Why does vacuum in steam condenser reduce or drop??

 

1-High exhaust temperature:

Vacuum drops or maintains at lower side due to high exhaust steam temperature flow into steam condenser. This high exhaust temperature is mainly due to

1-Operation of Turbine at lower loads

2-More clearance in labyrinth seals

3-Not operating exhaust hood sprays

4-More load on condenser

5-Breaking of ejector U loop

2-Low circulating cooling water flow

Vacuum in condenser reduces due to inadequate cooling water flow through steam condenser. This is mainly due to;

1-Problems associated with pumps

2-Air pockets in pipe line

3-Leakages in cooling water line

4-Stuck of discharge valve of pump

3-High cooling water temperature at condenser inlet

Higher cooling water temperature at condenser inlet results into reduction of vacuum due to poor heat transfer from steam to water

4-Poor heat transfer in condenser

Very less or poor heat transfer in steam condenser reduces vacuum to very low level resulting into high exhaust temperature & disturbances in hot well level.

Poor heat transfer is due to;

1-Fouling of condenser tubes due to poor water quality

2-High cooling water temperature

5-Air ingress in condenser & other vacuum pulling system

This is the most top reason for sudden drop of vacuum in steam condenser. Air ingress into the condenser is mainly from flange joints, gaskets, valves etc

Passing of vacuum breaker valve also be the one of the reason for maintaining low vacuum in condenser

6-Low steam temperature & pressure at ejector inlet:

Parameters lesser than rec recommended leads to reduction of vacuum pulling

7-Poor heat transfer in steam jet ejectors

This is mainly due to fouling of ejector tubes

steam condenser vacuum & calculations

8-Damages to air ejector nozzles

Increase in nozzle clearances leads to reduction of vacuum creating efficiency of ejectors

9-Breaking of ejector inter condenser U seal loop.

This creates escape of hot condensate directly into condenser leading to increase in exhaust temperature & reduction in vacuum

10-Low gland steam pressure:

Gland steam pressure lesser than design creates ingress of outside air into condenser through turbine glands.

11-Loading the condenser more than design

If the condenser steam load is more than design, vacuum drops slowly

12-High a temperature & low pressure in the atmosphere reduces vacuum pulling efficiency

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Steam Turbine SOPs

Why does load hunting occur in steam Turbines??

 If Turbine does not maintain the load as per set load, then this condition is called load hunting.Following are the some potential reasons for load hunting

1-Problems associated with actuators:

These are related to leakages in actuators, piston stuck up, oil holes elongation etc. Because of these issues there will interruption or fluctuation of secondary oil flow through actuators, this creates the problems of actuator miss-operation & eventually load hunting.

2-Improper calibration of actuators:

This results into mismatch of actuator opening & given set point or valve demand

3-Lower control oil pressure than required:

Actuators are designed for specific pressure of control oil, if the control oil pressure at actuator inlet becomes less, then there will be more chances of mal function of actuator.

4-Fluctuation of control oil pressure/flow:

Fluctuation of control oil pressure or flow due to malfunction of pump or line PRV may lead to actuator misoperation & hence creates load hunting.

5-Control oil line leakage:

Leakages in control oil line welding & flange joints will lead to fluctuation of flow & pressure causing actuator malfunction & load hunting.

6-Contamination in control oil

Foreign particles present in oil lead to improper functioning of actuators, which causes load hunting

7-Burs or scoring marks on actuator spindle & control valves spindles:

Burs or any rough scoring marks on spindles will lead to improper operation of actuator & valves

8-Passing of control valves:

This is the major reason for load fluctuation & Turbine over speed

9-Improperly set control valve cones:

During turbine HP valve assembly after maintenance, HP control valve cones should be set as per factory set readings, if it is disturbed, then there will be issues related to load hunting, low load at more HP demand or over speed.

10-Damage or broken control valves spindle & discs (cones):

Spindle damage or disc damage makes uncontrolled operation, as there will be wrong response from control valves to governor.

11-Wrongly tuned P&IDs in Governor: The disturbed values in P&ID tuning will result into heavy load hunting

12-Malfunction of Governor: This is very rare, but certainly results into load fluctuation of Turbine trip

13-Sudden changes in inlet steam pressure & extraction steam pressure

14-Fluctuation of grid frequency

15-Failure of Turbine inlet & extraction pressure sensor

For related articles on power plant Operation & Maintenance click here

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Calculated reasons for increase in Turbine specific steam consumption

1. Lower vacuum

Turbine consumes more steam, if vacuum in condenser is maintained on lower side.

Example:  Consider a 20 MW Steam Turbine having Inlet steam parameters 65 kg/cm2 & 490 Deg C & Vacuum maintained in condenser is -0.9 kg/cm2.

Calculate the steam consumption of turbine at vacuum -0.9 kg/cm2 & -0.85 kg/cm2

A-Steam consumption Q at -0.9 kg/cm2 to develop 20 MW power

P =Steam flow X( Enthalpy of inlet steam-Enthalpy of exhaust steam)/ 860

Enthalpy of inlet steam at inlet steam parameters =810 kcal/kg

Exhaust steam enthalpy at -0.9 kg/cm2 vacuum = 619 kcal/kg

Then, 20 = Q X (810-619)/860

Q1 = 90 MT

B- Steam consumption Q at -0.85 kg/cm2 to develop 20 MW power

Exhaust steam enthalpy at -0.85 kg/cm2 vacuum= 623 kcal/kg

Then, 20 = Q X (810-623)/860

Q 2= 90.9 MT

It is clear that, Turbine operating at -0.9 kg/cm2 vacuum consumes lesser steam as compared to turbine operating at vacuum-0.85 kg/cm2

2. Lower inlet main stream pressure& temperature

Turbine operating at higher main steam pressure consumes lesser steam as compared to turbines operating at lower pressure

Example: Consider a 20 MW Steam Turbine having Inlet steam temperature 490 Deg C & Vacuum maintained in condenser is -0.9 kg/cm2.

A-Inlet steam parameters: Pressure: 65 kg/cm2 & temperature 490 deg C , Enthalpy = 810 kcal/kg

Exhaust steam parameters P = 0.9 kg/cm2 & Enthalpy = 619 kcal/kg

Steam consumption of Turbine Q = P X 860 / (Enthalpy of inlet steam-Enthalpy of exhaust steam)

Q = 20 X 860 / (810-619)

Q1 = 90.05 MT

B-Inlet steam parameters: Pressure: 87 kg/cm2 & temperature 515 deg C , , Enthalpy = 818 kcal/kg

 

Steam consumption of Turbine Q = P X 860 / (Enthalpy of inlet steam-Enthalpy of exhaust steam)

Q = 20 X 860 / (818-619)

Q2 = 86.43 MT

It is clear that, Turbine operating at pressure 65 kg/cm2 & temperature 490 deg C consumes more steam as compared to turbine operating at 87 kg/cm2 & temperature 515 deg C

3. Higher extraction/bleed steam flow

Steam turbines consume more steam to develop same power on higher steam extraction as compared to lower extraction.

Example: A condensing & extraction steam turbine having Inlet steam flow 105 TPH at pressure 65 kg/cm2 & 490 Deg C & Vacuum maintained in condenser is -0.9 kg/cm2.

Here we can cross check the power generation by steam turbine by increasing the extraction flow keeping inlet steam constant.

A-Extraction pressure = 2 Kg/cm2 & Temperature = 150 Deg C, flow = 75 TPH, Exhaust steam to condenser = 30 TPH

Enthalpy of inlet steam, H1 = 810 kcal/kg

Main steam flow Q1 = 105 TPH

Enthalpy of extraction steam = H2 =660 kcal/kg

Extraction steam flow Q2 = 75 TPH

Enthalpy of exhaust team = 620 kcal/kg

Exhaust steam flow Q3 = 30 TPH

Power developed by steam Turbine P = (Q2 X (H1-H2) / 860) + (Q3 X (H1-H3) / 860 )

P = (75 X (810-660) / 860) + (30 X (810-620) / 860) = 19.7 MW

B- Extraction pressure = 2 Kg/cm2 & Temperature = 150 Deg C, flow = 65 TPH, Exhaust steam to condenser = 40 TPH

Enthalpy of inlet steam, H1 = 810 kcal/kg

Main steam flow Q1 = 105 TPH

Enthalpy of extraction steam = H2 =660 kcal/kg

Extraction steam flow Q2 = 65 TPH

Enthalpy of exhaust team = 620 kcal/kg

Exhaust steam flow Q3 = 40 TPH

Power developed by steam Turbine P = (Q2 X (H1-H2) / 860) + (Q3 X (H1-H3) / 860 )

P = (65 X (810-660) / 860) + (40 X (810-620) / 860) = 20.16 MW

It is clear that, Turbine power generation at same inlet main steam flow will increase as extraction flow gets decrease & vice versa

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Power plant & Calculations

4. Higher pressure/temperature of extraction & bleed steam

Higher pressure/temperature of extraction & bleed steam leads to increased steam consumption to generate same power or power consumption reduces at same inlet flow.

Example: A condensing , extraction & bleed steam turbine having Inlet steam flow 105 TPH at pressure 65 kg/cm2 & 490 Deg C & Vacuum maintained in condenser is -0.9 kg/cm2

 A-Bleed steam 10 kg/cm2 & Temperature 200 Deg C, flow =25 TPH, Extraction pressure = 2 Kg/cm2 & Temperature = 150 Deg C, flow = 60 TPH, Exhaust steam to condenser = 25 TPH

Enthalpy of inlet steam, H1 = 810 kcal/kg

Main steam flow Q1 = 105 TPH

Enthalpy of bleed steam = H2 =674 kcal/kg

Bleed steam flow Q2 = 25 TPH

Enthalpy of extraction steam = H3 =660 kcal/kg

Extraction steam flow Q3 = 60 TPH

Enthalpy of exhaust team H4= 620 kcal/kg

Exhaust steam flow Q4 = 20 TPH

Power developed by steam Turbine P = (Q2 X (H1-H2) / 860) + (Q3 X (H1-H3) / 860 ) +(Q4 X (H1-H4)/860)

P = (25 X (810-674) / 860) + (60 X (810-660)/860) + (20 X (810-620)/860)

P = 18.82 MW

B-Bleed steam 14 kg/cm2 & Temperature 260 Deg C, flow =25 TPH, Extraction pressure = 2.5 Kg/cm2 & Temperature = 170 Deg C, flow = 60 TPH, Exhaust steam to condenser = 25 TPH

Enthalpy of inlet steam, H1 = 810 kcal/kg

Main steam flow Q1 = 105 TPH

Enthalpy of bleed steam = H2 =704 kcal/kg

Bleed steam flow Q2 = 25 TPH

Enthalpy of extraction steam = H3 =669 kcal/kg

Extraction steam flow Q3 = 60 TPH

Enthalpy of exhaust team H4= 620 kcal/kg

Exhaust steam flow Q4 = 20 TPH

Power developed by steam Turbine P = (Q2 X (H1-H2) / 860) + (Q3 X (H1-H3) / 860) + (Q4 X (H1-H4)/860)

P = (25 X (810-704) / 860) + (60 X (810-669)/860) + (25 X (810-620)/860)

P = 18.43 MW

It is clear that, Turbine power generation reduces at higher extraction or bleed steam pressure &temperature

Note: Steam consumption of turbine increases if,

1-Bleed steam & extraction steam pressure increases

2-Bleed steam & extraction steam temperature increases

3-Bleed steam flow & extraction steam flow increases

4. Increase of exhaust steam temperature due to more clearance in labyrinth seals

Turbine steam consumption increases if exhaust steam temperature to condenser increases.

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How do you calculate the Power generation in steam Turbines??

Power in the steam Turbines produces at every stage where the steam is taken out, whether it may be bleed, extraction or exhaust steam. As the steam out from the turbine increases the power developed on that particular stage will increase.

Power generation phenomenon.

Power generation in steam Turbines is calculated based on difference between the heat content of inlet steam & extracted steam.

Factors affecting the power generation:

• Power generation at particular stage increases, when there is more steam flow &vice versa
• Power generation at particular stage increases when there is more difference between inlet & extraction steam & Vice versa
• Power develop at particular stage decreases if its extraction pressure increases & vice versa
• Power developed at particular stage decreases if its extraction temperature increases & vice versa
• Power developed in steam Turbine decreases if inlet live steam pressure & temperature decrease
• If steam vacuum decreases power generation reduces or else Turbine will consume more steam to develop same power
• If exhaust steam temperature increases then the power power generation reduces or else Turbine will consume more steam to develop same power
• If wheel chamber pressure increases, then the power generation capacity of the Turbine decreases

HOW DO YOU CALCULATE POWER GENERATION COST??

In which part of the Turbine higher power can be produced at lower steam consumption? And why?

It is at the exhaust stage. Because at the exhaust stage pressure & temperature of the steam is very lesser than bleed &extraction stages.

In which part of the Turbine lowest power is produced at higher steam consumption? And why?

It is at the bleed stage. Because at bleed steam pressure & temperatures are higher than extraction & exhaust stages

Calculation part:

1-Calculate the power generated in a back pressure steam Turbine, where 50 TPH steam enters the Turbine at 66 kg/cm2 & temperature 485 Deg C.And steam exhausts to process at pressure 2 kg/cm2 & temperature 180 Deg C.

For calculation of power we need to know the enthalpy of inlet & exhaust steam.

Refer steam table

Enthalpy of inlet steam at rated parameters H1 = 806.5 kcal/kg

Enthalpy of inlet steam at rated parameters H2 = 677 kcal/kg

Now power developed in steam turbine P = Q X (H1-H2) / 860

Where Q is steam flow

P = 50 X (806.5-677) / 860

P = 7.52 MW

Note: 860 kcal = 1 KWH

2. Calculate the power developed by a steam turbine by using following data

Sl No.	Particular	UOM	Value
1	Turbine inlet steam flow	TPH	145
2	Turbine inlet steam pressure	Kg/cm2	88
3	Turbine inlet steam temperature	C0	515
4	Bleed steam flow	TPH	20
5	Bleed steam pressure	Kg/cm2	12
6	Bleed steam temperature	C0	250
7	Extraction flow	TPH	100
8	Extraction steam pressure	Kg/cm2	1.8
9	Extraction steam temperature	C0	145
10	Exhaust flow to condenser	TPH	25
11	Exhaust pressure	Kg/cm2	0.08
12	Exhaust temperature	C0	44
13	Dryness fraction	%	90

Also calculate the specific steam consumption of this Turbine

Solution:

Note down the enthalpy of steam at various stages

Turbine inlet steam enthalpy H1 = 818 kcal/kg

Bleed steam enthalpy H2 =700 kcal/kg

Extraction steam enthalpy H3 = 657.2 kcal/kg

Exhaust enthalpy =Liquid heat + Dryness fraction X Vapour enthalpy = 41.77 + 0.9 X 615.5 = 595.72 kcal/kg

Also steam flow is given

Inlet steam flow Q1 = 145 TPH

Bleed steam flow Q2 = 20 TPH

Extraction steam flow Q3 =100 TPH

Exhaust steam flow Q4 = 25 TPH

Power generation at 1st stage (Bleed) P1 = Q2 X (H1-H2) / 860

P1 = 20 X (818-700) / 860 = 2.74 MW

Power generation at 2^nd stage (Extraction) P2 = Q3 X (H1-H3) / 860

P2 = 100 X (818-657.2) / 860 = 18.69 MW

Power generation at 3rd stage (Exhaust) P3 = Q4 X (H1-H4) / 860

P3 = 25 X (818-595.72) / 860 = 6.46 MW

SO total power developed at Turbine shaft P = P1+P2+P3 = 2.74+18.69+6.46 = 27.89 MW

Specific steam consumption SSC = Turbine inlet steam flow / Power generation = 145 / 27.89 =5.19 MT/MW

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3-By taking above example, explain how pure condensing steam Turbines have higher power generation & lower specific steam consumption (SSC)

Consider the above example, where Turbine inlet steam flow is 145 TPH & having 25 TPH exhaust steam flow to condenser.

If we condense 100% steam, then we will have reduced SSC

Let us see..

Q1=Q4=145 TPH..Where Q4 = Exhaust steam to condenser

Q2 & Q3 are considered zero (No flow)

Enthalpy of exhaust steam increases due to higher exhaust pressure

SO, Consider exhaust pressure = 0.1 kg/cm2 & Dryness fraction 0.9

Then, exhaust enthalpy becomes H1 = 46 + 0.9 X 617 = 601.9 kcal/kg

Total Power developed P= Q1 X (H1-H4) / 860

P =145 X (818-601.9) / 860 = 36.43 MW

So SSC = 145 / 36.43 = 3.98 MT/MW

This is very less as compared to bleed & extraction turbines

It's all about HP HEATERS

4-By taking an example No.2 explain how bleed steam flow will cause reduction in net power consumption

Solution:

We shall take the data of Example No.2

In this, we will increase bleed steam flow, pressure & temperature & parallel shall decrease extraction flow to match mass flow

Sl No.	Particular	UOM	Value
1	Turbine inlet steam flow	TPH	145
2	Turbine inlet steam pressure	Kg/cm2	88
3	Turbine inlet steam temperature	C0	515
4	Bleed steam flow	TPH	30
5	Bleed steam pressure	Kg/cm2	15
6	Bleed steam temperature	C0	285
7	Extraction flow	TPH	90
8	Extraction steam pressure	Kg/cm2	1.8
9	Extraction steam temperature	C0	145
10	Exhaust flow to condenser	TPH	25
11	Exhaust pressure	Kg/cm2	0.08
12	Exhaust temperature	C0	44
13	Dryness fraction	%	90

Solution:

Note down the enthalpy of steam at various stages

Turbine inlet steam enthalpy H1 = 818 kcal/kg

Bleed steam enthalpy H2 =716.63 kcal/kg

Extraction steam enthalpy H3 = 657.2 kcal/kg

Exhaust enthalpy =Liquid heat + Dryness fraction X Vapour enthalpy = 41.77 + 0.9 X 615.5 = 595.72 kcal/kg

Also steam flow is given

Inlet steam flow Q1 = 145 TPH

Bleed steam flow Q2 = 30 TPH

Extraction steam flow Q3 =90 TPH

Exhaust steam flow Q4 = 25 TPH

Power generation at 1st stage (Bleed) P1 = Q2 X (H1-H2) / 860

P1 = 30 X (818-716.63) / 860 = 3.53 MW

Power generation at 2^nd stage (Extraction) P2 = Q3 X (H1-H3) / 860

P2 = 90 X (818-657.2) / 860 = 16.82 MW

Power generation at 3rd stage (Exhaust) P3 = Q4 X (H1-H4) / 860

P3 = 25 X (818-595.72) / 860 = 6.46 MW

SO total power developed at Turbine shaft P = P1+P2+P3 = 3.53+16.82+6.46 = 26.81 MW

Specific steam consumption SSC = Turbine inlet steam flow / Power generation = 145 / 26.81 =5.41 MT/MW

So by increasing the bleed steam flow, the work done by the Turbine decreases & hence steam consumption will increase

Read Steam Turbine commissioning

5-A Turbine’s inlet steam enthalpy is 825 kcal/kg & Exhaust enthalpy is 590 kcal/kg. Calculate the work done by steam & specific steam steam consumption

We have,

H1 = 825 kcal/kg, H2 = 590 kcal/kg

Work done per kg of steam = (H1-H2) = 825-890 =235 kcal/kg

SSC = 860 / Work done = 860 / 235 =3.65 kg/kwh or 3.65 MT/MW

6-A steam Turbine inlet steam pressure & temperatures are 104 kg/cm2 & 540 C⁰ & exhausts at pressure 0.09 kg/cm2 & temperature 43 Deg C calculate the

a-      Work done per kg of steam

b-      Heat supplied per kg of steam

c-       Cycle efficiency

Enthalpy of inlet steam = 829 kcal/kg

Exhaust liquid enthalpy = 44 kcal/kg

Exhaust enthalpy by considering 90% dryness fraction = 44 + 0.9 X 616.44 =598.76 kcal/kg

A-Work done per kg of steam = (829-598.76) = 230.24 kcal/kg

B-Heat supplied per kg of steam = 829-44 = 785 kcal/kg

C-Cycle efficiency = Work done per kg of steam X 100 / Heat supplied per kg of steam

                                 = 230.24 X 100 / 785 = 29.32%

Note:

Power developed at Generator terminals = Power developed at Turbine Shaft X Reduction gear box efficiency X Alternator efficiency

For example:

Calculate the net power developed at Generator terminal if 100 TPH steam enters the Turbine at 811 kcal/kg enthalpy & leaves the Turbine at enthalpy 565 kcal/kg .Assume Gear box efficiency as 98% & Generator efficiency as 95%

Power developed on Turbine shaft = 100 X (811-565) / 860 = 28.0 MW

Net power developed at Generator output terminals = 28.0 X 0.98 X 0.95 = 26.06 MW

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