module-iv --- vapour power cycle applied thermodynamics
Transcript of 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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Dr.A.R.Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
MODULE-IV --- VAPOUR POWER CYCLE APPLIED THERMODYNAMICS 2014
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 APPLIED THERMODYNAMICS 2014
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:
http://courses.engr.illinois.edu/npre201/coursematerial/energy_cycles/lecture05figures/Figure5.2
b.html
6. Regeneration in a Rankine Cycle with Saturated Throttle Conditions
http://courses.engr.illinois.edu/npre201/coursematerial/energy_cycles/lecture05figures/Figure5.2
c.html
7. Rankine Cycle Regeneration Using Two Independent Closed Heaters:
http://courses.engr.illinois.edu/npre201/coursematerial/energy_cycles/lecture05figures/Figure5.2
d.html
8. T(S) with Two Independent Closed Heaters:
http://courses.engr.illinois.edu/npre201/coursematerial/energy_cycles/lecture05figures/Figure5.2
e.html
9. Rankine Cycle with Reheat:
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Dr.A.R.Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
MODULE-IV --- VAPOUR POWER CYCLE APPLIED THERMODYNAMICS 2014
http://courses.engr.illinois.edu/npre201/coursematerial/energy_cycles/lecture05figures/Figure5.2
f.html
10. T(S) and H(S) for Rankine Cycle with Reheat:
http://courses.engr.illinois.edu/npre201/coursematerial/energy_cycles/lecture05figures/Figure5.2
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:
http://www.youtube.com/watch?v=-Ot-SFAoTmE
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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Dr.A.R.Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
MODULE-IV --- VAPOUR POWER CYCLE APPLIED THERMODYNAMICS 2014
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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Dr.A.R.Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
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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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Dr.A.R.Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
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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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Dr.A.R.Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
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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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Dr.A.R.Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
MODULE-IV --- VAPOUR POWER CYCLE APPLIED THERMODYNAMICS 2014
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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Dr.A.R.Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
MODULE-IV --- VAPOUR POWER CYCLE APPLIED THERMODYNAMICS 2014
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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Dr.A.R.Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
MODULE-IV --- VAPOUR POWER CYCLE APPLIED THERMODYNAMICS 2014
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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Dr.A.R.Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
MODULE-IV --- VAPOUR POWER CYCLE APPLIED THERMODYNAMICS 2014
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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Dr.A.R.Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
MODULE-IV --- VAPOUR POWER CYCLE APPLIED THERMODYNAMICS 2014
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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Dr.A.R.Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
MODULE-IV --- VAPOUR POWER CYCLE APPLIED THERMODYNAMICS 2014
(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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Dr.A.R.Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
MODULE-IV --- VAPOUR POWER CYCLE APPLIED THERMODYNAMICS 2014
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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Dr.A.R.Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
MODULE-IV --- VAPOUR POWER CYCLE APPLIED THERMODYNAMICS 2014
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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Dr.A.R.Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
MODULE-IV --- VAPOUR POWER CYCLE APPLIED THERMODYNAMICS 2014
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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Dr.A.R.Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
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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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Dr.A.R.Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
MODULE-IV --- VAPOUR POWER CYCLE APPLIED THERMODYNAMICS 2014
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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Dr.A.R.Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
MODULE-IV --- VAPOUR POWER CYCLE APPLIED THERMODYNAMICS 2014
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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Dr.A.R.Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
MODULE-IV --- VAPOUR POWER CYCLE APPLIED THERMODYNAMICS 2014
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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