Power plant plays an important role for the continuous working of a steel industry. The Thermal power plant in a steel industry has 5 boilers each generating 330 tph steam at 101 atm and 5400 oC. The boilers are capable of firing combination of fuels namely coal, coke oven gas, blast furnace gas and oil. Normally 4 boilers are kept in full load operation to produce 247.5 MW of power, and supply steam to 2 turbo blowers. The outlet flue gas passes through electrostatic precipitators to control air pollution. The fly ash and bottom ash generated are pumped in slurry form to ash pond through ground pipelines. The clarified water is recirculated back to the ash system.

1.1 Working Principle of Thermal Power Plant

The high pressure steam is produced in the boilers by burning fuel in the furnace. The high pressure steam is expanded in the turbine that causes the turbine to rotate the prime rotor coupled with the generator or the alternator and generates power. The steam from the turbine is condensed in the condenser and re-circulated back to the steam generator (boiler). The outline model of a thermal power plant is shown in Figure 1.

Figure 1. Thermal Power Plant Outline

Different units of the thermal power plant are:

• Boilers
• Turbo generators
• Turbo blowers
• Chemical Water Treatment Plant
• Chilled Water Plant
• Coke Dry Cooling Plant (CDCP) Boilers
• Back Pressure Turbine Station
• Electrostatic Precipitators
• Auxiliaries of Thermal power plant

1.2 Importance of the Rankine Cycle

Rankine cycle is the theoretical cycle on which the steam turbine works. The common working fluid is water. It is a closed cycle that enables the working fluid to be used again and again. Water is usually the fluid of choice due to its favourable properties, such as non-toxic and nonreactive chemistry, abundance, and low cost, as well as its thermodynamic properties (Roy, 2015).

The cycle consists of 4 processes as shown in Figure 2.

Figure 2. Rankine Cycle

• Process 1-2: Reversible adiabatic expansion in turbine.
• Process 2-3: Constant pressure heat rejection in condenser.
• Process3-4: Reversible adiabatic pumping process in feed pump.
• Process 4-1: Constant pressure heat addition in boiler.

The thermal efficiency (η_th) is a dimensionless performance th measure of a thermal device such as an internal combustion engine, a boiler, or a furnace, for example. The input, Q_th to the device is heat, or the heat-content of a fuel in that is consumed. The desired output is mechanical work, W_out , or heat, Q_in , or possibly both. Because the input heat out out normally has a real financial cost, a memorable, generic definition of thermal efficiency is,

From the first and second law of thermodynamics, the output cannot exceed the input, so 0£ h£1.0. When th expressed as a percentage, the thermal efficiency must be between 0% and 100%. Due to inefficiencies such as friction, heat loss, and other factors, thermal efficiencies are typically much less than 100%.

1.3 Boilers Used in Thermal Power Plant of Steel Industry

The Thermal Power Plant in is equipped with 5 units of boilers. Normally 4 boilers are in running conditions at full load and one is kept idle for emergency cases.

1.3.1 Specifications of Boiler in TPP

Table 1 indicates the boiler design data. The salient design data of the boilers are,

Table 1. Boiler Data

Boiler Capacity - 330 tph

Fuels fired - Pulverized coal, BFG, COG, HSD, HFO

SH steam pressure - 100 kg/cm²

SH steam temperature - 540 ^oC

Feed water temperature - 170 ^oC

1.3.2 Efficiency of a Boiler

It is defined as the ratio of heat utilized in generation of steam to the heat supplied by the fuel in the same period (Roy, 2015; Ameri et al., 2009).

It can be calculated in two methods:

• Direct method
• Indirect (losses) method

Here we have calculated the efficiency of boiler by direct method.

Direct method:

The boiler efficiency can be obtained by calculating the output of the boiler (steam) and input of the boiler (water + coal + blast furnace gas (BFG) + coke oven gas (COG)) which can be expressed by

ηboiler = (Output/Input) ×100

Flow rate of steam (m) = 312.7 TPH

Output = Heat absorbed by working fluid

= Steam Enthalpy at boiler outlet – Water Enthalpy at boiler inlet

= 823.6 kcal/kg – 206.3 kcal/kg

= 617.3 kcal/kg

Heat generated (Output) by water

= m× (617.3) kcal/kg

=312700 × 617.3 kcal/kg

= 192939027 kcal/h

The heat input to the boiler is supplied by coal, blast furnace gas and coke oven gas. Hence the boiler is a multi fuel boiler. The consumption of each of the fuel may vary depending upon their availability. The fuel consumptions details are shown in Table 2.

Table 2. Fuel Consumption

Heat input by coal = 32000 kg/h × 28500 kcal/kg

= 91200000 kcal/h

Heat input by BFG = 107670 Nm³/h × 800 kcal/Nm3

= 86136000 kcal/h

Heat input by COG = 11600 Nm³/h × 4350 kcal/Nm³

= 50460000 kcal/h

Total heat input = 228096000 kcal/h

ηboiler = Heat generated (output) by water/Total heat input

= 192939027/228096000

= 84.5%

2.1 Efficiency of Thermal Power Plant Working on Simple Rankine Cycle:

Simple Rankine cycle in this thermal power plant,

η_Rankine = Wnet /Q₁ = (h₁ – h₂) / (h₁ – h_t4) (Rajput, 2015)

h₁ = Enthalpy of superheated steam at exit of boiler (at 530 ⁰C and 90 kg/cm² )

= 3459 kJ/kg

h2 = Enthalpy of saturated steam at the end of expansion in turbine (at 54 ⁰C)

= 2600 kJ/kg

h_f4 = Enthalpy of wet steam at which feed water is fed into boiler (at 170 ⁰C)

= 722 kJ/Kg

ηRankine = (h₁ – h2)/( h₁ – h₁₄ )

= (3459 kJ/kg -2600 kJ/kg)/(3459 kJ/kg -722 kJ/kg)

= (859 kJ/kg)/ (2737 kJ/kg)

= 31%

Thermal efficiency of thermal power plant with assumed simple Rankine cycle is 31%.

The thermal efficiency of the Rankine cycle can be improved by the following methods (Boonnasa & Namprakai, 2008).

• By regenerative feed heating
• By reheating of steam
• By combination of both regeneration and reheating.

2.2 Efficiency of Regenerative Rankine Cycle (Ameri et al., 2009)

Let 1 kg of steam enters the turbine. Let m₁ , m₂ , m₃ , m₄ , be the mass of steam that is bled at 4 intermediate stages from beginning of high pressure expansion to end of low pressure expansion in the turbine into the heaters (HPH4, HPH3, LPH2,LPH1) respectively.

(Note: HPH – High Pressure Heater, LPH – Low Pressure Heater)

From the diagram obtained from the control room the corresponding temperatures and pressures at the point of bleeding have been noted as shown in Table 3.

Table 3. Steam Tables Data Book Values (Jain, 2018)

The following values are taken from the control room of the power plant based on that the calculations are done (Murugan & Subbarao, 2008)

At the entrance of the turbine, at temperature 530 ⁰C and 90 kg/cm² ;

h₁ = 3459 kJ/kg ,

h_f1 =1352 kJ/kg

At the exit of the turbine at temperature 54 ⁰C and 0.86 2 kg/cm² ;

h₆ = 2600 kJ/kg,

h_f6 = 225 kJ/kg

Applying energy balance equations for each of the heaters (Rajput, 2015), we get,

For heater 4:

m₄ (h₅ –h_f5)=[1-(m₁ +m₂ +m₃ +m₄)](h_f5 –h_f6)

For heater 3:

m₃ (h₄ –h_f4 )=[1-(m₁ +m₂ +m₃)](h_f4 –h_f5 )

For heater 2:

m₂ (h₃ –h_f3 )=[1-(m₁ +m₂ )](h_f3 –h_f4 )

For heater 1:

m₁ (h₂ – h_f2 ) = [1- m₁ ] ( h_f2 – h_f3 )

Substituting the enthalpy values in the above energy balance equations, we get

m₁ (h₂ – h_f2 ) = [1- m₁ ] (h_f2 – h_f3 )

m₁ (3150 – 925) = (1- m₁) (925 – 747.6)

m₁ (2225) = (1- m₁) (177.4)

13.5m₁ =1 kg

m₁ = 0.0738 kg

m₂ ( h₃ – h_f3 ) = [(1- ( m₁ + m₂ )] (h_f3– h_f4)

m₂ (2976- 747) = [1- 0.0738 – m₂] (747 -425) 

m₂ (2229) = (0.926-m₂ ) (322)

m₂ = 0.1169 kg

m₃ (h₄ – h_f4) = [1- (m₁ + m₂ + m₃)] (h_f4 – h_f5)

m₃ (2829 – 425) = [1 – 0.0738 – 0.1169 –m₃] (425 -315)

m₃ (2404) = (0.809-m3) (110)

m₃ = 0.035 kg

m₄ (h₅ – h_f5 ) = [1- (m₁ + m₂ + m₃ +m₄ )] (h_f5 – h_f6 )

m₄ (2683 -315) = [1- 0.0738 – 0.1169 – 0.035 – m₄) (315- 225)

m4 (2368) = (0.77-m₄) (90)

m₄ = 0.0283 kg

Net Work done according to (Rajput, 2015), can be calculated as

= 1 (3459 -3150) + (1-0.0738) (3150 -2976) + (1 - 0.0738 - 0.1169) (2976 - 2829) +(1- 0.0738 - 0.1169 - 0.035)(2829 -2863) + (1 - 0.0738 - 0.1169 - 0.035 - 0.0283)(2683 -2368)

= 309+1610.15+128.96+113.04+254.99

= 965 kJ/kg

Heat Supplied per kg of steam = h₁ – h_f2

= 3459 – 925

= 2534 kJ/KG

= (965 kJ/Kg)/ (2534 kJ/Kg)

= 38%

The calculated thermal efficiency of plant working on regenerative cycle is 38%

Steam consumption with regenerative feed heating

= 3.73 kg/kWh

Steam Consumption without regenerative feed heating:

= 3.29 kg/kWh

Difference in steam consumption

= Steam consumption with feed heating – Steam consumption without feed heating.

= 3.73 kg/kWh – 3.29 kg/kWh

= 0.44 kg/kWh.

Quantity of steam passing through the last stage of a 60 MW turbine with regenerative feed heating

= 3.73 (1- m₁ – m₂ – m₃ – m₄)× 60000

= 3.73 (1-0.0738 -0.1169 –0.035-0.0283)× 60000

= 166954.8 kg/h

Same without regenerative arrangement

= 3.29 × 60000

= 197400 kg/h

2.3 Efficiency of Reheat Rankine Cycle (Murugan & Subbarao, 2008)

Thermal efficiency with 'Reheating' (neglecting pump work),

Heat supplied = (h₁ –h_f4) + (h₃-h₂)

Heat rejected = h₄ –h_f4

Work done by the turbine= Heat supplied – Heat rejected

= (h₁–h_f4) + (h₃ –h₂) - (h₄ –h_f4)

= (h₁ –h₂ ) + (h₃ –h₄)

Thus, theoretical thermal efficiency of reheat cycle is

The calculated thermal efficiency of Rankine Cycle with reheating is 34.4%

2.4 Efficiency of Combined Regenerative and Reheat Rankine Cycle (Haywood, 1949)

m₁, m₂, m₃, m₄ are the masses of steam that has been fed to the heaters at various stages from 1 kg of steam that has entered at high pressure turbine.

Applying energy balance equations at each of the heater (Rajput, 2015), we have

Substituting the values of h from the steam table, To calculate m1 :

m₁ (3150-925) = (1-m₁) (925-756)

m₁(2225) = (1-m₁) (169)

14.16 m₁ =1 kg

m1 = 0.070 kg

To calculate m₂ :

m₂ (h₅ –h_f5) = [(1 – (m₁+m₂)] [h_f5 –h_f6]

m₂ (3364 – 756) = [(1 – (0.070 + m₂) (756 -530)

m₂ (2608) = (0.93-m₂) (226)

m₂ (11.53) = 0.93 – m₂

12.53 m₂ = 0.93 kg

m₂ = 0.0742 kg.

To calculate m₃ :

m₃ (h₆ –h_f6) = [1 – (m₁ +m₂ +m₃ )] (h_f6 –h_f7)

m₃ (3100-530)= [1 – 0.070 – 0.074 – m₃](530-450)

m₃ (2570) = (0.856 – m₃) (80)

m₃ (33.12) = 0.856 kg

m₃ = 0.025 kg

To calculate m₄ :

m₄ (h₇ –h_f7) = [1 – (m₁ +m₂ +m₃ +m₄)] (h_f7 – h_f8)

m₄ (2945 -450) = [1 – 0.070 – 0.074 – 0.025 – m₄] (450 – 389)

m₄ (2495) = (0.83 – m₄)

m₄ (41.9) = 0.83 kg

m₄ = 0.0198 kg.

Total work done per kg of steam supplied by the boiler (Rajput, 2015),

Qtotal = Heat supplied by the boiler per kg of steam + Heat total supplied in reheating process

Qtotal = Q₁ + Q₂

Q₁ = Heat supplied by the boiler per kg of steam (Rajput, 2015)

Q₁ = h₁ – h_f1

= 3459 kJ/kg – 1352 kJ/kg

= 2107 kJ/kg

Q₂ = Heat supplied in reheating process (Rajput, 2015)

= (1 – m₁) (h₄ –h₃)

= (1 – 0.070) (3520 – 3128)

= 0.93 * 392

= 364.56 kJ/kg

Total heat supplied = Qtotal = Q₁ + Q₂ (Rajput, 2015)

= 2107 kJ/kg + 364.56 kJ/kg

= 2471.56 kJ/kg

The calculated thermal efficiency of plant with combined regenerative reheat cycle is 41.57%

The thermal efficiencies of the steel industry power plant are calculated using different improvement methods and tabulated. Only mathematical equations are used to determine the thermal efficiency of the power plant as it is already established in the plant. The efficiency obtained from these calculations are compared with the practically obtained values. There is an error of 6.45% to 9.67% with the obtained results and this is a case study of how the actual power plant works and the methods to be considered for the improvement of the efficiency.

Table 4 shows the calculated values of efficiency. From the results it is noticed that the thermal efficiency can be increased by using different methods. The regenerative and reheat Rankine cycle shows higher efficiency of 41.2%. The efficiency obtained from numerical analysis shows variations when practically implemented as the entire power plant used in a steel industry produces 4 million tonne steel. This can be considered as a case study approach in exact operating conditions. This study will be useful for future researchers in modifying and simplifying power plant. Conditions and this study will be useful for the future to modify and simplify the power plant.

Table 4. Steam Table Values (Jain, 2018)

Table 5. Entalphy Values (Jain, 2018)

Table 6. Calculated Values of Efficiency

The practical investigations and analysis of the various important factors affecting the thermal power plant and thermal efficiency is discussed. The calculations and analysis explains that efficiency of a simple Rankine cycle is improved by using a intermediate reheat cycle and enabling the improvement of thermal conditions. The regeneration is vital to improve the efficiency as it uses the sensible heat of exhaust steam for the preheating of feed water. Inclusion of a feed water heater also introduces an additional pressure level into the Rankine cycle. The condenser vacuum and supply of condenser tube cooling water is another parameter. It is observed that though reheating improves thermal efficiency of Rankine Cycle by a considerable percentage, it is lower than efficiency of cycle with Regenerative method. A combined Reheat and Regenerative method results in a significant improvement in thermal efficiency.