### Calculation of raw water requirement and Reservoir size for a power plant

In order to calculate the raw water requirement for a power plant, first of all we should know the losses in power plant.
As per new law raw water consumption should be 2.5 liters/kwh, presently it is around 4.5 liters/kwh.

Different losses in power plants :

A-Boiler

• Boiler blow down : 2% of steam generation
• Deaerator vent loss : 0.1% of Deaeration
• Soot blower steam : 0.8 to 1% of Boiler steam generation per day

B-Turbine & Process steam

• Process steam lines losses due to drain & traps : 1% of steam generation
• Cooling water evaporation loss : 2% of cooling water circulation
• Losses in process return condensate : 10% of steam given to process

C-Water treatment plant (WTP)

• Clarifier : 2% of total raw water input or 5% of DM water generation
• MGF back wash water loss : 40% of MGF inlet water (Back wash frequency : once in every 8 hours of operation)
• Reverse Osmosis (RO) plant reject water : 25% of total input water to RO plant or 35% of RO capacity
• Water losses in DM plant regeneration process : 30% of DM plant capacity for every regeneration process
D-Other miscellaneous losses & cunsumption

• Losses due to leakages : 10 MT/day
• Water used for drinking & service : 10 MT/day
• Fire fighting system : 40 MT/day...Considered Jockey pump is running for 4 hours in a day at 10 M3/hr flow
Let us discuss the above lossed by considering an example of 20 MW Cogeneration based power plant.This plant has Boiler of capacity 130 TPH,Steam given to process application is 65 TPH. Calculate the RO plant , DM plant capacity & Raw water consumption per day.

Calculation of total water requirement by considering various losses

Step-I

Calculate DM water requirement per hour
• Blow down loss = 130 X 2% = 2.6 TPH
• Deaerator vent loss = 130 X 0.1% = 0.13 TPH
• Soot blower steam loss = 130 X 1% / 24 = 0.054 TPH
• Process steam loss = 130 X 1% = 1.3 TPH
• Losses in process steam condensate = 65 X 10% = 6.5 TPH
•  SO total DM plant capacity = 2.6 + 0.13 + 0.054 + 1.3 + 6.5 =10.58 TPH
• DM plant is being stopped for 4 hours in a day for regeneration purpose,
Therefore, DM plant capacity = 10.58 X 24/20 = 12.696 TPH

Step-II
Calculate the cooling water make up.
• From the above example, quantity of steam going to the condenser is 130-65 = 65 TPH
• Cooling water circulation flow is = 65 X 60 = 3900 TPH
• Note: Considered 60 TPH of CW is required to condensate the 65 TPH of steam
• So total cooling tower loss = 3900 X 2% = 78 TPH

Step-III

Calculate WTP losses

• Clarifier blow down losses = 12.96 X 5% = 0.65 TPH
• RO plant reject water = 12.96 X 35% = 4.53 TPH
• DM water regeneration losses (SAC+SBA+MB) = (3 X (12.96 X 30% ))/24 = 0.486 TPH

Step-IV
Other miscellaneous
• Leakage losses = 10 TPD =10/24 =0.24 TPH
• Drinking & Service water = 10 TPD =10/24 = 0.24 TPH
• Fire fighting water = 10 TPD =10/24 =0.24 TPH
So total raw water requirement =12.696 + 78 + 0.65 + 4.53 + 0.486 +0.24 + 0.24 +0.24 = 97.08 =97 TPH

Total raw water required in a day = 97 X 24 =2328 TPD

Take 10% Extra Margin = 2328 X 110% = 2561 TPD

Size of raw water reservoir by considering water requirement for 7 days

Total storage capacity = 7 X 2561 =17927 Tonnes or M3.....As density of water is 1000 kg/m3

Take reservoir  depth of 6 meter
Then tank size
17927 M3 = Length X Breadth X 6

L X B =2987.3 M2

Take length of the reservoir = 60 meters
then breadth of the reservoir will be = 2987.3/60 = 49.8 =50 M3

So size of the reservoir will be = 60 X 50 X 6 meters
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1. Thanks more adds boe and powerplant as a whole