APPLIED THERMODYNAMICS 18ME42 Module 03: Vapour Power Cycles

Mr THANMAY J S, Asst Proff, Dept of Mechanical Engineering, VVIET Mysore Page 1
APPLIED THERMODYNAMICS
18ME42
Course Coordinator
Mr. THANMAY J. S
Assistant Professor
Department of Mechanical Engineering
VVIET Mysore
Module 03: Vapour Power Cycles
Course Learning Objectives
 To understand fundamentals of Vapour Power cycle, Construction and working Principle
and to calculate actual cycle Performance.
Course Outcomes
The students will understand the principle of Vapour Power cycle, applications and identify
methods for performance improvement.

Contents
Carnot vapour power cycle, drawbacks as a reference cycle.
Simple Rankine cycle; description, T-S diagram, analysis for performance.
Comparison of Carnot and Rankine cycles.
Effects of pressure and temperature on Rankine cycle performance.
Actual vapour power cycles.
Ideal and practical regenerative Rankine cycles, open and closed feed water heaters.
Reheat Rankine cycle.
Characteristics of an Ideal working fluid in vapour power cycles.

Introduction
Some of commonly used performance parameters in Vapour Power Cycle analysis are
described here.
a.
b.
c.
d.

Carnot vapour power cycle
Carnot cycle has already been defined earlier as an ideal cycle having highest thermodynamic
efficiency. Carnot vapour power cycle is as follows.
4– 1 = Reversible isothermal heat addition in the boiler
1 – 2 = Reversible adiabatic expansion in steam turbine
2 – 3 = Reversible isothermal heat rejection in the condenser
3 – 4 = Reversible adiabatic compression or pumping in feed water pump
𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑁𝑒𝑡 𝑊𝑜𝑟𝑘
𝐻𝑒𝑎𝑡 𝑎𝑑𝑑𝑒𝑑
=
𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝑊𝑜𝑟𝑘−𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑜𝑟 𝑃𝑢𝑚𝑝𝑖𝑛𝑔 𝑊𝑜𝑟𝑘
∴ 𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝜂(𝐶𝑎𝑟𝑛𝑜𝑡 𝑉𝑎𝑝𝑜𝑢𝑟) =
𝑊𝑇−𝑊𝐶
𝑄𝑆
𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝑊𝑜𝑟𝑘 𝑊𝑇 = (ℎ1 − ℎ2)
𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 𝑜𝑟 𝑃𝑢𝑚𝑝𝑖𝑛𝑔 𝑤𝑜𝑟𝑘 𝑊𝐶 = (ℎ4 − ℎ3)
𝐻𝑒𝑎𝑡 𝑎𝑑𝑑𝑒𝑑 𝑄𝑆 = (ℎ1 − ℎ4) = 𝑇1(𝑆1 − 𝑆4)
𝐻𝑒𝑎𝑡 𝑟𝑒𝑗𝑒𝑐𝑡𝑒𝑑 𝑄𝑅 = (ℎ2 − ℎ3) = 𝑇3(𝑆2 − 𝑆3)
∴ 𝜂(𝐶𝑎𝑟𝑛𝑜𝑡 𝑉𝑎𝑝𝑜𝑢𝑟) =
(ℎ1−ℎ2) −(ℎ4−ℎ3)
(ℎ1−ℎ4)
𝑹𝒆𝒂𝒓𝒓𝒂𝒏𝒈𝒊𝒏𝒈 (𝒉𝟏 − 𝒉𝟐) − (𝒉𝟒 − 𝒉𝟑) 𝒕𝒐 (𝒉𝟏 − 𝒉𝟒) − (𝒉𝟐 − 𝒉𝟑) 𝒘𝒆 𝒈𝒆𝒕
∴ 𝜼(𝑪𝒂𝒓𝒏𝒐𝒕 𝑽𝒂𝒑𝒐𝒖𝒓) =
(𝒉𝟏 − 𝒉𝟒) − (𝒉𝟐 − 𝒉𝟑)
(𝒉𝟏 − 𝒉𝟒)
= 𝟏 −
(𝒉𝟐 − 𝒉𝟑)
(𝒉𝟏 − 𝒉𝟒)
= 𝟏 −
𝑸𝑹
𝑸𝑺
∴ 𝜂(𝐶𝑎𝑟𝑛𝑜𝑡 𝑉𝑎𝑝𝑜𝑢𝑟) = 1 −
𝐻𝑒𝑎𝑡 𝑟𝑒𝑗𝑒𝑐𝑡𝑒𝑑
= 1 −
(ℎ2−ℎ3)
(ℎ1−ℎ4)
= 1 −
𝑇3(𝑆3−𝑆4)
𝑇1(𝑆2−𝑆1)
𝑏𝑢𝑡 𝑆1 = 𝑆4 𝑎𝑛𝑑 𝑆2 = 𝑆3
∴ 𝜼(𝑪𝒂𝒓𝒏𝒐𝒕 𝑽𝒂𝒑𝒐𝒖𝒓) = 𝟏 −
𝑻𝟑
𝑻𝟏
= 𝟏 −
𝑻(𝒎𝒂𝒙𝒊𝒎𝒖𝒎)
𝑻(𝒎𝒊𝒏𝒊𝒎𝒖𝒎)

The above Figure shows working of a Carnot Vapour cycle on T-s and p-V diagrams. It
consists of
(i) two constant pressure operations (4-1) and (3-2) and
(ii) two frictionless adiabatic (4-3) and (2-1).
These operations are discussed below:
1. Operation (4-1). 1 kg of boiling water at temperature T1 is heated to form wet steam of
dryness fraction x1. Thus heat is absorbed at constant temperature T1 and pressure p1 during
this operation.
2. Operation (1-2). During this operation steam is expanded isentropically to temperature T2
and pressure p2. The point ‘2’ represents the condition of steam after expansion.
3. Operation (2-3). During this operation heat is rejected at constant pressure p2 and
temperature T2. As the steam is exhausted it becomes wetter and cooled from 2 to 3.
4. Operation (3-4). In this operation the wet steam at ‘3’ is compressed isentropically till the
steam regains its original state of temperature T1 and pressure p1. Thus cycle is completed.
Drawbacks of Carnot vapour power cycle (Limitations of Carnot Cycle)
Though Carnot cycle is simple (thermodynamically) and has the highest thermal efficiency
for given values of T1 and T2, yet it is extremely difficult to operate in practice because of the
following reasons:
1. It is difficult to compress a wet vapour isentropically to the saturated state as required by
the process 3-4.
2. It is difficult to control the quality of the condensate coming out of the condenser so that
the state is exactly obtained.
3. The efficiency of the Carnot cycle is greatly affected by the temperature T1 at which heat
is transferred to the working fluid. Since the critical temperature for steam is only 374°C, there-
fore, if the cycle is to be operated in the wet region, the maximum possible temperature is
severely limited.
4. The cycle is still more difficult to operate in practice with superheated steam due to the
necessity of supplying the superheat at constant temperature instead of constant pressure (as it
is customary).
In a practical cycle, limits of pressure and volume are far more easily realized than limits of
temperature so that at present no practical engine operates on the Carnot cycle, although all
modern cycles aspire to achieve it.

Simple Rankine cycle; description, T-S diagram, analysis for performance.
The main change in Rankine cycle is in Rankin cycle complete condensation of water vapor in
the condenser, and then, pumping the water isentropically to boiler pressure is achieved.
(i) Boiler: In boiler the working fluid (water) at state ‘4’in sub cooled condition is converted
into dry saturated steam at state ‘1’ by receiving heat ‘Qs’ from high temperature heat source
through the following processes.
Process 4-5: As the water enters the boiler from pump in sub cooled condition state ‘4’ at
pressure PH, it is first heated up to the saturated state 5 at constant pressure (sensible heating).
Process 5-1: Then water at saturated condition 5 is further heated up at constant pressure PH and
constant saturation temperature to the saturated steam at state 1 (latent heat of vaporization).
Process 4-5-1: Total heat addition in boiler,
𝐐𝐬 = 𝒉𝟏 − 𝒉𝟒
(ii) Steam Turbine: In the steam turbine, the dry saturated steam from the boiler at state ‘1’ at
pressure ‘pH’ expands isentropically to wet steam at pressure ‘pL’ and thus produce mechanical
work, (WT).
Process 1-2: Isentropic expansion of steam in turbine (Steam turbine work(WT)), Steam
turbine work is given by,
𝑾𝑻 = 𝒉𝟏 − 𝒉𝟐

(iii) Condenser: In the condenser, the exhaust wet steam from turbine is condensed by
rejecting heat ‘Qr’ to the cooling water.
Process 2-3: Constant pressure (back pressure), constant temperature heat rejection in
condenser (Condensation), Total heat rejected in condenser,
𝐐𝐫 = 𝒉𝟐 − 𝒉𝟑
(iv) Feed pump: The feed pump is used to pump the condensate water (saturated water) from
the hot-well to the boiler at the boiler pressure, pH.
Process 3-4: Isentropic compression of water in pump (Pump work, (WP)), The Pump work is
given by,
𝑾𝑷 = 𝒉𝟒 − 𝒉𝟑
Pump work, (WP) is also given by 𝑾𝑷 = 𝑷𝟒 − 𝑷𝟑
𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 (𝑅𝑎𝑛𝑘𝑖𝑛𝑒) =
𝑁𝑒𝑡 𝑊𝑜𝑟𝑘
=
𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝑊𝑜𝑟𝑘−𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑜𝑟 𝑃𝑢𝑚𝑝𝑖𝑛𝑔 𝑊𝑜𝑟𝑘
∴ 𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝜂(𝑅𝑎𝑛𝑘𝑖𝑛𝑒) =
𝑊𝑇−𝑊𝐶
𝑄𝑆
𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝑊𝑜𝑟𝑘 𝑊𝑇 = (ℎ1 − ℎ2)
𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 𝑜𝑟 𝑃𝑢𝑚𝑝𝑖𝑛𝑔 𝑤𝑜𝑟𝑘 𝑊𝐶 = (ℎ4 − ℎ3)
𝐻𝑒𝑎𝑡 𝑎𝑑𝑑𝑒𝑑 𝑄𝑆 = (ℎ1 − ℎ4) = 𝑇1(𝑆1 − 𝑆4)
𝐻𝑒𝑎𝑡 𝑟𝑒𝑗𝑒𝑐𝑡𝑒𝑑 𝑄𝑅 = (ℎ2 − ℎ3) = 𝑇3(𝑆2 − 𝑆3)
∴ 𝜂(𝑅𝑎𝑛𝑘𝑖𝑛𝑒) =
(ℎ1−ℎ2) −(ℎ4−ℎ3)
(ℎ1−ℎ4)
𝑹𝒆𝒂𝒓𝒓𝒂𝒏𝒈𝒊𝒏𝒈 (𝒉𝟏 − 𝒉𝟒) 𝐚𝐬 (𝒉𝟏 − 𝒉𝟑) − (𝒉𝟒 − 𝒉𝟑) 𝒘𝒆 𝒈𝒆𝒕
∴ 𝜂(𝑅𝑎𝑛𝑘𝑖𝑛𝑒) =
(ℎ1 − ℎ2) − (ℎ4 − ℎ3)
(ℎ1 − ℎ3) − (ℎ4 − ℎ3)
=
(ℎ1 − ℎ2) − 𝑊𝑃
(ℎ1 − ℎ3) − 𝑊𝑃
In a Rankine cycle the pump work may be neglected as it is very small compared with other
energy transfers. Hence we have work done by pump as 𝑾𝑷 = 𝟎
∴ 𝜼(𝑹𝒂𝒏𝒌𝒊𝒏𝒆) =
(𝒉𝟏 − 𝒉𝟐)
(𝒉𝟏 − 𝒉𝟑)

Comparison of Carnot and Rankine cycles.
a) Rankine cycle without superheat: 1 - A - 2 - 3 - 4 - 1.
b) Rankine cycle with superheat: 1 - A - 2 - 2′ -3′ - 4 - 1.
c) Carnot cycle without superheat: A - 2 - 3 -4′ - A.
d) Carnot cycle with superheat: A - 2′′ - 3′ - 4′ - A.
 Heat addition process of Rankine cycle is reversible isobaric whereas heat addition process of
Carnot cycle is reversible isothermal.
 (η Rankine) < (η Carnot).
 The maximum efficiency of Rankine cycle (η Rankine) is the function of the mean
temperature of heat addition only.
 Efficiency of Rankine cycle increases with increase in superheat of the steam.
 The pressure at which heat is added in Rankine cycle increases, the moisture content at the
turbine exhaust increases.
 Increase in the pressure difference between which the Rankine cycle operates the chances of
corrosion of blades of turbine increase.

Effects of pressure and temperature on Rankine cycle performance.
(i) Decreasing the of Condenser Pressure (Lower TL)
Lowering the condenser pressure will increase the area enclosed by the cycle on a T-s diagram
which indicates that the net work will increase. Thus, the thermal efficiency of the cycle will
be increased.
Figure: Effect of lowering the condenser pressure on ideal Rankine cycle
(ii) Superheating the Steam to High Temperatures (Increase TH)
Superheating the steam will increase the net work output and the efficiency of the cycle. It also
decreases the moisture contents of the steam at the turbine exit. The temperature to which steam
can be superheated is limited by metallurgical considerations (~ 620°C).
Figure: The effect of increasing the boiler pressure on the ideal Rankine cycle.
(iii)Increasing the Boiler Pressure (Increase TH)
Increasing the operating pressure of the boiler leads to an increase in the temperature at
which heat is transferred to the steam and thus raises the efficiency of the cycle.
Figure: The effect of increasing the boiler pressure on the ideal cycle.

Actual vapour power cycles or Actual Rankine Cycle
Process 1-2: Turbine losses
When we consider the actual vapour cycle process, 1-2 process will not be vertical or 1-2
process will not be isentropic. Pressure drop because of friction and loss of heat energy to
surrounding are the most important causes.
In process 1-2, line will not be vertical but also it will move towards right side as shown in
figure below because according to the principle of increase in entropy there will be increment
in entropy during this process.
Actual work done by turbine, WT Actual = h1-h2
1
Ideal work done by turbine, WT,Ideal = h1-h2
As we know that friction will be present during the process of expansion through the turbine
and therefore friction will be converted in terms of intermolecular energy and this
intermolecular energy will increase the temperature and hence enthalpy will also be increased.
Therefore, h2
1
> h2
Process 3-4: Pump losses
Similarly, In process 3-4. Process 3-4 indicates the ideal process for working fluid flowing
through feed pump. In practical, process 3-41
will be the process for working fluid flowing
through feed pump. Now we will see the ideal work required by the feed pump and also actual
work required by the feed pump here.
Actual work required by the feed pump, WP, Actual = h4
1 – h3
Ideal work required by the feed pump, WP, Ideal= h4-h3
As we can easily observe that h4
1
> h4 and therefore actual work required by the feed pump will
be greater as compared to the ideal work required by the feed pump.
𝜼(𝑻𝒖𝒓𝒃𝒊𝒏𝒆/𝑹𝒂𝒏𝒌𝒊𝒏) =
𝑨𝒄𝒕𝒖𝒂𝒍 𝑻𝒖𝒓𝒃𝒊𝒏𝒆 𝑾𝒐𝒓𝒌
𝑰𝒅𝒆𝒂𝒍 𝑻𝒖𝒓𝒃𝒊𝒏𝒆 𝑾𝒐𝒓𝒌
=
(𝒉𝟏−𝒉𝟐′)
(𝒉𝟏−𝒉𝟐)
𝜼(𝑷𝒖𝒎𝒑/𝑹𝒂𝒏𝒌𝒊𝒏) =
𝑰𝒅𝒆𝒂𝒍 𝑷𝒖𝒎𝒑 𝑾𝒐𝒓𝒌
𝑨𝒄𝒕𝒖𝒂𝒍 𝑷𝒖𝒎𝒑 𝒘𝒐𝒓𝒌 𝑾𝒐𝒓𝒌
=
(𝒉𝟒−𝒉𝟑)
(𝒉𝟒′−𝒉𝟑)

Ideal and practical regenerative Rankine cycles, open and closed feed water heaters.
A regeneration process in steam power plants is accomplished by additional heating up steam
or water feed to the cycle, this produced more work by expanding further in the turbine. The
device where the feed water is heated by regeneration is called a regenerator, or a feed water
heater (FWH).
(a) Ideal Regeneration Rankine Cycle
𝑻𝒖𝒓𝒃𝒊𝒏𝒆 𝑾𝒐𝒓𝒌 𝑾𝑻 = (𝒉𝟐 − 𝒉𝟑)
𝑷𝒖𝒎𝒑 𝑾𝒐𝒓𝒌 𝑾𝑷 = (𝒉𝟓 − 𝒉𝟒)
𝑯𝒆𝒂𝒕 𝒂𝒅𝒅𝒆𝒅 𝑸𝑺𝟏 = (𝒉𝟐 − 𝒉𝟏)
𝑯𝒆𝒂𝒕 𝒓𝒆𝒋𝒆𝒄𝒕𝒆𝒅 𝑸𝑺𝟐 = (𝒉𝟏 − 𝒉𝟓)
𝑯𝒆𝒂𝒕 𝒓𝒆𝒋𝒆𝒄𝒕𝒆𝒅 𝑸𝑹 = (𝒉𝟑 − 𝒉𝟒)
𝑻𝒉𝒆𝒓𝒎𝒂𝒍 𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒄𝒚 𝜼(𝑹𝒂𝒏𝒌𝒊𝒏𝒆) =
𝑾𝑻−𝑾𝑪
𝑸𝑺
𝜼(𝑹𝒂𝒏𝒌𝒊𝒏𝒆) =
𝑾𝑻−𝑾𝑷
𝑸𝑺𝟏+𝑸𝑺𝟐
=
(𝒉𝟐−𝒉𝟑)−(𝒉𝟓−𝒉𝟒)
(𝒉𝟐−𝒉𝟏)+(𝒉𝟏−𝒉𝟓)
=
(𝒉𝟐−𝒉𝟓)−(𝒉𝟑−𝒉𝟒)
(𝒉𝟐−𝒉𝟓)
(𝒉𝟐−𝒉𝟓)−(𝒉𝟑−𝒉𝟒)
(𝒉𝟐−𝒉𝟓)
= 𝟏 −
(𝒉𝟑−𝒉𝟒)
(𝒉𝟐−𝒉𝟓)
= 𝟏 −
𝑸𝑹
𝑸𝑺𝟏+𝑸𝑺𝟐

Working
a) Theoretical arrangement shows that the steam enters the turbine at state 2 (temperature T2)
and expands to (temperature T3) state 3.
b) Condensate at state 5 enters the turbine casing which has annular space around turbine.
Feed water enters turbine casing at state 5 and gets infinitesimally heated up to state 1 while
flowing opposite to that of expanding steam.
c) This hot feed water enters into boiler where steam generation occurs at desired state, say 2.
Feed water heating in steam turbine casing is assumed
to occur reversibly as the heating of feed water occurs
by expanding steam with infinitesimal temperature
difference and is called “regenerative heating”. This
cycle is called regenerative cycle due to regenerative
heating employed in it.
Regenerative heating refers to the arrangement in
which working fluid at one state is used for heating
itself and no external heat source is used for this
purpose.
d) Here feed water picks up heat from steam expanding in steam turbine, thus the expansion
process in steam turbine shall get modified from 2-3' ideally to 2-3.
e) Heat picked up by feed water for getting heated up from state 5 to 1 is shown by hatched
area (1-7-6-5-1) on T-S diagram.
f) Under ideal conditions for cent per cent heat exchange effectiveness the two areas i.e. (2-
9-8-3-2) indicating heat extraction from steam turbine and (1-7-6-5-1) indicating heat
recovered by feed water shall be same.
Thus, T-S representation of regenerative cycle indicates that the cycle efficiency shall be more
than that of Rankine cycle due to higher average temperature of heat addition.
(b) Practical Regeneration Rankine Cycle
An open feed water heater is basically a mixing chamber, where the steam extracted from the
turbine mixes with the water exiting the pump. We have two types of feed water heater:
[1] Open Feed Water Heater
An open feed water heater which is known as a direct contact heat exchanger or mixing heat
exchanger. It is also called single – stage regenerative feed water heater.
[2] Closed Feed Water Heater
It is indirect heat exchanger or shell and tube heat exchanger.

(i) Rankine Cycle with Open Feed Heater
The schematic of a steam power plant with one open feed water heater is shown in Figure
below.
Energy entering regenerator = Energy
leaving regenerator
(𝒚). 𝒉𝟐 + (𝟏 − 𝒚)𝒉𝟓 = 𝒉𝟔
(𝒚). 𝒉𝟐 + 𝒉𝟓 − (𝒚)𝒉𝟓 = 𝒉𝟔
(𝒚). (𝒉𝟐 − 𝒉𝟓) = 𝒉𝟔 − 𝒉𝟓
(𝒚) =
(𝒉𝟔 − 𝒉𝟓)
(𝒉𝟐 − 𝒉𝟓)
Working
(i) In a Regenerative Rankine cycle with an open feed water heater, steam from the boiler
(state 1) expands in the turbine to an intermediate pressure (state 2).
(ii) At this state, some of the steam is extracted and sent to the feed water heater, while the
remaining steam in the turbine continues to expand to the condenser pressure (state 3).
(iii)Saturated water from the condenser (state 4) is pumped to the feed water pressure and send
to the feed water heater (state 5).
(iv)At the feed water heater, the compressed water is mixed with the steam extracted from the
turbine (state 2) and exits the feed water heater as saturated water at the heater pressure
(state 6).
(v) Then the saturated water is pumped to the boiler pressure by a second pump (state 7). The
water is heated to a higher temperature in the boiler (state 1) and the cycle repeats again.
The T-s diagram of this cycle is shown on the left.
Note that the mass flow rate at each component is different. If (1) kg steam enters the turbine,
(y) kg is extracted to the feed water heater and (1-y) kg continues to expand to the condenser
pressure. So if the mass flow rate at the boiler is (1) kg, then the mass flow rate from other
components are:
a. Condenser: (1-y)
b. Pump 01: (1-y)
c. Feed water Heater: [(y)+(1-y)]
d. Pump 02: (1)

For convenience, heat and work interactions for regenerative Rankine cycle is expressed per
unit mass of steam flowing through the boiler. They are:
Process 1-2 and 1-3: 𝑻𝒖𝒓𝒃𝒊𝒏𝒆 𝑾𝒐𝒓𝒌 𝑾𝑻 = (𝒉𝟏 − 𝒉𝟐) + (𝟏 − 𝒚)(𝒉𝟐 − 𝒉𝟑)
Process 4-5 and 6-7: 𝑷𝒖𝒎𝒑 𝑾𝒐𝒓𝒌 𝑾𝑷 = 𝑾𝑷𝟏 + 𝑾𝑷𝟐 = (𝟏 − 𝒚)(𝒉𝟓 − 𝒉𝟒) + (𝒉𝟕 − 𝒉𝟔)
𝑃𝑢𝑚𝑝 𝑊𝑜𝑟𝑘 𝑐𝑎𝑛 𝑎𝑙𝑠𝑜 𝑏𝑒 𝑚𝑒𝑛𝑡𝑖𝑜𝑛𝑒𝑑 𝑎𝑠 𝑾𝑷𝟏 = 𝑽𝟒(𝑷𝟓 − 𝑷𝟒) 𝑎𝑛𝑑 𝑾𝑷𝟐 = 𝑽𝟔(𝑷𝟕 − 𝑷𝟔)
Process 7-1: 𝑯𝒆𝒂𝒕 𝒂𝒅𝒅𝒆𝒅 𝑸𝑺 = (𝒉𝟏 − 𝒉𝟕)
Process 3-4: 𝑯𝒆𝒂𝒕 𝒓𝒆𝒋𝒆𝒄𝒕𝒆𝒅 𝑸𝑹 = (𝟏 − 𝒚)(𝒉𝟑 − 𝒉𝟒)
𝑻𝒉𝒆𝒓𝒎𝒂𝒍 𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒄𝒚 𝜼(𝑹𝒂𝒏𝒌𝒊𝒏𝒆)𝒊𝒔 𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝒂𝒔 =
𝑯𝒆𝒂𝒕 𝑺𝒖𝒑𝒑𝒍𝒊𝒆𝒅 −𝑯𝒆𝒂𝒕 𝑹𝒆𝒋𝒆𝒄𝒕𝒆𝒅
𝑯𝒆𝒂𝒕 𝑺𝒖𝒑𝒑𝒍𝒊𝒆𝒅
=
𝑸𝑺−𝑸𝑹
𝑸𝑺
𝑸𝑺 − 𝑸𝑹
𝑸𝑺
=
(𝒉𝟏 − 𝒉𝟕) − (𝟏 − 𝒚)(𝒉𝟑 − 𝒉𝟒)
(𝒉𝟏 − 𝒉𝟕)
Open feed water heaters are simple and inexpensive, and can also bring the feed water to
saturated state. However, each feed water needs a separate pump which adds to the cost.
(c) Rankine Cycle with Closed Feed Heater
(i) Closed FWH are shell-and-tube type heat exchanger in which feed water temperature
increases as the extracted steam condenses on the outside of the tubes carrying the feed
water.
(ii) The two streams can be at different pressures since the two streams do not mix.
(iii)The extracted stream condenses in the closed feed water while heating the feed water from
the pump.
(iv)The heated feed water is send to the boiler and condensate from the feed water heater.
There are two types of closed feed water heaters
Closed FWH with Drain Pumped Forward Closed FWH with Drain Cascaded Backward

Rankine Cycle with Closed FWH with Drain Pumped Forward System
𝑻𝒖𝒓𝒃𝒊𝒏𝒆 𝑾𝒐𝒓𝒌
𝑾𝑻 = (𝒉𝟔 − 𝒉𝟕) + (𝟏 − 𝒚)(𝒉𝟕 − 𝒉𝟖)
𝑷𝒖𝒎𝒑 𝑾𝒐𝒓𝒌 𝑾𝑷 = 𝑾𝑷𝟏 + 𝑾𝑷𝟐 = (𝟏 − 𝒚)(𝒉𝟐 − 𝒉𝟏) + (𝒚)(𝒉𝟒 − 𝒉𝟑)
𝑯𝒆𝒂𝒕 𝒂𝒅𝒅𝒆𝒅
𝑸𝑺 = (𝒉𝟔 − 𝒉𝟓)
𝑯𝒆𝒂𝒕 𝒓𝒆𝒋𝒆𝒄𝒕𝒆𝒅
𝑸𝑹 = (𝟏 − 𝒚)(𝒉𝟖 − 𝒉𝟏)
𝑾𝑻 − 𝑾𝑷
𝑸𝑺
=
(𝒉𝟔 − 𝒉𝟕) + (𝟏 − 𝒚)(𝒉𝟕 − 𝒉𝟖) − (𝟏 − 𝒚)(𝒉𝟐 − 𝒉𝟏) + (𝒚)(𝒉𝟒 − 𝒉𝟑)
(𝒉𝟔 − 𝒉𝟓)
𝑸𝑺
=
(𝒉𝟔 − 𝒉𝟓) − [(𝟏 − 𝒚)(𝒉𝟖 − 𝒉𝟏)]
(𝒉𝟔 − 𝒉𝟓)
Rankine Cycle with Closed FWH with Drain Cascaded Backward System
𝑬𝒏𝒆𝒓𝒈𝒚 𝒆𝒏𝒕𝒆𝒓𝒊𝒏𝒈 𝒓𝒆𝒈𝒊𝒏𝒊𝒓𝒂𝒕𝒐𝒓
= 𝑬𝒏𝒆𝒓𝒈𝒚 𝒍𝒆𝒂𝒗𝒊𝒏𝒈 𝒓𝒆𝒈𝒊𝒏𝒆𝒓𝒂𝒕𝒐𝒓
(𝒚)𝐡𝟓 + 𝐡𝟐 = (𝐲)𝐡𝟕 + 𝐡𝟑
(𝒚)( 𝐡𝟓 – 𝐡𝟕) = (𝐡𝟑 – 𝐡𝟐)
(𝒚) =
(𝐡𝟑 – 𝐡𝟐)
( 𝐡𝟓 – 𝐡𝟕)
Assignment Activity: Derive eqation for Work done by Turbine, Work done by Pump, Heat supply and
Thermal Efficiency (Heat Input: qin = h4 - h3; Heat Output: qout = (1 - y)(h1 - h6) + y(h8 - h1); Work Output:
Wturb,out = (h4 - h5) + (1 - y)(h5 - h6); Work input: Wpump,in = (h2 - h1))

Differentiate Between Open and Closed Feed Water Heater
The open and closed feed water heaters can be differentiated as follows:
Open feed water heater Closed feed water heater
Open and simple More complex in design
Good heat transfer characteristics Less effective heat transfer
Direct mixing extraction steam and
feed water temperature in a pressure
vessel
In-direct mixing feed water and steam in a shell and
tube type heat exchanger.
Pump is required to transfer the
water into next stage in the cycle.
Closed feed water pumps don’t require pump and
can operate with the pressure difference between the
various heaters in the cycle.
Requires more area Requires less area
Less expansive More expensive
All modern day power plants are employing the combination of open and closed feed water
heaters to maximize the thermal efficiency of the cycle.
𝑾𝑻 = 𝑾𝑻𝟏 + 𝑾𝑻𝟐 = (𝒉𝟏 − 𝒉𝟐) + (𝒉𝟑 − 𝒉𝟒)
𝑷𝒖𝒎𝒑 𝑾𝒐𝒓𝒌 𝑾𝑷 = (𝒉𝟔 − 𝒉𝟓)
𝑯𝒆𝒂𝒕 𝒂𝒅𝒅𝒆𝒅 𝑸𝑺 = (𝑸𝑺𝟏 + 𝑸𝑺𝟐)
𝑸𝑺𝟏 = (𝒉𝟏 − 𝒉𝟔) ; 𝑸𝑺𝟐 = (𝒉𝟑 − 𝒉𝟐)
𝑻𝒉𝒆𝒓𝒎𝒂𝒍 𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒄𝒚 𝜼(𝑹𝒂𝒏𝒌𝒊𝒏𝒆)
=
𝑸𝑺
=
𝑸𝑺
(𝑸𝑺𝟏 + 𝑸𝑺𝟐)
=
(𝒉𝟏 − 𝒉𝟐) + (𝒉𝟑 − 𝒉𝟒) − (𝒉𝟔 − 𝒉𝟓)
(𝒉𝟏 − 𝒉𝟔) + (𝒉𝟑 − 𝒉𝟐)

To take advantage of the increased efficiencies at higher boiler pressure without facing the
excessive moisture at the final stages of the turbine, reheating is used. In the ideal reheating
cycle, the expansion process takes place in two stages, that is at high-pressure and low-pressure
turbines. The total heat input and total turbine work output for a reheat cycle become:
𝑸𝑺𝟏 = (𝒉𝟏 − 𝒉𝟔) ; 𝑸𝑺𝟐 = (𝒉𝟑 − 𝒉𝟐) 𝑯𝒆𝒂𝒕 𝒂𝒅𝒅𝒆𝒅 𝑸𝑺 = (𝑸𝑺𝟏 + 𝑸𝑺𝟐) and total turbine work
output for a reheat cycle become:𝑾𝑻 = 𝑾𝑻𝟏 + 𝑾𝑻𝟐 = (𝒉𝟏 − 𝒉𝟐) + (𝒉𝟑 − 𝒉𝟒)
The incorporation of the single reheat in a modern power plant improves the cycle efficiency
by 4 % to 5 % by increasing the average temperature at which heat is transferred to the steam.
Characteristics of an Ideal working fluid in vapour power cycles
These factors suggest the properties of working fluids for a trouble free vapour power cycle.
(i) The fluid should have high critical temperature so that the saturation pressure at the
maximum permissible temperature is relatively low. It should have a large enthalpy of
evaporation at that pressure.
(ii) The saturation pressure at the temperature of heat rejection should be above atmospheric
pressure so as to avoid the necessity of maintaining vacuum in the condenser.
(iii)The specific heat of the liquid should be small so that little heat transfer is required to raise
the liquid to the boiling point.
(iv)The saturated vapour line of the T-s diagram should be steep, very close
to turbine expansion process so that excessive moisture does not appear during expansion.
(v) The freezing point of the liquid should be below room temperature, so that it does not get
solidified while flowing through the pipelines.
(vi)The fluid should be chemically stable and should not contaminate the material of
construction at any temperature.
(vii) The fluid should be nontoxic, noncorrosive, not excessively viscous, and low in cost.

List of Formulas
Carnot vapour power cycle
𝑾𝑻 = (𝒉𝟏 − 𝒉𝟐)
𝑾𝑷 = (𝒉𝟒 − 𝒉𝟑)
𝑸𝑺 = (𝒉𝟏 − 𝒉𝟒)
𝑸𝑹 = (𝒉𝟐 − 𝒉𝟑)
𝜼(𝑪𝒂𝒓𝒏𝒐𝒕 𝑽𝒂𝒑𝒐𝒖𝒓) =
𝑾𝑻−𝑾𝑷
𝑸𝑺
𝜼(𝑪𝒂𝒓𝒏𝒐𝒕 𝑽𝒂𝒑𝒐𝒖𝒓) =
(𝒉𝟏−𝒉𝟐) −(𝒉𝟒−𝒉𝟑)
(𝒉𝟏−𝒉𝟒)
𝜼(𝑪𝒂𝒓𝒏𝒐𝒕 𝑽𝒂𝒑𝒐𝒖𝒓) = 𝟏 −
(𝒉𝟐−𝒉𝟑)
(𝒉𝟏−𝒉𝟒)
= 𝟏 −
𝑸𝑹
𝑸𝑺
𝜼(𝑪𝒂𝒓𝒏𝒐𝒕 𝑽𝒂𝒑𝒐𝒖𝒓) = 𝟏 −
𝑻𝟑
𝑻𝟏
= 𝟏 −
𝑻(𝒎𝒂𝒙𝒊𝒎𝒖𝒎)
𝑻(𝒎𝒊𝒏𝒊𝒎𝒖𝒎)
Simple Rankine cycle
𝑾𝑻 = 𝒉𝟏 − 𝒉𝟐
𝑾𝑷 = 𝒉𝟒 − 𝒉𝟑
𝒐𝒓 𝑾𝑷 = (𝑷𝟒 − 𝑷𝟑)
𝐐𝐬 = 𝒉𝟏 − 𝒉𝟒
𝐐𝐫 = 𝒉𝟐 − 𝒉𝟑
𝜼(𝑹𝒂𝒏𝒌𝒊𝒏𝒆)
=
𝑾𝑻 − 𝑾𝑪
𝑸𝑺
(𝒉𝟏 − 𝒉𝟐) − (𝒉𝟒 − 𝒉𝟑)
(𝒉𝟏 − 𝒉𝟒)
(𝒉𝟏 − 𝒉𝟐) − 𝑾𝑷
(𝒉𝟏 − 𝒉𝟑) − 𝑾𝑷
(𝒉𝟏−𝒉𝟐)
(𝒉𝟏−𝒉𝟑)
𝒊𝒇 𝑾𝑷 = 𝟎
Actual Rankine Cycle
𝜼(𝑻𝒖𝒓𝒃𝒊𝒏𝒆) =
𝑨𝒄𝒕𝒖𝒂𝒍 𝑻𝒖𝒓𝒃𝒊𝒏𝒆 𝑾𝒐𝒓𝒌
𝑰𝒅𝒆𝒂𝒍 𝑻𝒖𝒓𝒃𝒊𝒏𝒆 𝑾𝒐𝒓𝒌
=
(𝒉𝟏 − 𝒉𝟐′)
(𝒉𝟏 − 𝒉𝟐)
𝜼(𝑷𝒖𝒎𝒑) =
𝑰𝒅𝒆𝒂𝒍 𝑷𝒖𝒎𝒑 𝑾𝒐𝒓𝒌
𝑨𝒄𝒕𝒖𝒂𝒍 𝑷𝒖𝒎𝒑 𝒘𝒐𝒓𝒌 𝑾𝒐𝒓𝒌
=
(𝒉𝟒 − 𝒉𝟑)
(𝒉𝟒′ − 𝒉𝟑)
Ideal Regeneration Cycle
𝜼(𝑹𝒂𝒏𝒌𝒊𝒏𝒆)
=
𝑸𝑺
(𝒉𝟐−𝒉𝟑)−(𝒉𝟓−𝒉𝟒)
(𝒉𝟐−𝒉𝟏)+(𝒉𝟏−𝒉𝟓)
(𝒉𝟐−𝒉𝟓)−(𝒉𝟑−𝒉𝟒)
(𝒉𝟐−𝒉𝟓)
𝜼(𝑹𝒂𝒏𝒌𝒊𝒏𝒆) = 𝟏 −
𝑸𝑹
𝑸𝑺𝟏 + 𝑸𝑺𝟐
𝜼(𝑹𝒂𝒏𝒌𝒊𝒏𝒆) = 𝟏 −
(𝒉𝟑 − 𝒉𝟒)
(𝒉𝟐 − 𝒉𝟓)
(i) Rankine Regeneration Cycle with Open Feed Heater
(𝒚) =
(𝒉𝟔 − 𝒉𝟓)
(𝒉𝟐 − 𝒉𝟓)
Process 1-2 and 1-3: 𝑻𝒖𝒓𝒃𝒊𝒏𝒆 𝑾𝒐𝒓𝒌
𝑾𝑻 = (𝒉𝟏 − 𝒉𝟐) + (𝟏 − 𝒚)(𝒉𝟐 − 𝒉𝟑)
Process 4-5 and 6-7: 𝑷𝒖𝒎𝒑 𝑾𝒐𝒓𝒌
𝑾𝑷 = 𝑾𝑷𝟏 + 𝑾𝑷𝟐
Process 7-1: 𝑯𝒆𝒂𝒕 𝒂𝒅𝒅𝒆𝒅
𝑸𝑺 = (𝒉𝟏 − 𝒉𝟕)
𝑾𝑷 = (𝟏 − 𝒚)(𝒉𝟓 − 𝒉𝟒) + (𝒉𝟕 − 𝒉𝟔)
𝑯𝒆𝒂𝒕 𝑺𝒖𝒑𝒑𝒍𝒊𝒆𝒅 −𝑯𝒆𝒂𝒕 𝑹𝒆𝒋𝒆𝒄𝒕𝒆𝒅
𝑯𝒆𝒂𝒕 𝑺𝒖𝒑𝒑𝒍𝒊𝒆𝒅
=
𝑸𝑺−𝑸𝑹
𝑸𝑺
𝑸𝑺−𝑸𝑹
𝑸𝑺
=
(𝒉𝟏−𝒉𝟕) −(𝟏−𝒚)(𝒉𝟑−𝒉𝟒)
(𝒉𝟏−𝒉𝟕)

(ii) a) Rankine Regeneration Cycle with Forward Closed Feed Heater
Forward
𝑾𝑻 = (𝒉𝟔 − 𝒉𝟕) + (𝟏 − 𝒚)(𝒉𝟕 − 𝒉𝟖)
𝑷𝒖𝒎𝒑 𝑾𝒐𝒓𝒌 𝑾𝑷 = 𝑾𝑷𝟏 + 𝑾𝑷𝟐
𝑾𝑷 = (𝟏 − 𝒚)(𝒉𝟐 − 𝒉𝟏) + (𝒚)(𝒉𝟒 − 𝒉𝟑)
𝑯𝒆𝒂𝒕 𝒂𝒅𝒅𝒆𝒅
𝑸𝑺 = (𝒉𝟔 − 𝒉𝟓)
𝑯𝒆𝒂𝒕 𝒓𝒆𝒋𝒆𝒄𝒕𝒆𝒅
𝑸𝑹 = (𝟏 − 𝒚)(𝒉𝟖 − 𝒉𝟏)
𝑾𝑻−𝑾𝑷
𝑸𝑺
(𝒉𝟔−𝒉𝟕)+(𝟏−𝒚)(𝒉𝟕−𝒉𝟖) −(𝟏−𝒚)(𝒉𝟐−𝒉𝟏)+(𝒚)(𝒉𝟒−𝒉𝟑)
(𝒉𝟔−𝒉𝟓)
𝑸𝑺−𝑸𝑹
𝑸𝑺
(𝒉𝟔−𝒉𝟕)+(𝟏−𝒚)(𝒉𝟕−𝒉𝟖) −(𝟏−𝒚)(𝒉𝟐−𝒉𝟏)+(𝒚)(𝒉𝟒−𝒉𝟑)
(𝒉𝟔−𝒉𝟓)
(ii) b) Rankine Regeneration Cycle with Backward Closed Feed Heater
Backward
Heat Input:
Qs = h4 - h3;
Heat Output:
Qr = (1 - y) (h1 - h6) + y (h8 - h1);
Work Output:
Wt = (h4 - h5) + (1 - y) (h5 - h6);
Work input:
Wp = (h2 - h1)
𝑾𝑻−𝑾𝑷
𝑸𝑺
(𝒉𝟒 − 𝒉𝟓) + (𝟏 − 𝒚) (𝒉𝟓 − 𝒉𝟔) −(𝒉𝟐 − 𝒉𝟏)
(𝒉𝟒−𝒉𝟑)
𝑸𝑺−𝑸𝑹
𝑸𝑺
𝑸𝑺−𝑸𝑹
𝑸𝑺
=
(𝒉𝟔−𝒉𝟓) −[(𝟏−𝒚)(𝒉𝟖−𝒉𝟏)]
(𝒉𝟔−𝒉𝟓)
𝑬𝒏𝒆𝒓𝒈𝒚 𝒆𝒏𝒕𝒆𝒓𝒊𝒏𝒈 𝒓𝒆𝒈𝒊𝒏𝒊𝒓𝒂𝒕𝒐𝒓 = 𝑬𝒏𝒆𝒓𝒈𝒚 𝒍𝒆𝒂𝒗𝒊𝒏𝒈 𝒓𝒆𝒈𝒊𝒏𝒆𝒓𝒂𝒕𝒐𝒓
(𝒚)𝐡𝟓 + 𝐡𝟐 = (𝐲)𝐡𝟕 + 𝐡𝟑
(𝒚)( 𝐡𝟓 – 𝐡𝟕) = (𝐡𝟑 – 𝐡𝟐)
(𝒚) =
(𝐡𝟑 – 𝐡𝟐)
( 𝐡𝟓 – 𝐡𝟕)
𝑾𝑻 = 𝑾𝑻𝟏 + 𝑾𝑻𝟐
𝑾𝑻 = (𝒉𝟏 − 𝒉𝟐) + (𝒉𝟑 − 𝒉𝟒)
𝑸𝑺𝟏 = (𝒉𝟏 − 𝒉𝟔) ; 𝑸𝑺𝟐 = (𝒉𝟑 − 𝒉𝟐)
𝑾𝑻−𝑾𝑷
𝑸𝑺
=
𝑸𝑺−𝑸𝑹
𝑸𝑺
𝑾𝑻−𝑾𝑷
(𝑸𝑺𝟏+𝑸𝑺𝟐)
=
(𝒉𝟏−𝒉𝟐)+(𝒉𝟑−𝒉𝟒)−(𝒉𝟔−𝒉𝟓)
(𝒉𝟏−𝒉𝟔)+(𝒉𝟑−𝒉𝟐)

Previous Year Solved Question Papers
Example Problem 01
5a In a steam power plant operating on ideal Rankine cycle, steam enters the turbine at 20 bar with an
enthalpy of 3248 kJ/kg and an entropy of 7.127 kJ/kg K. The condenser pressure is 0.1 bar. Find the
cycle efficiency and specific steam consumption in kg/kWh. Do not neglect pump work. You may make
use of the extract of steam table given below.
From
Steam
Table
Example Problem 2

Formulation of Rankine Cycle
Basic Nomenclature
Ideal Rankine Cycle
Actual Rankine Cycle

Formulation
The cycle has the following reversible processes:
(i) 1–2 or 1′–2′: Adiabatic reversible expansion through the turbine
(ii) 2–3 or 2′–3: A two-phase mixture constant temperature and pressure process. Heat is rejected in
the condenser at constant pressure.
(iii) 3–4: Adiabatic reversible compression. The pump increases saturated liquid at condenser pressure
at 3, to subcooled liquid at the steam generator pressure, 4. Line 3–4 is a vertical line as the liquid
is incompressible and pump work is adiabatic reversible.
(iv) 4–1 or 4–1′: Heat is added at constant pressure in the steam generator.

Modal Question Bank 01
5a
Sketch the flow diagram and corresponding T – s diagram of a reheat vapor power cycle and derive an
expression for the reheat cycle efficiency.
Reheat
Rankine
cycle.
𝑾𝑻 = 𝑾𝑻𝟏 + 𝑾𝑻𝟐
𝑾𝑻 = (𝒉𝟏 − 𝒉𝟐) + (𝒉𝟑 − 𝒉𝟒)
𝑸𝑺𝟏 = (𝒉𝟏 − 𝒉𝟔) ; 𝑸𝑺𝟐
= (𝒉𝟑 − 𝒉𝟐)
5b
A cyclic steam power plant is to be designed at turbine inlet temperature of 360°C and an exhaust pressure of
0.08 bar. After isentropic expansion of steam in the turbine, the moisture content at the turbine exhaust is not
exceeding 15%. Determine: (i) The greatest allowable steam pressure at the turbine inlet (ii) Efficiency of the
Rankine cycle and (iii) Specific steam consumption in kg/kW-hr
Interpolation Method: Using Superheated Steam Table where (s) is near
7.0833
20 bar x y 15bar x y
1 350 6.956 1 350 7.102
x= 360’C Y= 𝟔. 𝟗𝟗𝟎𝟐 X=360’C Y= 𝟕. 𝟏𝟑𝟓𝟒
2 400 7.127 2 400 7.269
𝑰𝒏𝒕𝒆𝒓𝒑𝒐𝒍𝒂𝒕𝒊𝒐𝒏 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 (𝒚 − 𝒚𝟏) =
(𝒚𝟐 − 𝒚𝟏)
(𝒙𝟐 − 𝒙𝟏)
× (𝒙 − 𝒙𝟏)
x y
1 7.1354 15
X=7.0833 y= 𝟏𝟔. 𝟕𝟗𝟒𝟎 ≈ 𝟏𝟔. 𝟖 𝒃𝒂𝒓
2 6.9902 20
For 20 bar
(𝑦 − 6.956) =
(7.127 − 6.956)
(400 − 350)
× (𝑥 − 350)
(𝑦 − 6.956) = 0.00342 × (360 − 350)
(𝑦) = 0.0342 + 6.956 = 𝟔. 𝟗𝟗𝟎𝟐
For 15 bar
(𝑦 − 7.102) =
(7.269 − 7.102)
(400 − 350)
× (𝑥 − 350)
(𝑦 − 7.102) = 0.00334 × (360 − 350)
(𝑦) = 0.0334 + 7.102 = 𝟕. 𝟏𝟑𝟓𝟒
For x= 7.0833
(𝑦 − 15) =
(20 − 15)
(6.9902 − 7.1354)
× (7.0833
− 7.1354)
(𝑦 − 15) = −34.4352 × (−0.0521)
(𝑦) = 1.79407 + 15 = 𝟏𝟔. 𝟕𝟗𝟒𝟎

Interpolation Method: Using Superheated Steam Table where
pressure (p1=16.8)
x y
1 15 3147.5
x= 16.8 Y=3,143.72
2 20 3137.0
𝑰𝒏𝒕𝒆𝒓𝒑𝒐𝒍𝒂𝒕𝒊𝒐𝒏 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 (𝒚 − 𝒚𝟏) =
(𝒚𝟐−𝒚𝟏)
(𝒙𝟐−𝒙𝟏)
× (𝒙 − 𝒙𝟏)
For x= 16.8 bar
(𝑦 − 3147.5) =
(3137.0 − 3147.5)
(20 − 15)
× (16.8 − 15)
(𝑦 − 3147.5) − 2.1 × (1.8)
(𝑦) = 3147.5 − 3.78 = 𝟑, 𝟏𝟒𝟑. 𝟕𝟐
Therefore h1=3143.72 kJ/kgK
6a With the help of a schematic and T-s diagram, explain the working of an ideal regenerative
vapor cycle and derive an expression for the overall efficiency.
𝑸𝑺
6b Steam enters the first stage of a reheat Rankine cycle at 8 MPa, 500°C and expands to 700
kPa. It is then reheated to 450°C before entering a second stage turbine, where it expands to
0.08 bar. The net power output is 100 MW. Determine: (i) Thermal efficiency of the cycle (ii)
Steam flow rate (iii) Quality of steam at the end of expansion, and (iv) Total heat rejected in
the condenser in MW.
Activity: To be completed by the students
5a With the help of a schematic and T –s diagram, explain the working of regenerative vapor power
cycle with one feed water heater.
(𝒚) =
(𝒉𝟔 − 𝒉𝟓)
(𝒉𝟐 − 𝒉𝟓)
Process 1-2 and 1-3: 𝑻𝒖𝒓𝒃𝒊𝒏𝒆 𝑾𝒐𝒓𝒌
𝑾𝑻 = (𝒉𝟏 − 𝒉𝟐) + (𝟏 − 𝒚)(𝒉𝟐 − 𝒉𝟑)
Process 4-5 and 6-7: 𝑷𝒖𝒎𝒑 𝑾𝒐𝒓𝒌
𝑾𝑷 = 𝑾𝑷𝟏 + 𝑾𝑷𝟐
Process 7-1: 𝑯𝒆𝒂𝒕 𝒂𝒅𝒅𝒆𝒅
𝑸𝑺 = (𝒉𝟏 − 𝒉𝟕)
𝑾𝑷 = (𝟏 − 𝒚)(𝒉𝟓 − 𝒉𝟒) + (𝒉𝟕 − 𝒉𝟔)
5b A steam power plant operates on theoretical reheat Rankine cycle. Steam enters the high
pressure turbine at 15 Mpa and 600°C and is condensed in the condenser at a pressure of 10
kPa. If the moisture content of the steam at the exit of the low pressure turbine is not to exceed
10.4 percent, determine (i) the pressure at which the steam should be reheated and (ii) the
thermal efficiency of the cycle. Assume the steam is reheated to the inlet temperature of the high
pressure turbine.
6a With the help of T-s diagrams, explain the effects of varying boiler pressure and condenser pressure on
the performance of a simple Rankine cycle.
varying boiler pressure varying condenser pressure varying Super heat
6b A steam power plant operates on simple ideal Rankine cycle. Steam enters the turbine at 3 MPa and
350°C and is condensed in the condenser at a pressure of 75 kPa. Determine the thermal efficiency of this
cycle.

APPLIED THERMODYNAMICS 18ME42 Module 03: Vapour Power Cycles

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