Topping cycle & calculations

  Topping cycle & calculations

 

A Co-generation system can be classified as either a topping cycle or a bottoming cycle on the basis of sequence of energy generated & use.

 In a topping cycle, the fuel supplied is used to first produce power and then thermal energy, which is the by-product of the cycle and is used to satisfy process heat requirements.

 In a topping cycle, a primary heat source, such as a gas turbine or an internal combustion engine, is used to drive a generator and produce electricity. The primary cycle typically operates at higher temperatures and generates high-pressure and high-temperature exhaust gases.

 The exhaust gases from the topping cycle are then directed to a waste heat recovery boiler or a heat exchanger, where their residual heat is captured. This waste heat is then used to produce steam, which drives a steam turbine or an organic Rankine cycle (ORC) turbine in the bottoming cycle.

 Topping cycles are commonly used in combined cycle power plants, where they offer improved efficiency and performance compared to standalone gas turbines or internal combustion engines. The integration of a bottoming cycle allows for the utilization of waste heat, maximizing the overall energy output of the system.

 In a bottoming cycle, the primary fuel used produces high temperature thermal energy and the heat rejected from the process is used to generate power through a heat recovery Boiler & Turbo generator.

 Bottoming cycles are suitable for manufacturing processes that require heat at high temperature in furnaces & kiln and reject heat at significantly high temperatures.

 The bottoming cycle operates at lower temperatures and utilizes the waste heat energy to generate additional power. By extracting energy from the waste heat, the topping cycle achieves higher overall efficiency compared to a single-cycle power generation system.

 Topping cycle calculation:

 A Co-generation facility is defined as one, which simultaneously produces two or more forms of useful energy such as electrical power and steam, electric power and shaft (mechanical) power, etc.” The project may qualify to be termed as a co-generation project, if it is in accordance with the definition and also meets the qualifying requirement outlined below:

 Topping cycle mode of co-generation – Any facility that uses non-fossil fuel input for the power generation and also utilizes the thermal energy generated for useful heat applications in other industrial activities simultaneously.

 For the co-generation facility to qualify under topping cycle mode, the sum of useful power output and one half the useful thermal output be greater than 45% of the facility’s energy consumption, during season.”

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Following inputs required for calculation of topping cycle:

• Fuel consumption
• Fuel GCV
• Steam given to processes & their heat content
• Power generation

Topping cycle is calculated by using following formula

 TC Eff = (Sum of total heat supplied to process in kcal X 50% + Total electricity generated in kcal) X 100 / Fuel energy

Example

 A 44 MW Co-generation plant is operating at 41 MW load and utilizing bleed & extraction steam for process heating. Calculate the topping cycle efficiency

The inputs required are as below

Sl No	Particular/Parameters	UOM	Value
1	Boiler fuel consumption	TPH	85
2	Fuel GCV	Kcal/kg	2250
3	Process-1 steam flow	TPH	12
4	Process steam-1 enthalpy	Kcal/kg	740
5	Process-2 steam flow	TPH	170
6	Process steam-2 enthalpy	Kcal/kg	653
7	Power generation	MWH	41

 Calculation:

Total heat content in input fuel = 85 X 1000 X 2250 =191250000 kcal

Heat content in process-1 steam = 12 X 1000 X 740 =8880000 kcal

Heat content in process-2 steam = 170 X 1000 X 653 =111010000 kcal

Power generation in kcal = 41 X 1000 X 860 = 35260000 kcal

 TC Eff = (Sum of total heat supplied to process in kcal X 50% + Total electricity generated in kcal) X 100 / Fuel energy

TC Eff = ((8880000+111010000) X 50% + 35260000) X 100 / 191250000

 TC eff = 49.78%

   

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Calculation of pressure drop in steam and water lines

 Calculation of pressure drop in steam and water lines

 The pressure drop in a water & steam lines refers to the decrease in pressure that occurs as water/steam flows through a pipe or conduit due to factors such as friction and flow resistance. Several factors influence the magnitude of pressure drop in a water line:

Pipe Characteristics: The diameter, length, and roughness of the pipe impact the resistance to flow and consequently the pressure drop. Smaller diameter pipes and longer pipe lengths tend to result in higher pressure drops. Additionally, rougher pipe surfaces create more friction and increase pressure drop compared to smoother surfaces.

Flow Rate: The rate at which water/steam flows through the pipe affects the pressure drop. Higher flow rates generally result in higher pressure drops due to increased frictional resistance.

Fluid Properties: The physical properties of the water/steam being transported, such as viscosity and density, can influence the pressure drop. However, for water at typical temperatures and pressures, these effects are usually negligible.

 Pipe Fittings and Valves: The presence of fittings, such as elbows, bends, valves, and other obstructions in the water line, can contribute to pressure drop. These components disrupt the flow and introduce additional resistance.

 It's important to note that pressure drop calculations for steam lines can be complex and require a comprehensive understanding of steam properties and fluid dynamics.

 Pressure drop in water line:

Head loss in water line for turbulent flow is given as

Head loss in meter = 4fLV² / (2gD)

 Where, f = Friction loss in pipe, generally varies from 0.005 to 0.007

L = Pipe length

D = Diameter of the pipe

g = Acceleration due to gravity, 9.81 m/s2

V = Velocity of the fluid

 Example:

A Boiler feed pump is delivering feed water flow 50 TPH to the boiler at a distance of 70 meter.The steam drum height is 38 meter from pump suction.Calculate the pressure drop in water line, assume pipe line size is 80 NB, water density 980 kg/m3 & neglect the other losses from pipe line fittings.

 Feed water flow in m3/sec = 50 000 kg/hr / 980 kg/m3 = 51.02 m3/hr =0.014 m3/sec

Area in side the pipe line = 3.142 X 0.08²/4 = 0.05 M²

Feed water velocity,V =  Flow / Area = 0.014 / 0.005 =2.78 m/sec

 Then, head loss, H = 4 X 0.005 X 38 X 2.78² / (2 X 9.81 X 0.08)

Head loss, H = 3.75 meter

 Minimum head required to lift the water up to steam drum, considering pressure drop in feed water control valve is 8 kg/cm2

 H = 3.75+80+38 =121.75 meter

 Pressure drop in steam line:

Head loss in meter = 2fLdV² / (500gD)

(Density of water is 500 times more than steam at atmospheric pressure)

Where, f = Friction loss in pipe, generally varies from 0.005 to 0.007

L = Pipe length

D = Diameter of the pipe

g = Acceleration due to gravity, 9.81 m/s2

V = Velocity of the fluid

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 Example: Turbine inlet steam flow is 100 TPH & the distance between Boiler MSSV & Turbine MSSV is 82 meter.The seam pressure & temperature are 65 kg/cm2 and 490 deg C respectively.Calculate the pressure drop in steam line.

 Density of steam at above parameters = 17 kg/m3

Steam flow in m3/sec = 100 X 1000 kg/hr / 17 kg/m3=5882.35 m3/hr = 1.63 m3/sec

Assume main steam velocity being 45 m/sec

Pipe inside area A = Flow / Velocity = 1.63/45 =0.0362 m2

Now, calculate pipe diameter , A = 3.142 X D²/4

D = SQRT (0.036 X 4/3.142) = 0.214 meter = 214 mm

Now, Pressure drop H =(2 X 0.005 X 82/0.224) X (17/500) X (45²/9.81) =25.7 m = 2.57 kg/cm2

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Why pressure reducing of steam is being done before de-super heating?

 Following advantages we can have, if pressure reducing is done before de-super heating

1.  Reducing the pressure before de-super heating helps in controlling the steam temperature with great accuracy.High-pressure steam entering the de-superheating station can be challenging to control accurately due to its high energy content. By reducing the pressure, the energy content decreases, allowing for better control of the de-superheating process and ensuring that the desired steam temperature is achieved.
2.  Heat transfer between water & steam is more better at lower pressure
3.  Helps is protecting the down stream equipment if desuper heating is done at lower pressure since there will not be no any much chances of low or high temperature steam entry into the Turbine or process.
4.  Erosion of down stream pipe lines can be minimized due to high pressure & high velocity steam
5.  Cost of high pressure lines can be reduced.
6.  Total head required for de-superheating water will be reduced, thereby reducing feed pump pumping power.
7.  Related cost of control valves & pipe lines will be reduced
8.  Damage to the de-super heating equipment can be reduced by avoiding thermal shock by avoiding direct mixing of high pressure steam and high pressure water
9.  Pressure reduction before the de-superheating station helps maintain safe operating condition.High-pressure steam carries a greater risk of causing damage or injury in the event of a malfunction or sudden pressure surge. By reducing the pressure beforehand, the risk of high-pressure steam reaching the de-superheating station is mitigated, enhancing overall system safety.
10.  Pressure reduction can also contribute to overall system efficiency. Lowering the pressure before the de-superheating station helps control the temperature more effectively, resulting in more efficient heat transfer and reduced energy consumption.
11. Reducing the pressure before de super heating will also helps in reducing the temperature of steam, which helps in reduction of water quantity required for desuper heating

 

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What are the various interlocks used in Boilers??

 

1.What do you mean by Interlocks?

 Interlocks are the programmed or hardwired control system made to protect the machine or system from damages or disturbances

Interlocks and protections involve sensors, cables, wires, local push buttons, logics, timers, probes etc.

 2.What is the significance of interlocks?

 Significance of interlocks;

• To protect the system against damages/disturbance
• To protect the equipment
• To avoid damages to the man and machine
• To avoid operation disturbances

 3.What are the various protections used in Boilers?

• High drum level trip
• Low drum level trip
• High main steam pressure trip
• High positive draught trip
• High negative draught trip
• High main steam temperature trip

 4.What are the various interlocks provided for Boiler feed pumps?

 Boiler feed pumps trip on acting following interlocks

• Low de-aerator level
• High bearing vibration
• High bearing temperature
• High feed water temperature at suction
• High drum level

 5.Write a brief note on Boiler interlocks

 Read Generator and Turbine inter tripping

Sl No.	Interlock	Significance
1	High drum level-FD fans trip followed by fuel feeding system & ID fans	To avoid carryover of water particles in steam To avoid thermal shock to super heater coils
2	Low drum level-FD fans trip followed by fuel feeding system & ID fans	To avoid over heating of pressure parts due to lack of water
3	FD fans trip-Fuel feeding system trip	To avoid jamming of fuel feeding system & grate
4	High furnace draught-FD & SA fans trip	To avoid furnace explosion
5	Low furnace draught-ID fans trip	To avoid explosion of ESP & related ducts to vacuum pressure
6	High main steam pressure-Boiler trips (FD & fuel feeding system)	To avoid failure of pressure parts due to high steam pressure
7	High main steam temperature-Boiler trips (FD & fuel feeding system)	To avoid failure of pressure parts due to high steam temperature
8	PA fan trips- Fuel feeding system trips	To avoid jamming of fuel feeding system
9	High main steam pressure-Start up vent CV auto open	To avoid failure of pressure parts due to high steam pressure

 

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Power plant and calculations

Read Practical approach to power plant O&M

Practical Approach to Power Plant Operation and Maintenance

100 + formulae for power plant calculations

A-Boiler

1-Boiler efficiency direct method

Boiler efficiency = (Mass of steam flow X Steam enthalpy-Feed water flow at economizer inlet X Enthalpy-Attemperator water flow X Enthalpy) / (GCV of fuel X Fuel consumption)

2-Boiler efficiency by indirect method

Boiler efficiency = 100-Various losses

3-Theoretical air requirement for combustion

Theoretical air T_hair = ((11.6 X C% + (34.8 X (H2-O2/8)) + (4.35 X %S))/100

Where C = % of carbon in fuel

H2 = % of Hydrogen present in fuel

S = % of sulphur present in fuel

4-Excess air requirement for combustion

%EA = O2% / (21-O2%)

Where O2 = % of oxygen present in flue gas

5-Mass of actual air supplied

AAS = (1 + EA / 100) X Theoretical air

6-Mass of flue gas

Mfg = Mass of Air + 1

6-Mass of dry flue gas

Mfg = Mass of Co2 in flue gas + Mass of Nitrogen in fuel + Mass of Nitrogen in combustion air + Mass of oxygen in flue gas + Mass of So2 in flue gas

Mfg =(Carbon % in fuel X Molecular weight of CO2 / Mol.weight of Carbon) + N2 in fuel + (Mass of actual air supplied X % of N2 in air i.e 77/100) + ((Mass of actual air – Mass of theoretical air) X 23/100) + S2 in fuel X Mol.weight of SO2 / Mol.weight of sulphur)

7-% of heat loss in dry flue gas

Heat loss = Mfg X Cp X (Tf-Ta) X 100 / GCV of fuel

Where,

Mfg = Mass of flue gas

Cp = Specific heat of flue gas in kacl/kg

Tf = Temperature of flue gas

Ta = Ambient air temperature

9-% of heat loss due to moisture in fuel

Heat loss = M X (584 + Cp X (Tf-Ta))  X 100 /  GCV of fuel

Where,

M = Moisture in fuel

Cp = Specific heat of flue gas in kcal/kg

10-% of heat loss due to moisture in air

Heat loss = AAS X humidity X Cp X (Tf-Ta) X 100/ (GCV of fuel)

Where,

AAS = Actual air supplied for combustion

Cp = Specific heat of flue gas in kcal/kg

Tf = Temperature of flue gas

Ta = Ambient air temperature

11-% of Boiler water blow down

Blow down % = (Feed water TDS X % of makeup water) X 100 / (Maximum permissible TDS in Boiler water –Feed water TDS)

12-Steam velocity in line

Velocity of steam in pipe line,V = Steam flow in m3/sec / Area of pipe line (A)

Steam flow in m3/sec = (Steam flow in kg/hr / Density of steam X 3600)

Area of pipe, A = Pi X D² / 4

Where D is pipe internal diameter

13-Condensate flash steam calculation

Flash steam % = (H1-H2) X 100 / Hfg)

Where, H1 = Sensible heat at high pressure condensate in kcal/kg

H2 = Sensible heat of steam at low pressure in kcal/kg

Hfg = Latent heat of flash steam

14-Calculation of amount of heat required to raise the water temperature

Heat required in kcal=Mw X Cp X (T2-T1)

Where, Mw = Mass of water

Cp = Specific heat of water in kcal/kg (1 kcal/kg)

T1 = Initial temperature of water in deg C

T2 = Final temperature of water in deg C

15-Calculation of heat required to raise air temperature

Heat required in kcal=Mair X Cp X (T2-T1)

Where, Mw = Mass of water

Cp = Specific heat of flue gas in kcal/kg (0.24 kcal/kg)

T1 = Initial temperature of air in deg C

T2 = Final temperature of air in deg C

16-Surface heat loss calculation

S = (10 + (Ts-Ta) / 20) X (Ts-Ta) X A

S = Surface heat loss in kcal/hr m2

Ts= Hot surface temperature in deg C

Ta = Ambient air temperature in deg C

17-Dryness fraction of steam

X = Mass of dry steam / (Mass of dry steam + Mass of water suspension in mixture)

18-Heat content in wet steam

h = hf + xhfg

h= Heat content in saturated steam

x = Dryness factor of steam

Hfg =Enthalpy of evaporation

19-Heat content in dry saturated steam

h = hf + hfg

h= Heat content in saturated steam

Hfg =Enthalpy of evaporation

20-Heat content in superheated  steam

h = hf + hfg + Cps (Tsup - Ts)

h= Heat content in super heated steam

hfg =Enthalpy of evaporation

Cps = Specific heat of super heated steam

Tsup= Superheated steam temperature in deg C

Ts = Saturated temperature of steam in deg C

21-Calculation of Equivalent evaporation

Me = Ms X (h-hf) / hfg

Ms = Mass of steam

h = Steam enthalpy

hf= Feed water enthalpy

22-Factor of evaporation

Fe = (h-hf) / 539

23-Ash (Total) generation calculation

Ash generation in TPH = Fuel consumption per hour X % of ash in fuel / 100

24-Fly ash generation calculation

Fly ash generation in TPH = Fuel consumption per hour X % of ash in fuel X 80% / 100

25-Bottom  ash generation calculation

Bottom ash generation in TPH = Fuel consumption per hour X % of ash in fuel X 20% / 100

26-Calculation of ash generation in ESP

Ash generation in ESP in TPH = Fuel consumption per hour X % of ash in fuel X 80% X 80% / 100

27-Boiler safety valve blow down calculation

Blow down % = (Set pressure - Re seat pressure) X 100 / Set pressure

28-Calculation of attemperator water flow

Attemperator water flow  in TPH= Steam flow in TPH X (h1-h2) / (h2-h3)

h1 = Enthalpy of steam before desuper heating in kcal/kg

29-Economiser efficiency calculation

ηEco. = (Economiser outlet feed water temperature-Economizer inlet feed water temperature )  X 100 / (Economizer inlet flue gas temperature - Economizer inlet feed water temperature)

30-APH efficiency calculation

APH air side efficiency

ηAPHa = (Air outlet temp-Air inlet temp)) X 100 / (Flue gas inlet temperature -Air inlet temperature)

APH gas side efficiency

ηAPHg = (Flue gas inlet temp.-Flue gas outlet temp) X 100 / (Flue gas inlet temperature -Air inlet temperature )

31-Calculation of steam cost

Steam cost per ton = Steam enthalpy  in kcal/kg X Fuel price per ton/ (Boiler efficiency % X GCV of fuel used in kcal/kg)

32-Travelling grate Boiler heating surface calculation

Boiler heating surface (Appx) = Boiler capacity in kg/hr / 18

33-AFBC Boiler heating surface calculation

Boiler heating surface (Appx) = Boiler capacity in kg/hr / 22

34-Travelling grate slop fired Boiler heating surface calculation

Boiler heating surface (Appx) = Boiler capacity in kg/hr / 12

35-AFBC  slop fired Boiler (Low pressure up to 10 kg/cm2 WP) heating surface calculation

Boiler heating surface (Appx) = Boiler capacity in kg/hr / 8.2

36-Calculation of draught produced in Chimney

Hw = 353 X H (1/Ta – 1/Tg (Ma+ 1)/Ma)

H = Chimney height in meters

Ta = Atmospheric temperature in K

Tg = Flue gas temperature in K

Ma = Mass of air & Mass of flue gas = Ma+1

 

Also given as;

 P = 176.5 X H / Ta

Hw = Chimney height in meters

Ta = Absolute atmospheric temperature in Kelvin

Hw = Draught in mmwc

37-Calculation of mass of flue gas flowing through chimney

Mg (kg/sec)= Density of gas (kg/m3) X Area of Chimney (m2) X Velocity of flue gas in Chimney (m/sec)

B-Turbine and Auxiliaries

1-Turbine heat rate calculation

a-Heat rate of Thermal power plant Turbine in kcal/kw =

Steam flow X (Steam enthalpy in kcal/kg-Feed water enthalpy in kcal/kg) / Power generation

b-Heat rate of Co-gen plant Turbine in kcal/kg =

Inlet steam flow X Enthalpy-(Sum of extraction steam flow X Their enthalpy + Exhaust steam X Enthalpy) / Power generation

i.e THR = ((Steam Flow x Steam Enthalpy)-(1St EXT Flow x Its Enthalpy + 2nd Ext flow x its Enthalpy + 3rd Ext flow x Its Enthalpy+ Exhaust Steam flow x its Enthalpy)) /Power Generation

Or

THR=((Steam Flow x Steam Enthalpy +Makeup Water flow x Its Enthalpy+ RC Flow x RC Enthalpy)-(Process-1 steam flow x its Enthalpy + Process-2 steam flow x Its Enthalpy+ FW Flow x FW Enthalpy)) /Power Generation

2-Turbine efficiency calculation

Efficiency = 860 X 100 /  Turbine heat rate

3-Steam condenser efficiency calculation

Condenser efficiency =Difference in cooling water inlet & outlet temperatures X 100/(Vacuum temperature-condenser Inlet temperature of cooling water)

Condenser efficiency = (T2 - T1) X 100/(T3 - T1)

T2: Condenser outlet cooling water temperature,

T1: Condenser inlet cooling water temperature,

T3: Temperature corresponding to the vacuum or absolute pressure in the condenser.

4-Vacuum efficiency calculation

Vacuum efficiency = (Actual vacuum in condenser X 100)/Max. Obtainable vacuum.

I.e

Vacuum efficiency in % =Actual vacuum X 100 / (Atmospheric pressure or barometric pressure-Absolute pressure)

5-Cooling tower range

Range = Cooling tower outlet water temperature-Cooling tower inlet water temperature

6-Cooling tower approach

Approach = Cooling tower outlet cold water temperature - Wet bulb temperature

7-Cooling efficiency calculation

Efficiency = Range X 100 / (Range + Approach)

8-Heat rejected or heat load of cooling towers

Heat load =Mass of circulating water X Specific heat of water Cp X Range

9-Cooling tower evaporation loss calculation

Evaporation loss in m3/hr = 0.00085 X 1.8 X Water circulation rate  m3/hr X Range 

Evaporation loss in % = Evaporation loss X 100 / Water circulation rate  m3/hr

10-Cooling tower blow down loss calculation

Blow down loss in % = Evaporation loss X 100 / (COC-1)

Where COC: Cycles of concentration

Its generally calculated as;

COC = Conductivity in circulation water / Conductivity in makeup water

OR

COC = Chloride in circulation water / Chloride in makeup water

11-Calculation of mass of cooling water required to condenser steam in surface condensers

Mw = (Ms X (hfg X dryness fraction(x) + Cpw (T3 - Tc)))/(Cpw X (T2 - T1))

Mw = Mass of cooling water required in TPH

Ms = Mass of exhaust steam to condenser in TPH

Hfg = Enthalpy of evaporation at exhaust pressure in kcal/kg

Cpw = Specific heat of cooling water in kcal/kg

T3= Temperature at exhaust pressure in deg C

Tc= Temperature of condensate in deg C

T1=Cooling water temperature entering condenser in deg C

T2 = Cooling water temperature leaving condenser in deg C

12-Steam turbine wheel chamber pressure calculation (Appx)

Turbine wheel chamber pressure (kg/cm2 ) = (Turbine inlet pressure (kg/cm2 ) X Turbine operating load (MW) X 0.6) / Turbine rated capacity (MW).

13-Calculation of power generation in steam Turbine

Power generation in MW= Turbine inlet steam flow  in TPH X (Inlet steam enthalpy in kcal/kg- Exhaust steam enthalpy in kcal/kg) / 860

14-Power generation calculation in multi stage Turbines

Power generation in MW= Steam flow from 1st stage X (Inlet steam enthalpy in kcal/kg- 1st stage extraction steam enthalpy in kcal/kg) + Steam flow from 2nd  stage X (Inlet steam enthalpy in kcal/kg- 2nd  stage extraction steam enthalpy in kcal/kg) + Exhaust steam flow to condenser X (Inlet steam enthalpy in kcal/kg- Exhaust  steam enthalpy in kcal/kg) / 860

15-Calculation of work done per kg of steam in Turbine

Work done/kg of steam = Inlet steam enthalpy in kcal/kg-Exhaust steam enthalpy in kcal/kg

16-Calculation of steam required per per KWH

Steam required per KWH = 860 / (Work done per kg of steam)

Or

Steam required per KWH =860/(Inlet steam enthalpy in kcal/kg-Exhaust steam enthalpy in kcal/kg)

17-Thermal power plant efficiency calculation

Efficiency = 860 X Power generation / Heat input

Efficiency = 860 X PG X 100 / (Fuel consumption X Fuel GCV)

18-Co gen-plant efficiency calculation

Efficiency = 860 X Power generation X 100 / (Fuel consumption X GCV + Make up water X Make up water enthalpy + Return condensate water X Enthalpy-Process steam flow X Enthalpy)

19-HP heater steam consumption calculation

Steam flow in TPH = FW flow in TPH X (HP heater outlet FW temperature-HP heater inlet FW temperature) /(Steam enthalpy in kcal/kg-HP heater outlet condensate water enthalpy in kcal/kg)

Where, FW =  Feed water

20-Deaerator steam consumption

Mass of steam in TPH = (Deaeraor outlet Feed water flow in TPH X Enthalpy –CEP flow X Enthalpy-Makeup water X Enthalpy) / (Enthalpy of steam-Enthalpy of deaerator outlet water)

Note: Enthalpy in kcal/kg

Feed water, CEP water & Make up water flow in TPH

C-Pumps fans and air compressors

1-Centrifugal pumps hydraulic power consumption

Hydraulic power in Kwh = Flow (M3/sec) X Total head (m) X Fluid density (kg/m3) X g (m/s2) / 1000

2-Pump shaft power calculation

Pumps shaft power Ps = Hydraulic power / Pump efficiency

3-Pump electrical power calculation

Electrical power = Pump shaft power / Motor efficiency

4-Pump affinity laws

a-Q1/Q2 = N1/N²

Q1 & Q2 are pump flow at different RPM

N1 & N2 are pump speed

b-H1 / H2 = (N1)² / (N2)²

Where H1 & H2 are pump head at different RPM

N1 & N2 are pump speed

c-P1 / P2 = (N1)³ / (N2)³

P1 & P2 are pump flow at different RPM

N1 & N2 are pump speed

5-Relation between head,flow ,power and impeller diameter

Q = D (Flow is directly proportional to impeller diameter)

H = D2 (Head is directly proportional to square root of impeller diameter)

P =D3 (Power is directly proportional to impeller diameter)

6-Relation between NPSHR, flow and speed

NPSHR = N2 (NPSHR is directly proportional to square root of speed)

That is as the flow and speed increase, pump need more NPSH

7-Fans volumetric flow calculation

Flow (m3/sec) = Velocity of fluid (m/sec) X Area of duct (m2)

8-Fans Mechanical efficiency calculation

Mechanical efficiency % = Flow (m3/sec) X Total pressure (mmwc) X 100 / (102 X power input to fan shift in KW)

9-Fans Technical efficiency calculation

Static efficiency % = Flow (m3/sec) X Static pressure (mmwc) X 100 / (102 X power input to fan shift in KW)

11-Fans affinity laws

a-Q1/Q2 = N1/N2

Q1 & Q2 are fans flow in m3/sec at different RPM

N1 & N2 are fans speed

b-SP1 / SP2 = (N1)² / (N2)²

Where SP1 & SP2 are static pressure in mmwc of fan at different RPM

N1 & N2 are pump speed

c-P1 / P2 = (N1)³ / (N2)³

P1 & P2 are flow flow at different RPM

N1 & N2 are pump speed

11-Isothermal power of reciprocating compressors

P = P1 X Q X log_e r/36.7

Where, P = Isothermal power

Q = Free air delivered by compressor in m3/hr

R = P2/P1

P1 & P2 are absolute inlet & deliver pressure respectively

10-Isothermal efficiency of reciprocating compressors

Isothermal efficiency = Isothermal power X 100 / Actual measured input power

11-Isothermal efficiency of reciprocating compressors

Isothermal efficiency = Free air delivered (m3/min) X 1007 / Compressor displacement (m3/min)

12-Compressor displacement = 3.142 X D2 X L X S X n X N

Where, D = Cylinder bore in meter

L = Cylinder stroke in meter

S = Compressor speed in rpm

n= 1 for single acting & 2 for double acting cylinders

N=No.of cylinders

13-Compressed air leakage calculation

% of leakage = T X 100 /  (T+t)

Or air leakage in m3/min = T X Compressor capacity (m3/min) / (T+t)

T = Loading time in minutes, t = Unloading time in minutes

14-Calculation of intermediate pressure of reciprocating air compressors

P2 =SQRT ( P1 X P3)

Where P1,P2 & P3 are suction pressure, intermediate pressure and discharge absolute pressure respectively

D-Fuel handling

1-Minimum length of belt conveyor take up

Total belt length X 1.5%

2-Conveyor belt speed calculation

Belt speed (m/sec) = 3.142 X D X N / 60

Where, D= Head pulley diameter in meter

N = Head pulley speed in RPM

Head pulley speed = Moto RPM / Gear box reduction ratio

3-Conveyor start up tension calculation

T (N/mm)= P X 3.2 / (V X W)

Where,P = Power consumption of conveyor

V = Belt speed in m/sec

W = Belt width in mm

4-Tensile strength of belt

Initial tension of the belt X 5.4 / Belt splicing efficiency

E-Mechanical Maintenance

1-V pulley speed calculation

N1/N2 = D2/D1

N1 & N2 are speed in RPM of drive and non drive end pulleys

D1 & D2 are diameter (inch or mm) of drive and non drive end pulleys

2-Gear box output speed calculation

Gear box out put shaft speed = Motor RPM / Reduction ratio

3-Speed of pulley calculation

Pulley speed (m/sec) = 3.142 X D X N / 60

Where, D = Diameter of pulley in meter

N = Pulley rotary speed in RPM

4-V belt length calculation

 L =∏ (R1+R2) + 2C + (R2+R1)2/C

Where, R1 & R2 are radius of DE & NDE pulleys

5-C is centre distance between two pulleys

6-Torque developed on a shaft

T = P X 60 / (2 X ∏ X N)

Where, P = Power consumption, N is speed of the shaft

7-Calculation of hub diameter of a shaft

Hub diameter = 2 X Shaft diameter

8-Bearing lubrication quantity calculation

Bearing lubrication quantity in grams = Bearing OD (D) X Width (B) X 0.05

9-Safe working load of steel wire rope of size D (inches)

SWL = 8D²

10-Safe working load of Chain block having Load link diameter D in mm

 Safe working load of chain block = 80 X 0.4 X D²

Where 80 is the grade of chain block steel material

D is the link diameter in mm

12-Safe working load of Nylon lifting rope of  diameter D in mm

Safe working load of chain block = D²

13-Calculation of weight of steel plate

Weight in kg = Plate length (m) X width (M) X Thickness (m) X density of steel (kg/m3)

14-Calculation of weight of round steel plate

Weight in kg = (∏D²/4) X plate thickness in meter

D is diameter of plate in meter

15-Calculation of weight of hollow cylinder

Weight in kg = (∏R²₂X h- ∏R²₁X h) X steel density in kg/m3

Where, R1 & R2 are outer radius & inner radius of tank in meter

H is height of tank in meter

F-Electrical and Instrumentation

1-Single phase power consumption calculation

P = VI Cos θ

Where,Cos θ P is power consumption in Watts, V&I are Voltage in volts and current in Amps receptively

Cos θ is power factor

2-Three  phase power consumption calculation

P = √3VI Cos θ

3-Apparent power = V X I (Voltage X Current)

4-Real power = √3VI Cos θ

5-Reactive power of 3-phase power =√3VI Sin θ

Reactive power of single-phase power =VI Sin θ

 

 

6-Resistance of a conductor

R = ƍ X L/A

R = Resistance in Ohm-meter or Ohm-cm

L & A are Length and area of conductor measured in meter & m2

7-Voltage, current and resistance relation

Voltage = Current X Resistance, i.e. V = IR

8-Resistance of a conductor at temperature t °C is given by

Rt = R0 (1 + α t) Where α is temperature coefficient at t °C.

8-Electrical Power calculation

   Power= Work done watt-sec./Time in second

   P = VI X t/t i.e. VI Watts

   P = I²R… Watts

9-Power factor calculation

Power factor (cosՓ) = Active power (KW) / Apparent power (KVA)

10-Synchronous speed calculation

Synchronous speed/RPM = (120 X frequency)/No. of poles.

11-Absolute Pressure = Gauge Pressure + Atmospheric Pressure

12-Relation between flow and differential pressure (DP). Flow (Q) = √(DP

Q1/Q2 = (√∆P1/∆P2)

13-Compensated flow calculation

Compensated flow= Raw value from flow meter X Sqrt ((Absolute Design temperature/Absolute Operating temperature) X (Absolute operating pressure/Absolute design pressure))

Atmospheric pressure goes on decreasing as altitude goes on increases.

It is 1.033 kg/cm2 at sea level & goes on decreasing at higher elevation from sea level.

It is calculated as below;

Formula for calculation of atmospheric pressure at height more than sea level

Atmospheric pressure at particular elevation = Atmospheric pressure at sea level X (1–2.255 X (10^-5) X Elevation in meter)^5.25585

Note: Pressure is in Kg/cm2

Elevation is in meters.

For more read>>> Power plant and calculations

Practical Approach to Power Plant Operation and Maintenance