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Course of Energy Conversion A 2014/2015

Prof. G. Valenti

_________________________________________________

4th project

Advanced steam cycles

Index:

1. Introduction;

2. Entropic analysis for all configurations ;

3. General results and comments

1. Introduction

Steam cycle are really important in electric power generation nowadays.

In this project we will analyze six different configuration, starting from a basic configuration and varying

some of the plant parameters such as:

• Fuel

• Regenerators number

• Reheaters number

• Condensing pressure

• Economizer water inlet

• Maximum temperature of the cycle

The starting configuration is typical for large scale steam power plant:

• Superheater outlet mass flow rate: 1˙850˙000 kg/h

• Superheater outlet steam pressure: 270 bar

• n re-heaters: 1

• n pre-heaters (deaerator included) : 8 (4 BP, 4 AP)

• n intermediate-pressure two-flow rotors: 2

• n low-pressure two-flow rotors: 4

• last stage blading: 33.5 inches – mean diameter 99.5 inches

• steam temperatures: 580 °C / 580 °C

• condensing pressure: 0.05 bar

• fuel: coal

• exhaust gases stack temperature 150 °C

• O2 concentration in exhaust gases at the stack: 5%

• economizer water inlet temperature: ~ 315 °C

Thanks to CICVAP we perform easily the analysis of the plant both in terms of first law and second law.

We report in a table and with some graphs the entropy analysis of the plant dividing the irreversibilities:

• combustion related losses (including air and fuel pre-heating);

• cycle heat introduction losses: heat exchange between exhausts and isothermal line at Tmax ;

• cycle heat introduction losses: heat exchange between isothermal line at Tmax – steam;

• regeneration related losses (deaerator included);

• cycle waste heat losses;

• working fluid compression losses;

• working fluid expansion losses (including outlet kinetic energy losses);

• stack losses;

• various

2. Entropic analysis for all configurations

Configuration A

We perform the entropic analysis for this plant taken as a reference, hence we vary some parameters to

get some information about the losses of exergy in a power plant in relation with the plant configuration.

Entropic analysis A

second law efficiency 41,83%

combustion related losses 27,46%

heat exchange between exhausts and Tmax 10,32%

heat exchange between Tmax – steam 7,43%

regeneration related losses 1,82%

cycle waste heat losses 2,94%

working fluid compression losses 0,40%

working fluid expansion losses 4,60%

stack losses 1,33%

various 1,86%

TOT 100,00%

0,00% 10,00% 20,00% 30,00% 40,00% 50,00%

Percentual

Second law analysis A

various

stack losses

working fluid expansion losses

working fluid compression losses

cycle waste heat losses

regeneration related losses

heat exchange between Tmax – steam

heat exchange between exhausts and

Tmaxcombustion related losses

second law efficiency

Configuration B: Natural gas, lower temperature at the stack

• fuel: natural gas

• exhaust gases stack temperature: 120 °C

• stack O2 concentration in exhausts: 2.5%

Entropic analysis B









stack losses 0,75%

various 1,87%

TOT 100,00%

0,00% 10,00% 20,00% 30,00% 40,00% 50,00%

1

Second law analysis Bvarious

stack losses









Configuration C: higher condensing pressure

Entropic analysis C









stack losses 1,33%

various 1,84%

TOT 100,00%

Condensing pressure: 0.1 bar

The choice of the minimum pressure of the cycle is really important. Thermodynamically speaking, the lower is

the pressure the better is the cycle. Reality is different due to some factors. The main reason is the availability

of the refrigerant medium.

Where there’s the possibility to exploit cold sea water the pressure decreases to values of 0.028 bar. In other

cases, as the cooling medium is air or water, the pressure is forced to increase to guarantee the right

temperature of condensation.

Moreover the difference between the condensing temperature and temperature of the cooling medium is

important in the choice of the pressure.

For economic reasons the higher this difference (and so the higher the pressure) the lower are the costs

related to the surfaces of the heat exchangers and to the pumps or fans to move the coolant medium .

For this reasons, in reality, especially in power plants with evaporative tower is not so unusual to find pressures

of 0.01bar. The heat exchange with a high temperature difference increases the rate of entropy generation,

and obviously this fact decreases the efficiency.

0,00% 10,00% 20,00% 30,00% 40,00% 50,00%

1

Second law analysis Cvarious

stack losses







Tmax

combustion related losses


Configuration D: higher temperature of SH and RH

Steam temperature 640°C, 640°C

Increasing the maximum temperature of the cycle bring the efficiency to grow significantly due to:

• Increase of the mean temperature of introduction of heat in the cycle

• Increase in the efficiency of turbine expansion : the liquid fraction in the turbine decreases

So it’s a good ideal strategy for increasing efficiency. Reality is affected by some problems in the real plant

physical configuration:

• SH and RH are at high pressure, the stress they are forced to support is not negligible and brings to

increase the thickness of tubes

• The temperatures are high for common materials so special alloys or special steels are required.

Their cost is predictably wide.

For this reason nowadays technology uses temperature of 610-620°C. In the future with the development

of new materials these temperature are expected to grow.

Entropic analysis D









stack losses 1,33%

various 1,86%

TOT 100,00%

0,00% 10,00% 20,00% 30,00% 40,00% 50,00%

1

Second law analysis D various

stack losses







Tmax



Configuration E: 4 preheaters, lower water temperature in economizer • N° regenerators: 4 (2 LP deaereator included,

2HP)

• Economizer water inlet temperature: ~ 280 °C

Regeneration is thermodynamically good. The idea is to use a low temperature heat source to preheat a

low temperature fluid. In a modern power plant there are usually many regenerators (8-10). The aim is to

provide the lower temperature difference between the water and the steam bled from the turbine. The

higher the number of regenerators, the higher is the number of condensing pressure of the steam

bleedings. This allows lower temperature differences in the surface countercurrent heat exchangers.

Actually the steam bled is not able to provide power in the turbine, but from a total point of view the

efficiency still grows. The working fluid arrives in the boiler at a higher temperature, so the heat required

from the boiler itself, keeping fixed the maximum temperature, is lowered. Accordingly the efficiency

fraction sees the denominator decreasing more rapidly than the numerator, thus increases.

Entropic analysis E









stack losses 1,33%

various 0,23%

TOT 100,00%

0,00% 10,00% 20,00% 30,00% 40,00% 50,00%

1

Second law analysis Evarious

stack losses





heat exchange between Tmax –

steam


Tmax



Configuration F: double RH

• N° re-heaters: 2

• Steam temperatures: 580°C/580°C/580°C

Thermodynamically the two most attracting perspectives are here done together. Reheating is done

twice.

Entropic analysis E









stack losses 1,33%

various 2,34%

TOT 100,00%

0,00% 10,00% 20,00% 30,00% 40,00% 50,00%

1

Second law analysis Fvarious

stack losses









3. General results and comments It’s shown the comparison among all the possible variation of the technology, and after having explained

mainly the role of all inefficiencies we analyze the weight of plant configuration and eventually the

technical-economical compromises.

• Second law efficiencies of steam power plant are between 40% and 45% ;

• The mayor loss is combustion. Unfortunately this loss is unavoidable using combustion as a source

of heat for the system. For the future, researchers are interested in electrochemical processes to

avoid this huge term : it about 26% of the reversible work entering the cycle.

• Boiler architecture plays a fundamental role or heat exchange. The mechanism used are radiation

and convection. The biggest issues are caused by material thermal resistances. As thermodynamics

suggests the research focuses on the increase of the temperature. Materials keep the temperature

of the boiler’s part much lower than the exhaust from combustion. This is the cause for the

irreversibility. Moreover being a steam power plant an external combustion engine, the

temperature of the fluid is lower than the temperature of the tubes and this is another source of

irreversibility.

• The losses are divided between heat exchange between exhaust and a fictitious isothermal line at T

max and heat exchange between this line and the steam this is done in order to simplify the view

and this simplified the thermal exchange of combustion.

• It’s noticeable that only small differences are present in the trends (2-3%), the least variable is

combustion related losses.

• Configuration D and F with Re-heaters, and higher temperature are the most performing. This was

clearly predictable from thermodynamics. The reason why not all the plants are like those is that it

is not economically profitable to build such steam cycles. It is important to keep in mind that re-

heater work at high pressure and high temperature, therefore they need carefully selected alloys.

The second re-heater works with moderate pressure, this implies large specific volume for the flow.

0,00%

5,00%

10,00%

15,00%

20,00%

25,00%

30,00%

35,00%

40,00%

45,00%

50,00%

second law efficiency combustion related losses heat exchange between

exhausts and Tmax

heat exchange between

Tmax – steam

Major Losses

A B C D E F

Hence according to Mariotte’s formula large volumetric flow rates require both large diameters

and high thickness. In other words they are really expensive.

• Case C has the worst efficiency. Condensing pressure must be kept as low as possible.

Here are reported the minor losses, they do not reach high values and can be maximized working on some

cycle parameter.

• Regeneration as we have already said is thermodynamically good. On the technical side it requires

large amount of materials and careful study of heat exchangers. Case E uses only 4 Preheaters while

all other configuration use 8 preheaters. In the picture is highlighted the usual configuration of a

preheating line.

0,00%

1,00%

2,00%

3,00%

4,00%

5,00%

6,00%

7,00%

regeneration

related losses

cycle waste heat

losses

working fluid

compression

losses

working fluid

expansion losses

stack losses various

Minor losses

A B C D E F

The loss of efficiency is given by :

It is clear how the loss in case E is larger than the others. Case E has just 4 preheaters so

temperature differences inside the heat exchangers are increased very much. Moreover the mean

temperature of the water decreases. On the other side from an economic stand point this practice

is definitely a simplification of the plant.

• Cycle waste losses are given by:

The pedex 2 means the quantity of heat given to the ambient while the pedex 1 is used for the

power entering the cycle.

Actually in rejecting thermal power there are two exchanges, the first is between the water and the

cooling medium, the other is between the cooling medium and the environment. The formula

keeps into account this two terms.

Cycle C, having a higher pressure, has a higher Temperature of condensation, so the losses are

quite relevant in the overall balances. On the other side cycle E has a low term.

In the condenser not only the working fluid gives heat to the environment, also the condensate

used in the preheater line is brought back to the saturated liquid condition. For this reason case E

has a low losses.

• Stack losses are low for almost all cases (case B has a lower T at the stack)

• Various are not very relevant to the overall balance, being made by electrical losses and losses in

the auxiliaries which are not big.

• Steam turbine losses stop at a low value of 5%, this is good for the plant. We notice how case 6 is

the best expander of all due to the absence of a strong fraction of liquid in the turbomachines

avoided by the double reheating.

We must also report net powers and First law efficiencies.

A; 582,81

B; 571,65

C; 553,98

D; 629,95

E; 584,62

F; 660,25

500

520

540

560

580

600

620

640

660

680

MW

Net electrical power

• Case C is strongly penalized on the side of power produced, while on the other side the double

reheated cycle and the cycle D with high temperatures are naturally favored.

• Case F is the most efficient.

• It is important to see how for a given amount of fuel these are the scale of power production for

the technologies considered.

• The differences are after all in the order of 3-4%. Despite the numerical values in reality one

percentage point can do really the difference in a power plant business plant, and above all in the

budget planning of the plant

A; 42,07%

B; 43,11%

C; 40,79%

D; 43,46%

E; 41,02%

F; 43,64%

39,00%

39,50%

40,00%

40,50%

41,00%

41,50%

42,00%

42,50%

43,00%

43,50%

44,00%

1

Efficiency

4. Plant configuration for case A

This is the plant configuration for a typical ultra-super critical steam cycle. In our case the configuration is

very similar. The differences are the presence of one only RH, and the values of temperature and pressure

in the cycle.

This values are reported:

T [°C] P [bar]

HP in 578 270 HP out 341,903 57 MP in 578 53 MP out 321,726 8 LP in 321,726 8

LP out 32,898 0,05

5. First and second law Sankey diagrams

Sankey diagrams are really useful to view the relative weight of losses inside a power cycle. The width of an

arrow is proportional to the amount of power of the relative voice. For the second law it is listed the

reversible work . Here are reported first and second law Sankey diagrams.

First law Sankey diagram

Second law Sankey diagram

Some interesting consideration can be drawn:

• Condenser losses are huge in first law analysis, but as the loss is occurring at low temperature they

affect for a little percentage the second law analysis.

• Heat transfer and combustion are really relevant in second law analysis as occurring at high

temperature.

• Thermal losses are less relevant in a first principle analysis.

6. T-Q diagram for high pressure preheaters Here is reported the T-Q diagram for the 5th, 6th, 7th and 8th preheater for water.

In the diagram is highlighted the temperature of the various steam flows.

In the following table some important numerical values are highlighted, in the following part we replace the

numerical calculation we have used to build the graph.

427,0

310,0

271,3

571,0

261,7

246,5

508,0

236,7

214,3

422,0

204,3167,6

0; 154,49

1; 316,62

0,0

100,0

200,0

300,0

400,0

500,0

600,0

0 0,2 0,4 0,6 0,8 1 1,2

High Pressure Preheaters

8th

7th

6th

5th

WATER

High pressure preheaters

Mass flow P H T S M Q

kg/h bar kJ/kg °C kJ/kg/K kg/s kW Dearator

Water out 1850000 5,361 651,55 154,49 1,887 513,8889

Preheater N° 5 Water out 1850000 333 894,3 206,31 2,342 513,8889 124746,5

Steam 111117,3 17,895 3300,86 422 7,256 30,86592

Out 510370,5 17 709,01 167,57 2,016 141,7696

Preheater N° 6

Water out 1850000 332 1037,09 238,54 2,631 513,8889 73378,19 Steam 84832,2 33,243 3471,31 508 7,208 23,5645 Out 399253,3 31,581 917,79 214,31 2,463 110,9037

Preheater N° 7

Water out 1850000 331 1150,4 263,29 2,847 513,8889 58228,75 Steam 71241,8 50,736 3597,1 571 7,173 19,78939 Out 314421,1 48,199 1069,02 246,54 2,759 87,33919

Preheater N° 8

Water out 1850000 330 1413 316,62 3,314 513,8889 134947,2 Steam 228175,1 103,895 3174,89 427 6,312 63,38197 Out 243179,3 98,7 1190,39 271,29 2,975 67,54981

For the construction of the diagram you need to follow this path.

8th Preheater

This is the situation for the 8th preheater. Calculation must start from this we can apply the energy balance

neglecting thermal losses.

�� ℎ��, � − ℎ��, �� = �� ℎ�,�� + �� ℎ�� − �� + �� ℎ� ��

Moreover mass balances hold.

�� = ��

Water in Water out

Steam Leakage

Outlet

�� + �� = � ��

In the CICVAP output many information are given about this preheating line.

In this balance everything is known except for the enthalpy of the leakage that can be computed starting

from this equation.

m leakage h leakage

kg/s kJ/kg

4,167833 3389,542

5th, 6th ,7th Preheater

This is the situation for the other three preheaters. Being them at lower pressure than the previous preheater

the condensate from it must be laminated. This brings the flow in the bi-phase region. Lamination is

considered an isoenthalpic process.

Also here the energy balance holds.

�� ℎ�, � − ℎ�, �� = �� ℎ�,�� + �� ,!�"�ℎ� ��,!�"� − �� ,!�" + �� ℎ� ��

�� ,!�" + �� = �� , �

The calculation of the fraction of heat exchanged is performed in this way.

We perform the calculation of the total heat exchanged using water properties, we obtain Qtot.

# = $#�� =�� %�& �!�'�� − &��!�'��

Water in Water out

Outlet

Steam

Laminated condensate from the

previous preheater

For the 8th preheater we divide the exchange in 3 parts.

In the C part

(# = �� ℎ�,�� − ℎ��"�!@!�� + �� ℎ��,�� − ℎ��"�!@!��

In the B part

(# = �� + �� ℎ��"�!@!�� − ℎ��*@!��

In the A partq

(# = �� + �� ℎ��*@!�� − ℎ� ��, ��

All the terms are known and knowing Q0 from the water heating process we can compute

%#% = 1 − (#%/# �

%#. = %#% − (#./# �

The procedure holds true going through the heat exchangers and the fraction of heat exchanged can be

computed with the previous equation.

A

B

C

5th ,6th, 7th Preheater

For the 5th ,6th ,7th preheater we divide the exchange in 3 different parts as we have done for the 8th.

The condensate is laminated and hence is a bi-phase fluid .

In the C part

(# =�� ℎ�,�� − ℎ��"�!@!��

In the B part

(# = �� ℎ��"�!@!�� − ℎ��*@!�� + �� ℎ� ��,�� − ℎ��*@!��

In the A part

(# = �� + �� ℎ��*@!�� − ℎ� ��, ��

Than we can compute the percentage of heat

%#% = %#/012��345 − (#%/# �

%#. = %#% − (#./# �

The diagram has been plotted following this ideal path.

The total entropy generation is not so big as it could seem from a first inspection of the diagram. Huge

differences of temperature are relevant in the system. The overall entropy generation depends also on

mass flow rates and the values of this parameter for steam are low.

7. H-s diagram of turbine The steam turbine is the actual protagonist of the power production. In our plant the turbine is very large

and is composed by High , Medium and Low pressure sections. Tecnically there are some parameters which

confirm that the turbine examinated is a large scale one. This parameters are:

• 2 intermediate pressure rotors with two flows • 4 rotors in the low pressure size with two flows • The dimension of the machine are big: the last stage has 33,5 inches blades and a big mean

diameter(99,5 inches) The calculation of efficiency should deal with one big problem: mass bleedings.

The preheaters and the dearator require steam from the turbine, moreover the feed water pump is driven

by a small turbine which operates with bled steam.

The idea we have used in calculation is the following.

• Compute for every turbine section the mass flow rate. This can be done with the CICVAP output. In

fact inside this file are present all the mass flow rates of bleedings going to preheaters.

• Determine the thermodynamic conditions of every point, using the pressure enthalpy and entropy

values.

• Compute for each section in which the mass flow rate is constant the efficiency, given by the

formula:

6 = ℎ�� − ℎ �ℎ�� − ℎ ��

Where the inlet and the outlet condition are known from the Cicvap output and the isoentropic

enthalpy can be easily computed. This value is the enthalpy of the thermodynamic state with the

outlet pressure and the inlet entropy.

This allows to calculate the efficiency. In the following table is reported a summary of the calculation

previously explained. We have reported ℎ �� and 6 referred to a section in the same line of the inlet

condition.

m [kg/s] h[kJ/kg] s [kJ/kgK] p [bar] hiso[kJ/kg] ad. Eff

Hpin 504,9303889 3403,6 6,23 264,6 3136,88015 0,793342

bleed_8 431,9285833 3192 6,3 111,2 3008,46956 0,877729

Hpout 431,9285833 3030,91 6,337 56,842 3013,22834 ------------

Mpin 429,9195278 3610,07 7,16 52,9 3586,50192 0,549896

bleed_7 410,5861111 3597,11 7,173 50,736 3444,40652 0,823229

bleed_6 387,5404722 3471,4 7,208 33,243 3267,84135 0,836909

bleed_5 357,4690278 3301,04 7,256 17,895 3291,53869 0,825149

bleed_turb_fwp 324,4535556 3293,2 7,258 17,37 3070,44772 0,852561

Mpout 324,4535556 3103,29 7,314 8 3103,33949 ------------

Lpin 323,7637222 3103,29 7,314 8 3012,76167 0,864812

bleed_4 314,7431667 3025 7,337 5,643 2838,95547 0,873769

bleed_3 294,9056944 2862,44 7,387 2,506 2663,48763 0,889761

bleed_2 278,127 2685,42 7,446 0,878 2499,96924 0,890371

bleed_1 263,5971667 2520,3 7,506 0,272 2289,04129 0,782167

Lpout 263,5971667 2339,417 7,593 0,05 .---------------

Notice how the mass flow rate is nearly reduced by a factor of two. Preheating of the water is a loss of

power produced. On the otherside as the water comes to the boiler with a quite high temperature the fuel

required to heat up the working fluid is lowered. It is possibile to compute that if we lose one kW in terms

of power generation , we save up to 3kW in termal power. Moreover as the fluid reaches the boiler with

high T , the entropy generation inside the boiler is lower.

Preheating is thermodynamically good.

Here are reported the efficiencies of the single stages.

Adiabatic efficiency

Hpin /Bleed8 0,793342 Bleed8 /Hpout 0,877729 Mp in / Bleed7 0,549896 Bleed7/ Bleed6 0,823229 Bleed6/ Bleed5 0,836909 Bleed5/ Bleed turbfw 0,825149 Bturb/Mpout 0,852561 LP in/ Bleed4 0,864812 Bleed4/ Bleed3 0,873769 Bleed3/ Bleed2 0,889761 Bleed2/ Bleed1 0,890371 Bleed1/ Lpout 0,782167

The medium pressure stages are the ones with the best efficiencies.

Every pressure level has his own requirements to work with high efficiency: if the rotation speed is not

fixed it is a powerful tool to improve the efficiency of a stage; in particular, high pressure stages tend to

have higher optimal rotation speed than low pressure stages.

All this considerations come from the existence of a maximum peripheral velocity. Usually, the higher this

parameter the higher the power extracted. In reality the peripheral velecity has a limit caused by the

mechanical stress on the blades. We should aim at going as close as possible to this maximum value, which

for modern turbines is about 400 m/s.

HP stages have short blades so in order to keep the peripheral velocity as high as possibile should work at

high rotational speed. In low pressure the situation is exactly the opposite. Low pressure stages would work

properly at low rotational speed.

The efficiency is a useful parameter to take into account many geometrical and operational aspects. We

can list some of the most important losses in a turbine stage.

• Angles of the blades: in particular the absolute velocity deflection in the inlet rotor section 71, and

the outlet relative angle β2. To keep losses reasonably low this angles must be higher than 15

degrees.

• Deflection of the flow: this parameter is generated by the presence of boundary layers on the

blades. The most critic section is on the suction side as the flow encounters an opposite pressure

gradient.

• Ratio between relative velocities: decelerating too much brings irreversibilities in the expander.

• Solidity of blades: there is an optimal value for solidity, the ratio of the distance between two

blades and the axial length of the blade. Low solidities mean large boundary layer influence in the

rotor. On the other side high solidities mean difficulties in deflecting properly the fluid vein.

• Roughness of the blade: obviously low roughness are related to low losses.

• Thickness of trailing edge: the higher this value the higher the influence of the blade on the wake.

This value has an inferior limit caused by technological issues.

• Blade height with respect to the width of the channel: short have big losses as the boundary layer

on the casing and on the shaft have a strong weight on the flow

• Blade height with respect to the mean diameter of the flow: if the height of the blade is too high

there is a pressure gradient along the radial direction. This causes vortexes and so irreversibilities.

This parameter for a good design should be kept under 0.3.

• Clearances between the casing and the rotor: the lowest value is limited by technological limits.

In a second moment it is reported the graph of the efficiency of the stages. Here the calculation is quite

simpler as we choose to neglect the mass influence. The efficiency computed is valid just for enthalpies but

it does not affect the power produced. The path is very similar to the previous one but larger turbine

section are investigated.

There are some graphical errors occurring as our calculator is not able to compute enthalpy and entropy in

that zone of the diagram, nevertheless this are the numerical values.

P H S T

bar kJ/kg kJ/kg/K °C

HP in 270 3403,632 6,23 576,6643

MP in 54 3610,067 7,16 577,0521

LP in 8 3103,027 7,313 321,8971

HP out 56,842 3030,914 6,337 342,4571

MP out 8 3103,027 7,313 321,8971

LP out 0,05 2343,577 7,592 36,85594

HP out iso 56,842 2966,416 6,23 320,7281

MP out iso 8 3014,949 7,16 280,1493

LP out iso 0,05 2229,979 7,313 32,87542

2000

2200

2400

2600

2800

3000

3200

3400

3600

3800

4000

5 5,5 6 6,5 7 7,5 8

H

S

H-S diagram expansion

From the table above using this equation we compute efficiencies for high and medium pressure

6 = ℎ�� − ℎ �ℎ�� − ℎ ��

For low pressure section we get a value of efficiency which is not in accordance with the theoretical results,

due to the liquid fraction inside the turbine this efficiency should be lower than the other two. This is a

problem but there could be some issues in the definition of the efficiency. In fact the CICVAP output reports

also the kinetic energy of the fluid outside the turbine, this could affect the definition of the efficiency.

Efficiency

HP 0,852481

MP 0,851999

LP 0,865572

1500

2000

2500

3000

3500

4000

4500

5 5,5 6 6,5 7 7,5 8

H-S expansion

Serie1 HP IDEAL HP MP IDEAL MP

LP IDEAL LP Espo. (ISO HP) Espo. (ISO MP) Espo. (ISO OUT)

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