Overview of Obra Thermal and Hydro Power Plant

Overview of Obra Thermal and Hydro Power Plant

Amrit Shankar Srivastava

Lecturer Mechanical Engineering, Banaras Institute Of Polytechnic And Engineering, Varanasi

Introduction

Obra Hydel Power Plant is one of the earliest large hydroelectric projects in eastern Uttar Pradesh, located near Obra town in Sonebhadra district, on the Son River. It works in coordination with the nearby Obra Thermal Power Plant, forming a classic hydrothermal power complex.

Basic Overview

• Type- Hydroelectric Power Plant

• Water source- Son River

• Commissioned- 1960s (Indias early planned hydro projects, turbines in 1965)

• Owner/Operator- Uttar Pradesh Power Corporation Limited (UPPCL)

  Importance

• First major hydel project in UP

• Provides peak load support

• Stabilizes power supply for-

  • Obra Thermal Plant

  • Nearby industrial zones

• Helps in grid frequency control (fast START/STOP)

  Installed Power Capacity

• The Obra Hydel plant has a total installed capacity of about 99 MW.

• It has 3 generating units (all commissioned in the mid-1960s).

  Working Principle

  Step 1 Water Storage

  Water is stored at height in the Rihand reservoir behind the dam this stores potential energy.

  Step 2 Penstock Flow

  At Obra, for each turbine 1 Penstocks of 89m diameters and length of 200250m arranged as Dam integrated/ tunnel type steel-lined conduit with gentle slope are used. When Power is needed

• Intake gates open

• Water flows through penstocks

  Step 3 Turbine Action

  High-head water strikes the turbine blades, converting Hydraulic energy Mechanical energy

  Type Kaplan Vertical turbine

  Step 4 Generator

• Turbine shaft rotates generator

• Mechanical energy Electrical energy (AC)

  Step 5 Power Transmission

• Voltage stepped up using transformers

• From generator 11kV 33kV 132kV 220kV

• Power sent to grid

  River, Reservoir and Level Terminology

  River system (Natures view) Son River tributary Rihand River

  Near Pipri, an artificial stream is created. This stream is an engineered waterway which feeds Obra Hydel turbines.

  Main Water body is Son River.

  River Bed The river bed is the natural bottom of the river channel. It is irregular and variable due to erosion and siltation and is never used as a reference level in dam or hydropower design.

  Datum and Reduced Levels (RL)

  All elevations are measured from a fixed survey datum, usually-

• Mean Sea Level (MSL), or

• A permanent benchmark tied to MSL

• Here, Reservoir water level, RL = 192 ft, this means the water surface elevation is 192 ft above datum, not above the river bed.

  Full Reservoir Level (FRL) FRL is the maximum normal operating water level

  At Obra, FRL 202 ft, FRL is decided based on hydrology, flood studies, and storage requirements.

  Schematic longitudinal section of a river-bed before (a) & during (b) the propagation

  Dam

  Dam Height

  Dam Height = Crest Level Foundation Level

  Dam Crest level = FRL + freeboard

  Foundation level lies well below the river bed on hard rock

  Dam height is a structural parameter, not related to power generation.

  This is a gravity dam which resists water pressure by its own weight.

  It has-

• Huge base width

• Massive concrete volume

• Internal galleries, drains, and conduits

  During construction-

• The penstock passage is planned first

• Concrete is poured around it

• It becomes a monolithic part of the dam

  Penstock diameter 8 m, length 200m

  Since a typical gravity dam has base width of 60m 100m and relative size of diameter is kept around 10%

  Location-

• Near the neutral zone of the dam cross-section

• Not near upstream face

• Not near downstream toe

  This zone has-

• Minimum bending stress

• Maximum concrete mass around

  So stress distribution is hardly disturbed.

  Concrete lining + steel liner- Inside the conduit-

• Steel liner (2025 mm) handles water pressure

• Surrounding concrete handles structural load

• Water pressure does not act on dam concrete directly, it is carried by steel liner

  Internal galleries- Gravity dams have-

• Inspection galleries

• Drainage galleries

• Grouting galleries

  Penstock conduits are just like the bigger versions of these planned spaces.

  Structural design check Engineers check-

• Sliding

• Overturning

• Shear stress

• Principal stresses

with conduits included in the model so the dam is designed assuming these voids exist.

Low Head Dam

Calculations

Upstream water level (Reservoir side) This is measured as- Reservoir Water Level (RWL), taken near intake opening.

Reservoir water level = 192 ft 58.5 m

This is gross available height or Gross Head

Downstream water level (Tail race) This is water level after turbine in tail race / river

At Obra, Tail race level = 125 ft 38.1 m

Gross head Gross Head is the vertical difference between the reservoir water surface level and the tail-race water surface level. It is denoted by Zg.

Zgross = Zreservoir Ztail race = 58.5m 38.1m Zg = 20.4 m

Losses Losses occur in Penstock due to friction, bends, turbulence. Typically in Hydel systems head losses are taken around 15% of Gross head.

Total losses 15% 20% of Zg 3.4m

Net Head Net Head is calculated between two water levels

Net Head = Gross Head Head Loss

Net Head = 20.4m 3.4m = 17 m **

**These Losses may slightly vary**

Discharge Discharge per turbine is calculated by

P=gQH

Where, P = power = 33MW = 33×106 W

= density = 1000kg/m3 g = 9.81m/sec2

H = Net Head = 17m

= turbine efficiency = 90% = .9 then, Q = discharge = P ÷ gH

= 220m3/sec

Total Discharge = 3Q = 660 m3/sec

Penstock velocity

Discharge, Q = Area x velocity = D2/4 x velocity 220 = x 82/4 x velocity

Velocity, V = 4.3767 m/sec

Penstock thickness

Max internal pressure inside penstock, p = gH = 1000 x 9.81 x 17.34 = 170105.4 N/m2

= 0.170 MPa

For steel, FOS = 3, then, design pressure = 0.170 x 3 = 0.510 MPa = 0.510 N/mm2 For structural steel, allowable = 120MPa

circumferential allowable = pd/2t

Then, thickness, t = 0.510 x 8000 / (2 x 120) = 17mm Therefore, available thickness 20mm 25mm

Friction loss The loss of head (or energy) in pipes due to friction is calculated from Darcy-Weisbach equation

Friction loss, hf = 4 f L V2 / 2gD

Where, L = length of pipe (or penstock) = 200 m

V = velocity of water inside pipe = 4.2882 m/sec D = diameter of pipe = 8 m

g = 9.81 m/sec2

f = co-efficient of friction, which is a function of Reynolds number, Re

= 16/Re for Re < 2000 (viscous flow)

= .079/ Re1/4 for Re varying from 4000 to 106

Reynolds number, Re = VxD / where, = kinematic viscosity

= 4.3767 x 8 / (.01 x 10-4) [for water = .01 poise or cm2/sec]

= 35013600 = 35.0136 x 106

f = 079/ Re1/4 = .079/ (35.0136 x 106)1/4

= .00102699

Now, hf = 4fLV2/2gD

= .100267 m

Spillways They protect Reservoir level which indirectly ensures continuous hydel operations.

Purpose of spillway gates-

• To control excess reservoir water (Flood control)

• To protect the dam

• To maintain safe reservoir level They operate-

• during monsoon

• during floods

• when reservoir exceeds Full Reservoir Level (FRL)

  Spillway gates are meant for flood control and dam safety, they do not participate in power generation.

  Intake Gate It is large vertical steel gate which moves UP & DOWN housed in a concrete intake tower. It is located on the reservoir side at lower elevation than Spillway behind trash rack and before Penstock.

  Reservoir Trash rack Intake gate Penstock Turbine

  It is operated slowly to START / STOP water supply to the turbine and also act as Emergency shut- off.

  Guide Vanes or Wicket Gates Wicket gates are located inside the turbine casing and continuously control power output. These look like many curved blades arranged in a circle moving together connected to a governor ring.

  Penstock Spiral casing Wicket gates (guide vanes) Turbine runner Draft tube

  It act as real throttle of the turbine and has following purposes-

  1. control how much water enters the runner

  2. control angle of attack

  3. maintain constant speed

  4. regulate power output

     Simple Logic

     • Reservoir fuel tank

     • Spillway gate overflow pipe

     • Intake gate fuel cock (ON / OFF)

     • Wicket gate accelerator

     • Turbine runner engine

  Intake trash rack

  A trash rack is a steel bar screen installed in front of the intake opening to prevent debris from entering the penstock / turbine.

  At Obra it protects against-

• tree branches

• floating vegetation

• plastic / debris

• fish (to some extent)

  Trash rack is always submerged usually and not visible from the dam top.

  Location at Obra- Intake is below FRL which is separated from spillway and Penstock is embedded inside dam So the sequence is-

  Reservoir

  ~~~~~~~~~~~~~

  | Trash rack | submerged, vertical

  | |

  | Intake gate|

  | |

  | Pressure conduit (penstock)

  Discharge passing through ONE intake = 220m3/sec Total Discharge = 3 x 220 = 660m3/sec

  Turbines

  Obra Hydel Power plant uses Reaction type Kaplan turbine.

  Turbine dimensions (runner size)-

  For Kaplan turbines, runner diameter depends mainly on head.

• Low head large runner

• High head small runner

For Head = 17 m

and Power = 33 MW Size of intake openings

Intake opening velocity is kept low to avoid clogging, vibration, fish injury

Typical intake velocity, V = 1 m/sec 1.5 m/sec 1.2 m/sec [remarkably less than penstock velocity]

Discharge per turbine, Q = 220 m3/sec

Required intake flow area, A= Q ÷ V = 220 ÷ 1.2 = 183.33m2

Overall Efficiency of the turbine, o = Shaft Power ÷ Water Power = Power developed ÷ gQH

Given, P = 33 MW

Water Power = 1000 x 9.81 x 220 x 17 = 36689400 watt = 36.68MW

o = P ÷ gQH

o = 33 ÷ 36.68 = 0.89944 90%

at Obra, Runner diameter, Do = 5.17m Diameter of hub, Db = 35% of Do = 1.8095 m In Kaplan turbine,

Velocity of flow at inlet, Vf1 = Velocity of flow at outlet, Vf2

Q = (Do2 Db2) Vf1 ÷ 4

Vf1 = 11.942756m/sec 12m/sec

Flow ratio = Vf1 ÷ (2gH) = 11.942756 ÷ (2×9.81×17)

= 0.65

Peripheral velocity at inlet, u1 = Peripheral velocity at outlet, u2

u1 = Do N ÷ 60

for Kaplan turbine, peripheral velocity, u1 = (2gH)

where, = speed ratio = 1.6 to 2.0 [it is different from flow ratio]

u1 = 1.7x(2×9.81×17) [taking 1.7]

= 31.047 31m/sec

Speed of the Turbine, N = u1x60 ÷ Do = 31×60 ÷ 5.17

= 114.5176 115rpm

Specific speed of the turbine, Ns = N P ÷ H5/4

Where, N = turbine rpm / rotational speed (rpm) P = power (kW)

H = net head (m) Ns = 115×33000 ÷ 171.25

= 605.1926

C:—————— llvllrnull< M11d1inf".. T\Jrb1mt1 8751

8;,; ,peed of turbine1,263.72. And') n hmnou, peed (,Y.) ". cquolio 2 0.I fence. the peed of 1urb1nc

,,n..:hnlf'loo, TllC ,peed ot turbine ,hould 250r p.m

,,.9 AXIAL FLOW REACTION TURr! N .

JJ "Jtcr flo"., parallel.to the .t\1, of the rorauon of the shnn. the turbine i known as axial ow turbine.

Ifth<' he.1d .11 the inlet of the lurbme 1, 1hr ,um ofprc\surc energy nnd kincuc energy and dunng the now

l,,nf_J.Jfrt: rhn>Ugh nmnerJ pan of prc,,urc energy i convened intokinetic energy the turbine is known os

,..,J,,."{k)IJ rurbine.

,. fl)( rh< J.uol no" reacuon turt>ine. the ,haft of the turbine is vertical. Thelower end of theshaft is made

hich 1, kno"n :t> ·hub· or ·oo»· The vanes arc fixed on the hub and hence hub acts asa runner for Jo n,Jcuon rurtnne. The following are the imponan1 rypc of axial now reaction turbines:

f. Propeller Turbine. and

: Kapl..n Turt>ine.

lll>en rlH! ,..nc, an, fixed ro the hub and rhey an, nor adjus1-

lt.the rurbme ,, lnov. n as p pc!ler turbine. But if the vanes on

!btbub are adJu,rable. the rurbme rs known as a Kaplan Turbine. j,rrh< nl/llc of\ Kaplan. an Austnan Engineer. This turbineis uble "herea large qu:muty of warer ar low head is available.

:r 1 15 ,.holl) the runner of a Kaplan turbine, which consists

-, bob fixed 10 tbe shaft. On the hub. the adJustable vanes are

,e,Ja,,bo"nmFig. 1825.

n,, nwn p.ut> of a Kapl:m turbine are :

1 Scroll casmg.

1 Guide \ ane.!. mechanism,

J Hub 11.11h ,:me; or runner of rhe turbine. and

Draft rube.

rg 18.26 show; all mam pans of a Ian 1urb1ne. The warer from rock enter; !he ;croll casing and mo,e; 10 the guide vanes. From rhe 1ane,. the warer rums through 90° flo"; axiall> through rhe runner as

·o rn Fig. 18 26. The discharge

gb rbe runner is obtained as

Q=f(D;-Dnxvt,

…(18.25)

D0 = Ou1er diameter of !he

runner,

Fig. 18.25 Kapl,,n t11Tbine rwnner.

SHAFT SCROLL CASING

-,

DRAFT

TUBE

Db = Diameter of hub. and

= Velocity of flow at inlet. inlet and ou1le1 velocity triangles

11.n at the extreme edge of the vane corresponding 10 the points as shown in Fig. 18.26.

'

'

Fill. 18.26 Main components of Kaplan turbie.

Switch Yard

POWER GENERATION

99MW it is the maximum instantaneous output when the plant is fully loaded. Number of turbines = 3

3 hydro turbinegenerator units

Each unit 33 MW

total = 3x33MW = 99MW (installed capacity) Energy is calculated as:

Energy (MWh)=Power (MW)×Time (hours)

If, Case-1 all 3 turbines ON- each at 33 MW- running for 1 hour Energy generated = 99×1=99 MWh

That is 99,000 units (kWh) in 1 hour

Case-2 if run on full load for 10 hours Energy generated = 99×10=990 MWh That is 9.9 lakh units in 10 hours

Case-3 Typical real operation (not ideal)

Hydel plants rarely run full load all day.

If 2 turbines ON = 66 MW, running for 6 hours Energy generated = 66×6=396 MWh

That is 3.96 lakh units

Why Obra Hydel does NOT run 24×7 at 99 MW?

Hydel plants are used mainly for-

• Peak load

• Morning & evening demand

• Grid frequency control They depend on-

• reservoir level

• irrigation demand

• thermal plant coordination So actual operation is-

• few hours at high load

• few hours at partial load

• sometimes shutdown

  Actual electrical path at Obra Hydel-

  Turbine (mechanical)

  Generator output 11 kV

  Step-up transformer

  33 kV (plant level)

  Switchyard

  Further step-up (132 / 220 kV) Grid

  Generator

  Each turbine has its own 11kV- hydro umbrella shaped Vertical-shaft synchronous generator connected via shaft

  Generator power rating (per unit)

  Since Turbine power = 33 MW, Generator is always rated slightly higher.

  Typical margin 510%

  So generator rating 3537 MW per unit

  Generator voltage, Vg 11 kV Why not higher?

  • insulation limits

  • rotor stress

  • cooling constraints

    Calculations

    Poles count Since generator is directly coupled to turbine via shaft so formula-.

    N = 120f ÷ p

    Where, N = shaft rpm = turbine rpm f = Grid frequency

    p = .number of poles

    p = 120×50÷115 = 52.1739

    = 52

    Rotor Diameter In umbrella type hydro generator, rotor diameter is usually larger than runner diameter.

    Drotor 1.5 to 1.8 times runner diameter Drotor = 1.6×5.17 = 8.3m

    Pole Pitch

    = Drotor ÷ p = 8.3÷52

    = .50144m 500mm

    This value is comfortable, optimum for field coil placement and mechanically strong

    Generator Torque

    Power, P = T

    Angular speed, =2N÷60 = 2 x 115 ÷ 60

    = 12.04277 rad/sec

    Torque, T = P÷ = 33×106 ÷ 12.04277 = 2.74023 x106 N-m

    = 2.74 MN-m

    This Torque is very high torque because speed is low (115 rpm)

    Low speed High torque High speed Lower torque

    Thats why hydro generators (low rpm) have-

• Large shaft diameter

• Massive thrust bearing

• Huge rotor diameter

• Salient pole construction

  Generator height (axial length)

  Generator height depends on-

  1. Rotor rim + pole height

  2. Stator core height

  3. Air gap

  4. Thrust bearing arrangement

  5. Top cover + spider

  6. Brake system

  7. Excitation system In umbrella type-

• Thrust bearing is above the rotor

• Generator is compact vertically compared to conventional vertical type

  Typical Proportions for Low-Speed Hydro Generators- For large low-speed salient pole machines

• Rotor diameter- 79 m

• Rotor axial length (stack height)- 1.21.8 m

• Pole height- 11.5 m

• Air gap- 2040 mm

• Total generator height- usually 0.6 to 0.8 × rotor diameter

  Here, Rotor diameter = 8.3 m

  Practical height ratio 0.7

  Height = 0.7×8.3 = 5.8m

  Height Reference

  Similar Kaplan unit of Obra was manufactured by H.E. (I) ltd [which is now BHEL] in 1965

  Presently For 3040 MW, 100150 rpm turbine BHEL & NHPC (National Hydroelectric Power Corporation) keep the generator height around 5 m to 7 m, which matches the calculation.

  Generatorturbine alignment

  Vertical stack (top to bottom): Generator rotor

  Generator stator

  Thrust bearing carries weight of runner + shaft

  Main shaft

  Kaplan runner

  Draft tube

  This explains- very tall powerhouse, deep foundations, heavy crane capacity

  Where the transformer is placed at Obra?

  Physical location- Just outside the powerhouse, each generator has its own step-up transformer called Generator Transformer (GT).

  Connection-

  Turbine Generator (11 kV) Generator Transformer 33 kV bus Transmission lines

  Voltage levels are staged-

  11 kV 33 kV 132 / 220 kV

  Why specifically 33 kV (and not directly 132 kV)? It is because 33 kV is used for- nearby substations, local/regional transmission, inter-connection with thermal plant network. Further step-up is done at the switchyard, not at the generator.

  Technically we can but practically there are reasons.

  Reason 1 Switchyard & Local Distribution Often 33 kV is used for-

• 33 kV is used for local distribution

• Plant auxiliaries may operate at 6.6 kV / 11 kV / 33 kV

• Nearby industries use 33 kV So 33 kV bus is useful.

  If you directly go 132 kV:

• You still need a 33 kV transformer for local supply

• So intermediate voltage becomes necessary anyway

  Reason 2 Transformer Rating & Flexibility

  Instead of one large 11/132 kV transformer, Plants often use-

• 11/33 kV generator transformer

• Then 33/132 kV grid transformer

  It provides- better protection coordination, easier maintenance, modular expansion, fault isolation

  Reason 3 Fault Level Control In direct 11/132 kV transformer

• High short circuit

• Higher fault currents on 132 kV side

• Protection complexity increases Intermediate 33 kV bus helps in system segmentation.

  Reason 4 Insulation Stress

  11 kV 132 kV is large ratio (1:12), such transformer requires-

• Larger insulation requirement

• More expensive

• Bigger bushings

• Higher impulse stress

  Two-step transformation reduces insulation gradient per stage.

  One clean diagram

  Kaplan Turbine

  Generator (11 kV)

  Generator Transformer (11/33 kV)

  33 kV switchyard

  Power Transformer (33/132 kV)

  Grid

  Final Summary

  RIHAND RESERVOIR

  (Water)

  HYDRO TURBINE

  (Mechanical Power)

  GENERATOR (~11 kV)

  11 kV BUS

  STEP-UP TRANSFORMER (11 kV / 33 kV)

  33 kV BUS

  33 kV FEEDER 33 kV FEEDER

  (Local Grid) (Sub-station/ Fields)

  33 kV SUB-STATION

  11 kV 132 / 220 kV

  (Local) (Long distance)

  OBRA THERMAL POWER PLANT

  Introduction

  Obra Thermal Power Plant (OTPP) is a coal-based thermal power station located in Obra, Sonbhadra district, Uttar Pradesh. It is one of the major power-generating stations of UP and plays a crucial role in supplying electricity to the state.

• Owner & Operator Uttar Pradesh Rajya Vidyut Utpadan Nigam Limited (UPRVUNL)

• Type Coal-fired thermal power plant

• Boiler Water tube boiler, firing by- pulverized coal

• Turbine Steam turbine

• Primary fuel Coal

• Cooling system Water-based cooling

Installed Capacity

The plant consists of multiple units developed in phases. Obra A 5x200MW = 1000MW

+ 5×50 = 250 MW

+ 3×100 = 300 MW

Total Capacity = 1550 MW Present Capacity = 1288MW

ObraB units are supercritical technology based, making them more efficient and environment-friendly

compared to older units

Location advantage

• Located near coal belt of Eastern India

• close to railway (DHN Division, ECR) connectivity for coal transport

• availability of cooling water

• Sonbhadra is known as the Energy Capital of Uttar Pradesh

Working Principle

1. Coal is burned in the boiler

2. Heat converts water into high-pressure steam

3. Steam rotates the steam turbine

4. Turbine drives the alternator (generator)

5. Electricity is stepped up by transformers

6. Power is sent to the grid

   Major Systems in Obra TPS

   • Coal Handling Plant (CHP)

   • Boiler (Pulverized coal fired)

   • Steam Turbine

   • Generator

   • Condenser

   • Cooling Tower

   • Ash Handling Plant

   • ESP (Electrostatic Precipitator) for pollution control

     Environmental Control Measures

   • ESP to control fly ash

   • Ash ponds for ash disposal

   • High efficiency boilers

   • Supercritical units to reduce coal consumption

   • Plantation around plant premises

PLANT LAYOUT

1. Boiler side (Right side of sketch)

   Boiler (Water-tube, Pulverized coal fired)

   • Coal is burnt in the furnace

   • Water flows inside tubes

   • Steam generated is high pressure + high temperature

     Economiser (Before boiler)

   • Feed water from pump Economiser

   • uses waste heat of flue gas

   • raises water temperature before entering boiler

     Purpose Fuel saving + increasing efficiency

     Superheater (After steam generation)

   • Saturated steam superheated steam

   • Temperature raised to ~ 535°C

     Why needed? Dry steam prevents turbine blade damage.

     Reheater (Between HP & IP turbine)

   • After HP turbine, steam pressure drops

   • Steam sent back to boiler reheater

   • Temperature raised again

   • Sent to IP turbine

     Purpose avoid moisture + increase turbine life

2. Turbine section (Upper middle of sketch)

   The drawing shows three turbines in series-

   HP Turbine High Pressure Turbine

   • Receives superheated steam

   • High pressure, lower volume

   • Produces first stage power

     IP Turbine Intermediate Pressure Turbine

   • Steam comes from reheater

   • Medium pressure Further expansion

     LP Turbine Low Pressure Turbine

   • Low pressure, very large volume steam

   • Long blades

   • Extracts last usable energy

     All turbines are on the same shaft connected to Generator

3. Generator & Electrical side (Left / bottom)

   Generator

   • Turbine shaft generator rotor

   • Mechanical energy electrical energy

     Station Transformer

   • Generator voltage ~ 1115 kV

   • Step-up to- 220 / 400 kV

   • Power sent to- Switchyard

   • Then to- Grid

     one of the reasons why Switchyard is outside the plant boundary.

4. Condenser system (Right bottom of sketch)

   Condenser

   • Exhaust steam from LP turbine

   • Cooled by circulating water

   • Steam water (condensate)

     Why condenser is critical?

   • Creates vacuum

   • Increases turbine efficiency

   • Allows reuse of water

     Condensate Extraction Pump (CEP)

   • Takes water from condenser

   • Sends to LP heaters

5. Feed water heating system

   LP (Low Pressure) Heaters

   • Heated using extraction steam from turbine

   • Raise water temperature gradually

     Deaerator (DA)

   • Removes O and CO

   • Prevents corrosion

   • Also acts as storage tank Boiler Feed Pump (BFP)

   • Raises pressure to boiler pressure

   • One of the highest power-consuming pumps HP Heaters

   • Final heating before economiser

   • Uses turbine extraction steam

     This whole process is called Regenerative Feed Heating

6. Cooling tower & circulating water (Left side)

   Cooling Tower

   • Hot water from condenser

   • Cooled by air

   • Sent back to condenser

     This is a separate loop from boiler water.

7. Air & flue gas path (Lower middle of sketch)

   FD Fan

   • Pushes air into boiler

   • Supports combustion

     PA Fan

   • Carries pulverized coal

   • Primary air for combustion

     ID Fan

   • Pulls flue gas out

   • Maintains negative pressure in furnace

     ESP (Electrostatic Precipitator)

   • Removes fly ash

   • Protects environment

   • Gas exits through chimney

8. Coal handling path

Coal yar Crusher Pulveriser PA fan carries coal-air mixture Boiler furnace

Complete cycle summary

Coal heats water

steam runs turbines

power generated

steam condensed

water reheated

reused

cycle continues

l Stearn Flow

Water l

Boiler Feed Pumps Boiler Feed Pumps Condensate Extraction Pumps

Wat@r & Cooling Water " Coal,AlraGas -ii,, Coal Flow Cooling water _., Coal, Air, & Flue Gas Flow

PLANT WALKTHROUGH

A thermal power plant can be considered as a very advanced steam engine running in a loop. The plant layout lies in 4 parts-

1. How do we make steam?

2. How do we use steam to make electricity?

3. How do we convert steam back into water?

4. How do we reuse that water safely and efficiently?

   STEP 1 Making fire and heat (Coal Boiler)

   • Coal of size 250-300mm is brought from the coal yard via Conveyor belt

   • It is crushed to 20mm bowl mill which makes it upto 70-75

   • This fine powdered coal is blown into the boiler furnace

   • Coal burns huge heat is produced

     STEP 2 Raw Water System

     Reservoir / River Intake It is the source of water for-

   • Cooling tower

   • DM plant

   • Service water

     Working-

   • Pumps lift water from reservoir

   • Sent to raw water treatment plant

     Raw Water Treatment Plant (Clarifier + Filters) Its main function is to remove mud, suspended solids, organic matters

     Working-

   • Alum dosing particles settle

   • Clarifier removes sludge

   • Sand filters polish water

     Now water is clean but still contains dissolved salts.

     DM (Demineralization) PLANT DM Plant (RO + Ion Exchange) Working-

   • RO system- removes 9598% dissolved salts

   • Cation exchanger- removes positive ions (Ca², Mg²)

   • Anion exchanger- removes negative ions (Cl, SO²)

   • This prevents- Scale formation, Turbine blade deposits

     Output Ultra pure water (conductivity < 0.2 µS/cm) goes to boiler system.

     STEP 3 Feed water System

     Deaerator (DA) Remove dissolved oxygen and CO

     Working-

   • Steam sprayed over water

   • Heating drives out gases

   • Vent removes oxygen Why? Because oxygen causes:

     4Fe+2H2O+O2 4Fe(OH)2 Rust formation

     Even 510 ppb oxygen can cause pitting in high-pressure boilers, deaerator reduces oxygen to less than 7 ppb

     Chemical Dosing (Final Protection) After DM + Deaeration, chemicals are added

     Hydrazine or oxygen scavengers Removes trace oxygen NH3 maintains pH around 99.5

     Why alkaline? Because steel is stable in alkaline conditions, Acidic water causes corrosion.

     Why doesnt the boiler corrode continuously? Because inside the boiler-

   • High temperature

   • High pressure

   • Controlled pH

   • Very low oxygen

   • Protective magnetite (FeO) layer forms This thin oxide layer protects tubes.

     If minerals are present scale forms overheating tube burst

     If oxygen present pitting corrosion

     Between these two Oxygen is far more dangerous than pure water.

     Real Corrosion Scenario in Plants Corrosion usually happens due to-

   • Oxygen ingress during shutdown

   • Improper layup

   • Condenser tube leakage

   • Low pH excursion

   • Poor deaerator performance

     Boiler Feed Pump (BFP) Raise pressure of water to boiler pressure (~130170 bar)

     Working-

   • Multi-stage centrifugal pump

   • Motor or turbine driven

   • Pushes water through heaters to boiler Very high power consumption.

     Low Pressure (LP) Heaters Preheat feedwater using extracted turbine steam

     Working-

   • Steam from LP turbine heats water

   • Improves efficiency

     High Pressure (HP) Heaters Final heating before boiler

     Working-

   • Steam extracted from IP turbine

   • Raises water temperature further

This process is called Regenerative Heating.

STEP 4 Turning water into very powerful steam (Inside the Boiler)

Inside and around the boiler, three important things happen-

1. Economiser Preheat feedwater using flue gas heat

   Working-

   • Flue gases pass over tubes

   • Water temperature increases

   • Fuel saving

2. Boiler furnace convert water into steam

   Working-

   • Pulverized coal burns

   • Water flows inside tubes

   • Heat transfer occurs

   • Steam forms

3. Superheater Increase steam temperature above saturation

   Working-

   • Steam passes through hot gas path

   • Temperature raised to ~535°C

   • Prevents moisture in turbine

4. Reheater Reheat steam after HP turbine

   Working-

   • Steam returns from HP turbine

   • Heated again

   • Sent to IP turbine

   • Improves turbine life.

     STEP 5 Turbine System

     HP Turbine High Pressure turbine, First stage energy extraction

     Working-

   • High-pressure steam enters

   • Rotates blades

   • Pressure drops

     IP Turbine Intermediate expansion

     Working-

   • Reheated steam expands

   • More energy extracted

     LP Turbine Low Pressure turbine, Final expansion stage

     Working-

   • Low-pressure high-volume steam

   • Long blades

   • Maximum energy extraction

     All turbines connected to same shaft

     STEP 6 Generator & Electrical

     Generator (Alternator) converts mechanical energy electrical energy

     Working-

   • Shaft rotates rotor

   • Magnetic field rotates

   • Electricity induced in stator

     Output 1115 kV

     Step-up Transformer increase voltage to transmission level

     Working-

   • 11 kV 220/400 kV

   • Reduces transmission losses

     Switchyard Send electricity to grid

     Working-

   • Circuit breakers

   • Isolators

   • Busbars Transmission lines STEP 6 Condensing & Cooling Condenser convert steam back to water Working-

   • Steam contacts cold tubes

   • Condenses

   li data-list-text=””>

   Vacuum maintained

   • Improves efficiency.

     Circulating Water Pump (CWP) Pump cooling water through condenser

     Cooling Tower Cool hot water from condenser

     Working-

   • Hot water sprayed

   • Air flows upward

   • Some water evaporates

   • Remaining water cools

   • Recycled back

STEP 7 Air & Flue Gas System

FD Fan (Forced Draft) Pushes air into boiler

PA Fan (Primary Air) Carries pulverized coal

ID Fan (Induced Draft) Pulls flue gas out and maintains negative pressure

ESP (Electrostatic Precipitator) Removes fly ash particles

Chimney Releases cleaned gases to atmosphere

Working of ESP

Induced Draft Cooling Tower- work-diagram

STEP 8 Ash Handling

Bottom Ash System Collects heavy ash from furnace

Fly Ash System Collects fine ash from ESP Ash pond Cement industry

Final Simplified Flow

Raw water DM plant Deaerator Pump Heaters Boiler Turbine Generator Transformer

Grid

Steam Condenser Cooling tower Recycle

Calculations

1. Water Requirement Water requirement can be breakdown in 4 parts-

   1. How much water is needed initially?

   2. How much steam is produced?

   3. Where losses happen?

   4. How much steam/water is actually lost?

   1. How much water is needed for one 200 MW boiler?

      A 200 MW subcritical coal unit typically requires-

      Steam generation rate 650750 tonnes per hour (TPH) = 700TPH

      which means- 700 tonnes/hour = 700,000 kg/hour of water boiler converts into steam every hour

      Initial filling requirement Before starting from cold condition-

      • Boiler drum + water walls + piping + economiser

      • Condenser hotwell

      • Deaerator storage tank

      • Feedwater lines

        Roughly, 1,500 2,000 tonnes of water initially

        This is one-time filling, not continuous consumption. Initial filling depends on- Total internal water holding volume

        + Size of condenser hotwell + Deaerator storage capacity + Boiler circulation volume

        It is not directly equal to steam generation rate, but proportional to plant size.

   2. What happens to 700 TPH steam?

      Steam path Boiler HP Turbine Reheater IP LP Condenser back to boiler

      In ideal condition, almost all 700 TPH is reused but there are losses (real plant ideal plant)

   3. Where losses happen (step-by-step)

   In the direction of flow

   1. Boiler losses Blowdown loss

      To remove dissolved solids typically 13% of steam generation 2% (assume) 700×0.02=14 TPH

      Loss = 14 tonnes/hour

   2. Steam leakages From- Valve glands + Flanges + Start-up vents + Drains 0.5 1% = 1% (assume) 700×0.01=7 TPH

      Loss = 7 tonnes/hour

   3. Turbine gland steam loss Steam used for sealing turbine shaft 0.5%

      700×0.005=3.5 TPH

      Loss = 3.5 tonnes/hour

   4. Auxiliary system drains Sampling + Chemical dosing + Heater drains 12% = 1% (assume) 700×0.01=7 TPH

      Loss = 7 tonnes/hour

   5. Cooling tower evaporation loss This steam is NOT loss directly from boiler, but cooling water evaporation. For 200 MW unit, Cooling water circulation 25,00030,000 m³/hr

   Evaporation loss formula, E= 0.00085×C×T

   Where, C = Circulating water (m³/hr) =30000 m³/hr (assuming)

   T = Temperature rise (°C) 10 °C (outlet water may rise by 810°

   C)

   Evaporation loss, E = 0.00085×30000×10 = 255 m³/hr

   1% 300m³/hr water loss *this is makeup water requirement

   Total steam/water loss per hour = 30.5 TPH 30 tonnes/hour 4%

   Therefore, Steam reused 670 TPH 96%

   & Makeup water requirement 30 tonnes per hour

   + Cooling tower evaporation 300400 m³/hr

   Therefore, totals plant water makeup = 300m³/hr + 30TPH (300 + 30) m³/hr

   350 m³/hr *for 200 MW unit

   Important Concept The boiler does NOT consume 700 tonnes of water every hour, it circulates 700 tonnes per hour in a loop. Actual fresh water consumption is only ~4% of steam flow.

2. Steam

   Steam turbine works on Rankine Cycle

   Main processes Standard numbering:

   1. Turbine inlet (superheated steam)

   2. Turbine exit

   3. Condenser exit (saturated liquid)

   4. Pump exit (compressed liquid entering boiler)

STEP 1 Realistic Steam Conditions

For 200 MW unit

Main Steam Pressure = 130 bar (safe for steel structures) Temperature = 535°C

Condenser pressure= 0.1 bar (~45°C saturation)

STEP 2 Steam Table & MOLLIER DIAGRAM

At Turbine Inlet 130 bar & 535°C

by MOLLIER DIAGRAM, Specific Enthalpy, p = 3460 kJ/kg

Specific Entropy, s1 = 6.58 kJ/kg K

At Condenser Pressure (0.1 bar)

According to STEAM TABLE 1, saturated water and steam (temperature) table Absolute pressure, p = 0.10080 bar

Temperature = 46°C

Specific enthalpy of water, hf = 192.5 kJ/kg Specific enthalpy of steam, hg = 2585.1 kJ/kg Specific entropy of water, sf = 0.651 kJ/kg K Specific entropy of steam, sg = 8.148 kJ/kg K

STEP 3 Find Turbine Exit State (Isentropic Expansion)

Since ideal expansion, s2 = s1 = 6.58

Quality equation, s2 = sf + x(sg sf)

6.58 = .651 + x(8.148 .651)

Therefore, x = 0.790849

So steam at exit is 79% dry

For exit enthalpy, p = hf + x(hghf)

p = 192.5 + .790849(2585.1 192.5)

Therefore, p = 2084.685317 kJ/kg

STEP 4 Turbine Work

Wt = p p = 3460 2084.685317

= 1375.314683 kJ/kg

STEP 5 Pump Work

Pump work, Wp 10 kJ/kg

Net work, Wnet = 1375.314683 10 = 1365.314683

1365 kJ/kg

STEP 6 Heat Supplied in Boiler

Qin = p h4

Where, p = Enthalpy of water at turbine inlet = 3460 kJ/kg h4 = Enthalpy of water at pump outlet = p + Wp

p = Enthalpy of saturated water leaving the condenser

According to STEAM TABLE 2, saturated water & steam (pressure) table At Absolute pressure, p = 0.100 bar

p = 191.8 kJ/kg Therefore, h4 = p + Wp = 191.8 + 10 = 201.8

200 kJ/kg

Finally Qin = p h4 = 3460 200

= 3260 kJ/kg

STEP 7 Rankine Cycle Efficiency

Efficiency, = Wnet ÷ Qin

= 1365 ÷ 3260 = 41.8711%

Therefore, 41.87%

This is Ideal Rankine efficiency

Now apply real-world losses

• Turbine efficiency 85%

• Generator efficiency 98%

• Boiler efficiency 8590%

• Mechanical losses

Actual overall plant efficiency = 41.87% × 0.85 × 0.9 = 32.03055%

Overall Plant efficiency = 32%

Conclusion

Ideal thermodynamic efficiency 42%

Actual plant efficiency 32 33%

Loss difference 910% due to real-world irreversibility

A thermal plant is not inefficient because Engineering failed it is limited by the Second Law of Thermodynamics.