Skip to main content

Solved Numerical Problems from Boiler Operation Engineer Exams

Properties of Steam

Problem (Gujarat BOE 2021, 2024, 5 marks)

Steam is generated in a boiler at 110 kg/cm2 at 5200C. Assume drum pressure is 118 kg/cm2. Using steam table, find:

1. Saturated steam temperature.

2. Degree of superheat.

3. Enthalpy of steam.

Solution

(i) From the steam table, saturated temperature corresponding to 118 kg/cm2 (=118/1.02 = 116 bar) (drum pressure) is 322 °C.


 (ii) Degree of super heat = Superheated steam temperature – Saturated temperature

 = 520 – 322 = 198 °C.

(iii) Enthalpy of the superheated steam as found from steam table corresponding to 110 kg/cm2 (=110/1.02 = 108 bar) and 520 °C is 815.8 kcal/kg.

Problem (Gujarat BOE 2021, 5 marks0

Calculate the heat required to be added to 1000 kg of steam at 15 kg/cm2 in order to superheat to 400 centigrade.

Solution

msteam = 1000 kg

Psteam = 15 kg/cm2

Final temperature of superheated steam, T2 = 4000C

Saturation temperature of steam at 15 kg/cm2 = 198.30C

Specific heat capacity of superheated steam, Cp = 2.09 kJ/kg0C

T = 400 - 198.3 = 201.70C

Now using 

\(Q = m{C_p}\Delta T\) 

= 1000 x 2.09 x 201.7 

= 421543 kJ

Boiler Mechanics

Problem (Jharkhand BOE 2024)

A solid round bar 3 m long 5cm in diameter is used as a strut with both ends hinged. Determine the crippling load. Take E=2 x 105 N/mm2

Solution

Moment of inertia

\(I = \frac{\pi }{{64}}x{5^4} = 30.68\,c{m^4} = 30.68x{10^4}\,m{m^4}\)                        

Both ends of the bar are hinged. Crippling load for this condition is

\(P = \frac{{{\pi ^2}EI}}{{{l^2}}} = \frac{{{\pi ^2}x2x{{10}^5}x30.68x{{10}^4}}}{{{{3000}^2}}}\)                        

 = 67288 N = 67.288 kN

These questions are taken from study material for  Boiler Operation Engineer Exam by amiestudycircle.com. Our study material is prepared by scholars and retired faculty from IIT, Roorkee.

Problem (Jharkhand BOE 2022, 3 marks)

A hollow shaft is to transmit 300 kW at 80rpm. If shear stress is not to exceed 60 N/mm2 a diameter is 0.6 of the external diameter, find external and internal diameters maximum torque is 1.4 times the mean torque. 

Solution   

\(P = \frac{{2\pi NT}}{{60}}\)         

\(300x1000 = \frac{{2(3.14)(80)(T)}}{{60}}\)                         

Giving T = 35828 Nm

\(T = \frac{{\pi {f_s}}}{{16}}\left[ {\frac{{{D_e}^4 - {D_i}^4}}{{{D_e}}}} \right]\) 

\(35828 = \frac{\pi }{{16}}x60x\left[ {\frac{{{D_e}^4 - {{(0.6{D_e})}^4}}}{{{D_e}}}} \right]\) 

Giving De = 14.24 mm

Fuel, Combustion and Fuel Gases

Example (Karnataka BOE 2023)

If one ball of coal of 250 c.c. volume breaks in to small 250 pcs. of ball having a volume of 1 cc of each ball, calculate how many times the surface area exposed to heat for combustion will increase?

Solution

Volume           

 \(\begin{array}{l}{V_L} = \frac{4}{3}\pi {r_L}^3\\ \Rightarrow 250 = \frac{4}{3}{r_L}^3\\ \Rightarrow {r_L} = {\left( {\frac{{750}}{{4\pi }}} \right)^{1/3}}\,cm\end{array}\)

\(1 = \frac{4}{3}\pi {r_S}^3 \Rightarrow {r_s} = {\left( {\frac{3}{{4\pi }}} \right)^{1/3}}cm\)  

[Volume of one ball = 250/250 = 1 cc] 

Now area        

 \({A_L} = 4\pi {\left[ {{{\left( {\frac{{750}}{{4\pi }}} \right)}^{1/3}}} \right]^2} = 4\pi {\left( {\frac{{750}}{{4\pi }}} \right)^{2/3}}c{m^2}\)

\({A_S} = 4\pi {\left[ {{{\left( {\frac{3}{{4\pi }}} \right)}^{1/3}}} \right]^2} = 4\pi {\left( {\frac{3}{{4\pi }}} \right)^{2/3}}c{m^2}\) 

Ratio

\(\frac{{{A_{S,total}}}}{{{A_L}}} = \frac{{250(4\pi ){{\left( {\frac{3}{{4\pi }}} \right)}^{2/3}}}}{{4\pi {{\left( {\frac{{750}}{{4\pi }}} \right)}^{2/3}}}} = 6.3\)

Problem (Gujarat BOE 2023, 5 marks)

Following parameters are noted from the ultimate analysis of coal sample:

            Carbon: 40%

            Sulphur: 2%

            Hydrogen: 4%

Calculate the theoretical quantity of air required in kilograms. If the boiler is operated at 4% excess oxygen, then calculate the actual air quantity in kilograms.

Solution

Theoretical air

Method 1

Total O2 required for complete combustion = 1.41 - 0 = 1.41 kg/kg of fuel

Air required = 1.41 x (100/23) = 6.12 kg/kg of fuel

Method 2

Quantity of theoretical air required (by weight) per kilogram of fuel is given by

\(4.35\left[ {\left( {\frac{8}{3}C + 8{H_2} + S} \right) - {O_2}} \right]kg\)

= \(4.35\left( {\frac{8}{3}x0.4 + 8x0.04 + 0.02} \right) = 6.12\,kg\) 

Actual air quantity

Excess air percentage = \(\frac{{Oxygen\,percentage\,{\mathop{\rm i}\nolimits} n\,flue\,gas}}{{21 - Oxygen\,percentage}}x100\)

Boiler is operated at 4% excess oxygen.

Hence, excess air percentage = \(\frac{4}{{21 - 4}}x100 = 23.5\% \)

\(Actual\,air\,qty = Theoretical\,air\left( {1 + \frac{{excess\,air\% }}{{100}}} \right)\)

= 6.12 x 1.235 = 7.35 kg/kg of fuel

These questions are taken from study material for  Boiler Operation Engineer Exam by amiestudycircle.com. Our study material is prepared by scholars and retired faculty from IIT, Roorkee.

Problem (Kerala BOE 2023, 5 marks)

The height of a chimney of a steam generation plant is 40 m. Determine (a) the draught produced by the chimney (b) the available draught from the following parameters: Flue gas temperature = 300°C, Ambient air temperature = 27°C, Mass of air supplied = 22 kg/kg of fuel, Available draught = 0.82 times natural draught.

Solution

Chimney draught theoretical

H = 40 m

T1 = 27 + 273 = 300 K

T = 300 + 273 = 573 K

W = 22 kg/kg of fuel

\(h = 353H\left[ {\frac{1}{{{T_1}}} - \left( {\frac{{W + 1}}{W}} \right)\frac{1}{T}} \right]\) mm of H2O

= \(353x40\left[ {\frac{1}{{300}} - \left( {\frac{{22 + 1}}{{22}}} \right)} \right]\frac{1}{{573}}\)

= 21.18 mm of H2O

Available draught

Available draught

= 0.82 x natural draught

= 0.82 x 21.18

= 17.37 mm of H2O

Problem (Maharashtra BOE 2025, 6 marks)

A boiler fitted with a forced draught has the following particulars

1. Mass of air required = 18 kg/kg of fuel

2. Mass of fuel used = 1400 kg/hr

3. Temperature of outside air = 400C

4. Temperature of chimney gas = 1700C

5. Draught pressure = 40 mm of water

6. Efficiency of fan = 70%

Determine the power required to drive the fan in HP. If the boiler is equipped with induced draught fan instead of forced draught fan with same efficiency what will be the power required driving it?

Solution

h = 40 mm

Tgas = 170 + 273 = 443 K

mf = (1400/60) kg/min

m = 18 kg/kg of fuel

Tair = 40 + 273 = 313 K

ηf = 0.7

For forced draught, the capacity of the motor is

\({P_{fan,forced}} = \frac{{h(m)({m_f}){T_{air}}}}{{60x36x0.7}} = \frac{{40(18)(1400/60)(313)}}{{60x36x0.7}}\)= 3477 W = 3.48 kW

For induced draught, the capacity of the motor is

\({P_{fan,induced}} = \frac{{h(m)({m_f}){T_{gas}}}}{{60x36x0.7}} = \frac{{40(18)(1400/60)(443)}}{{60x36x0.7}}\) = 4922 W = 4.92 kW

These questions are taken from study material for  Boiler Operation Engineer Exam by amiestudycircle.com. Our study material is prepared by scholars and retired faculty from IIT, Roorkee.

Problem (Gujarat BOE 2024, 10 marks)

A solid fuel contains 74% carbon and 16% ash. The ash discharged from furnace contains 20% carbon. Calculate:

1. The weight of carbon lost in the ash per kg of fuel.

2. The percentage carbon burned.

3. The heat lost by the incomplete combustion.

Solution

Carbon Associated with Ash

If it is assumed that the ratio of carbon associated with ash is the same both in the solid fuel and the ash discharged then,

\(\frac{x}{{16}} = \frac{{20}}{{(100 - 20)}}\)

where x is the carbon associated with ash in 100 kg solid fuel.

Hence,  x = 4 kg C/100 kg of fuel

Carbon lost in ash

We see that when fuel is 100 kg then carbon lost in ash = 4 kg

Hence, when fuel is 1 kg then carbon lost will be 4/100 = 0.04 kg

Percent carbon burned

Effective carbon = 74 - x = 74 - 4 = 70 kg

Hence, % carbon burnt = \(\frac{{70}}{{74}}x100 = 94.59\)

Heat Lost Due to Incomplete Combustion

The loss of heat due to incomplete combustion is due to slippage of carbon into ash.

Thus heat lost = (4/100) (8137.5) kcal/kg of fuel = 325.5 kcal/kg of fuel.

(The calorific value of carbon is 8137.5 kcal/kg)

Therefore, per cent heat loss due to incomplete combustion

= (325.5/8137.5) (100) = 4%

Problem (Gujarat BOE 2023, 15 marks)

Determine the height and diameter of the chimney used to produce a draught for a boiler which has an average coal consumption of 1800 kg/hr and flue gases formed per kg of coal fired are 14 kg. The pressure losses through the system are given below:

Pressure loss in fuel bed =7 mm of water

Pressure loss in boiler tubes =7 mm of water

Pressure loss in bends = 3 mm of water

Pressure loss in chimney = 1.3 mm of water

Pressure head equivalent to velocity of the gases passing through the chimney =1.3 mm of water

The temperature of ambient air and flue gases are 350C and 3100C respectively.

Assume actual draught is 80% of theoretical one.

Solution

Height of the chimney

ma + 1 = 14

Hence, ma = 13 kg/kg of fuel

Ta = 35 + 273 = 308 K

Tg = 310 + 273 = 583 K

Draught required is equivalent to overcome the losses and velocity head

= 7+ 7 + 3 + 3+1.3 = 21.3 mm of water

Actual draught produced

\({h_w} = \frac{{21.3}}{{0.8}} = 26.62\,mm\,of\,water\)                       

Now using equation

\({h_w} = 353H\left( {\frac{1}{{{T_a}}} - \left( {\frac{{{m_a} + 1}}{{{m_a}}}} \right)\frac{1}{{{T_g}}}} \right)\)                        

\(26.62 = 353H\left( {\frac{1}{{308}} - \frac{{14}}{{13}}.\frac{1}{{583}}} \right)\)                        

Solving, H = 53.88m

Diameter of the chimney

\({\rho _g} = \frac{{353}}{{{T_g}}}\left( {\frac{{{m_a} + 1}}{{{m_a}}}} \right) = \frac{{353}}{{583}}x\frac{{14}}{{13}} = 0.652\,kg/{m^3}\)                        

Flue gases formed per second 

= \(\frac{{1800x14}}{{3600}} = 7\,kg\)

mg = ACρg       (1)

But 

 \(C = \sqrt {2g{H_1}} \)

where H1 is the equivalent velocity expressed in m of gas.

\({H_1}{\rho _g} = {h_w}{\rho _w}\)                       

where hw is the water head equivalent to velocity head responsible for giving velocity to the gas.

Hence              

 \({H_1} = \frac{{{h_w}{\rho _w}}}{{{\rho _g}}} = \frac{{1.3x1000}}{{1000x0.652}}\)

Hence              

\(C = \sqrt {2x9.81x1.993}  = 6.25\,m/s\)

Puttint this value in eq. (1)

\(7 = \frac{\pi }{4}{D^2}(6.25) \Rightarrow D = 1.478\,m\)   

These questions are taken from study material for  Boiler Operation Engineer Exam by amiestudycircle.com. Our study material is prepared by scholars and retired faculty from IIT, Roorkee.   

Problem (Jharkhand BOE 2024, 6 marks)

Propane Gas is reacted with air in such as ratio that an analysis of the dry products of combustion gives CO2 = 11.5%, O2 = 2.7% and CO = 0.7%. What is the percentage excess air used?

Solution

The right way to begin the analysis is to write down the chemical equation for 100 moles of the dry product gases. The total of the product is 14.9%, hence 85.1% is the nitrogen in the product. With this data, the chemical equation may be written as

\(\begin{array}{*{20}{l}}{x{C_3}{H_8} + a{O_2} + 3.76a{N_2}}{ \to 11.5C{O_2} + 0.7CO + 2.7{O_2}}{ + 85.1{N_2} + b{H_2}O}\end{array}\)

Since N2 is inert, it is written down as a side product. On the basis of the composition of air

a = 85.1/3.76 = 22.63 moles of air

Carbon balance gives

3x = 11.5 + 0.7 = 12.2

or x = 4.066

O2 balance gives 

2 x 22.63 = 2 x 11.5 + 0.7 + 2 x 2.7 + b

or b = 16.16

Now, employing the hydrogen balance, we check the computation

8x = 2b

or 8 x 4.066 = 2 x 16.16

or  32.28 = 32.32

This check is fairly accurate and the actual reaction may be written as

\(\begin{array}{*{20}{l}}{4.066{C_3}{H_8} + 22.63{O_2} + 85.1{N_2}}{ \to 11.5C{O_2} + 0.7CO + 85.1{N_2}}{ + 16.16{H_2}O}\end{array}\)

Dividing throughout by 4.066, the equation could be reduced to the basis of fuel. The stoichiometric equation for propane is

\(\begin{array}{l}{C_3}{H_8} + 5{O_2} + 18.8{N_2} \to C{O_2} + 4{H_2}O + 18.8{N_2}\end{array}\)

The moles of oxygen used theoretically per mole of fuel is 5. For the actual combustion, the ratio is 22.63/4.066 = 5.565. Hence, the percentage of theoretical air used in the actual combustion process to percentage of theoretical air

= \(\frac{{5.565x100}}{5} = 111.3\% \)

The percentage of excess air is 111.3%.

Problem (Assam BOE 2022, 8 marks)

A fuel C10H22 is burnt using an air fuel ratio of 12:1 by mass. Air contains 77 per cent of nitrogen and 23 per cent of oxygen by mass. If the whole amount of hydrogen burns to form water, vapour and there is neither any free oxygen nor any free carbon, determine the complete volumetric analysis of the products of combustion.

Solution

2C10H22 + 31O2 = 20CO2 + 22H2O

2x 142 + 31 x 32 = 20 x 44 + 22x 18

or 284 + 992 = 880 + 396

Air required for complete combustion

\(\frac{{992x100}}{{284x23}} = 15.2\,kg/kg\,of\,fuel\)

Air actually supplied = 12 kg/kg of fuel

Deficiency of air = 15.2 -12 = 3.2 kg/kg of fuel

Also, 1 kg of C requires \(\frac{4}{3}x\frac{{100}}{{23}} = 5.8\)kg of less air to burn to CO instead of CO2.

Hence, \(\frac{{3.2}}{{5.8}} = 0.55\) kg C is burnt to CO

and \(\frac{{12x10}}{{142}} - 0.55 = 0.23\)kg of C is burnt to CO2.

Weight of CO2 formed = 0.23 x (11/3) = 0.84 kg

Weight of CO formed = 0.55 x (7/3) = 1.28 kg

Weight of H2O formed = 9 x (22/142) = 1.394 kg

Weight of N2 from air = 12 x 0.77 = 9.24 kg

The percentage composition of products of combustion is given below in a table.

Performance of Boiler

Problem (Maharashtra BOE 2024, 6 marks)

Efficiency of 300 TPH coal fired boiler is 84%. Calculate quantity of coal required per day. Feed water temperature is 1400C, GCV of coal 3600 kcal/Kg. The enthalpy of superheated steam is 803 kcal/kg.

Solution

Use the formula

\({\eta _{boiler}} = \frac{{steam(kg)[total\,heat\,in\,steam(kcal/kg) - total\,heat\,in\,feedwater(kcal/kg)]}}{{coal\,consumption(kg)\,x\,GCV\,of\,coal\,(kcal/kg)}}x100\) 

Taking all the data on per hour basis,

Coal consumption per hour 

= \(\frac{{300x1000(803 - 140)}}{{0.84x3600}}x100\) 

= 65774 kg/hr

Hence, coal consumption per day 

= 65774 x 24 = 1578572 kg i.e. 1579 t/day

Problem (Maharashtra BOE 2023, 6 marks)

200 TPH steam is generated in a Boiler at 100 kg/cm2 pressure and 500°C temperature. If feed water inlet temperature is 135°C and 52 Tons of coal is consumed per hour then calculate

(i) Equivalent evaporation from and at 1000C.

(ii) Equivalent evaporation from and at 100°C per ton of coal.

(iii) Boiler HP.

Solution

Equivalent evaporation from and at 1000C

Equivalent evaporation is the quantity of water evaporated from and at 100 °C to produce dry saturated steam at 100 °C by absorbing the same amount of heat equal to the actual operating condition.

\({M_{eq}} = \frac{{{M_{act}}(H - {H_{fw}})}}{{539}}\)

where, Meq is the Equivalent evaporation from and at 100 °C, Mact is the actual steam generation in the boiler, H is the heat of steam and Hfw is the heat of feedwater.

Latent heat of dry saturated steam at 100°C is 539 kcal/kg

Heat in steam at 100 kg/cm2 pressure and 500 °C temperature is 809 kcal/kg.

\({M_{eq}} = \frac{{200(809 - 135)}}{{539}} = 250\,TPH\)

Equivalent evaporation from and at 100 °C per ton of coal.

To find equivalent evaporation from and at 100 °C per ton of coal, Mact is taken as actual steam generation per unit quantity of fuel.

Mact = 200/52 = 3.85 t per ton of coal

 \({M_{eq}} = \frac{{3.85(809 - 135)}}{{539}} = 4.814\,t\,of\,steam/t\,of\,coal\)

Boiler Horse Power (HP)

One boiler horsepower is the capacity to evaporate 15.653 kg of feedwater per hour at 100°C into dry saturated steam at 100 °C.

\(HP = \frac{{Eq.\,evaporation\,from\,and\,at\,{{100}^0}C(kg/hr)\,}}{{15.663}}\)

= \(\frac{{250x1000}}{{15.663}} = 15971\)

 


 





Comments

Popular posts from this blog

Energy Systems (Solved Numerical Problems)

Wind at 1 standard atmospheric pressure and \({15^0}C\) has velocity of 15 m/s, calculate (i) the total power density in the wind stream (ii) the maximum obtainable power density (iii) a reasonably obtainable power density (iv) total power (v) torque and axial thrust Given: turbine diameter = 120 m, and turbine operating speed = 40 rpm at maximum efficiency. Propeller type wind turbine is considered. (AMIE Winter 2023) Solution For air, the value of gas constant is R = 0.287 kJ/kg.K 1 atm = 1.01325 x 105 Pa Air density \(\rho  = \frac{P}{{RT}} = \frac{{1.01325x{{10}^5}}}{{287}}(288) = 1.226\,kg/{m^3}\) Total Power \({P_{total}} = \rho A{V_1}^3/2\) Power density \(\begin{array}{l}\frac{{{P_{total}}}}{A} = \frac{1}{2}\rho {V_1}^3\\ = \frac{1}{2}(1.226){(15)^3}\\ = 2068.87{\mkern 1mu} W/{m^2}\end{array}\) Maximum power density \(\begin{array}{l}\frac{{{P_{\max }}}}{A} = \frac{8}{{27}}\rho A{V^3}_1\\ = \frac{8}{{27}}(1.226){(15)^3}\\ = 1226{\mkern 1mu} W/{m^2}\end{array}\) Assuming eff...

Mechanics of Fluids (Solved Numerical Problems)

Numerical The surface Tension of water in contact with air at 20°C is 0.0725 N/m. The pressure inside a droplet of water is to be 0.02 N/cm² greater than the outside pressure. Calculate the diameter of the droplet of water. (7 marks) (AMIE Summer 2023) Solution Surface tension, σ = 0.0725 N/m Pressure intensity, P = 0.02 N/m 2 P = 4σ/d Hence, the Diameter of the dropd = 4 x 0.0725/200 = 1.45 mm Numerical Find the surface tension in a soap bubble of 40 mm diameter when the inside pressure is 2.5 N/m² above atmospheric pressure. (7 marks) (AMIE Summer 2023) Answer: 0.0125 N/m Numerical The pressure outside the droplet of water of diameter 0.04 mm is 10.32 N/cm² (atmospheric pressure). Calculate the pressure within the droplet if surface tension is given as 0.0725 N/m of water. (AMIE Summer 2023, 7 marks) Answer: 0.725 N/cm 2   Numerical An open lank contains water up to a depth of 2 m and above it an oil of specific gravity 0.9 for a depth of 1 m. Find the pressure intensity (i) at t...

Analysis and Design of Structures (Solved Numerical Problems)

Structural Analysis