Power plant and calculations: How to calculate the capacity of Sugar based Cogeneration plant??

How do you design a Thermal power plant?????

Capacity of the Cogen plant is nothing but the power and thermal energy (heat) generation capacity of the particular plant.

Cogen plant capacity calculation involves deciding the capacity of the Boiler, steam Turbine, WTP and other auxiliaries like cooling tower, fuel handling plant etc.

Inputs required for Cogen plant capacity calculation.

1-Sugar plant capacity

2-Sugar plant electrical power demand

3-Sugar plant thermal load (Steam) demand

4-Percentage of fuel (bagasse) generated from the Sugar plant

5-No. of operation days of the Sugar plant.

6-Price per unit for excess power export to the Grid.

7-Area available

8-Future expansions in sugar plant

Let us consider the Sugar plant capacity 5000 TCD for demonstration.

A-Steam demand

1-Steam for sugar plant.

Sugar plant capacity: 5000 TCD or TPD (Tonnes per day)

Usually sugar process needs low pressure saturated steam for juice heating.

Required Steam parameters are;

Pressure: 1.5 to 2 Kg/cm2

Temperature: 125 deg C to 140 deg C

Consider thermal load i.e steam demand for Boiling house (Juice heating) is 40% of Sugar cane crushing.

Then steam required to sugar process = 5000 X 40% = 2000 TPD

Steam flow per hour = 2000 / 24 = 83.34 = 84 TPH

Rounding the figure, steam flow to sugar Boiling house = 85 TPH

2-Steam for distillery plant.

Generally, molasses-based distillery plant consumes around 3 to 3.5 kg of steam to generate 1 litre of ethanol.

So, total steam required for distillery process plant =Distillery plant capacity X 3.5

= 60 KLPD X 3.5 = 210 TPD = 8.75 TPH

Therefore, total steam demand for sugar and distillery process = 85 + 8.75 = 93.75 TPH

B-Power demand

The auxiliary power consumed per hour by any sugar plant as a Thumb rule = Cane crushing capacity X 80%

i.e total power required for sugar plant process = 5000 X 80% = 4 MW/day

Total power required for sugar plant considering 10% extra margin = 4 MW X 110% = 4.4 MWH

Other power consumption units related to sugar process are;

Employee & labour Colony power consumption = Sugar plant power consumption X 4 to 5%

= 4400 X 4 to 5% = 170-220 KWH = 0.17 to 0.22 MW

Consider Distillery plant of capacity = 60 KLPD for 5000 TCD sugar process.

Power consumption of distillery plant = Sugar plant power consumption X 20-25%

Power consumption of distillery plant = 4400 X 20-25% = 0.88 to 1.1 MW

Therefore, total power consumption of Sugar process and related auxiliaries is = 4.4 + 0.22 +1.1=5.72 MWH

C-Bagasse or fuel generated from sugar mill

Generally, an average of 28 to 29% bagasse is generated from sugar mill at various cane crushing conditions.

Therefore, bagasse generated from Sugar mill =5000 X 29% = 1450 TPD = 60.42 TPH.

D-No. of operation days of the Sugar plant

Based on current availability of sugar crop, sugar plant runs around 120 days in a year.

Cogeneration plant Capacity calculation

Let us decide the capacity of the power plant major equipment / auxiliaries.

Major equipment auxiliaries in the power plant/cogeneration plant are;

1-Boiler

2-Steam Turbine

3-Alternator

4-Lube oil system

5-Export transformer

6-Auxilary Transformer

7-Boiler feed pumps

8-Boiler fans

8-Air compressor units

9-Bagasse handling plant

10-Water treatment plant

11-Cooling tower and cooling water pumps

12-ESP

13-Chimney/stack

Let us decide the Boiler operating pressure and temperature for this plant.

Now a days high pressure and high temperature Boilers are selected for economical point of view.

High pressure boilers are more efficient and consume less fuel and also steam Turbines operating on high pressure steam consume lesser steam to generate unit power.

1-Calculation of Boiler Capacity.

Let us consider the operating pressure be 110 Kg/cm2 and temperature 540 deg C.

In design let us add HP heaters to increase the cycle efficiency.

Assumptions:

Return condensate from Sugar and distillery process = 95%

Total steam load = Process steam + Deaerator steam + HP heater steam + Exhaust steam

Note: During cane crushing season exhaust steam to condenser is very minimum, it is around 12 to 15% of Turbine inlet steam & during off season operation i.e. during condensate mode operation, turbine exhaust flow will be around 70% of Turbine inlet steam.

During cane crushing operation, steam flow to the Deaerator is least and is around 10% of total steam extracted from Turbine.

As a thumb rule,

TG inlet steam = Extraction steam / 70%

Extraction steam flow = Process steam flow + Deaerator steam flow

Extraction steam flow= 85 + Process steam flow X 10% = 85 + 85 X 10% = 93.5 TPH

Therefore, TG inlet steam = 93.5 / 0.7 = 133.5 TPH

Boiler capacity = 135 TPH-------------------------------------------------------1

Mass balance & recalculation of Boiler capacity

Total steam load = 93.75 TPH + Process steam X 10% (Thumb rule) + HP-1-Turbine inlet steam X 6% + HP-2-Turbine inlet steam-2 X 10% + Exhaust steam to Condenser-15%

Extraction steam flow = TG inlet steam X 70% = 133.5 X 70% =93.45 TPH

Deaerator steam flow = Process steam flow X 10% = 85 X 10% =8.5 TPH

Bleed-1 steam/HP heater-1 steam flow = TG inlet steam flow X 6% = 133.5 X 6% =8.01 =8.0 TPH

Bleed-2 steam flow (HP heater-2 & Distillery process) = TG inlet steam X 15% = 133.5 X 15% = 20.02=20 TPH

Exhaust steam flow = TG inlet steam flow X 15% = 133.5 X 10% = 13.35=14 TPH

TG inlet steam flow = 93.45+ 8 + 20 + 14 = 135.45 =135 TPH

Based on above thumb rules, Boiler of capacity matches 135 TPH as per Equation---I.

Boiler fuel (Bagasse) consumption:

At the average SFR (steam to fuel ratio) 2.4 Boiler bagasse consumption = 135 / 2.4 =56.25 TPH

So, bagasse saving per day = (Sugar mill bagasse generation-Boiler fuel consumption) X 24

= (60.42-56.25) X 24 = Appx. 100 MT/day.

% of bagasse saving = 100 X 100/(60.42 X 24) =6.9%

Ideally, bagasse saving will be around 5 to 8% based on moisture in the fuel and steam consumption of the sugar process.

2-Turbine capacity calculation

Let us consider the following parameters for Turbine power generation calculation

Sl No.	Particular	UOM	Value
1	Turbine inlet steam flow	TPH	140
2	Main steam pressure	Kg/cm2	105
3	Main steam Temperature	Deg C	540
4	Main steam Enthalpy	Kcal/kg	828
5	Bleed-1 steam flow	TPH	8
6	Bleed-1-steam pressure	Kg/cm2	21
7	Bleed-1 steam Temperature	Deg C	330
8	Bleed-1 steam Enthalpy	Kcal/kg	738
9	Bleed-2 steam flow	TPH	20
10	Bleed-2 steam pressure	Kg/cm2	12
11	Bleed-2 steam Temperature	Deg C	250
12	Bleed-2 steam Enthalpy	Kcal/kg	703
13	Extraction steam flow	TPH	93.45
14	Extraction steam pressure	Kg/cm2	3
15	Extraction steam Temperature	Deg C	140
16	Extraction steam Enthalpy	Kcal/kg	645
17	Exhaust steam flow	TPH	14
18	Exhaust steam pressure	Kg/cm2	0.087
19	Exhaust steam Temperature	Deg C	43
20	Exhaust steam Enthalpy	Kcal/kg	611

Assumptions:

Gearbox efficiency = 98%

Alternator efficiency = 95%

Power generation in Turbine = (Bleed-1 steam flow X (Main steam Enthalpy-Bleed-1 enthalpy) + Bleed-2 steam flow X (Main steam Enthalpy-Bleed-2 enthalpy) + Extraction steam flow X (Main steam Enthalpy-Extraction enthalpy) + Exhaust steam flow X (Main steam Enthalpy-Exhaust enthalpy)) / 860

Power generation in Turbine = (8 X (828-738) + 20 X (828-703) + 93.45 X (828-645) + 14 X (828-611)) / 860

Power generation in Turbine = 720+2500+17110.5+3038 =27.17 MW

Net Power generation = 27.17 X 98% X 95% = 25.29 MW = 25 MW

Therefore, Turbine power generation capacity = 25 MW--------------------II

Specific steam consumption (SSC) = Steam consumption / Power generation = 135 / 25 = 5.4

Specific fuel consumption = SSC / SFR = 5.4 / 2.4 = 2.25 Kg/Kw

3-Alternator capacity/size

Alternator capacity = Turbine Power generation capacity/Power factor

= 25/0.8 = 31.25 MVA----------------III

4-Lube oil system capacity calculation.

Lube oil system majorly consists of;

Main Oil Tank (MOT)
Over Head Oil Tank (OHOT)
Lube oil pumps-2 Nos
Control oil pumps
Emergency oil pump
Oil cooler and filters
Oil vapour extractions fans.

Let us decide the capacity of major system like Oil pumps, OHOT & Oil coolers.

A-Lube & control oil pumps capacity calculation.

As a thumb rule, Lube oil pump capacity in M3/Hr = Turbine capacity in MW X 2.0

That is AOP/MOP capacity = 25 MW X 2.0 = 50 M3/hr each

I.e. capacity of MOP and AOP is 50 M3/Hr.

Capacity of EOP pump = Lube oil pump (AOP or MOP) capacity X 25% = 50 X 25% = 12.5 M3/hr

Capacity of COP pump = Lube oil pump capacity (AOP or MOP) X 10% = 50 X 10% = 5 M3/hr

B-Capacity of the main oil tank:

As a thumb rule, MOT storage capacity = Quantity of oil flow (Lube & control oil) X 225

= (50+5) X 225 =12375 litre, can be rounded up to 12500 Litres

Or else MOT tank capacity can also be calculated as;

MOT storage capacity in M3 = Total maximum possible oil flow from lube oil and control oil pumps X air release time / 60

Air releasing time in MOT is around 4 to 6 minutes. If total oil quantity being handled by MOP, AOP & COP is 110 M3/hr then MOT storage capacity will be;

Consider air releasing time is 6 minutes.

Then, main oil tank (MOT) hold up capacity = (50 X 2 + 5 X 2 + 12.5 X 1) X 6 /60 =12.25 M3 =12500 liters.

Let’s us decide the Overhead oil tank (OHOT) capacity.

As a thumb rule, Storage capacity of Overhead oil tank (OHOT) is = Main Oil tank (MOT) tank capacity X 35% = 12500 X 35% = 4375 liters, can be taken up to 4500 liters

C-Oil cooler capacity calculation.

Generally, oil coolers capacity is measured by heat load.

So, as a thumb rule, oil cooler heat load in KW= Lube oil flow X 10

Note: Consider only the oil that flows through the cooler.

In some Turbines, where only one pump is used for both lube & control oil, then both Lube & Control oil are cooled in oil coolers & in some Lube oil system Control oil is not cooled in oil coolers, as there will be separate pumps for handling control oil.

So, for this case, Oil cooler heat load = 50 M3/hr X 10 = 500 KW = 500 X 860 = 430000 kcal/hr

Oil flow through the Lube oil cooler = Lube oil pump flow

Cooling water required for oil cooler to cool the lube oil to desired temperature in M3/hr = Lube oil flow through oil cooler X 2.1

Therefore, cooling water circulating through the lube oil cooler in M3/hr = 50 X 2.1 =105 M3/hr

Also, cooling water circulation through the oil cooler is calculated as below;

Quantity of Cooling Water circulation = Heat load / Cooling water temperature difference

= 500 X 860 kcal/hr / 4 deg C X 1000 = 107.5 M3/Hr.

Note: Cooling water temperature difference in oil cooler is considered as 3 to 4 deg C

There should be two numbers of oil coolers, one working and one standby. In some areas where atmospheric temperature is more especially in summer season, in that case both oil coolers should be taken into line.

D-Oil Vapour extraction fans capacity calculation

As a Thumb rule Oil vapour extraction fan capacity in m3/sec = Total oil flow (Lube & Control oil) X 9

OVEF capacity = (50+5)X9 = 495 M3/sec and static pressure will be 250-300 mmwc--IV

5-Capacity of Export transformer and Auxiliary Transformers.

Generally, the auxiliary power consumption (APC) of any power plant is in the range of 8 to 12% of power generation. And process auxiliary power consumption is 18 to 22%cof total power generation.

However, we need to decide the maximum power export during off season for calculation of Export Transformer capacity.

Let us consider the APC of Cogen is 8% during off season operation to select the transformer size.

So, Power available for export to grid = Total maximum power generated-Cogen Auxiliary power consumption

= 25 MW-(25 X 8%) = 23 MW---------V

Therefore, export transformer capacity will be 23 MW/Power factor =23/0.8 =28.75 or 30 MVA

Auxiliary Transformer capacity = Cogeneration Maximum Home Load X 120% / Power factor

=25 X 12% X 120% / 0.8 = 4.5 =4.5 MVA X 2 Nos

Note: Maximum APC considered is 12% and total transformers considered -2 Nos (2 X 100%)

Capacity of Transformers= 4.5 MVA……………………….VI

Out of two transformers one transformer can be used as converter (For VFD) and one as distribution transformer for other auxiliaries.

6-Boiler feed pumps capacity calculation.

Capacity of Operating Boiler feed pump = Boiler MCR X 135%

So, total capacity of the feed pumps = 135 X 135% = 182.25 M3/hr.

So, it is good to operate the Boiler with Three feed pumps (2-Running + 1-Standby)

So, capacity of each pump = 182.25/2 = 91.12 or 92 M3/hr……………….VII

Head of the pump = Boiler operating pressure X 140% = 110 X 140% = 154 Kg/cm2 or 1540 meter (Appx).

7-Calculation of Boiler fans capacity.

Bagasse fired Boiler usually have three types of fans, ID fans, FD fans and SA fans.

Let us decide the capacity of each fan.

As a thumb rule,

Capacity of the ID fan in m3/sec = Boiler capacity / 2 = 135 / 2 = 67.5 m3/sec

Capacity of FD & SA fans = ID fan capacity X 80%

Therefore, capacity of FD & SA fans = 67.5 m3/sec X 80% = 54 m3/sec

Capacity of FD fan = 60% of total combustion air required

Capacity of FD fan = 60% X 54 m3/sec = 32.4 m3/sec

Capacity of SA fan = 40% of total combustion air required

Capacity of SA fan = 40% X 54 m3/sec = 21.6 m3/sec

Therefore, capacity of ID fans = 2 X 67.5 m3/sec & Static pressure: 250-300 mmwc

capacity of FD fans = 2 X 32.4 m3/sec & Static pressure: 250-280 mmwc

capacity of SA fans = 2 X 21.6 m3/sec & Static pressure: 600-650 mmwc---------VII

8-Calculation of Water treatment plant capacity.

Water treatment plant consists of Raw water storage tank, Clarifier (HRSCC), MGF, UF, RO and DM plants.

In this case let us decide the capacity of DM plant only. The capacity of DM plant depends on the water make up required for Boiler due to various losses

Major losses in Power plant are;

1.Boiler blow down: 2% of steam generation

2.Deaerator vent loss: 0.1% of Deaeration.

3.Soot blower steam: 0.8 to 1% of Boiler steam generation per day

4.Process steam lines losses due to drain & traps: 1% of steam generation

5.Losses in process return condensate: 10% of steam given to process

6.Losses due to leakages: 10 MT/day

Let us discuss the above losses for this 25 MW Cogeneration plant.

Calculation of total DM water requirement by considering above listed various losses

Blow down loss = 135 X 2% = 2.7 TPH

Deaerator vent loss = 135 X 0.1% = 0.135 TPH

Soot blower steam loss = 135 X 1% / 24 = 0.056 TPH

Process steam loss due to drain and trap condensate = 135 X 1% = 1.35 TPH

Losses in process steam condensate = 93.75 X 10% = 9.37 TPH

Note: For DM plant capacity calculation, recovery of condensate from process has been considered 90% against above assumed 95%.

Other miscellaneous losses: 10 Tonnes/day = 0.42 TPH

So, DM plant capacity = 2.7 + 0.135 + 0.056 + 1.35 + 9.37 + 0.42 =14.031 TPH or 15 TPH

DM plant is being stopped for 4 hours in a day for regeneration purpose,

So, DM plant capacity considering 20% extra margin= 15 X 24 X 120% / 20 = 21.6 M3/hr or 22 M3/hr.

Select two streams for DM plant i.e 2 X 22 M3/hr as one plant working and 1 plant standby

So, Capacity of RO plant should be = DM plant capacity X 110%

Therefore, RO plant capacity = 22 X 110% = 24.2 or 25 M3/hr---------VIII

Select two streams for RO plant also i.e 2 X 25 M3/hr as one plant working and 1 plant standby.

9-Cooling tower and cooling water pumps capacity calculation.

Let us decide the total cooling water required for power plant operation;

A-Main circulating cooling water

MCW water = Maximum exhaust from condensate X 70

i.e Around 70 M3 of cooling water is required to condense 1 Ton of steam in surface condenser.

Maximum exhaust from Turbine= Turbine inlet steam in season operation X 75% X 70%

i.e Turbine inlet steam reduces from 100% to 75-77% during off season operation and exhaust flow to condenser will be 70% of the inlet steam.

Therefore, Maximum exhaust = 135 X 75% X 70% = 70.87 = 71 TPH

Then, Main circulating water (MCW) = 71 X 70 = 4970 M3/hr

Auxiliary cooling water required for various heat exchangers like Generator air cooler, oil cooler, Boiler feed pumps, SWAS panel, Air conditioning system, Ash handling system, bearing cooling systems, air compressors is around 8 to 10% of the total main circulating water.

So, Auxiliary cooling water required = 4970 X 8% =397.6 = 400 M3/hr.

So, Total circulating water = MCW + ACW water = 4970 + 400 = 5370 M3/hr.

Now, let us decide the cooling water size and storage capacity;

Cooling tower storage capacity = Total circulating water X 25% = 5370 X 25% =1342.5 = 1345 M3.

Total no.of cooling towers cells may be 3 or 4. Let us consider 3 in this case.

Water circulation in each cell = 5370/3 = 1790 M3/hr.

Capacity of MCWP pumps.

Let us consider 3 Nos of pumps (2-working + 1 stand by) for above main cooling water circulation flow (4970 M3/hr)

Capacity of each main cooling water pumps (MCWP) : 4970 X 50% = 2485, can take 2500 M3/each pump

Capacity of ACWP pumps = 400 M3/hr -2 Nos (1-working + 1-standby)-------IX

10-Capacity calculation of air compressors.

As a Thumb rule: Capacity of instrumentation air compressor in m3/min = Power plant capacity in MW/4

So, Compressor capacity = 25/4 =6.25 M3/min X 2 Nos (1-Working + 1-Stand by) ----X

Similarly, capacity of the Service compressor, where high pressure air is being used for pneumatic ash handling system for conveying ash into silo = Power plant capacity in MW/4.2

So, Compressor capacity = 25/4.2 =5.8 M3/min X 2 Nos (1-Working + 1-Stand by)----X

11-Capacity of fuel handling plant.

Capacity of the fuel handling plant would be = Bagasse generated from mill X 120%

Therefore, capacity of the fuel handling plant = 60.42 TPH X 120% =72.50, can be rounded up to 75 TPH.

Or Bagasse handling capacity = Boiler fuel consumption at SFR 2.4 X 130%

= (135/2.4) X 130% = 73.125, can be rounded up to 75 TPH………XI

12- Stack or Chimney size:

Let us consider the stack is of RCC construction.

Stack height is decided based on sulphur dioxide emission & local statutory compliances.

Stack height = 14 X Q^0.3

Where Q is the Sulphur dioxide emission in Kg/hr.

Generally, bagasse has no sulphur content in it. However, we shall assume 0.1% of sulphur content in bagasse for calculating Stack height.

Sulphur dioxide generated per hour = Bagasse consumption/hour X sulphur content in fuel/100

Q = 56.25 X 1000 X 0.1/100 =56.25 Kg.

Sulphur dioxide generation = S + O2 = SO2

SO2 generation/hour = 56.25 X 2 = 112.5 Kg/hr

Therefore, minimum stack height in meter, H = 14 X 112.5^^0.3

H = 57.74 meter, can be rounded up to 60 meters.

However, considering local statutory compliances like airport permissions. The total height of chimney should be maintained around 75 meters---------XII

Calculation of the Diameter of the Chimney

Flue gas flow through Chimney = Mass of bagasse fired X Total combustion air given.

= 56.25 X 3.7 = 208.12 TPH

Let us consider 10% extra margin for calculation of chimney internal diameter

So, total flue gas generated per hour = 208.12 X 110% =228.9, can be rounded up to 230 TPH

Convert the above quantity in M3/sec

Flue gas volume = 230 X 1000 kg/hr / (Flue gas Density at temperature 140 Deg C X 3600)

Density of flue gas at 140 Deg C temperature is 0.85 Kg/m3

Therefore, Flue gas volume = 230000 / (0.85 X 3600) = 75.16 m3/sec (270576 M3/hr)

We have, Flue gas volume, Q = Area of chimney, A X Velocity, V

Assume velocity of flue gas in chimney is around 12 m/sec.

So, 75.16 = 3.142 X D²/4 X 12

Internal diameter of the Chimney, D =2.82 meter, can be rounded up to 3 meters

Internal diameter of the chimney considering 10% extra margin = 3 X 110% = 3.3 meter

Outside diameter of the Chimney D1 = 3.3 X 0.3 X 2 =3.9 meter (0.3 meter is the concrete wall thickness) ----------XII

13-ESP sizing

Aa a thumb rule, ESP collection area = Flue gas flow in M3/hr / 67

ESP collection area = 270576 M3/hr /67 = 4038 M2

Note: convert Flue gas flow from 230000 kg/hr to M3/hr by dividing by density of flue gas i.e o.85 kg/m3 and again convert it into m3/sec by dividing by 3600

Specific collection area = 4038 / 75.16 =53.73, can be rounded up to 54 m2/m3/sec

Flue gas velocity in ESP = 0.9 to 1 m/sec

Migration velocity =8 to 9 cm/sec

Let’s design 3 fields for this ESP with Height of collecting plates 11 meter & width 500 mm.

Read>>>>>> 40-Thumb rules for Cogeneration plant design

Power plant and calculations

How to calculate the capacity of Sugar based Cogeneration plant??

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