Power plant and calculation site basically includes the detailed study of power plant operation and maintenance, its related all calculations and thumb rules. It also involves detailed troubleshooting guides for operation and maintenance of power plant system/equipments like Boiler, fans, compressors, belt conveyors, ash handling system, ESP, steam turbine, cooling tower, heat exchangers, steam ejectors, condensers WTP. etc. Heat rate, efficiency
Monday, 31 August 2026
Online Steam Table
Friday, 10 April 2026
Online calculator Deaerator steam consumption
What are the inputs required to calculate the deaerator steam consumption?
Read>>>How do you calculate the Deaerator & HP heaters steam consumption???
Inputs required for the calculation of Deaerator steam consumption
Check for Best Dissolved Oxygen Meters for Boiler Feedwater Testing
- Steam pressure
- Steam temperature
- Steam enthalpy
- Feed water temperature at Deaerator inlet
- Feed water temperature at Deaerator outlet
- Feed water flow
- Any other make up water and its flow and temperature
DA Steam Consumption Calculator
Fuel GCV & NCV Online Calculator
Monday, 10 November 2025
Online steam cost calculator
In power plant, calculation of cost
of steam is very vital in commercial point of view. Following are the
parameters which affect the cost of steam.
1. Steam pressure
2. Steam temperature
3. GCV of fuel
4. Price of fuel
5. And Boiler efficiency
Following gives you the relation among steam cost & above all
parameters & vice versa
- Steam cost increases as the
enthalpy or heat content in steam increases and vice versa
- Steam cost increases as the
GCV of fuel decreases and vice versa
- Steam cost increases as the
fuel price increases & vice versa
- Steam cost increases as the
Boiler efficiency decreases & vice versa
Understanding with examples.
1.Calculate the cost of steam per kg,
which is been using for Steam turbine having pressure 121 kg/cm2 &
temperature 550 deg C.The boiler of efficiency 75% uses coal of GCV 4200
kcal/kg to produce this steam.Cosnsider the price of coal is Rs 5000/MT
Enthalpy of steam at above pressure & temperature H = 830.43 kcal/kg
Boiler efficiency â´–b= 75%
GCV of coal = 4200 kcal/kg
Now, cost of steam = Heat content in steam in kcal/kg X Fuel
price / (GCV of fuel in kcal/kg X Boiler efficiency â´–b)
=
830.43 X 5000 / (4500 X 0.75)
=
1230.26 rupees / MT of steam or Rs 1.23 / kg of steam
2.Calculate the cost of steam per kg, which is been using for chemical process plant having pressure 5 kg/cm2 & temperature 180 deg C.The boiler of efficiency 65% uses biomass of GCV 2800 kcal/kg to produce this steam.Cosnsider the price of biomass is Rs 2400/MT
Boiler efficiency â´–b= 65%
GCV of coal = 2800 kcal/kg
Now, cost of steam = Heat content in steam in kcal/kg X Fuel price /
(GCV of fuel in kcal/kg X Boiler efficiency â´–b)
=
670 X 2400 / (2800 X 0.65)
=
883.51 rupees / MT of steam or Rs 0.88 / kg of steam
Read more>>>>Powerplant and calculations
Steam Cost Calculator
Note: This tool estimates steam energy and cost based on simplified enthalpy equations. It does not consider feedwater temperature or full steam tables.
Sunday, 24 August 2025
What is Entropy??
In heat engine theory, the term ENTROY plays a very
vital role and leads to important results which by other methods can be
obtained much more laboriously.
It may be noted that, all the heat is equally not
available for converting into work. Heat that is supplied to a substance at
higher temperature has greater possibility of conversion into work rather than
heat supplied to a substance at lower temperature.
Meta Description: Confused by entropy? This guide breaks down the law of
entropy in simple terms. Learn how this fundamental concept of thermodynamics
affects everything from the cosmos to your daily life.
We’ve
all experienced it. An ice cube melts in your drink. Your once-organized desk
slowly descends into chaos. A hot cup of coffee gets colder, never hotter, on
its own. These everyday events seem unrelated, but they are all governed by a
single, powerful, and often misunderstood scientific law: the Second Law of Thermodynamics, often called the Law of Entropy.
In
simple terms:
·
Entropy
is a measure of how energy is distributed in a system.
·
It tells us how much of the system’s energy is unavailable to do
useful work.
·
It also represents the degree of disorder or randomness
of particles within the system.
Entropy is a function of quantity of heat which shows
the possibility of conversion of that heat into work. The increase in entropy
is small when heat is added at higher temperature and is greater when heat
addition is made at a lower temperature.
Thus, the for maximum ENTROPY there is minimum availability
for conversion into work and for minimum entropy there is maximum availability for
conversion into work.
As per the Third law of Thermodynamics: When a system
is at zero absolute temperature, the entropy of a system is zero. That is The
Entropy of all perfect crystalline solids is zero at absolute zero temperature.
Read more>>>>>Boiler and Turbine hub
Entropy may also be defined as the thermal property of a substance which remains constant when substance is expanded or compressed adiabatically in a cylinder.
In the simplest sense, entropy is a measure of disorder
or randomness in a system. In thermodynamics, however, it is
more precisely defined:
Entropy is
a measure of the amount of energy in a system that is unavailable to do useful
work.
This makes entropy a central
concept in the Second Law of Thermodynamics, which states that in any
natural process, the total entropy of an isolated system always increases or
remains constant.
In an isolated system, entropy never decreases.
Processes naturally move toward states of higher entropy.
Characteristics
of Entropy:
1-It increases when the heat is supplied irrespective
of the fact whether temperature changes or not.
2-Entropy decreases when heat is removed whether
temperature changes or not
3-It remains unchanged in all adiabatic frictionless processes
4-Entropy increases if temperature of heat is lowered
without work being done as in a throttling processes.
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Entropy and Irreversibility:
One of the most practical consequences
of entropy is that it explains why certain processes are irreversible.
A cup of hot coffee placed on a table
will cool down, but the table will never spontaneously heat the coffee back up.
An inflated balloon can burst, but the
air inside will not naturally return into the balloon.
Entropy is
more than an abstract thermodynamic term — it is a guiding principle of nature.
It tells us why processes have direction, why machines have limits, and why
time seems to move forward. From a melting ice cube to the vast fate of the
universe, entropy governs the flow of energy and the evolution of systems.
Understanding entropy is not only essential for physicists and
engineers but also provides us with a deeper appreciation of the hidden order
behind everyday phenomena.
Two
Sides of the Same Coin: Thermodynamic vs. Informational Entropy
The concept of entropy has also revolutionized the field of information
theory, thanks to Claude Shannon.
·
Thermodynamic
Entropy: Deals with the physical dispersal of energy.
·
Informational
Entropy: Measures uncertainty or the surprise factor
in a message. A string of random letters has high informational entropy (it's
very surprising/unpredictable). A meaningful sentence in English has low
informational entropy (it's predictable and ordered).
Saturday, 2 August 2025
40-Thumb rules for sugar Cogeneration plant design
Objective: Design of Cogeneration
plant for 5000 TCD Sugar and 60 KLPD Molasses based Distillery plant
Thumb rules for power plant design
|
Sl.No. |
Particular |
Capacity as per Thumb rule |
|
1 |
Capacity of the Sugar plant |
5000
TCD |
|
2 |
Distillery plant capacity |
60
KLPD |
|
3 |
Steam required to Sugar process |
Cane
crushing capacity X 40% = 5000X 40% =2000 MT/day =85 TPH |
|
4 |
Steam required for distillery
process |
Steam
required for distillery process plant =Distillery plant capacity X 3.5
= 60 KLPD X 3.5 = 210 TPD = 8.75 TPH |
|
5 |
Total steam demand for sugar and
distillery process |
85 +
8.75 = 93.75 TPH |
|
6 |
Sugar plant auxiliary Power
consumption |
The
auxiliary power consumed per hour by any sugar plant as a Thumb rule = Cane
crushing capacity X 80% X 110% =5000X80%110%=4.4 MWH |
|
7 |
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 |
|
8 |
Power consumption of distillery
plant |
Sugar
plant power consumption X 20-25% = 4400 X 20-25% = 0.88 to 1.1 MW |
|
9 |
Total power consumption of Sugar
process and related auxiliaries |
Total
power consumption= 4.4 + 0.22 +1.1=5.72 MWH |
|
10 |
Turbine extraction steam flow |
Extraction
steam flow = Process steam flow + Deaerator steam flow Deaerator
steam flow=Process steam flow X 10% |
|
11 |
Boiler capacity |
Boiler
capacity = Turbine extraction steam/0.7 = = 85 + 85 X 10% = 93.5 TPH/0.7
=133.5 =135 TPH |
|
12 |
Boiler fuel (Bagasse) consumption |
At the
average SFR (steam to fuel ratio) 2.4 Boiler bagasse consumption = 135 / 2.4
=56.25 TPH |
|
13 |
Bagasse saving per day |
=
(Sugar mill bagasse generation-Boiler fuel consumption) X 24 = (60.42-56.25) X 24 = Appx. 100 MT/day |
|
14 |
% of Bagasse saving |
% of
bagasse saving = 100 X 100/(60.42 X 24) =6.9% |
|
15 |
Turbine capacity |
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 |
|
16 |
Specific steam consumption (SSC) |
SSC=Steam
consumption / Power generation = 135 / 25 = 5.4 |
|
17 |
Specific fuel consumption (SFC) |
SFC=
SSC / SFR = 5.4 / 2.4 = 2.25 Kg/Kw |
|
18 |
Alternator capacity/size |
Alternator
capacity = Turbine Power generation capacity/Power factor =
25/0.8 = 31.25 MVA |
|
19 |
Lube oil system capacity
calculation |
|
|
19a |
Lube oil pump capacity |
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 |
|
19b |
Capacity of EOP pump |
Capacity
of EOP pump = Lube oil pump (AOP or MOP) capacity X 25% = 50 X 25% = 12.5
M3/hr |
|
19c |
Capacity of COP pump |
Capacity
of COP pump = Lube oil pump capacity (AOP or MOP) X 10% = 50 X 10% = 5 M3/hr |
|
19d |
MOT storage capacity |
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 |
|
19e |
OHOT storage 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 |
|
19f |
Oil cooler capacity |
Oil
cooler heat load = 50 M3/hr X 10 = 500 KW = 500 X 860 = 430000 kcal/hr. 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. |
|
19g |
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) X 9 = 495 M3/sec and static pressure will be 250-300 mmwc |
|
20 |
Capacity of export Transformer |
Capacity = (Total maximum power
generated-Cogen Auxiliary power consumption) = 25 MW-(25 X 8%) = 23 MW Therefore,
export transformer capacity will be 23 MW/Power factor =23/0.8 =28.75 or 30
MVA |
|
21 |
Auxiliary Transformer capacity | 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 Out of two transformers one transformer can be used as converter (For VFD) and one as distribution transformer for other auxiliaries. |
|
22 |
Boiler feed pumps capacity |
Capacity
of Operating Boiler feed pump = Boiler MCR X 135% So, total
capacity of the feed pumps = 135 X 135% = 182.25 M3/hr Capacity
of each pump = 182.25/2 = 91.12 or 92 M3/hr Head
of the pump = Boiler operating pressure X 140% = 110 X 140% = 154 Kg/cm2 or
1540 meter (Appx). |
|
23 |
ID fans capacity |
Capacity
of the ID fan in m3/sec = Boiler capacity / 2 = 135 / 2 = 67.5 m3/sec
& Static pressure: 250-300 mmwc |
|
24 |
Total combustion air required |
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 |
|
25 |
Capacity of FD fan |
Capacity
of FD fan = 55% of ID fan flow Capacity
of FD fan = 55% X 67.5 m3/sec = 37.12 m3/sec & Static pressure: 250-280 mmwc |
|
26 |
Capacity of SA fan |
Capacity
of SA fan = 25% of ID fan flow Capacity
of SA fan = 25% X 67.5 m3/sec = 16.87 =17 m3/sec & Static pressure: 600-650 mmwc |
|
27 |
DM plant capacity |
Capacity
= Total losses in Power plant = 2 X 22 M3/Hr. |
|
28 |
RO plant capacity |
RO plant capacity = 22 X 110% = 24.2 or 25 M3/hr 2 X 25
M3/Hr |
|
29 |
Main circulating cooling water |
MCW water = Maximum exhaust from condensate X 70 Maximum exhaust from Turbine= Turbine inlet steam
in season operation X 75% X 70% Maximum exhaust = 135 X 75% X 70% = 70.87 = 71
TPH Main circulating water (MCW) = 71 X 70 = 4970
M3/hr |
|
30 |
Auxiliary Cooling water |
Auxiliary cooling water required = 4970 X 8% =397.6 = 400 M3/hr. |
|
31 |
Total circulating water |
Total circulating water = MCW + ACW water = 4970 + 400 = 5370 M3/hr |
|
32 |
Cooling tower storage capacity |
Cooling tower storage capacity = Total
circulating water X 25% = 5370 X 25% =1342.5 = 1345 M3. |
|
33 |
Capacity of MCWP pumps |
Capacity of each (3 Nos-2W+1S) main cooling water
pumps (MCWP) : 4970 X 60% = 2982, can take 2900 M3/each pump |
|
34 |
Capacity of ACWP pumps |
Capacity of ACWP pumps = 400 M3/hr -2 Nos (1-working + 1-standby) |
|
35 |
Capacity of instrumentation air
compressor |
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) |
|
36 |
Capacity of Service air
compressor |
Service air Compressor capacity = 25/4.2 =5.8 M3/min X 2 Nos
(1-Working + 1-Stand by) |
|
37 |
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 |
|
38 |
Stack or Chimney size |
Stack
height = 75 ID :
3.3 M OD:
3.9 M |
|
39 |
ESP sizing |
AS a
thumb rule, ESP collection area = Flue gas flow in M3/hr / 67 ESP
collection area = 345960 M3/hr /67 = Specific
collection area = 5164 / 96.1 =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 |
Read more>>>>Sugar based Cogeneration plant complete design
Read more>>>>>Powerplant and calculations
Read more>>>>Top-36 Thumb rules for steam Turbine
Read more>>>>Boiler and Turbine hub
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