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3 Steam turbines

A steam turbine is a thermal machine, in which the thermal internal energy of the

working substance (steam) converts into mechanical work.

Produced mechanical energy is mainly used for the production of electricity for the

propulsion of large ships, heavy machinery and rarely to power vehicles or other

machinery. The working substance is usually steam with high pressure and

temperature, heat for producing steam can be produced from fossil fuels in steam

boilers, from nuclear reactions or other heat sources.

A Thermodynamic properties of water and steam

In general, the changes of all state variables are interrelated, but usually these

connections cannot be expressed by simple empirical terms. Instead of that, the state

variables can be provided in tabular or graphical format. For water and water vapour there

are several resources where are collected the thermodynamic properties of water and

water vapour, for example: Engineering handbook1 and Tables of thermodynamic

properties of water and water vapour2.

When dealing with water as the working substance is required to identify three specific

areas in which the working substance responds differently to the effects of ambient

accordingly while also changing its properties. These regions are:

1 Liquid region (water),

2 Vapour region (superheated) steam and

3 Saturated region (mixture of water and steam, wet steam).

Properties of the first two regions are given as separate tables, each of which is

characteristics of a certain (constant) pressure. At the known pressure and temperature

we can from appropriate table find values (for a known pressure) for specific volume

(v), the specific enthalpy (h) and the specific entropy (s) of water or steam.

1 Kraut, B.; Puhar, J.; Stropnik, J.: Krautov strojniški priročnik, 14. slovenska izd., predelana; Ljubljana:

Littera Picta, 2007

2 Kuštrin, I.; Senegačnik, A.: Tabele termodinamičnih lastnosti vode in vodne pare, po modelu IAPWS-IF97;

Ljubljana: Fakulteta za strojništvo, 2001

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For the area of wet steam (mixture of water and steam) is subject to a separate table,

which contains the same information, but in one table arranged by steadily increasing

temperatures and in the second table by pressure.

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A mixture of liquid water and gaseous steam can exist only in very specific conditions;

they called the state of saturation. At a given temperature, there is a defined pressure

(saturation pressure) at which the phase change can take place, which takes place over

the area of the wet steam. In this area are given only the two end points - the boiling

(water) point and saturated steam, for all the intermediate states may, however,

calculate properties of the mixture with the use of additional parameters - dryness

steam fraction (x), sometimes called the quality. X gives the fraction (by mass) of

gaseous substance in the wet region and it has a value between 0 and 1.

steamwater

steam

mm

mx

State variables of wet steam define with following expressions:

v = v' + x (v'' – v')

h = h' + x (h'' – h')

s = s' + x (s'' – s')

The values v', h' in s' are for the boiling water and v’’, h'' in s’’ for saturated steam at

given pressure or temperature.

Example: 1. Known data: pressure p = 0,6 bar, temperature T = 85 °C

Calculate the specific volume of water/steam? 2. Find the water/steam table for a given pressure 3. Because the known temperature is not in the table, find the first lower (Tp = 80 °C) and first

higher (Tn = 100 °C) 4. Determine the corresponding values for specific volumes (vp in vn) 5. Calculate

/kgm7118,0001029,0844,280100

8085001029,0

bar0,6 C;85

3

pn

pn

pp vv

TT

TTvv

Because between the temperatures of 80 and 100 ° C, there is a phase change (indicated by a line in the

table, also evident from the major changes in volume, enthalpy and entropy), it is necessary to perform

interpolation with respect to the parameters of the saturation.

6. In the appropriate table find for pressure 0,6 bar the proper boiling point and limit values of searched variable Ts = 85,95 °C v' = 0,001055 m3/kg v'' = 1,091 m3/kg

7. Because the T < Ts, please note as the ''next'' point Tn = Ts and vn = v' 8. Calculate

/kgm001051,0001029,0001055,08095,85

8085001029,0

bar0,6 C;85

3

80

pn

pn

pvv

TT

TTvv

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In the event that the table does not include data for a given pressure or temperature, we have to help with the linear interpolation to get proper properties of the water or steam. In doing so, considering the general expression for the equation of a line through two known points:

pnpn

pp yy

xx

xxyy

We replaced parameter y with search variable and parameter x with known variable, of

which value is not in the table. We find in the tables data for previous value (xp) and next

value (xn) of a known variable and their corresponding values yp and yn of a searched

variable.

Example: 1. Known data: pressure p = 1,2 bar, temperature T = 45 °C

Calculate the specific enthalpy? 2. Find the water/steam table for a given temperature 3. Because the known temperature is not in the table, find the first lower (Tp = 40 °C) and first

higher (Tn = 60 °C) 4. Determine the corresponding values for enhalpies(hp in hn) 5. Calculate

kJ/kg5,1886,1672,2514060

40456,167

C45

pn

pn

pp hh

TT

TThh

Example: 1. Known data: pressure p = 1,6 bar, specific entropy s = 7,3 kJ/kgK

Calculate the temperature for known data? 2. Find the water/steam table for a given pressure 3. Because the known entropy is not in the table, find the first lower (sp = 7,237 kJ/kgK) and first

higher (sn = 7,340 kJ/kgK) 4. Determine the corresponding values for temperatures (Tp in Tn) 5. Calculate

C2,132120140237,7340,7

237,73,7120

kJ/kgK3,7

pn

pn

pp TT

ss

ssTT

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By interpolation we must always be careful to never (!) interpolate over the area of the

phase change.

Another way to determine the thermodynamic properties of water and steam is the use

of Moliere diagram, also called h-s diagram. This diagram shows the relationship

between the pressure, temperature, enthalpy, entropy, and dryness of the steam (the

area of the wet steam). Usually diagram only shows the area of superheated steam and

part of the area wet steam.

Primer: 1. Known data: pressure p = 0,6 bar, temperature T = 85 °C

Calculate the specific volume of water/steam? 2. Find the water/steam table for a given pressure 3. Because the known temperature is not in the table, find the first lower (Tp = 80 °C) and first

higher (Tn = 100 °C) 4. Determine the corresponding values for specific volumes (vp in vn) 5. Calculate

/kgm7118,0001029,0844,280100

8085001029,0

bar0,6 C;85

3

pn

pn

pp vv

TT

TTvv

Because between the temperatures of 80 and 100 ° C, there is a phase change (indicated by a line in the

table, also evident from the major changes in volume, enthalpy and entropy), it is necessary to perform

interpolation with respect to the parameters of the saturation. 6. In the appropriate table find for pressure 0,6 bar the proper boiling point and limit values of

searched variable Ts = 85,95 °C v' = 0,001055 m3/kg v'' = 1,091 m3/kg

7. Because the T < Ts, please note as the ''next'' point Tn = Ts and vn = v' 8. Calculate

/kgm001051,0001029,0001055,08095,85

8085001029,0

bar0,6 C;85

3

pn

pn

pprej vv

TT

TTvv

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In Moliere diagram are drawn curves of constant parameters:

— Saturation line (x = 1) – is the boundary between the areas of superheated steam

and wet steam and provides the point where it begins or ends phase change

— Isobaric lines (p = const.) – are the curves on which the pressure do not change,

but other state variables changes along the line; in the area of wet steam along the

isobaric lines the temperature remains constant (saturation temperature)

— Isothermal lines (T = const.) – are the curves at which the temperature does not

change but other state variables changes along the line; in the area of wet steam

isotherms they are not drawn because coincide with isobaric lines

— Curves of constant moisture (dryness fraction) of a steam (x =const.), or quality –

are lines which represents the quantity of saturated vapor in unit mass of wet

vapor in the field of wet steam in addition to pressure (and corresponding

saturation temperature), accurately define the state of a substance

— Isentropic lines (s=const.) – are the vertical lines in the diagram and represents

constant entropy, other state variables changes along the lines

— Isenthalpic lines (h=const.) – are the horizontal lines in the diagram and

represents constant enthalpy, other state variables changes along the lines

In the area of superheated steam (above the saturation line) we can determine state

properties of the steam with only two parameters, for example: pressure and

temperature, pressure and entropy, temperature and enthalpy. In the region of wet

steam (under the saturation line), the combination of the pressure-temperature is

insufficient, it is needed to add an additional parameter - the dryness of the steam (x).

Based on known parameters, of which can be read from the diagram very simply, we can

determine unknown state variables, but the accuracy of the readings is usually less

accurate than with the use of the tables.

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B The steam expansion in a turbine

The power of the steam turbine is determined by the amount of steam flowing through

the turbine, and the enthalpy difference between the inlet and outlet state of steam.

hmP st

The turbine inlet state of the steam (point 4), it is typically constant and

predetermined. In such a case, we can change the power of turbine with the changing of

steam flow through the turbine, which is called quantitative regulation. The power of

turbine can be changed also by changing the input parameters of the steam (pressure),

whereby, accordingly, be adapted to the steam generator (e.g. boiler).

The turbine outlet state of the steam (point 5s

and 5) according to the turbine inlet state and

pressure after expansion in turbine can be

determined in several ways. First we find ideal

endpoint of (isentropic) expansion.

a) With the use of h-s diagram:

From point 4 we pulled line vertically

(isentropic line) to the pressure line at the

end of expansion in turbine and we get

endpoint of ideal (isentropic) expansion.

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b) With the use of thermodynamic tables of water and water vapour: Firstly read the

entropy of saturated steam at outlet pressure 5s and compared with the entropy

of the point 5s, (s5s = s4). If:

55 ss s (Example B and C)

Point 5s lies in the saturated region

and we first determine dryness of the

steam (x).

55

545

ss

ssx s

Then we can determine the enthalpy

at point 5s.

55555 hhxhh ss

55 ss s (Example A)

Point 5s lies in the superheated steam

(vapour) region and an enthalpy is

determined by interpolation.

55

55

5455 hh

ss

sshh s

Entropy s5– is first lower and s5+ is first

higher value according to the s4 at

pressure p5. Enthalpy h5– and h5+

corresponding entropy s5– and s5+.

With the known enthalpy in a theoretical point 5s we should determine the actual enthalpy in point 5 with respect to the turbine efficiency. For the turbine without steam extraction (the whole steam expands in one step) is

st

hh

hhη

54

54

and

st hhηhh 5445

Temperature at point 5 is determined depending on the outlet state of the steam. If is

55 hh (Example C)

Point 5 lies in the saturated region and

the temperature is the same as

saturation temperature at point 5

(pressure p5).

T5 = Ts(p5)

We determine the dryness of the

steam at point 5.

55

555

hh

hhx

55 hh (Examples A and B)

Point 5s lies in the superheated steam

(vapour) region and the temperature

is determined by interpolation.

55

55

5555 TT

hh

hhTT

Enthalpy h5– is first lower and h5+ is

first higher value according to the h5

at pressure p5. Temperatures T5– and

T5+ corresponding enthalpy h5– and

h5+.

Overheated steam means the temperature difference between the actual temperature

and the saturation temperature at selected pressure. For point 5 apply:

ΔT5 = T5 – Ts(p5)

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Power and efficiency of the turbine

Internal turbine power generally calculated as

N

i

iiit hhmP1

1

Where N is the number of steam extractions in the

turbine, mi is the steam mass flow through the turbine

between the extractions i – 1 and i.

Turbine efficiency is defines as the ratio between the actual and theoretical turbine

power.

teor

act

tP

P

For turbine without steam extractions applies

sss

s

thh

hh

hhm

hhm

21

21

21

21

For turbine with steam extractions (for example,

with two extractions, see sketch turbines and

expansion scheme), It applies similarly to its

individual parts (high pressure, medium pressure in

low pressure)

s

HPhh

hh

21

21

s

MPhh

hh

32

32

s

LPhh

hh

43

43

For overall (combined) turbine efficiency applies:

ssssssss

thhmhhmhhm

hhmhhmhhmη

433322211

433322211

Overall turbine efficiency therefore cannot be expressed only by specific enthalpy before

and after the turbine, not even with the efficiency of each turbine parts!

sst

hh

hhη

41

41

NTSTVTt ηηηη

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C Steam extraction and turbine outlet

With the steam extractions out of turbine at certain points the steam are discharged

from the turbine and we use extracted steam as a heat source in appropriate heat

exchangers. The generated heat can be returned to the cycle (regenerative heating of

water) or use it for other purposes (district heating, heat for technical processes). Also,

in the condenser, where the steam flows from the end of the expansion in the turbine,

the steam transferred the heat, which is discarded into the environment. In the

condenser flows superheated or wet steam depending on the expansion in turbine and

extraction point. From condenser flows boiling water or water with temperature lower

than boiling point. If the state of steam is not known at outlet of the condenser, it

assumes that it is boiling water at a pressure of steam at the inlet to the heat exchanger

(condenser). The major part of the heat generated by the steam, named condensation

heat, is released during the phase change from steam to water. The overall heat flow in

each case is calculated as the product of the mass flow rate and the change in enthalpy.

21 hhmQ s

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D Exercise task

Steam boiler produce 20 kg/s superheated steam of 500 °C

temperature and pressure 60 bar. Steam is lead in turbine

without repeated superheating and internal efficiency

81,5 %. Steam in turbine expands to pressure 4 bar and

then it is lead to heat consumer where condensate to 70 °C

and pressure drop to 3 bar.

Calculate:

a) Steam properties at the turbine outlet (p2, T2) and steam

overheated,

b) Turbine power and

c) Heat flow, which is usefully extracted in heat consumer.

We lead in Condensing steam turbine with

steam extraction 50 kg/s superheated steam

with following properties: 500 °C and 60 bar.

Steam extraction is at 4 bar and 200 °C and mass

flow of the extracted steam is 20 kg/s. Extracted

steam is lead in heat consumer where the steam

condensate and cooled down to 70 °C. The

pressure of extracted steam drops to 3 bar. The

rest of the steam flow is expanded to pressure

0,05 bar, which is the same as in condenser. The

efficiency of the low pressure turbine is 77%. Define:

a) Steam properties at the turbine outlet (p4, T4, x4),

b) Internal efficiency of the high pressure turbine,

c) Internal turbine power,

d) Internal turbine efficiency,

e) Heat flow, which is usefully extracted in heat consumer (exchanger) and

f) Heat flow in the condenser.