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

m_steam = 1000 kg

P_steam = 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, C_p = 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

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 D_e = 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

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

T_gas = 170 + 273 = 443 K

m_f = (1400/60) kg/min

m = 18 kg/kg of fuel

T_air = 40 + 273 = 313 K

η_f = 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

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

m_a + 1 = 14

Hence, m_a = 13 kg/kg of fuel

T_a = 35 + 273 = 308 K

T_g = 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

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

m_g = ACρ_g (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 h_w 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)

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

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, M_eq is the Equivalent evaporation from and at 100 °C, M_act is the actual steam generation in the boiler, H is the heat of steam and H_fw 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, M_act is taken as actual steam generation per unit quantity of fuel.

M_act = 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.

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

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

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

C_p 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)

H_w = 213.1 kcal/kg

Latent heat, L = 455.1 kcal/kg

Saturation temperature, θ_s = 208.8⁰

Degree of superheat

Total heat of the seam = h_w + L + C_p∆θ

Specific enthalpy of feedwater = h_fw

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

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, m_a = 7.5 kg

Steam pressure, p = 11 bar

Temperature of feed water = 70⁰C

Efficiency of boiler = 75%

Factor of evaporation, F_e = 1.15

Specific heat of steam, c_ps = 2.3 kJ/kg K

Degree of superheat and temperature of steam generated

At 11 bar, from steam tables (p, t_s, h_f, h_fg and h_g)

h_f = 781.1 kJ/kg

h_fg = 1998.5 kJ/kg

t_s = 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, t_sup = 504.3 K

Degree of superheat

= t_sup - t_s = 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, m_e

\({m_e} = \frac{{{m_a}(h - {h_{f1}})}}{{2257}} = \frac{{7.5[(2888.16 - 1x4.18x(70 - 0)]}}{{2257}}\)

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

h_f1 = 1 x 4.18 x (t_e2 - t_e1)

= 1 x 4.18 x (115 - 32) = 346.9 kJ/kg

Heat utilised in boiler per kg of feed water

h_boiler = (h_f + xh_fg) - h_f1

At 14 bar, we get following data from steam table (p, t_s, h_f, h_fg and h_g)

t_s = 195⁰C, h_f = 830.1 kJ/kg

h_fg = 1957.7 kJ/kg

∴ h_boiler = (830.1 + 0.96 x 1957.7) - 346.9 = 2362.6 kJ/kg

Heat utilised in superheater by 1 kg of feed water

h_superheater = (1 - x)h_fg + c_p(T_sup - T_s)

= (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

= h_f1 + h_boiler + h_superheater

= 346.9 + 2362.6 + 227.8 = 2937.3 kJ/kg

Now, \({\eta _{overall}} = \frac{{8.8x2937.3}}{{31600}} = 0.81 = 81\% \)

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 = h_ss

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

h_f4 = 163.4 kJ/kg at 0.07 bar

h_f5 = 561.4 kJ/kg at 3 bar

Amount of steam bled/kg of steam entering the turbine

Energy balance

m(h₂ - h_f4) = (h_f5 - h_f4)

∴ \(m = \frac{{({h_{f5}} - {h_{f4}})}}{{({h_2} - {h_{f4}})}} = \frac{{561.6 - 163.4}}{{2684.98 - 163.4}} = 0.158\,kg\)

Steam generated per hour

Net work done by the turbine

W_turbine = (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

\({\eta _{overall}} = \frac{{Net\,available\,power}}{{Heat\,\sup plied}} = \frac{{25.5}}{{93.293}} = 0.2733 = 27.33\% \)

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

P_abs = p_s + p_air = 0.1087 kgf/cm²

∴ p_air = 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

h_w (at 35⁰C) = 35.06 kcal/kg

Latent heat, L = 576.4 kcal/kg

Dryness fraction of steam

m_s(h_w + xL - h_cond) = mc_p∆T_cw

24[35.06 + (x)(576.4) - 29.06] = 1100 x 1 x 10

∴ x = 0.79

Vacuum efficiency

\({\eta _{cond}} = \frac{{condenser\,vacuum}}{{ideal\,vacuum}}\)

=\( = \frac{{680}}{{760 - (0.0563/0.001359)}}x100 = 94.63\% \)

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%

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