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%.
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 (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.
\(HP = \frac{{Eq.\,evaporation\,from\,and\,at\,{{100}^0}C(kg/hr)\,}}{{15.663}}\)
= \(\frac{{250x1000}}{{15.663}} = 15971\)
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 (Maharashtra BOE 2024, 6 marks)
The performance of a boiler plant are as given below:
A boiler generated 6.5 ton of steam per ton of coal fired.
The boiler feed water temperature is 1100C downstream of deaerator.
The steam generated is at 18 kg/cm2(g)
Boiler Efficiency is 75%
Factor of evaporation is 1.15
Cp of the steam is 0.55 kcal/kg0C
Determine:
(i) The temperature of the steam and degree of superheat,(if any).
(ii) The equivalent evaporation per ton of coal burned.
(iii) The calorific value of coal.
Solution
Steam parameters for 18 kgf/cm2 (gauge)
Pressure = 19 kgf/cm2 (abs)
Hw = 213.1 kcal/kg
Latent heat, L = 455.1 kcal/kg
Saturation temperature, θs = 208.80
Degree of superheat
Total heat of the seam = hw + L + Cp∆θ
Specific enthalpy of feedwater = hfw
Hence, the factor of evaporation
\(\frac{{{h_w} + L + {C_p}\Delta \theta - {h_{fw}}}}{{539}}\)
\(1.15 = \frac{{213.1 + 455.1 + 0.55(\Delta \theta ) - 110}}{{539}}\)
Giving, ∆θ = 112.090C
Superheated steam temperature
∆θ = 112.090C
Or \(\theta - {\theta _s} = 112.09\)
\(\theta = 208.8 + 112.09 = {321^0}C\)
Heat output
Heat required to generate steam
= 6.5 x 103(213.1 + 455.1 + 0.55∆θ - 110) kcal/t of coal
=\(6.5x{10^3}(558.2 + 0.55x112.09)\)
= 4029.021 x 103 kcal/t of coal
Heat input
Calorific value of coal = CV kcal/kg
Energy released per ton of coal burnt = 103 x CV of coal
Boiler efficiency
\({\eta _{boiler}} = \frac{{Heat\,output}}{{Heat\,input}} = \frac{{4029.021x{{10}^3}}}{{{{10}^3}(CV)}}\)
\(0.75 = 4029.021/cV\)
Giving CV = 5372 kcal/kg of coal
Equivalent evaporation
\({M_{eq}} = \frac{{6.5(213.1 + 455.1 + 0.55x112.09 - 110)}}{{539}}\)
= 7.474 t/t of coal burnt
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 2021, 12 marks)
A boiler generates 7.5 kg of steam per kg of coal burnt at a pressure of 11 bar, from feed water having a temperature of 70°C. The efficiency of the boiler is 75%, factor of evaporation is 1.15 and specific heat of steam at constant pressure is 2.3. Determine (i) Degree of superheat and temperature of steam generated (ii) Calorific value of coal in kJ/kg and (iii) Equivalent evaporation in kg of steam per kg of coal.
Solution
Steam generated per kg of coal, ma = 7.5 kg
Steam pressure, p = 11 bar
Temperature of feed water = 700C
Efficiency of boiler = 75%
Factor of evaporation, Fe = 1.15
Specific heat of steam, cps = 2.3 kJ/kg K
Degree of superheat and temperature of steam generated
At 11 bar, from steam tables (p, ts, hf, hfg and hg)
hf = 781.1 kJ/kg
hfg = 1998.5 kJ/kg
ts = 184.10C = 184.1 + 273 = 457.1 K
Factor of evaporation
\({F_e} = \frac{{\left[ {\left\{ {{h_f} + {h_{fg}} + {c_{ps}}({t_{\sup }} - {t_s}} \right\} - {h_{f1}}} \right]}}{{2257}}\)
Hence
\(1.15 = \frac{{[781.1 + 1998.5 + 2.3({t_{\sup }} - 457.1)] - 1x4.18x(70 - 0)}}{{2257}}\)
Solving, tsup = 504.3 K
Degree of superheat
= tsup - ts = 504.3 - 457.1 = 47.20C
Calorific value of coal, C
Boiler efficiency =\(\frac{{{m_a}(h - {h_{f1}})}}{C}\)
Hence, \(0.75 = \frac{{{m_a}\left[ {\left\{ {{h_f} + {h_{fg}} + {c_{ps}}({t_{\sup }} - {t_s})} \right\} - {h_{f1}}} \right]}}{C}\)
\(0.75 = \frac{{7.5[(781.1 + 1998.5 + 2.3(504.3 - 457.1) - 1x4.18x(70 - 0)]}}{C}\)
Solving
\(0.75 = \frac{{7.5(2888.16 - 292.6)}}{C}\)
Giving, C = 25955 kJ/kg
Equivalent evaporation, me
\({m_e} = \frac{{{m_a}(h - {h_{f1}})}}{{2257}} = \frac{{7.5[(2888.16 - 1x4.18x(70 - 0)]}}{{2257}}\)
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 (Punjab BOE 2021, 10 marks)
The following data refer to a boiler plant consisting of an economiser, a boiler and a superheater.
Mass of water evaporated per hour = 5940 kg.
Mass of coal burnt per hour = 675 kg,
LCV of coal - 31600 kJ/kg,
Pressure of steam at boiler stop valve = 14 bar,
Temperature of feed water entering the economizer = 32 °C
Temperature of feed water leaving the economizer = 115 °C
Dryness fraction of steam leaving the boiler and entering superheater = 0.96.
Temperature of steam leaving the superheater = 260 °C.
Specific heat of superheated steam = 2.3 kJ/kg K.
Determine
(a) Percentage of heat in coal utilized in economizer, boiler and superheater.
(b) Overall efficiency of boiler plant.
Solution
Heat utilized by 1 kg of feed water in economiser
hf1 = 1 x 4.18 x (te2 - te1)
= 1 x 4.18 x (115 - 32) = 346.9 kJ/kg
Heat utilised in boiler per kg of feed water
hboiler = (hf + xhfg) - hf1
At 14 bar, we get following data from steam table (p, ts, hf, hfg and hg)
ts = 1950C, hf = 830.1 kJ/kg
hfg = 1957.7 kJ/kg
∴ hboiler = (830.1 + 0.96 x 1957.7) - 346.9 = 2362.6 kJ/kg
Heat utilised in superheater by 1 kg of feed water
hsuperheater = (1 - x)hfg + cp(Tsup - Ts)
= (1 - 0.96) x 1957.7 + 2.3(260 - 195) = 227.8 kJ/kg
Also, mass of water evaporated per hour per kg of coal burnt
= 5940/675 = 8.8 kg
Percentage of heat utilised in economiser
= (346.9/31600) x 8.8 x 100 = 9.66%
Percentage of heat utilised in boiler
= (2362.6/31600) x 8.8 x 100 = 65.7%
Pertcentage of heat utilised in superheater
= (227.8/31600) x 8.8 x 100 = 6.34%
Overall efficiency of boiler plant
Total heat absorbed in kg of water
= hf1 + hboiler + hsuperheater
= 346.9 + 2362.6 + 227.8 = 2937.3 kJ/kg
Now, \({\eta _{overall}} = \frac{{8.8x2937.3}}{{31600}} = 0.81 = 81\% \)
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, 12 marks)
A stoker - fired water tube boiler burns coal at the rate of 4 ton/hr to generate steam of 30 kg/cm2 abs and 430°C at the rate of 30 ton/hr. Evaluate the boiler performance from the following data:
(i) Components by proximate analysis of the coal: Ash - 12.7% by weight, Moisture - 7.9% by weight
(ii) Gross calorific value of coal - 6250 kcal/kg,
(iii) Components by flue gas analysis: CO2 = 12.85%, O2 = 6.5%, N2 =rest
(iv) Carbon present in the cinder as unburnt combustible - 2.75%,
(v) The feed water temperature - 90’C,
(vi) Flue gas temperature at economizer outlet - 1500C, Flue gas pressure at economizer outlet - 755 mm Hg,
(vii) Air temperature at burner inlet: 30°C DB and 22°C WB. Ignore the presence of sulphur arid oxygen in the coal.
Solution
The boiler performance, i.e. the overall thermal efficiency of the boiler is to be evaluated.
Heat Input Rate
Rate of fuel burning = 4 ton coal/h
Gross clorific value of fuel = 6250 kcal/kg
Heat input rate = 4 x 1000 x 6250 = 25000000 kcal/h
Ignoring the heat input of air (at 30°C), the net heat imput rate = 25000000 kcal/h
Heat Output Rate
Heat absorbed in generating superheated steam
= 30,000 x (787.8 - 90.04) kcal/h
= 20932800 kcal/h
where 787.8 kcal/kg = enthalpy of steam at 30 kgf/cm2 abs and at 430°C = hss
90.04 kcal/kg = enthalpy of water at 90°C
Overall thermal efficiency of boiler
= \(\frac{{Heat\,output(Steam\,generation)}}{{Heat\,input(fuel\,combustion)}}\)
=\(\frac{{20932800}}{{25000000}}x100 = 83.73\% \)
Note: The other data is superfluous as this will not be used in calculating overall efficiency of the boiler. It may however be used in doing a heat balance.
Turbine
Problem (Assam BOE 2022, 10 marks)
A steam turbine gets steam at 60 bar and 450°C from a steam generator that expands to a condenser pressure of 0.07 bar. Some amount of steam is bled from the turbine at 3 bar to heat the feed water from the condenser. Steam turbine generates 30 MW through a directly coupled generator with 95 per cent efficiency. Assuming turbine efficiency of 90 per cent determines
(i) the amount of steam bled/kg of steam entering the turbine.
(ii) the steam generation per hour.
(iii) the overall efficiency of the plant if boiler efficiency is 92 per cent and mechanical efficiency is 98 per cent.
Assume the pump work is negligible and 15 per cent of generated power is used to run auxiliaries.
Solution
The actual expansion in the turbine is shown by dotted lines on the T-s diagram. From steam tables at 60 bar, 450°C
h1 = 3301.8 kJ/kg
s1 = 6.7193 kJ/kgK
At 3 bar for process 1-2'
s1 = s2'
6.7193 = 1.6717 + x2' x 5.3192
∴ x2' = 0.95
h2' = 561.4 + 0.95 x 2163.2 = 2616.44 kJ/kg
∴ h1 - h2 = turbine efficiency x (h1 - h2')
= 0.90 x (3301 - 2616.44) = 616.82 kJ/kg
Also,
h2 = h1 - 616.82
= 3301.8 - 616.82
= 2684.98 kJ/kg
= 561.4 + x2 x 2163.2 at 3 bar
∴ x2 = 0.982
s2 = 1.6717 + 0.982 x 5.3192 = 6.895 kJ/kg K
Considering the process 2-3' (isentropic)
s2 = s3' at 0.07 bar
6.895 = 0.5591 + x3' x 7.7176
Giving
x3' = 0.821
∴ h3' = 163.4 + 0.821 x 2409.2 at 0.07 bar = 2141.35 kJ/kg
∴ h2 - h3 = 0.9 x (h2 - h3') = 0.9 x (2684.98 - 2141.35) = 489.27 kJ/kg
Also
h3 = h2 - 489.27 = 2684.98 - 489.27 = 2195.71 kJ/kg
hf4 = 163.4 kJ/kg at 0.07 bar
hf5 = 561.4 kJ/kg at 3 bar
Amount of steam bled/kg of steam entering the turbine
Energy balance
m(h2 - hf4) = (hf5 - hf4)
∴ \(m = \frac{{({h_{f5}} - {h_{f4}})}}{{({h_2} - {h_{f4}})}} = \frac{{561.6 - 163.4}}{{2684.98 - 163.4}} = 0.158\,kg\)
Steam generated per hour
Net work done by the turbine
Wturbine = (h1 - h2) + (1 - m)(h2 - h3)
= 616.82 + (1 - 0.158) x 489.27 = 1028.785 kJ/kg
Actual turbine work
=\(\frac{{30}}{{0.95x0.98}} = 32.22\,MW\)
Steam generated in kg/s
\({m_s} = \frac{{{W_{actual}}}}{{{W_{turbine}}}} = \frac{{32.22x{{10}^3}\,kJ/s}}{{1028.785\,kJ/kg}}\)
= 31.32 kg/s = 112.752 t/hr
Heat supplied to the boiler
\(q = \frac{{{m_s}({h_1} - {h_{f5}})}}{{0.92}} = \frac{{31.32(3301.8 - 561.4)}}{{0.92}} = 93.293\,MW\)
As 15 per cent of turbine work is utilized for running auxiliaries, actual generated power available is
= 30 x 0.8 = 25.5 MW
Overall plant efficiency
\({\eta _{overall}} = \frac{{Net\,available\,power}}{{Heat\,\sup plied}} = \frac{{25.5}}{{93.293}} = 0.2733 = 27.33\% \)
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 2023, 5 marks)
Calculate the gross turbine heat rate and Net turbine heat rate with the help of parameters given below:
Feed water Inlet temperature to Boiler = 1250C
Boiler feedpump load = 1.6 MW
Steam generation = 160 TPH
Steam pressure = 100 Kg/cm2
Steam temperature = 5150C
Enthalpy of steam (at 100 kg/cm2 & 5150C) = 816 kcal/kg
Generator capacity = 40 MW
Auxiliary Power Consumption at Full load = 4 MW
Solution
Gross turbine heat rate (considering generation for one hour)
= [Steam flow to turbine (kg) (total heat of steam - total heat of feedwater at boiler inlet)]/Kilowatt output
= [160 x 1000(816 - 125)]/(40 x 1000)
= 2764 kcal/kWh
Net turbine heat rate
= [Steam flow to turbine (kg) (total heat of steam - total heat of feedwater at boiler inlet)]/[Kilowatt hour generated - Kilowatt hour consumed by feedpump)
= [160 x 1000(816 - 125)]/[(40 x 1000) - (1.6 x 1000)]
= 2879.17 kcal/kWh
Problem (Kerala BOE 2021, 2023, 8 marks)
The following parameters were recorded for a surface condenser:
Condenser vacuum = 680 mm Hg
Barometric pressure = 760 mm Hg
Mean condenser temperature = 35°C
Condenser exit temperature = 29°C,
Rise in temperature of the cooling water = 10°C
Rate of steam condensation - 24 ton/hr
Rate of cooling water flow = 1100 ton/hr
Determine (i) the mass of air present per unit volume of the condenser (b) the dryness fraction of steam at the inlet to condenser. (c) the vacuum efficiency of the condenser.
Solution
Absolute Pressure in the Condenser
Condenser vacuum = 680 mm Hg
Barometric pressure = 760 mm Hg
Absolute pressure in the condenser
= 760 - 680 = 80 mm Hg = 80 x 0.001359
= 0.1087 kgf/cm2
Mass of air/m3 of condenser volume
Partial pressure of steam at 350C = 0.0563 kgf/cm2 (from steam table)
Absolute pressure in the condenser
Pabs = ps + pair = 0.1087 kgf/cm2
∴ pair = 0.1087 - 0.0563 = 0.0524 kgf/cm2
Applying the characteristic gas equation
PV = mRT
\(\frac{m}{V} = \frac{P}{{RT}} = \frac{{0.0524x{{10}^4}}}{{(29.27)(273 + 35)}} = 0.0581\,kg/{m^3}\)
Dryness fraction of steam
hw (at 350C) = 35.06 kcal/kg
Latent heat, L = 576.4 kcal/kg
Dryness fraction of steam
ms(hw + xL - hcond) = mcp∆Tcw
24[35.06 + (x)(576.4) - 29.06] = 1100 x 1 x 10
∴ x = 0.79
Vacuum efficiency
\({\eta _{cond}} = \frac{{condenser\,vacuum}}{{ideal\,vacuum}}\)
=\( = \frac{{680}}{{760 - (0.0563/0.001359)}}x100 = 94.63\% \)
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)
For a cooling tower, the TDS of cooling water and make up water are 1500 ppm and 250 ppm respectively. If evaporation loss is 2% with windage loss negligible . Calculate COC (Cycles of concentration) & blow down percentage.
Solution
COC = TDS of cooling water/TDS of make-up water
= 1500/250 = 6
As windage loss is negligible, so
Blowdown% = Evaporation%/(COC - 1)
= 2/(6 - 1) = 0.4%
Comments