1/2+0.150x1/1=6.53 kg moles

O2 theoretical= O2 required O2 in fuel= 6.53-0.265=6.265 kg moles

Air theoretical= O2 theoretical x100/21= 6.265x100/21= 29.833 kg moles

Theoretical air fuel ratio (by weight) = 29.833x29/ 100= 8.652

O2 supplied with 20% excess= 6.265x1.2= 7.518 kg moles

N2 from air= 7.518x79/21= 20.282 kg moles

Gases leaving- CO2, H2O, SO2, O2, N2

Moles of CO2 leaving= Moles of CO2 produced=5.612 kg moles

Moles of H2O leaving= Moles of H2O produced= 2.335 kg moles

Moles of SO2 leaving= Moles ofSO2 produced =0.150 kg moles

Moles O2 leaving = O2 supplied- O2 consumed= O2 in fuel + O2 from air- O2 consumed

=0.265+7.518- 6.53 = 1.253 kg moles (O2

utilized = O2

required if the reaction goes

to completion)

Moles N2 leaving= N2 in fuel + N2 from air= 0.045+20.282=20.327 kg moles

Composition of flue gases:

Component Moles Mole %

CO2 5.612 18.910

H2O 2.335 7.868

SO2 0.150 0.505

O2 1.253 4.222

N2 20.327 68.494

29.677 99.999


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2. A furnace is to be designed to burn coke at the rate of 500 kg/hr. The composition of

the coke used in the furnace is C-89%, ash 11%. The efficiency of the grate is such

that only 90% of the carbon charged is burnt.25% excess air is supplied. Of the carbon

burnt only 95% is oxidized to dioxide and 5% goes to carbon monoxide. Calculate the

composition of flue gases from the furnace. Calculate the volumetric flow of the gases at

a temperature of 250 C and 750 mm Hg pressure for the purpose of designing the

stack.

25% excess air

500 kg/hr coke

C-89%, ash-11%

Flue gases

Ash

Basis: 100 kg of coke

Carbon present= 89 kg= 89/12= 7.417 kg moles

C+O2 CO2Mole O2 required= 7.417 kg moles

Moles O2 supplied= 7.417x1.25=9.271 kg moles

N2 supplied with O2 = 9.271x79/21= 34.877 kg moles

Gases leaving: CO2, CO, O2, N2

Moles of CO2 leaving= moles of CO2 produced= 7.417x0.9x0.95= 6.342 kg moles

Moles CO leaving= moles CO produced= 7.417x0.9x0.05= 0.334 kg moles

Moles O2 leaving= moles O2 entering O2 consumed= O2 from air- O2 consumed

=9.271- 6.342x1/1-0.334x1/2= 2.762 kg moles

Moles N2 leaving= moles N2 from air= 34.877 kg moles

Composition of gases:

Component Moles Mole %

CO2 6.342 14.311


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CO 0.334 0.754

O2 2.762 6.233

N2 34.877 78.702

44.315 100.000

3. A furnace uses coke containing 80% carbon, 0.5 % hydrogen and rest ash. The

furnace operates with 50% excess air. The ash contains 4% unburned carbon. Of the

carbon burnt 5% goes to monoxide. Calculate the composition of the flue gases. What

is the quantity of ash produced per 100 kg of coke burned?

100 kg coke 50% excess air

80% carbon

0.5% Hydrogen

19.5% ash Flue gases

96% ash, 4% carbon

Basis: 100 kg coke burnt

Carbon present= 80 kg= 80/12= 7.5 kg moles

Hydrogen present=0.5 kg= 0.5/2= 0.25 kg moles

C+O2 CO2

H2+1/2 O2 H2O

O 2 required= 7.5x1/1+0.25x1/2= 7.625 kg moles

O2 supplied= 7.625x1.5= 11.4375 kg moles

N2 with O2= 11.4375x79/21= 43.027 kg moles

Gases leaving: CO2, CO, H2O, O2, N2

Ash in fuel= ash in residue

Carbon lost in residue=4/96x19.5= 0.8125 kg= 0.8125/12= 0.0677 kg moles

Carbon burnt= 7.5- 0.0677= 7.4323 kg moles

CO2 leaving=CO2 produced= 7.4323x0.95= 7.061 kg moles

CO leaving= 7.4323x0.05= 0.3716 kg moles

H2O leaving= 0.25x1/1=0.25 mole


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O2 leaving= O2 entering- O2 consumed= 11.4375- 7.061x1/1-0.3716x1/2-0.25x1/2=

4.0657 kg moles

N2 leaving= 43.027 kg moles

Composition of gases leaving:

Component Moles Mole%

CO2 7.061 12.891

CO 0.3716 0.678

H2O 0.25 0.456

O2 4.0657 7.423

N2 43.027 78.552

54.7753 99.999

Ash produced per 100 kg of coke= (100/96) x19.5= 20.3125

4. A producer gas made from coke has the following composition CO-28%, CO2- 3.5%,

O2-0.5 % and the rest N2. The gas is burnt with such a quantity of air that the O 2 from air

is 25% in excess of that required for combustion. The combustion is 95% complete.

Calculate the composition of the flue gases by volume and by weight for 100 kg of gas

burnt. What is the density of the flue gas?

Producer gas 25% excess air 95% completion

CO-28%

CO2-3.5% Flue gases

O2-0.5%, N2-68%

Basis: 100 moles of producer gas

CO+ O2 CO2

Moles O2 required= 28x1/2=14 moles

Moles O2 theoretical = O2 required- O2 in fuel= 14-0.5= 13.5 moles

Mole O2 supplied= 13.5x1.25= 16.875 moles

Moles N2 supplied= 16.875x79/21= 63.482 moles

Gases leaving: CO2, CO, O2, N2

CO2 leaving= CO2 produced+ CO2 entering= 28x0.95+ 3.5= 26.6+3.5= 30.1 moles

CO leaving = CO unreacted= 28x0.05= 1.4 moles

O2 leaving= O2 entering-O2 consumed=0.5+16.875-26.6x1/2= 17.375-13.3=4.075 moles


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Moles N2 leaving= 68+ 63.482= 131.482 moles

Composition of flue gases

Component moles Mole% Mole wt weight Weight %

CO2 30.1 18.018 44 1324.4 25.590

CO 1.4 0.838 28 39.2 0.757

O2 4.075 2.439 32 130.4 2.520

N2 131.482 78.705 28 3681.496 71.133

167.057 100.000 5175.496 100.000

Composition of flue gases will be same irrespective of the quantity burnt.

Density of the gases leaving at NTP= mass/volume at NTP = 5175.496/

167.057x22.414= 1.382 kg/ m3

5.A petroleum refinery burns a gas mixture containing C5H12-7.5%, C4H10-10%, C3H8-

15%, C2H6 -9%, CH4-55%, N2-3.5% at the rate of 200m3 measured at 4bars and 25C.

The air rate is adjusted to have 20% excess air. The flue gases contain CO2 and CO in

the ratio 20:1. Calculate the volume of air used at 1.5 bars and 30 C and composition of

flue gases.

Gas mixture 20% excess air

C5H12-7.5% Flue gases

C4H10-10% CO2: CO=20:1

C3H8-15%

C2H6-9%

CH4-55%

N2-3.5%

Basis: 100 moles of gas mixture entering

Reactions


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C5H12 +8O2 5CO2+ 6H2O

C4H10 +13/2O2 4CO2+5H2O

C3H8 +5O2 3CO2+4H2O

C2H6 +7/2O2 2CO2+3H2O

CH4 +2O2 CO2+2H2O

O2 required =7.5x8+10x13/2+15x5+9x7/2+55x2= 60+65+75+31.5+110=341.5 moles

O2 supplied=341.5x1.2 = 409.8 moles

N2 supplied= 409.8x79/21= 1541.629 moles

Gases leaving CO2, CO, H2O, O2, N2

Carbon in fuel= 5x7.5+4x10+3x15+2x9+1x55=195.5

Moles CO2 leaving = moles CO2 produced= 195.5x20/21= 186.19

Moles CO leaving= moles CO produced= 195.5x1/21=9.31

Water leaving= water produced=6x7.5+5x10+4x15+3x9+2x55= 45+50+60+27+110=292

O2 leaving= O2 supplied-O2 utilized= 409.8-186.19x1/1-9.31x1/2-292x1/2=72.955

Moles N2 leaving= moles N2 entering=3.5+1541.629=1545.129

Volume of gases entering at 4 bars and 25C= 100x22.414x1/4x298/273= 611.664 m3

Volume of air used at 1.5 bar and 30C=

(409.8+1541.629)x22.414x1/1.5x303/273=32363.90 m3

Volume of air used per 200 m3 of gases entering= (32363.9/ 611.664)200= 10582.25 m3

Composition of gases leaving

Components Moles Mole %

CO2 186.19 8.843

CO 9.31 0.442

H2O 292 13.868

O2 72.955 3.465

N2 1545.129 73.382

2105.584 100.000

Orsat analysis:


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Orsat analysis is the analysis with out water and SO3 free. The flue gases contain water

and SO3 if sulphur is present along with other products of combustion. In the presence

of H2O vapor and SO3 the analysis will be difficult .Therefore H2O and SO3 are removed

before analysis and the composition of other constituents are presented in flue gases.

Such analysis is called orast analysis. Orsat apparatus is used for the analysis.

Type II problems:

Air in excess quantity

Fuel Orsat analysis of flue gases given

Fuel

Residue

Select a suitable basis

Carry out elemental balance for N, O, C or C, N, O depending on the problem

Calculate theoretical O2

Calculate excess O2 and Calculate the other quantities asked for

6. A furnace is fired with a fuel oil. The orsat analysis of the flue gases indicated 10.6%

CO2,

2.6%O2

and rest N2

by volume. Find C-H ratio of the fuel assuming that fuel oil

does not contain nitrogen.

Air excess

Fuel oil Flue gases

CO2-10.6%,O2-2.6%, N2-86.8%

Basis: 100 moles of dry flue gases

N2 balance:

N2 in flue gases= 86.8 moles

N2 from air= 86.8 moles


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O2 balance:

O2 from air= 86.8x21/79 = 23.07 moles

O 2 accounted in flue= 10.6x1/1+2.6= 13.2 moles

O2 un accounted=23.07 -13.2= 9.87 moles, this O2 must have combined with H2 In flue

to form water and has been eliminated before analysis

H2 balance:

Moles H2 in fuel = 9.87x2=19.74 moles

C balance:

Moles carbon in flue= 10.6 = moles of carbon in fuel

C to H ratio (mole) in fuel= 10.6/ 19.74= 0.537

C to H ratio (weight) in fuel= 10.6x12/ 19.74x2= 3.222

7. A combustion chamber is fed with butane and excess Air. The combustion of butane

is complete. The composition of combustion gases by volume is given below.CO2-

9.39%, H2O 11.73%, O2-4.5% and N2-74.38%. Find the % excess air used and mole

ratio of air to butane used

Excess air

Flue gases

Butane

CO2-9.39%,H2O-11.73%,O2-4.5%,N2-4.38%

Basis: 100 moles of flue gases

C4H8 + 6O2 4CO2+ 4H2O

N2 balance:

Moles N2 in flue= 74.38

Moles N2 in fuel= 74.38

O2 balance:

O2 from air= 74.38x21/79=19.772 moles

Carbon balance:

Moles of carbon in flue= 9.39 moles


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Moles butane entering= 9.39/4 =2.3475 moles (formula of butane C4H8)

O2 theoretical= 2.3475x6 = 14.085 moles

% excess air= [(O2 supplied- O2 theoretical)/O2 theoretical] x100

= [(19.772-14.085)/14.085]x100=40.38%

Air to butane ratio (mole) = (19.772x100/21)/ 2.3475= 40.108

8. The orsat analysis of flue gases is CO2-12.7% O2-7.1 %, N2-80.2%. Determine the %

excess air used in the combustion. The N2 present in the flue gas is contributed by air

only.

O2 in excess

Fuel Flue gases

CO2-12.7%,O2-7.1%, N2-80.2%

Basis: 100 moles of dry flue gases

N2 balance:

N2 in flue= 80.2 moles =N2 from air

O2 balance:

O2

from air= 80.2 x21/79=21.319 moles

O2 excess= 7.1 moles

O2 theoretical= 21.319- 7.1=17.219

% excess= (7.1/17.219)x100= 41.234 %

9. A coal containing C-67.9% H2-4.4% S-0.8% N2-13.21% O2- 7.9%, ash 4.5%, H2O-

1.29% is burnt in a furnace. The combustion dry gas has CO2-14.5%, O2-4.7% and rest

N2. Calculate a. theoretical volume of air per 100 kg coal. b. % excess air.

Excess air

C-67.9%

H2-4.4% Coal dry flue gas

S-0.8% CO2-14.5 %, O2-4.7%, N2-80.8%

N2-13.21%

O2- 7.9%, ash 4.5%, H2O-1.29%


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Basis: 100 kg coal

Carbon in fuel= 67.9 kg = 67.9/12=5.658 kg moles

Hydrogen in fuel= 4.4 kg= 4.4/2=2.2 kg moles

Nitrogen in fuel= 13.21 kg= 13.21/28= 0.472 kg moles

Oxygen in fuel=7.9 kg= 7.9/32=0.247 kg moles

Water in fuel= 1.29 kg= 1.29/18= 0.072 kg moles

Ash in fuel= 4.5 kg

Carbon balance:

Carbon in fuel= 5.658 kg mole

Carbon in flue= 5.658 kg moles

Moles of carbon in 100 moles of dry flue gases= 14.5 kg moles

Moles of dry flue gases produced= (100/14.5)x5.658= 39.021 kg moles

Moles CO2 in flue= 39.021x0.145=5.656 moles

Moles of O 2 in flue= 39.021x0.047=1.838

Moles of N2 in flue= 39.021x0.808=31.527 moles

N2 balance:

N2 in fuel + N2 in air= N2 in flue 0.472+ N2 from air= 31.527

Therefore N2 from air= 31.527-0.472= 31.055 moles

O2

balance:

O2 from air= 31.055x21/79= 8.255

O2 theoretical= 5.658x1/1+2.2x1/2-0.247= 6.511

% excess air= [(O2 supplied-O2 theoretical)/O2 theoretical] x100= [(8.255-

6.511)/6.511]x100=26.765%

Theoretical volume of air/100 kg coal= (6.511x100/21)22.414= 694.94 m3 at NTP

10. The gas obtained from the furnace fired with a hydrocarbon fuel analyses 10.2%

CO2, 7.9% O2 and 81.9% N2 on dry basis. Calculate a. the % excess air b. carbon to

hydrogen ratio c. kg of air supplied per kg of fuel burned.

Excess air

Hydrocarbon fuel dry flue gas

CO2- 10.2 %, O2-7.9%

N2-81.9%


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Basis: 100 kg moles dry flue gas

N2 balance:

N2 in flue= 81.9 moles

N2 from air= 81.9 moles

O2 balance:

O2 from air= 81.9x21/79= 21.771 moles

O2 accounted in flue= 10.2+7.9= 18.1 moles

O2 un accounted= 21.771- 18.1= 3.671 which would have combined with H2 to form

water

Hydrogen in fuel= 3.671x2= 7.342 moles

Carbon in fuel= 10.2 moles

O2 theoretical= 10.2+ 3.671= 13.871

O2 excess= [(21.771-13.871)/ 13.871] x100= 56.953

Carbon to hydrogen ratio (mole) = 10.2/ 7.342= 1.389

Carbon to hydrogen ratio (weight) = 10.2x12/ 7.342x2= 122.4/14.684= 8.336

Kg air supplied per kg fuel burnt= (21.771+81.9)29/ (122.4+14.684) = 21.932

11. A sample of coal is found to contain 65% carbon, 12.7% ash by weight and the

remaining H2 and water vapor. The refuse obtained after burning the fuel is found to

have 8.6% carbon. Assume negligible oxygen is present in the coal. Flue gas analysis

showed CO2-10.6%, O2-8.7% and N2-80.7% by volume. Calculate a. actual weight of

flue gas produced by burning coal b. actual weight of dry air used for burning the coal c.

amount of combustible H2 present in fuel d. % excess air.

Excess air

Coal Flue gas

C-65% CO2-10.6%

Ash-12.7% O2- 8.7%

H2 and H2O-22.3% N2-80.7%

Refuse- 8.6% C, 91.4% ash

Basis: 100 kg of coal

Ash balance:


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Ash in fuel =12.7 kg = ash in refuse

Carbon balance:

Carbon present in fuel = 65 kg

Carbon lost in refuse= (8.6/91.4)12.7= 1.195 kg

Carbon burnt= 65-1.195= 63.805 kg= 63.805/12=5.317 kg moles

Carbon present per 100 kg mole dry flue=10.6 moles

Moles of dry flue gas produced = (100/10.6)5.317 = 50.16 moles

CO2 in flue=50.16x0.106= 5.317 moles

O2 in flue= 50.16x0.087=4.364 moles

N2 in flue= 50.16x0.807=40.479 moles

N2 balance: 20.3-

N2 from air= N2 in flue= 40.479 moles

O2 balance:

O2 from air= 40.479x21/79= 10.760 moles

O2 accounted in flue= 5.317x1/1+4.364= 9.681 moles

O2 un accounted= 10.760-9.681= 1.079 moles which would have combined with H2 to

form water and eliminated before analysis

Therefore moles H2 available in fuel= 1.079 x2= 2.158 moles

Weight of H2

in fuel= 2.518x2= 5.036 kg

Therefore weight of water in fuel= 20.3- 5.036 = 15.264 kg

Water produced=2.158 moles

Actual weight of flue produced= dry flue gases + water

=5.317x44+4.364x32+40.479x28+2.158x18+15.264

= 233.948+139.648+1133.418+38.844+15.264=1561.122 kg

Weight of dry air used= (10.760+40.479)29=1485.931 kg

Weight of combustible H2 in fuel= 5.036 kg

O2 theoretical=65/12x1/1+1.079=6.496 moles

% excess= (10.76-6.496)/6.496x100= 65.64%

12. A solid fuel with composition by weight of C-78%, H2-6%, O2-9.8%, N2-1.2%, ash-

5% and is burnt in air. The products of combustion give an orsat analysis CO 2-12.5%,


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CO-0.09%, O2-5.6% and rest N2. The refuse removed from ash pit contains 30% carbon

by weight. Find the actual air fuel ratio and % excess air.

Solid fuel C-78% Flue

H2-6% gas

O2-9.8% CO2- 12.5%

N2-1.2% CO-0.09%

Ash-5% O2-5.6%

N2-81.81%

Refuse 30% C, 70% ash

Basis: 100 kg of solid fuel

Moles of carbon entering= 78/12= 6.5 kg moles

Carbon lost in refuse from an ash balance= (30/70)5= 2.143 kg= 2.143/12=0.179 kg

moles

Therefore carbon burnt= 6.5-0.179=6.321 kg moles

Carbon present per 100 kg mole dry flue gas= 12.5+0.09= 12.59 kg mole

Therefore dry flue produced= (100/ 12.59)x 6.321= 50.207 kg moles

Moles CO2= 50.207x 0.125=6.276 kg moles

Moles CO = 50.207x0.0009= 0.047 kg moles

Moles of O2= 50.207x0.056= 2.8116 kg moles

Moles N2= 50.207x0.8181= 41.074 kg moles

N2 from air= N2 in flue-N2 in fuel= 41.074- 1.2/28=41.031

O2 from air= 41.031x21/79=10.907 kg moles

Air fuel ratio= (41.031+10.907)29/ 100= 15.06

O2 theoretical= 6.5x1/1+6/2x1/2-9.8/32=6.5+1.5-0.306=7.694

% excess air= (10.907-7.694)/ 7.694x100= 41.76%


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13. Wood having the analysis C-46%, O223%, ash-5% and rest being moisture and

H2 is burnt in a furnace. The orsat analysis of flue gas obtained is CO2-14.9%, CO-

1.65%, O2-3.45% and N2-80%. Calculate a. complete analysis of wood used b. fuel to

air ratio by weight c. % excess air used d. complete composition of stack gases.

Wood excess air

C-46% flue

O2-23% gas

Ash-5%

Moisture&H2= 26%

CO2-14.9%

CO-1.65%

O2-3.45%

N2-80 %

Basis: 100 kg of wood:

Carbon in wood= 46 kg=46/12= 3.833 kg moles

O2 in wood=23 kg=23/32=0.719 kg moles

Moles carbon per 100 kg mole dry flue gas= 14.9+1.65= 16.55 kg moles

Moles dry flue produced by a carbon balance= (100/16.55)x3.833=23.23 kg moles

Moles of CO2 in flue= 23.23x0.149= 3.361 kg moles

Moles CO in flue= 23.23x0.0165= 0.383 kg moles

Moles O2 in flue= 23.23x 0.0345= 0.801 kg moles

Moles of N2 in flue= 23.23x0.8=18.584 kg moles

N2 from air= N2 in flue= 18.584 kg moles

O2 from air = 18.584x21/79= 4.940 kg moles

O2 accounted in flue= 0.801 kg moles

O2 un accounted= 4.940- 0.801= 4.139 kg moles

Therefore combustible H2 present= 4.139x2= 8.278 kg moles

Weight of H2 present in wood= 8.278x2= 16.556 kg

Moisture present in wood= 26 -16.556 = 9.444 kg

Fuel to air ratio (weight) = 100/ (18.584+4.940)29=0.147


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O2 theoretical= 3.833x1/1-0.719=3.114

% excess= (4.94-3.114)/3.114x100= 58.64 %

Composition of stack gases:

Component moles Mole%

CO2 3.361 10.700

CO 0.383 1.219

H2O 8.278 26.357

O2 0.801 2.550

N2 18.584 59.172

Total 31.407

14. The analysis of the flue gas from a combustion reaction is as follows. CO2 -10.8%,

CO-4%, O2-9.2% and N2-76% by volume. The producer gas burnt had the composition

CO2-9.2%, C2H4-0.4%, CO-20.9%, H2-19.6%, CH4-1.9% and N2-48% on volume basis.

Compute a.m3 of air used in combustion of 1 m3 of producer gas b. % excess air.

Producer gas excess air Flue

gas

CO2-9.2% CO2-10.8

C2H

4-0.4% CO-4%

CO-20.9% O2-9.2%

H2-19.6% N2-76%

CH4-1.9%

N2-48%

Basis: 100 moles of producer gas.

Carbon balance:

Carbon in fuel= 9.2x1/1+0.4x2+20.9x1+1.9x1= 32.8 moles

Carbon per 100 moles of dry flue= 10.8+4= 14.8 moles

Moles of dry flue gas produced= (100/ 14.8)32.8= 221.622 moles

CO2 in flue= 221.622x0.108= 23.935 moles

CO in fuel= 221.622x0.04=8.865 moles

O2 in flue= 221.622x0.092=20.389 moles


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N2 in flue= 221.622x0.76=168.433 moles

N2 balance:

N2 from air=N2 in flue-N2 in fuel= 168.433-48= 120.433 moles

O2 balance:

O2 from air= 120.433x21/79= 32.014 moles

m3 air used per m3 of producer gas= (120.433+32.014)22.414/ 100x22.414= 1.524

O2 theoretical= 0.4x3+20.9x1/2+19.6x1/2+1.9x2= 25.25 moles

% excess air= (32.014-25.25)/25.25x100= 26.788%

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