steam cycle1

7/25/2019 Steam Cycle1

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STEAM CYCLE

For the steam cycle below, calculate the total electrical efficiency. The following is

given:

Mechanical efficiency turbine 0.96Efficiency generator 0.95

Efficiency boiler 0.92

Isentropic efficiency turbine 0.75

The rest of the given values are given on the figure

Turbine G

m =120 kg/s

p =110 bar

t = 525 deg C

m = 19 kg/s

p = 20 bar

p = 0.8 bar

p = 1.5 bar

p = 6.0 bar

h = 900 kJ/kg h = 450 kJ/kg

h= 390 kJ/kg

h = 480 kJ/kg

h= 640 kJ/kg

Boiler


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HINTS

Hint 1 (-2p):The total electrical efficiency is defined as the power output dividedwith the thermal power obtained from the boiler. Start calculate the power output

from the steam turbine

Hint 2 (-5p):In order to find the steam turbine power output, you have to calculate

the enthalpies and mass flows from the steam extractions. The enthalpies can all be

found in an h-s diagram on the expansion line drawn from turbine from inlet to the

outlet to the condenser.

h2s

h2

0,8 bar

1,5 bar

6 bar

20 bar110 bar

525 deg C h1

a

b

c

Hint 3 (-3p): The steam extraction mass flows can be found via heat balances on the

feed water preheaters. Heat balance on the high pressure preheater gives data to

calculate the boiler power.


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SOLUTION

Start calculating the power output from the turbine.

First we draw the expansion line, from the inlet conditions (1) to the condenserpressure. The enthalpy h2 can be found, as we know the isentropic efficiency of the

turbine, as follows:

h2 = 3427 0.75*(3427-2370) = 2634 kJ/kg, (where 2370 is the enthalpy of the

isentropic expansion process).

h2s

h2

0,8 bar

1,5 bar

6 bar

20 bar

110 bar

525 deg C h1

a

b

c

Enthalpies at extractions can be read on the expansion line at the given pressures on

an enthalpy entropy diagram.

ha = 3115 kJ/kg

hb = 2920 kJ/kg

hc = 2720 kJ/kg

The water enthalpy into the boiler is found through a heat balance over the high

pressure preheater, and then we calculate the power in the boiler Pboiler:

MWP

kgkJh

h

Boiler

in

in

4.292)7.9903427(*120

/7.990

120*64019*311519*900120*

=

+=+

The fuel input to the boiler becomes:


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MWP

Qboiler

boilerfuel 8.317

92.0

4.292===

&

The flows

bm and

cm are given through heat balances over the feed water tank andover the low pressure feed water preheater:

Feed water tank:

640*120900*19480*)19120(2920* =++

bb mm

skgmb /6.4

Low pressure preheater:

skgm

mm

c

cc

/82.3

480*4.96450*390*4.962720*

+=+

The electrical power generated is:

MW

Pel

9.7695.0*96.0*

)]2.26342720(*58.92)27202929(4.96)29203115(*101)31153427(*120[

+++=

The total efficiency of the plant can be calculated as:

24.08.317

9.76==

fuel

eltot

Q

P

&

steam cycle1

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