module-iv --- vapour power cycle applied thermodynamics

Dr.A.R.Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM

MODULE-IV --- VAPOUR POWER CYCLE APPLIED THERMODYNAMICS 2014

VTU-NPTEL-NMEICT

Project Progress Report

The Project on Development of Remaining Three Quadrants to NPTEL

Phase-I under grant in aid NMEICT, MHRD, New Delhi

DEPARTMENT OF MECHANICAL ENGINEERING,

GHOUSIA COLLEGE OF ENGINEERING,

RAMANARAM -562159

Subject Matter Expert Details

SME Name : Dr.A.R.ANWAR KHAN

Prof & H.O.D

Dept of Mechanical Engineering

Course Name:

Applied Thermodynamics

Type of the

Course

web

Module

IV

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CONTENTS

Sl.

No. DISCRETION

1. Quadrant -2

a. Animations.

b. Videos.

c. Illustrations.

2. Quadrant -3

a. Wikis.

b. Open Contents

3. Quadrant -4

a. Problems.

b. Assignments

c. Self Assigned Q & A.

d. Test your Skills.

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MODULE-IV

VAPOUR POWER CYCLE

QUADRANT-2

Animations:

(Animation links related to Vapour Power Cycle)

1. Vapour power systems:

http://me.queensu.ca/Courses/230/Lect24.pdf

2. Vapour power systems:

http://www.saylor.org/site/wp-content/uploads/2013/01/ME103-6.1_Ciccarcelli_Introduction-to-

Thermodynamics_Lecture-24.pdf

3. Rankine cycle:

http://courses.engr.illinois.edu/npre201/coursematerial/energy_cycles/lecture05notes/lecture05te

xt/Section5.2.html

4. Rankine Cycle Energy Flow and Process Unit Diagram:

http://courses.engr.illinois.edu/npre201/coursematerial/energy_cycles/lecture05figures/Figure5.2

a.html

5. Rankine Cycle Temperature-Entropy Diagram:


b.html

6. Regeneration in a Rankine Cycle with Saturated Throttle Conditions


c.html

7. Rankine Cycle Regeneration Using Two Independent Closed Heaters:


d.html

8. T(S) with Two Independent Closed Heaters:


e.html

9. Rankine Cycle with Reheat:

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http://me.queensu.ca/Courses/230/Lect24.pdf

http://www.saylor.org/site/wp-content/uploads/2013/01/ME103-6.1_Ciccarcelli_Introduction-to-Thermodynamics_Lecture-24.pdf

http://www.saylor.org/site/wp-content/uploads/2013/01/ME103-6.1_Ciccarcelli_Introduction-to-Thermodynamics_Lecture-24.pdf

http://courses.engr.illinois.edu/npre201/coursematerial/energy_cycles/lecture05notes/lecture05text/Section5.2.html

http://courses.engr.illinois.edu/npre201/coursematerial/energy_cycles/lecture05notes/lecture05text/Section5.2.html

http://courses.engr.illinois.edu/npre201/coursematerial/energy_cycles/lecture05figures/Figure5.2a.html



http://courses.engr.illinois.edu/npre201/coursematerial/energy_cycles/lecture05figures/Figure5.2b.html

http://courses.engr.illinois.edu/npre201/coursematerial/energy_cycles/lecture05figures/Figure5.2b.html

http://courses.engr.illinois.edu/npre201/coursematerial/energy_cycles/lecture05figures/Figure5.2c.html

http://courses.engr.illinois.edu/npre201/coursematerial/energy_cycles/lecture05figures/Figure5.2c.html

http://courses.engr.illinois.edu/npre201/coursematerial/energy_cycles/lecture05figures/Figure5.2d.html



http://courses.engr.illinois.edu/npre201/coursematerial/energy_cycles/lecture05figures/Figure5.2e.html

http://courses.engr.illinois.edu/npre201/coursematerial/energy_cycles/lecture05figures/Figure5.2e.html




f.html

10. T(S) and H(S) for Rankine Cycle with Reheat:


g.html

Videos:

(Animation links related to Vapour Power Cycle)

1.Vapour power cycle:

http://www.youtube.com/watch?v=4-BI22Wx4Pc

http://www.youtube.com/watch?v=vt1_7f5l3hI

2. Rankine cycle:

http://www.youtube.com/watch?v=_yx-O4r0Ymc

http://www.youtube.com/watch?v=Lg_tpEsk-0s

http://www.youtube.com/watch?v=CTUefqBpBI0&list=PL4B6F0F6ACE7466D5

http://www.youtube.com/watch?v=-Ot-SFAoTmE

3. Ideal rankine cycle:


4. Regenerative rankine cycle:

http://www.youtube.com/watch?v=uLlQBoSvwxw

5. Open feed water heater:

http://www.youtube.com/watch?v=lcAU6Rel41Y

6. Closed feed water heater:

http://www.youtube.com/watch?v=QOjcTUR_5fI\

7. Reheat rankine cycle:

http://www.youtube.com/watch?v=yAySar5gYOk

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http://courses.engr.illinois.edu/npre201/coursematerial/energy_cycles/lecture05figures/Figure5.2f.html

http://courses.engr.illinois.edu/npre201/coursematerial/energy_cycles/lecture05figures/Figure5.2f.html

http://courses.engr.illinois.edu/npre201/coursematerial/energy_cycles/lecture05figures/Figure5.2g.html

http://courses.engr.illinois.edu/npre201/coursematerial/energy_cycles/lecture05figures/Figure5.2g.html

http://www.youtube.com/watch?v=4-BI22Wx4Pc

http://www.youtube.com/watch?v=vt1_7f5l3hI

http://www.youtube.com/watch?v=_yx-O4r0Ymc

http://www.youtube.com/watch?v=Lg_tpEsk-0s

http://www.youtube.com/watch?v=CTUefqBpBI0&list=PL4B6F0F6ACE7466D5



http://www.youtube.com/watch?v=uLlQBoSvwxw

http://www.youtube.com/watch?v=lcAU6Rel41Y

http://www.youtube.com/watch?v=QOjcTUR_5fI/

http://www.youtube.com/watch?v=yAySar5gYOk



Illustrations

1. A simple rankine cycle works between pressures 28 bar and 0.06 bar, the initial condition of

steam being dry saturated. Calculate the cycle efficiency, work ratio and specific steam

consumption.

Solution:

From steam table,

At 28 bar: h1 = 2802 kJ/kg, s1 = 6.2104 kJ/kg K

At 0.06 bar: hf2 = hf3 = 151.5kJ/kg. hfg2 = 2415.9

kJ/kg

Sf2 = 0.521 kJ/kgK, sfg2 = 7.809 kJ/kg K

Vf = 0.001 m3/kg

Considering turbine process 1-2, we have: S1 = S2

6.2104 = sf2 + x2 sfg2 = 0.521 + x2 x7.809

X2 =

h2 = hf2 + x2 hfg2 = 151.5 + 0.728 x 2415.9 = 1910.27 kJ/kg

Turbine work, Wturbine = h1 – h2 = 2802 – 1910.27 = 891.73 kJ/kg

Pump work = Wpump = hf4 – hf3 = vf ( p1 – p2) =

kJ/kg

Net work, Wnet = Wturbine - Wpump = 891.73 – 2.79 = 888.94 kJ/kg

Cycle efficiency =

=

=

= 0.3357 or 33.57%

Work ratio =

==

= 0.997

Steam specific consumption =

=

= 4.049 kg/kWh

2. In a steam power cycle, the steam supply is at 15 bar and dry and saturated. The condenser

pressure is 0.4 bar.Calculate the carnot and rankine efficiencies of the cycle.Neglect pump work.

Solution: Steam supply pressure, p1 = 15 bar, x1 = 1

Condenser pressure, p2 = 0.4 bar

Carnot and rankine efficiencies:

From steam table:

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At 15 bar: ts = 198.30C, hg = 2789.9 kJ/kg Sg = 6.4406 kJ/kg K

At 0.4 bar ts = 75.90C, hf = 317.7 kJ/kg, hfg = 2319.2 kJ/kg

Sf = 1.0261 kJ/kg K, Sfg = 6.6448 kJ/kg K

T1 = 198.3 + 273 = 471.3 K

T2 = 75.9 + 273 = 348.9 K

=

=

= 0.259 or 25.9 %

=

=

Where, h2 = hf2 + x2 hfg2 = 317.7 + x2 x 2319.2

To find x2? As the steam expands isentropically,

6.4406 = Sf2 +x2 Sfg2 = 1.0261 + x2 x 6.6448

x2 = 0.815

h2 = 317.7 + 0.815 x 2319.2 = 2207.8 kJ/kg

Hence, =

= 0.2354 or 23.54%

3. Steam is supplied to a turbine at a pressure of 30 bar and temperature of 4000C and is

expanded adiabaticacallly to a pressure of 0.04 bar.At a satge of turbine where the pressure is 3

bar a connection is made to surface heater in which the feed water is heated by bled steam to a

temperature of 1300C.The condensed steam from the feed feed heater is cooled in a drain cooler

to 270C.the feed water passes through the drain cooler before entering the feed heater.The cooled

drain water combines with the condensate in the well of the condenser. Assuming no heat losses

in the steam, calculate the following:

(i) Mass of steam used for feed heating per kg of steam entering the turbine.

(ii) Thermal efficiency of the cycle.

Solution:

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From steam table:

At 3 bar: ts = 133.50C, hf = 561.4 kJ/kg

At 0.04 bar: ts = 290C, hf = 121.5 kJ/kg

From Mollier chart :

Ho = 3231 kJ/kg (at 30 bar, 4000C)

h1 = 2700 kJ/kg (at 3 bar)

h2 = 2085 kJ/kg (at 0.04 bar)

(i) Mass of steam used, m1:

Heat lost by the steam = Heat gained by water.

Taking the feed-heater and drain-cooler combined, we have:

m1 (h1 – hf2) = 1x 4.0186 (130-27)

m1 (2700 – 121.5) = 4.186 (130-27)

m1 = 0.1672 kg.

(ii) Thermal efficiency of the cycle:

Work done per kg of steam = 1 ( h0 – h1) + (1 – m1) (h1 – h2)

= 1 ( 323.1 – 2700) + ( 1-0.1672) (2700 – 2085) = 1043.17 kJ/kg

Heat supplied per kg of the steam = h0 – 1x4.186 x 130

= 3231 – 544.18 = 2686.82 kJ/kg

=

=

= 0.3882 or 38.82%

4. A steam power plant operates on a theoretical reheat cycle.Steam at boiler at 150bar, 5500C

expands through the high pressure turbine.It is reheated at a constant pressure of 40 bar to 5500C

and expands through the low pressure turbine to a condenser at 0.1 bar.Draw T-s and h-s

diagrams.Find:

(i) Quality of steam at turbine exhaust. (ii) Cycle efficiency and (iii) Steam rate in kg/kWh.

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Solution: Refer to figs. Given below.

From Mollier diagram (h-s diagram):

h1 = 3450 kJ/kg: h2 = 3050 kJ/kg : h3 = 3560 kJ/kg: h4 = 2300 kJ/kg: hf4 = (From steam table,

at 0.1 bar) = 191.8 kJ/kg

(i) Quality of steam at turbine exhaust, x4:

X4 = 0.88 ( form Mollier diagram)

(ii) Cycle efficiency, cycle: =

=

= 0.4405 or 44.05%

(iii) Steam rate in kg/kWh: Steam rate =

=

= 2.17 kg/kWh

5. A steam power plant equipped with regenerative as well as reheat arrangement is supplied

with steam to the H.P turbine at 80 bar 4700C. For feed heating , a part of steam is extracted at 7

bar and remainder of the steam is reheated to 3500C in a reheater and then expanded in L.P

turbine down to 0.035 bar.Determine:

(i) Amount of steam bled-off feed heating.

(ii) Amount of steam supplied to L.P turbine

(iii) Heat supplied in the boiler and reheater.

(iv) Cycle efficiency

(v) Power developed by the steam

The steam supplied by the boiler is 50 kg/s

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Solution:

From h-s chart and steam tables, we have enthalpies at different points as follows:

h1 = 3315 kJ/kg: h2 = 2716 kJ/kg; h3 = 3165 kJ/kg: h4 = 2236 kJ/kg

hf6 = hf2 = 697.1 kJ/kg: hf5 = hf4 = 101.9 kJ/kg( from h-s chart and steam table)

(i) Amount of steam bled-off for feed heating:

Considering energy balance at regenerator, we have:

Heat lost by steam = heat gained by water.

m (h2 – hf6) = (1 – m) (hf6 – hf5)

m(h2 – hf2) = (1 – m) (697.1 – 111.9)

m = 0.225 kg of steam supplied.

Hence amount of steam bled-off is 22.5 % of steam generated by the boiler.

(ii) Amount of steam supplied to L.P turbine = 100 – 22.5 = 77.5% of the steam

generated by the boiler.

(iii) Heat supplied in the boiler and reheater: = h1 – hf6 = 3315 – 697.1 = 2617.9 kJ/kg

Heat supplied in the reheater per kg of steam generated = ( 1- m) ( h3 – h2)

= (1 – 0.225) (3165 – 2716 ) = 347.97 kJ/kg

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Total amount of heat supplied by the boiler and reheater per kg of steam generated,

Qs = 2617.9 + 347.97 = 2965.87 kJ/kg

(iv) Cycle efficiency ; =

where,

W = 1 ( h1 –h2) + (1 – m) (h3 – h4)

= 1(3315 -2716) + ( 1-0.225) ( 3165 – 2236) = 1319 kJ/kg

=

=

= 0.4447 = 44.47%

(v) Power developed by the system: = ms x W = 50 x 1319 = 65.9

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QUADRANT-3

Wikis : (This includes wikis related to Vapour Power Cycle contains practical application and

research trends in Vapour power cycle)

1. Vapour power cycle:

http://www.powershow.com/view/109b32ZGI1Y/Vapor_Power_Cycles_powerpoint_ppt_presen

tation

2. Rankine cycle:

http://www.slideshare.net/akashdjkid/rankine-cycle-12868084

3. Vapour power cycle:

http://www.eng.fsu.edu/~shih/eml3015/lecture%20notes/rankine%20cycle.ppt

Open Contents: (This includes wikis related to introduction to Vapour power cycle contains

practical application and research trends)

1. Modern Engineering Thermodynamics By Robert T. Balmer / Chapter -13 / Vapor and Gas

power cycles / Pages-447 to 525

2.Engineering Thermodynamics By P. K. Nag / Chapter-13 / Gas Power Cycles / Pages-482 to

540

3. Thermal Engineering By R.K. Rajput / Chapter-21 / Gas Power Cycles / Pages-932 to 1003

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http://www.powershow.com/view/109b32ZGI1Y/Vapor_Power_Cycles_powerpoint_ppt_presentation

http://www.powershow.com/view/109b32ZGI1Y/Vapor_Power_Cycles_powerpoint_ppt_presentation

http://www.slideshare.net/akashdjkid/rankine-cycle-12868084

http://www.eng.fsu.edu/~shih/eml3015/lecture%20notes/rankine%20cycle.ppt



QUADRANT-4

Problems:

1. A simple rankine cycle works between pressures 28 bar and 0.06 bar, the initial condition of

steam being dry saturated. Calculate the cycle efficiency, work ratio and specific steam

consumption.

Solution: From steam table,

At 28 bar: h1 = 2802 kJ/kg, s1 = 6.2104 kJ/kg K

At 0.06 bar: hf2 = hf3 = 151.5kJ/kg. hfg2 = 2415.9

kJ/kg

Sf2 = 0.521 kJ/kgK, sfg2 = 7.809 kJ/kg K

Vf = 0.001 m3/kg

Considering turbine process 1-2, we have: S1 = S2

6.2104 = sf2 + x2 sfg2 = 0.521 + x2 x7.809

X2 =

h2 = hf2 + x2 hfg2 = 151.5 + 0.728 x 2415.9 = 1910.27 kJ/kg

Turbine work, Wturbine = h1 – h2 = 2802 – 1910.27 = 891.73 kJ/kg

Pump work = Wpump = hf4 – hf3 = vf ( p1 – p2) =

kJ/kg

Net work, Wnet = Wturbine - Wpump = 891.73 – 2.79 = 888.94 kJ/kg

Cycle efficiency =

=

=

= 0.3357 or 33.57%

Work ratio =

==

= 0.997

Steam specific consumption =

=

= 4.049 kg/kWh

2. In a steam power cycle, the steam supply is at 15 bar and dry and saturated. The condenser

pressure is 0.4 bar.Calculate the carnot and rankine efficiencies of the cycle.Neglect pump work.

Solution: Steam supply pressure, p1 = 15 bar, x1 = 1

Condenser pressure, p2 = 0.4 bar

Carnot and rankine efficiencies:

From steam table:

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At 15 bar: ts = 198.30C, hg = 2789.9 kJ/kg Sg = 6.4406 kJ/kg K

At 0.4 bar ts = 75.90C, hf = 317.7 kJ/kg, hfg = 2319.2 kJ/kg

Sf = 1.0261 kJ/kg K, Sfg = 6.6448 kJ/kg K

T1 = 198.3 + 273 = 471.3 K

T2 = 75.9 + 273 = 348.9 K

=

=

= 0.259 or 25.9 %

=

=

Where, h2 = hf2 + x2 hfg2 = 317.7 + x2 x 2319.2

To find x2? As the steam expands isentropically,

6.4406 = Sf2 +x2 Sfg2 = 1.0261 + x2 x 6.6448

x2 = 0.815

h2 = 317.7 + 0.815 x 2319.2 = 2207.8 kJ/kg

Hence, =

= 0.2354 or 23.54%

3. Steam is supplied to a turbine at a pressure of 30 bar and temperature of 4000C and is

expanded adiabaticacallly to a pressure of 0.04 bar.At a satge of turbine where the pressure is 3

bar a connection is made to surface heater in which the feed water is heated by bled steam to a

temperature of 1300C.The condensed steam from the feed feed heater is cooled in a drain cooler

to 270C.the feed water passes through the drain cooler before entering the feed heater.The cooled

drain water combines with the condensate in the well of the condenser. Assuming no heat losses

in the steam, calculate the following:

(i) Mass of steam used for feed heating per kg of steam entering the turbine.

(ii) Thermal efficiency of the cycle.

Solution: From steam table:

At 3 bar: ts = 133.50C, hf = 561.4 kJ/kg

At 0.04 bar: ts = 290C, hf = 121.5 kJ/kg

From Mollier chart :

Ho = 3231 kJ/kg (at 30 bar, 4000C)

h1 = 2700 kJ/kg (at 3 bar)

h2 = 2085 kJ/kg (at 0.04 bar)

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(i) Mass of steam used, m1:

Heat lost by the steam = Heat gained by water.

Taking the feed-heater and drain-cooler combined, we have:

m1 (h1 – hf2) = 1x 4.0186 (130-27)

m1 (2700 – 121.5) = 4.186 (130-27)

m1 = 0.1672 kg.

(ii) Thermal efficiency of the cycle:

Work done per kg of steam = 1 ( h0 – h1) + (1 – m1) (h1 – h2)

= 1 ( 323.1 – 2700) + ( 1-0.1672) (2700 – 2085) = 1043.17 kJ/kg

Heat supplied per kg of the steam = h0 – 1x4.186 x 130

= 3231 – 544.18 = 2686.82 kJ/kg

=

=

= 0.3882 or 38.82%

4. A steam power plant operates on a theoretical reheat cycle.Steam at boiler at 150bar, 5500C

expands through the high pressure turbine.It is reheated at a constant pressure of 40 bar to 5500C

and expands through the low pressure turbine to a condenser at 0.1 bar.Draw T-s and h-s

diagrams.Find:

(i) Quality of steam at turbine exhaust. (ii) Cycle efficiency and (iii) Steam rate in kg/kWh.

Solution: Refer to figs. Given below.

From Mollier diagram (h-s diagram):

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h1 = 3450 kJ/kg: h2 = 3050 kJ/kg : h3 = 3560 kJ/kg: h4 = 2300 kJ/kg: hf4 = (From steam table,

at 0.1 bar) = 191.8 kJ/kg

(i) Quality of steam at turbine exhaust, x4:

X4 = 0.88 ( form Mollier diagram)

(ii) Cycle efficiency, cycle: =

=

= 0.4405 or 44.05%

(iii) Steam rate in kg/kWh: Steam rate =

=

= 2.17 kg/kWh

5. A steam power plant equipped with regenerative as well as reheat arrangement is supplied

with steam to the H.P turbine at 80 bar 4700C. For feed heating , a part of steam is extracted at 7

bar and remainder of the steam is reheated to 3500C in a reheater and then expanded in L.P

turbine down to 0.035 bar.Determine:

(i) Amount of steam bled-off feed heating.

(ii) Amount of steam supplied to L.P turbine

(iii) Heat supplied in the boiler and reheater.

(iv) Cycle efficiency

(v) Power developed by the steam

The steam supplied by the boiler is 50 kg/s

Solution:

From h-s chart and steam tables, we have

enthalpies at different points as follows:

h1 = 3315 kJ/kg: h2 = 2716 kJ/kg;

h3 = 3165 kJ/kg: h4 = 2236 kJ/kg

hf6 = hf2 = 697.1 kJ/kg: hf5 = hf4 = 101.9

kJ/kg( from h-s chart and steam table)

(i) Amount of steam bled-off for feed

heating:

Considering energy balance at

regenerator, we have:

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Heat lost by steam = heat gained by water.

m (h2 – hf6) = (1 – m) (hf6 – hf5)

m(h2 – hf2) = (1 – m) (697.1 – 111.9)

m = 0.225 kg of steam supplied.

Hence amount of steam bled-off is 22.5 % of steam generated by the boiler.

(ii) Amount of steam supplied to L.P turbine = 100 – 22.5 = 77.5% of the steam

generated by the boiler.

(iii) Heat supplied in the boiler and reheater: = h1 – hf6 = 3315 – 697.1 = 2617.9 kJ/kg

Heat supplied in the reheater per kg of steam generated = ( 1- m) ( h3 – h2)

= (1 – 0.225) (3165 – 2716 ) = 347.97 kJ/kg

Total amount of heat supplied by the boiler and reheater per kg of steam generated,

Qs = 2617.9 + 347.97 = 2965.87 kJ/kg

(iv) Cycle efficiency ; =

where,

W = 1 ( h1 –h2) + (1 – m) (h3 – h4)

= 1(3315 -2716) + ( 1-0.225) ( 3165 – 2236) = 1319 kJ/kg

=

=

= 0.4447 = 44.47%

(iv) Power developed by the system: = ms x W = 50 x 1319 = 65.9

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Frequently asked Questions.

(1) Discuss the effect of i) Boiler pressure, ii) Condenser pressure, iii) Superheating o f the steam on

the performance of a Rankine cycle.

(2) Why Carnot cycle is not practical for a steam power plant?

(3) What are the methods which lead to increase in thermal efficiency of Rankine cycle explain

briefly.

(4) What are the effects of reheating of steam on (a) specific power output, (b) cycle efficiency?

(c) Steam rate?

(5) What are the effects of regenerative method of steam on (a) specific power output, (b) cycle

efficiency, (c) Steam rate?

(6) Why ideal regeneration is not possible? Explain in brief.

(7) With the help of a schematic diagram and T-S diagram, explain the working of a regenerative

vapour

Power cycle with one feed heater and derive an expression for its overall efficiency.

(8) With the help of a schematic diagram and T-S diagram, explain the working of a reheating vapour

power

Cycle and derive an expression for its overall efficiency.

(9) A 40 MW steam power plant working on Rankine cycle operates between boiler pressure of 4MPa

and condenser pressure of 10 KPa. The steam leaves the boiler and enters the steam turbine at 4000C.

The isentropic efficiency of the turbine is 85%. Determine: i) Cycle efficiency, ii) quality of the stem

from the turbine, iii) steam flow rate in term of Kg/hr. Consider the pump work.

(10) In a reheat cycle, steam at 5000C expands in a H.P. turbine till it is saturated vapour. It is then

reheated at constant pressure to 4000C and then expanded on an L.P. turbine to 40

0C. If the maximum

moisture content at the turbine exhaust is limited to 15%. Find i) the reheat pressure, ii) pressure of the

steam at the inlet to H.P. turbine, iii) the net specific work output, iv) cycle efficiency and v) the steam

rate. Assume all ideal processes.

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Assignments:

1) Steam is supplied to a turbine at 30 bar and 350

0C. The turbine exhaust pressure is 0.08

bar. The mail condensate is heated regenerative in two stages by steam bled from the

turbine at 5 bar and 1 bar respectively. Calculate i) bled steams mass per Kg of steam

entering, ii) cycle efficiency.

Answers: i) bled steams mass per Kg of steam entering -0.107 Kg at 5bar & 0.085 Kg at

1 bar, ii) cycle efficiency – 36.48%.

2) In a Rankine cycle, the steam at inlet to turbine is saturated at a pressure of 35 bar and the

exhaust pressure is 0.2 bar. Determine i) pump work, ii) turbine work, iii) cycle

efficiency, iv) condensate heat flow, the dryness fraction at the end of expansion.

Assume flow rate of 9.5 Kg/Sec.

Answers: i) pump work – 33.63 KW, ii) turbine work – 7495.5 KW, iii) cycle efficiency

– 30.93%, iv) condensate heat flow – 16734.25 KW,v) the dryness fraction at the end of

expansion – 0.747.

3) Steam at a pressure of 15 bar and 2500C is expanded through a turbine at first to a

pressure of 4 bar. It is then reheated at constant pressure to the initial temperature of

2500C and finally expanded to 0.1 bar. Estimate work done per Kg of steam and heat

supplied during reheat.

Answers: work done per Kg of steam – 885 Kj/Kg and heat supplied during reheat – 300

Kj/Kg.

4) Discuss the effect of i) Boiler pressure, ii) Condenser pressure, iii) Superheating o f the steam

on the performance of a Rankine cycle.

5) Why Carnot cycle is not practical for a steam power plant?

6) What are the methods which lead to increase in thermal efficiency of Rankine cycle explain

briefly.

7) What are the effects of reheating of steam on (a) specific power output, (b) cycle efficiency?

(c) Steam rate?

8) What are the effects of regenerative method of steam on (a) specific power output, (b) cycle

efficiency, (c) Steam rate?

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Self Answered Question & Answer

1) In a reheat cycle, steam at 5000C expands in a H.P. turbine till it is saturated vapour. It is then

reheated at constant pressure to 4000C and then expanded on an L.P. turbine to 40

0C. If the maximum

moisture content at the turbine exhaust is limited to 15%. Find i) the reheat pressure, ii) pressure of the

steam at the inlet to H.P. turbine, iii) the net specific work output, iv) cycle efficiency and v) the steam

rate. Assume all ideal processes.

Answers: the reheat pressure – 20 bar, pressure of the steam at the inlet to H.P. turbine – 150

bar, the net specific work output – 1529.8 Kj/Kg, cycle efficiency – 42.7%, the steam rate –

2.353 Kg/Kw hr.

2) The net power output of an ideal regenerative – reheat steam cycle is 80 MW. Steam enters

the HP turbine at 80 bar, 5000C and expands till it becomes saturated vapour. Some of the steam

then goes to an open feed heater and the balance is reheated to 4000C, after which it expands I

the LP turbine to 0.07 bar. Compute i) the reheat pressure, ii) steam flow rate to HP turbine, iii)

cycle efficiency.

Answers: The reheat pressure – 6.5 bar, steam flow rate to HP turbine – 60 Kg/Sec,

cycle efficiency – 43.15%

3) Stem at 20 bar,3600C is expanded in a steam turbine to a pressure of 0.08 bar. It then enters a

condenser, where it is condensed to saturated liquid water. Assuming the turbine and pump

efficiencies are 60% and 90% respectively. Determine per Kg of steam, the net work, the heat

transfer to the working fluid and cycle efficiency.

Answers: The net work – 580.79 Kj/Kg, the heat transfer to the working fluid – 2983 Kj / Kg,

cycle efficiency – 19.46%

4) A steam turbine operates on simple regenerative Rankine cycle. The steam is supplied dry and

saturated at 40 bar and is exhausted to condenser at 0.07 bar. The condensate is pumped at a

pressure of 3.5 bar at which it is mixed with bled steam from the turbine at 3.5 bar. The mixture is

saturated water at 3.5 bar is then pumped to the boiler. Neglecting the pump work, for ideal cycle,

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calculate amount of bled steam per Kg of steam supplied, cycle efficiency and specific steam

consumption.

Answers: Amount of bled steam per Kg of steam supplied – 0.19 Kg/Kg, cycle efficiency – 37% and

Specific steam consumption – 4.39Kg/Kw hr.

5) Steam expands in a turbine from 25 bar and 3000C to a condenser pressure of 20 KPa. Calculate

Rankine efficiency. (i) What would be the efficiency if the temperature of the steam to be 5000C

instead of 3000C?, (ii) If the boiler pressure is increased to 60 bar maintain the steam temperature at

5000C, calculate cycle efficiency. Assume condenser pressure remain constant in all cases.

Answers: Rankine efficiency – 27.64%, if the temperature of the steam to be 5000C instead of 300

0C,

Rankine efficiency – 32.26%, If the boiler pressure is increased to 60 bar and 5000C - 36.63%.

Test Your Skills:

Fill up the blanks 1) In Carnot cycle, expansion of steam in turbine takes at ………………… process

2) Efficiency of Carnot cycle increases with ………… boiler pressure.

3) Efficiency of Rankine cycle increases with ………… of condenser pressure.

4) Unit of specific steam consumption ………………..

5) In rankine cycle heat supply takes at ……………. Process.

6) Regenerative method is ……………. The heat input.

7) In steam turbine corrosion of turbine blade is reduced by ………………… the input steam.

8) Rankine cycle efficiency of a good steam power plant in the range of ………..

9) ………… cycle is maximum thermal efficiency cycle for given range of temperature.

10) Rankine cycle with multiple regeneration can be approximated as ……………..

Answers :

1) Isentropic, 2) Increases, 3) Decreasing, 4) Kg / Kw hr, 5) Constant pressure,

6) Reduces, 7) Super heating, 8) 35-45%, 9) Carnot, 10) Carnot vapour cycle.

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Match the following Part A Part B

1. Carnot vapour power cycle a. By increasing degree of super heat

2. In Rankine cycle heat supply takes at b. By using feed water heater

3. Efficiency of Rankine cycle increases c. Change in enthalpy between inlet and

outlet

4. Reheating d. Difficult to achieve practically

5. Thermal stress on the boiler surface reduces e. Simple Rankine cycle

6. Rankine cycle work output of turbine f. Modified Rankine cycle

7. Regenerative cycle thermal efficiency always g. Constant pressure process

higher than

8. Steam engine operates on h. Increases the output of the turbine

Answers :

1-d, 2-g, 3-a, 4-h, 5-b, 6-c, 7-e, 8-f.

TRUE (T)/ FALSE (F) 1) For the same temperature limits Rankine cycle produce higher specific work output than a

carnot cycle. ---------- (T)

2) Practically achieving of Carnot vapour power cycle is possible. ------------- (F)

3) For given boiler pressure, efficiency of Rankine cycle is always lower than Carnot cycle. --(T)

4) Vapour power cycle efficiency is the ratio of net work output to the heat supplied. ------ (T)

5) Regenerative feed heater is reduces the thermal stress on the boiler surface. -------- (T)

6) For same higher and lower pressure limits, Carnot cycle efficiency is higher than Rankine

cycle.—(F)

7) Rankine vapour power cycle is most popular cycle (T)

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module-iv --- vapour power cycle applied thermodynamics

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