An Autonomous Institution,

Affiliated to VTU, Belagavi, Approved by AICTE, New Delhi,

Recognised by UGC with 2(f) & 12 (B), Accredited by NBA & NAAC

APPLIED THERMODYNAMICS (18ME42)

MODULE 4 -VAPOR POWER CYCLE
Prepared by: Dr. Vivekanand Huddar

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 1
CONTENTS RBT Level : L2, L3, L4
Time : 12 hrs
• Rankine Cycle ideal and actual.
• Mean temperature of heat addition.
• Reheat Cycle,
• Ideal Regenerative Cycle, and Regenerative Cycle with feed
water heaters.
• Binary Vapour Cycle.
• Problems

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 2
IMPORTANCE OF VPCs - BEHAVIOR OF TWO-PHASE SYSTEMS

Phase is “a quantity of matter that is homogeneous throughout”.

Example: Water, pure substance
It’s important to know the relations between phases and the
relations that describe the change of phase (from solid to liquid, or
from liquid to vapor) of a pure substance, including the work done
and the heat transfer.
Consider a system consisting of a liquid and its vapor in
equilibrium, which are enclosed in a container under a moveable
piston, as shown in Figure.
The system is maintained at constant temperature through contact
with a heat reservoir at temperature T, so there can be heat
transfer to or from the system.
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 3
 Water vapor

Water vapor

Liquid water Liquid water

Figure : Two-phase system in contact with constant temperature heat reservoir & p-T diagram

For a pure substance, as shown at the right, there is a one-to-one

correspondence between the temperature at which vaporization
occurs and the pressure.
These values are called the saturation pressure and saturation
temperature

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 4
The behavior shown is found for all the
isotherms that go through the vapor dome.
At a high enough temperature, specifically
at a temperature corresponding to the
pressure at the peak of the vapor dome,
there is no transition from liquid to vapor
and the fluid goes continuously from a
liquid-like behavior to a gas-type behavior.

This behavior is unfamiliar, mainly because

the temperatures and pressures are not
ones that we typically experience; for
water the critical temperature is 374oC and
the associated critical pressure is 220
atmospheres. Figure: P-v diagram for two-phase system showing isotherms

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 5
 STEAM POWER PLANT
• Vapour power cycles are used
in steam power plants.
• In a power cycle heat energy
(released by the burning of
fuel) is converted into work
(shaft work), in which a
working fluid repeatedly
performs a succession of
processes.
• In a vapour power cycle, the
working fluid is water, which
undergoes a change of phase.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 6
Figure shows a simple
steam power plant
working on the vapour
power cycle.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 7
• Heat is transferred to the water in
the boiler (QH) from an external
source. (Furnace, where fuel is
continuously burnt) to raise steam.
• The high pressure high temperature
steam leaving the boiler expands in
the turbine to produce shaft work
(WT).
• The steam leaving the turbine condenses into water in the
condenser (where cooling water circulates), rejecting heat (QL),
and
• Then the water is pumped back (WP) to the boiler.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 8
Since the fluid is undergoing a cyclic process, the net energy
transferred as heat during the cycle must equal the net energy
transfer as work from the fluid.

By FLTD

෍ 𝑄𝑛𝑒𝑡 = ෍ 𝑊𝑛𝑒𝑡
𝑐𝑦𝑐𝑙𝑒 𝑐𝑦𝑐𝑙𝑒

Or
QH – QL = WT - WP

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 9
Where
QH = heat transferred to the working fluid (kJ/kg)
QL = heat rejected from the working fluid (kJ/kg)
WT = work transferred from the working fluid (kJ/kg)
WP = work transferred into the working fluid (kJ/kg)

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 10
 IDEALIZED STEAM POWER CYCLES: CARNOT CYCLE

We know that the efficiency of a Carnot Cycle

Carnot engine is maximum and it
does not depend on the working
Heat addition
fluid. Q in
Compression
It is, therefore, natural to examine W in
Work Done
of a steam power plant can be WT
operated on the Carnot cycle.

Figure shows the Carnot cycle on the Condensation Q out

T-S diagram.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 11
• Heat addition at constant pressure P2,
can be achieved isothermally in the
process 1-2 in a boiler. Carnot Cycle
• The decrease in pressure from P2 to P3
in the process 2-3 can also be attained Heat addition
through the performance of work in a Compression Q in
steam turbine. W in
Work Done
• But in order to bring back the saturated WT

liquid water to the boiler at the state 1,

the condensation process 3-4 in the Condensation Q out
condenser must be terminated at the
state 4, where the working fluid is a
mixture of liquid water and vapour.
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 12
• But it is practically impossible
to attain a condensation of this Carnot Cycle
kind.
• Difficulty is also experienced in
Heat addition
compressing isentropically the Q in
Compression
binary mixture from state 4 to W in
Work Done
the initial state 1, where the WT
working fluid is entirely in the
liquid state.
Condensation Q out
• Due to these inherent practical
difficulties, Carnot cycle remains
an ideal one.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 13
IDEAL RANKINE CYCLE:

• The simplest way of overcoming the inherent practical

difficulties of the Carnot cycle without deviating too much
from it is to keep the processes 1-2 and 2-3 of the latter
unchanged and to continue the process 3-4 in the condenser
until all the vapour has been converted into liquid water.
• Water is then pumped into the boiler up to the pressure
corresponding to the state 1 and the cycle is completed. Such a
cycle is known as the Rankine cycle.
• This theoretical cycle is free of all the practical limitations of
the Carnot cycle.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 14
Figure (a) shows the schematic diagram for a simple steam power
cycle which works on the principle of a Rankine cycle.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 15
 Figure (b) represents the T-S diagram of the cycle.

The Rankine cycle comprises the following processes.

Process 1-2: Constant pressure heat transfer process in the boiler
Process 2-3: Reversible adiabatic expansion process in the steam turbine
Process 3-4: Constant pressure heat transfer process in the condenser and
Process 4-1: Reversible adiabatic compression process in the pump.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 16
• For any given pressure, the steam approaching
the turbine may be dry saturated (state 2), wet
(state 21) or superheated (state 211), but the
fluid approaching the pump is, in each case,
saturated liquid (state 4).
• Steam expands reversibly and adiabatically in
the turbine from state 2 to state 3 (or 21 to 31 or
211 to 311), the steam leaving the turbine
condenses to water in the condenser reversibly
at constant pressure from state 3 (or 31, or 311)
to state 4.
• Also, the water is heated in the boiler to form
steam reversibly at constant pressure from
state 1 to state 2 (or 21 or 211)
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 17
Applying SFEE to each of the processes on the basis of unit mass
of fluid and neglecting changes in KE & PE, the work and heat
quantities can be evaluated.

For 1kg of fluid, the SFEE for the boiler as the CV, gives,
h1 + QH = h2 i.e., QH = h2 – h1 --- (1)
SFEE to turbine, h2 = WT + h3 i.e., WT = h2 – h3 ---(2)
SFEE to condenser, h3 = QL + h4 i.e., QL = h3 – h4 --- (3)
SFEE to pump, h4 + WP = h1 i.e., WP = h1 – h4 --- (4)
QH
𝑊𝑛𝑒𝑡 𝑊𝑇 − 𝑊𝑃 Boiler
∴ 𝜂𝑐𝑦𝑐𝑙𝑒 = =
𝑄𝐻 𝑄𝐻 h1
h2
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 18
 𝑊𝑇 −𝑊𝑃 ℎ2 −ℎ3 − ℎ1 −ℎ4 ℎ2 −ℎ1 − ℎ3 −ℎ4
𝑖. 𝑒. , 𝜂𝑅 = = =
𝑄𝐻 ℎ2 −ℎ1 ℎ2 −ℎ1

The pump handles liquid water which is incompressible i.e., its density
or specific volume undergoes little change with an increase in pressure.
For reversible adiabatic compression, we have Tds = dh – vdp; since ds = 0
We have, dh = vdp
Since change in specific volume is negligible,
Δh = v ΔP or (h1 – h4) = v4 (P2 – P3)

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 19
Usually the pump work is quite small compared to the turbine
work and is sometimes neglected. In that case, h1 = h4
ℎ2 − ℎ3 ℎ2 − ℎ3
∴ 𝜂𝑐𝑦𝑐𝑙𝑒 = =
ℎ2 − ℎ1 ℎ2 − ℎ4
The efficiency of the Rankine cycle
is presented graphically in the T-S
diagram
QH α area 1-5-6-4-1,
QL α area 3-2-5-6-3
Wnet = (QH – QL)
= area 1-2-3-4-1 enclosed by the cycle.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 20
EXTRA INFORMATION
The capacity of the steam plant is expressed in terms of steam rate
defined as the rate of steam flow (kg/h) required to produce unit
shaft output (1kW)

1 𝑘𝑔
∴ 𝑠𝑡𝑒𝑎𝑚 𝑟𝑎𝑡𝑒 =
𝑊𝑇 −𝑊𝑃 𝑘𝐽

1𝑘𝑔/𝑠
𝑆𝑝. 𝑠𝑡𝑒𝑎𝑚 𝑐𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛(𝑠𝑠𝑐) =
1𝑘𝑊
1 𝑘𝑔 3600 𝑘𝑔
∴ 𝑠𝑡𝑒𝑎𝑚 𝑟𝑎𝑡𝑒 = =
𝑊𝑇 −𝑊𝑃 𝑘𝑊𝑠 𝑊𝑇 −𝑊𝑃 𝑘𝑊ℎ

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 21
The cycle efficiency also expressed alternatively as heat rate which
is the rate of heat input (QH) required to produce unit work output
(1kW)
3600𝑄𝐻 3600 𝑘𝑔
∴ ℎ𝑒𝑎𝑡 𝑟𝑎𝑡𝑒 = =
𝑊𝑇 − 𝑊𝑃 𝜂𝑐𝑦𝑐𝑙𝑒 𝑘𝑊ℎ
‫𝑊ׯ‬ ℎ2 − ℎ3 − ℎ1 − ℎ4
𝐿𝑎𝑠𝑡𝑒𝑙𝑦, 𝑊𝑜𝑟𝑘 𝑅𝑎𝑡𝑖𝑜𝑟𝑤 = =
𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑤𝑜𝑟𝑘 ℎ2 − ℎ3

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 22
ACTUAL VAPOUR POWER CYCLES- DEVIATION FROM IDEAL CYCLE

The actual Vapour power cycle differs from the ideal Rankine
cycle, as shown in figure, as a result of irreversibilities in
various components mainly because of fluid friction and heat
loss to the surroundings.
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 23
Pressure drop as irreversibility

• Fluid friction causes pressure drops in the boiler, the

condenser, and the piping between various components.
• As a result, steam leaves the boiler at a lower pressure.
• Also the pressure at the turbine inlet is lower than that at
the boiler exit due to pressure drop in the connecting pipes.
• The pressure drop in the condenser is usually very small.
• To compensate these pressure drops, the water must be
pumped to sufficiently higher pressure which requires the
larger pump and larger work input to the pump.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 24
Heat Transfer as irreversibility

• The other major source of irreversibility is the heat loss from the
steam to the surroundings as the steam flows through various
components.
• To maintain the same level of net work output, more heat needs to
be transferred to the steam in the boiler to compensate for these
undesired heat losses.
• As a result, cycle efficiency decreases.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 25
• As a result of irreversibilities, a pump requires a greater work
input, and a turbine produces a smaller work output.
• Under the ideal conditions, the flow through these devices are
isentropic.
• The deviation of actual pumps and turbines
from the isentropic ones can be accounted
for by utilizing isentropic efficiencies,
defined as

𝑊𝑠 𝐼𝑑𝑒𝑎𝑙 𝑤𝑜𝑟𝑘 𝑖𝑛𝑝𝑢𝑡 ℎ1𝑠 − ℎ4

∴ 𝜂𝑃 = = =
𝑊𝑎 𝐴𝑐𝑡𝑢𝑎𝑙 𝑤𝑜𝑟𝑘 𝑖𝑛𝑝𝑢𝑡 ℎ1𝑎 − ℎ4
𝑊𝑎 𝐴𝑐𝑡𝑢𝑎𝑙 𝑒𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛 ℎ2 − ℎ3𝑎
& 𝜂𝑡 = = =
𝑊𝑠 𝐼𝑑𝑒𝑎𝑙 𝐸𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛 ℎ2 − ℎ3𝑠
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 26
Example: Dry saturated steam at 17.5 bar enters the turbine of a
steam power plant and expands to the condenser pressure of 0.75
bar. Determine the Carnot and Rankine cycle efficiencies. Also find
the work ratio of the Rankine cycle.
Solution: P1 = 17.5 bar P2 = 0.75 bar
ηCarnot = ? ηRankine = ?
•Carnot cycle: At pressure 17.5 bar from
steam tables, (THERMODYNAMICS DATA
HAND BOOK by B.T.Nijaguna &
B.S.Samaga)
P tS hf hfg hg Sf Sfg Sg
17 204.3 871.8 1921.6 2793.4 2.3712 4.0246 6.3958
18 207.11 884.5 1910.3 2794.8 2.3976 3.9776 6.3751
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 27
For P = 17.5 bar, using linear interpolation
(Y2-Y1)(X3-X1) = (X2-X1)(Y3-Y1)

P tS
17 204.3
17.5 Y2
18 207.11
Similarly,
hf = 878.15 kJ/kg hfg = 1915.95 kJ/kg hg = 2794.1 kJ/kg
Sf = 2.3844 kJ/kg K Sfg = 4.0011 kJ/kgK Sg = 6.3855 kJ/kg K

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 28
Also at pressure 0.75 bar from steam tables
P tS hf h fg hg Sf S fg Sg
0.7 89.96 376.8 2283.3 2660.1 1.1921 6.2883 7.4804
0.8 93.51 391.7 2274.0 2665.8 1.233 6.2022 7.4352

For 0.75 bar, using linear interpolation,

tS = 91.740C =364.74K
hf = 384.25 hfg = 2278.65 hg = 2662.95
Sf = 1.2126 Sfg = 6.2453 Sg = 7.4578

𝑇2 364.74
𝐶𝑎𝑟𝑛𝑜𝑡 𝐶𝑦𝑐𝑙𝑒 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝜂𝑐 = 1 − =1− =0.238
𝑇1 478.71

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 29
 1 1
𝑆𝑡𝑒𝑎𝑚 𝑟𝑎𝑡𝑒 𝑜𝑟 𝑆𝑆𝐶 = =
‫ 𝑇𝑊 𝑊 ׯ‬− 𝑊𝑃
Since the expansion work is isentropic, S2 = S3
But S2 = Sg = 6.3855 and S3 = Sf3 + x3 Sfg3
i.e., 6.3855 = 1.2126 + x3 (6.2453)
∴x3 = 0.828

∴ Enthalpy at state 3, h3 = hf3 + x3hfg3

= 384.25 + 0.82(2278.65) = 2271.63 kJ/kg
∴ Turbine work or expansion work or positive work = h2 – h3
= 2794.1 – 2271.63 = 522.47 kJ/kg
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 30
Again since the compression process is isentropic i.e.,
S4 = S1 = Sf1 = 2.3844
Hence 2.3844 = Sf4 + x4 Sfg4 = 1.2126 + x4 (6.2453) ∴ x4 = 0.188

∴ Enthalpy at state 4 is h4 = hf4 + x4 hfg4

= 384.25 + 0.188 (2278.65)= 811.79 kJ/kg

∴ Compression work, WP = h1 – h4 = 878.15 – 811.79 = 66.36 kJ/kg

∴ Heat addition, QH = h2 – h1 = 2794.1 -878.15 = 1915.95 kJ/kg

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 31
 1
∴ 𝑆𝑡𝑒𝑎𝑚 𝑟𝑎𝑡𝑒 𝑜𝑟 𝑆𝑆𝐶 =
𝑊𝑇 − 𝑊𝑃
1
= = 2.192 x 10−3 kg / kJ
522.47 − 66.36

‫𝑊ׯ‬
𝑊𝑜𝑟𝑘 𝑅𝑎𝑡𝑖𝑜 𝑟𝑤 =
𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑤𝑜𝑟𝑘

𝑊𝑇 − 𝑊𝑃 456.11
= = = 0.873
𝑊𝑇 522.47

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 32
 RANKINE CYCLE:

𝑊𝑇 −𝑊𝑃 ℎ2 −ℎ3 − ℎ1 −ℎ4 ℎ2 −ℎ1 − ℎ3 −ℎ4

𝑖. 𝑒. , 𝜂𝑅 = = =
𝑄𝐻 ℎ2 −ℎ1 ℎ2 −ℎ1

Since the change in volume of the saturated liquid water during

compression from state 4 to state 1 is very small, v4 may be taken
as constant.
In a steady flow process, work W = -v∫dp
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 33
∴ WP = h1S – h4 = vfP2 (P1 – P2)
= 0.001037 (17.5 – 0.75) x 105 x (1/1000)
= 1.737 kJ/kg
∴ h1S = 1.737 + 384.25 = 385.99 kJ/kg
Hence, turbine work = WT = h2 – h3 = 522.47kJ/kg
Heat supplied = QH = h2 – h1S = 2.794.1 – 385.99 = 2408.11 kJ/kg

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 34
 𝑊𝑇 −𝑊𝑃 522.47−1.737
∴ 𝜂𝑅 = = = 0.2162
𝑄𝐻 2408.11
1
∴ 𝑆𝑡𝑒𝑎𝑚 𝑟𝑎𝑡𝑒 𝑜𝑟 𝑆𝑆𝐶 =
𝑊𝑇 − 𝑊𝑃
1
= = 19204 x 10−3 kg / kJ
522.47 − 1.737
‫𝑊ׯ‬
𝑊𝑜𝑟𝑘 𝑅𝑎𝑡𝑖𝑜 𝑟𝑤 =
𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑤𝑜𝑟𝑘

𝑊𝑇 − 𝑊𝑃 522.47 − 1.737
= = = 0.9967
𝑊𝑇 522.47

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 35
 2) Steam at 20 bar, 360 0C is expanded in a steam turbine to 0.08 bar. It then enters a condenser, where
it is condensed to saturated liquid water. The pump feeds back the water in to boiler. a) Assuming ideal
processes, find per kg of steam the net work and the cycle efficiency. b) If the turbine and pump have
each 80% efficiency find the percentage reduction in the network and cycle efficiency.
Ans : Need to use Steam tables

Saturation temperature at 20bar is 212.4 0C. Given steam is at 360 0C, thus it is in the super heated
condition.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 36
 Referring to the superheated steam tables

At 0.08 bar, entropy sg= 8.229 kJ/kgK

Sf = 0.593.
Since s4 = 6.992 kJ/kgK, steam after expansion is in
Enthalpy before expansion, h3 = 3159kJ/kg wet condition
Entropy, s3 = 6.992 kJ/kg.K
s4 = sf + x*(sg-sf)
During the expansion process, 3-4, Entropy is constant at 0.08 bar
Thus s4 = s3 = 6.992 kJ/kg.K 6.992 = 0.593 + x(8.229-0.593)
Pressure in the condenser = 0.08 bar x, steam quality after expansion = 0.838

Need to refer the saturated steam tables to h4 = hf + x*(hg-hf)

at 0.08 bar
determine the quality of steam after expansion.
h4 = 173.8+0.838*(2577-173.8) = 2187.7 kJ/kg
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 37
Calculation of Pump work
Pumping process 1-2 is isentropic compression process. Thus ds=0
Tds = dh -vdp
h2-h1 = vf(p2-p1)
At 0.08 bar, vf = 1.008 X 10-3 m3/kg

h2-h1 = 1.008 X 10-3 (20-0.08)*100 =2.008 kJ/kg

Turbine work, WT = h3 – h4 =3159 -2187.7 = 971.3 kJ/kg
Pump work input, Wp = 2.008 kJ/kg
Network output = 971.3 – 2.008 = 969.3 kJ/kg
h1 = hf at 0.08 bar =173.8 kJ/kg
h2 =2.008 + 173.8 = 175.8 kJ/kg
Heat Input to the cyle, Q1= h3-h2 = 3159 – 175.8 = 2983 kJ/kg
Efficiency of the cycle = Net work output / Heat Input = 969.3 / 2983 =
0.3249 = 32.49%
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 38
 Net efficiency calculation with the turbine and pump efficiency of 80%

Irreversibilities decreases the efficiencies T

of compression and expansion processes
and entropy increases (does not remain
constant)

Pump efficiency = (h2-h1)/(h2’-h1) 2’

0.8=( 175.8-173.8)/(h2’-173.8)
h2’ = 178.32 kJ/kg 4’

Turbine efficiency = (h3-h4’)/(h3-h4)

0.8=( 3159-h4’)/(3159-2187.7)
h4’ = 2382 kJ/kg
Heat input, Q1 = h3-h2’ = 3159-178.32 = 2980.7 kJ/kg
Turbine work output = h3-h4’ = 3159 – 2382 = 777 kJ/kg
Pump work input = h2’-h1 = 178.32 - 173.8 = 4.52 kJ/kg

Net work output = 777 – 4.52 =772.5 kJ/kg Efficiency= 772.5 / 2980.7 *100 = 25.9%
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 39
 Work output Work output
% reduction in with Isentropic - with 80%
work output = process efficiency
X 100
Work output
with Isentropic
process

969.3 - 777
= X 100 = 20.3%
969.3

Efficiency with Efficiency with

% reduction in Isentropic - 80% efficiency
Efficiency = process 32.49 - 25.9
X 100 = X 100 = 20.23%
Efficiency with 32.49
Isentropic
process

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 40
 MEAN TEMPERATURE OF HEAT ADDITION

In the Rankine cycle, heat is added

reversibly at a constant pressure,
but at infinite temperatures. Let
Tm1, is the mean temperature of
heat addition, so that area under 1s
and 2 is equal to the area under 5-6.

Heat added, QH = h2 – h1S

= Tm1 (S2 – S1S)

T3 limited by atmospheric temperature

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 41
 ℎ2 − ℎ1𝑠
𝑀𝑒𝑎𝑛 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 ℎ𝑒𝑎𝑡 𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛 𝑇𝑚 =
𝑆2 − 𝑆1𝑠
heat rejected QL = h3S – h4 = T3 (S2 – S1S)
𝑄𝐿 𝑇3 𝑆2 − 𝑆1𝑠 𝑇3
𝜂𝑅 = 1 − =1− =1−
𝑄𝐻 𝑇𝑚1 𝑆2 − 𝑆1𝑠 𝑇𝑚1
Where T3 is temperature of heat rejection
As T3 is lowered for a given Tm1, the h R . But the lowest practical
temperature of heat rejection is the ambient temperature T0
i.e., h R = f (Tm1 ) only.
0r higher the mean temperature of heat addition, the higher will
be the cycle efficiency.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 42
 The effect of increasing the initial temperature at constant pressure
on cycle efficiency

The effect of increasing the initial

temperature at constant pressure
on cycle efficiency is shown in
Figure. When the initial state
changes from 2 to 2’, Tm1, between 2
and 2’ is higher than Tm1 between 1s
and 2.

So an increase in the superheat at constant pressure increases the

mean temperature of heat addition and hence the cycle η.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 43
But the maximum temperature of steam that can be used is fixed from
metallurgical considerations (i.e., materials used for the manufacture of
the components which are subjected to high pressure, high temperature
steam such as super heaters, valves, pipelines, inlet stages of turbines
etc).
When the maximum temperature Metallurgical Limit
is fixed, as the operating steam
pressure at which heat is added in
the boiler increases from P1 to P2,
the mean temperature of heat
addition increases (since Tm1
between 5S and 6 higher than
between 1S and 2).

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 44
 The effect of increasing the initial pressure at turbine inlet

But when the turbine

inlet pressure increases
from P1 to P2, the ideal
expansion line shifts to
the left and the moisture
content at the exhaust
increases ( x 7S < x 3S )

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 45
If the moisture content of steam
in the turbine is higher the
entrained water particles along
with the vapour coming out from
the nozzles with high velocity
strike the blades and erode their
surfaces, as a result of which the
longevity of the blades decreases.
From this consideration,
moisture content at the turbine
exhaust is not allowed to exceed
15% or x < 0.85. Steam quality at Turbine Exit >15%

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 46
 Fixing of exhaust quality, maximum temperature & maximum
pressure in Rankine cycle efficiency

With the maximum steam temperature at the

turbine inlet, the minimum temperature of
heat rejection and the minimum quality of
steam at the turbine exhaust fixed, the
maximum steam pressure at the turbine inlet
also gets fixed.
The vertical line drawn from 3S, fixed by T3
and x3S, intersects the Tmax line, fixed by
material, at 2, which gives maximum steam
pressure at the turbine inlet.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 47
3) A cyclic steam power plant is to be designed for a steam temperature at turbine inlet of 360 0C 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 to exceed 15%. Determine the greatest allowable steam pressure at the turbine inlet
and calculate the Rankine cycle efficiency for these steam conditions. Estimate also the mean temperature
of heat addition.
Ans :
Turbine inlet temperature, T3 = 360 0C = 633 K
Exhaust pressure, P4 = 0.08 bar
Max moisture content after expansion ie at 4 = 0.15
Steam quality at station 4, i.e. x4 = 1- 0.15 = 0.85

Entropy after expansion (station 4) = Entropy before expansion (station 3)

s4 = 0.593 +0.85(8.229 – 0.593) =7.08 kJ/kgK

h4 = 173.8 +0.85(2577 – 173.8) =2216.5 kJ/kgK
h1 = 173.8 kJ/kg
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 48
At 360 0C saturation vapor entropy is 5.053 kJ/kgK which is lower than s4. Hence steam at turbine inlet is in super
heated condition

From super heated tables

At 360 0C
15 bar  entropy is 7.136 kJ/kgK
20 bar  entropy is 6.992 kJ/kgK
Pressure corresponding to entropy of 7.08
kJ/kgK is  7.08  7.136 20  15  15
6.992  7.136
 16.94 bar
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 49
Enthalpy at station 1 ie h3 = Enthalpy at 16.94 bar and 360 0C

15 bar  enthalpy = 3169 kJ/kg

20 bar  enthalpy = 3159 kJ/kg

Enthalpy at 16.94 bar  16.94  15 3159  3169  3169

20  15
 3165.2 kJ / kg

h3 = 3165.2 kJ/kg

Enthalpy h2
For process 1-2
Tds = dh -vdp
h2-h1 = 1.008 X 10-3(16.94-0.08) X 100 =1.7 kJ/kg
h1 = 173.8 kJ/kg
h2 = 173.8 + 1.7 = 175.5 kJ/kg

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 50
Turbine work output, WT = h3 – h4 =3165.2 – 2216.5 = 948.7 kJ/kg
Work input to pump, WP = h2 –h1 =1.7 kJ/kg
Net work output = 948.7 – 1.7 = 947 kJ/kg
Heat input in the boiler = h3 –h2 = 3165.2 – 175.5 = 2989.7 kJ/kg

Ranking cycle efficiency = Net work output / Heat input *100

=947/2989.7 *100 = 31.7 %
Mean Temperature of Heat Addition
Tmean = (h3-h2)/(s4-s1) = 2989.7/(7.08-0.593) = 460.8 K = 187.8 0C

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 51
Benson Boiler

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 52
IDEAL REHEAT CYCLE

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 53
We know that, the efficiency of the Rankine cycle could be increased
by increasing steam pressure in the boiler and superheating the
steam.
But this increases the moisture content of the steam in the lower
pressure stages in the turbine, which may lead to erosion of the
turbine blade.
∴The reheat cycle has been developed to take advantage of the
increased pressure of the boiler, avoiding the excessive moisture of
the steam in the low pressure stages.
In the reheat cycle, steam after partial expansion in the turbine is
brought back to the boiler, reheated by combustion gases and then
fed back to the turbine for further expansion.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 54
In the reheat cycle the expansion of steam from the initial state (2)
to the condenser pressure is carried out in two or more steps,
depending upon the number of reheats used.
In the first step, steam expands in HP turbine from state 2 to
approximate the saturated vapour line (process 2-3s).
The steam is then reheated (or resuperheated) at constant pressure
in the boiler (or in a reheater) process 3s-4 and the remaining
expansion process 4s-5 is carried out in the LP turbine.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 55
Note: 1) To protect the reheater tubes, steam is not allowed to
expand deep into the two-phase region before it is taken for
reheating, because in that case the moisture particles in steam
while evaporating would leave behind solid deposits in the form
of scale which is difficult to remove.
Also a low reheat pressure may bring down Tm1 and hence cycle
η. Again a high reheat pressure increases the moisture content at
turbine exhaust.
Thus reheat pressure is optimized. Optimum reheat pressure is
about 0.2 to 0.25 of initial pressure.

We have for 1 kg of steam

QH = (h2 – h1S) + (h4 – h3S); QL = h5S – h6
WT = (h2 – h3S) + (h4 – h5S); WP = h1S – h6
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 56
 𝑊𝑇 − 𝑊𝑃 3600 𝑘𝑔
∴ 𝜂𝑅 = & ℎ𝑒𝑎𝑡 𝑟𝑎𝑡𝑒 =
𝑄𝐻 𝑊𝑇 − 𝑊𝑃 𝑘𝑊ℎ

Since higher reheat pressure is used, WP work is appreciable.

2) In practice, the use of reheat gives a marginal increase in

cycle η, but it increases the net work output by making possible
the use of higher pressures, keeping the quality of steam at
turbine exhaust within a permissible limit. The quality
improves from x’5S to x5S by the use of reheat.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 57
3) A steam power station uses the following
cycle :
Steam at the boiler outlet is 150 bar, 550 0C
Reheat at 40 bar to 550 0C
Condenser at 0.1 bar
Assuming ideal processes, find the a) quality
at the turbine exhaust b) cycle efficiency and
c) steam rate

Ans :
Steam at boiler outlet, P3 = 150 bar
T3 = 550 0C =823 K

Reheat at 40 bar i.e., P4 = 4.0 bar Condenser at 0.1 bar ie

T5 = 550 0C =823 K P6 = P1 = 0.1 bar

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 58
At Station 5: at 40 bar, 550 0C

From superheated tables

Enthalpy, h5 = 3560.2 kJ/kg
Entropy, s5 = 7.233 kJ/kgK

At condenser, P = 0.1 bar

From saturated steam tables
sg =8.15 kJ/kg K & sf = 0.649 kJ/kg K

hf hg sf sg

s5 = s6 = sf + x(sg-sf)
7.233 = 0.649 + x(8.15-0.649) Thus x6= 0.877

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 59
Enthalpy at station h6 = hf + x(hg-hf)
=191.8 + 0.877(2585-191.8) = 2290.6 kJ/kg
h1 = 191.8 kJ/kg

h2-h1 = vf(p2-p1) =1.01 X 10-3 (150-0.1) X 100 = 15.14 kJ/kg

h2 = 191.8 + 15.14 = 206.94 kJ/kg

Enthalpy at HP Turbine inlet (150 bar, 550 0C)

From super heated tables, h3 = 3448.5 kJ/kg, s3 = 6.521 kJ/kgK

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 60
After HP turbine expansion , pressure is 40 bar, P4 = 40 bar
From 40 bar superheated tables,
By interpolation, temperature corresponding to entropy of 6.521 kJ/kgK
6.521  6.455
 360  320  320
6.621  6.455
 335.9 0C

By interpolation, Enthalpy at 40 bar & 335.9 0C

335.9  320
h4  3117  3015  3015
360  330
 3069.1 kJ / kg

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 61
Total Heat Input QH = h3-h2 + h5 – h4
= 3448.5 – 206.9 + 3560.2 – 3069.1 = 3732.7 kJ/kg
Turbine work output WT = h3 – h4 + h5 –h6
= 3448.5 – 3069.1 + 3560.3 – 2290.6 = 1649.1 kJ/kg
Efficiency η= 1649.1/3732.7 ×100 = 44.17%

Steam rate = Steam flow required to produce unit power output

= 3600 / 1649 = 2.183 kg/kWh

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 62
 IDEAL REGENERATIVE CYCLE:
The mean temperature of heat addition can
also be increased by decreasing the amount
of heat added at low temperatures. In a
simple Rankine cycle (saturated steam
entering the turbine), a considerable part of
the total heat supplied is in the liquid phase
when heating up water from 1 to 1’, at a
temperature lower than T2, the maximum
temperature of the cycle.
For maximum η, all heat should be supplied at T2, and feed water should
enter the boiler at 1’. This may be accomplished in what is known as an
ideal regenerative cycle as shown in figures (a) and (b).
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 63
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 64
The unique feature of the ideal regenerative cycle is that the
condensate, after leaving the pump circulates around the
turbine casing, counter-flow to the direction of vapour flow in
the turbine. Thus it is possible to transfer heat from the vapour
as it flows through the turbine to the liquid flowing around the
turbine.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 65
Let us assume that this is a reversible heat transfer i.e., at each
point, the temperature of the vapour is only infinitesimally higher
than the temperature of the liquid. ∴The process 2-3’ represents
reversible expansion of steam in the turbine with reversible heat
rejection. i.e., for any small step in the process of heating the
water ΔT(water) = - Δ T(steam) and (Δ S)water = (Δ S)steam. Then the
slopes of lines 2-3’ and 1’-4 will be identical at every temperature
and the lines will be identical in contour. Areas 1-1’-b-a-1 and 3’-
2-d-c-3’ are not only equal but congruous. ∴ all heat added from
external source (QH) is at constant temperature T2 and all heat
rejected (QL) is at constant temperature T3, both being reversible.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 66
Then Qs=h2-h1’=T1(s2-s1’)
Qr=h3’-h4 = T3(s2-s1’) ∵ s3’-s4=s2-s1’
∴ Efficiency η= 1-(Qr/Qs) = 1-(T3/T1)
This implies that;
• The efficiency of the ideal regenerative cycle is equal to the
Carnot cycle efficiency.
• The net work out put is less and hence steam flow rate will be
more.
However, the cycle is not practicable for the following reasons;
• Reversible heat transfer cannot be obtained in finite time
• Heat exchange in the turbine is mechanically impracticable
• The moisture content of the steam in the turbine will be high

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 67
REGENERATIVE CYCLE WITH FEED WATER HEATERS:

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 68
In a practical regenerative cycle, the feed water enters the boiler
at a temperature between 1 and 1’ (previous article figure), and it
is heated by steam extracted from intermediate stages of the
turbine. The flow diagram of the regenerative cycle with
saturated steam at the inlet to the turbine and the corresponding
T-S diagram are shown in figure.
For every kg of steam entering the turbine, let m1 kg steam be
extracted from an intermediate stage of the turbine where the
pressure is P2, and it is used to heat up feed water [(1 – m1) kg at
state 9] by mixing in heater (1).

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 69
The remaining (1-m1) kg of steam then expands in the turbine from
pressure P2 (state 3) to pressure P3 (state 4) when m2 kg of steam is
extracted for heating feed water in heater (2). So (1 – m1 – m2)kg of
steam then expands in the remaining stages of the turbine to pressure
P4, gets condensed into water in the condenser, and then pumped to
heater (2), where it mixes with m2 kg of steam extracted at pressure
P3. Then (1-m1) kg of water is pumped to heater (1) where it mixes
with m1 kg of steam extracted at pressure P2. The resulting 1kg of
steam is then pumped to the boiler where heat from an external source
is supplied. Heaters 1 and 2 thus operate at pressure P2 and P3
respectively. The amounts of steam m1 and m2 extracted from the
turbine are such that at the exit from each of the heaters, the state is
saturated liquid at the respective pressures.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 70
∴ Turbine work,
WT = 1(h2 – h3) + (1 – m1) (h3 – h4) + (1 – m1 – m2) (h4 – h5)
Pump work, WP = WP1 + WP2 + WP3
= (1 – m1 – m2) (h7 – h6) + (1 – m1) (h9 –h8) + 1 (h11 – h10)
QH = (h2 – h11); QL = (1 – m1 – m2) (h5 – h6)

𝑄𝐻 − 𝑄𝐿 𝑊𝑇 − 𝑊𝑃
∴ 𝜂𝑐𝑦𝑐𝑙𝑒 = =
𝑄𝐻 𝑄𝐻

3600 𝑘𝑔
∴ 𝑠𝑡𝑒𝑎𝑚 𝑟𝑎𝑡𝑒 =
𝑊𝑇 − 𝑊𝑃 𝑘𝑊ℎ

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 71
FEED WATER HEATERS (FWH)

A practical Regeneration process in steam power plants is

accomplished by extracting or bleeding, steam from the turbine at
various points. This steam, which could have produced more work
by expanding further in the turbine, is used to heat the feed water
instead. The device where the feed water heated by regeneration is
called a Regenerator or a Feed water Heater (FWH).

A feed water heater is basically a heat exchanger where heat is

transferred from the steam to the feed water either by mixing the
two streams (open feed water heaters) or without mixing them
(closed feed water heaters).

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 72
REGENERATIVE RANKINE CYCLE WITH OPEN & CLOSED FEED
WATER HEATERS

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 73
Open Feed water Heaters
An open (or direct-contact) feed water heater is basically a mixing chamber, where
the steam extracted from the turbine mixes with the feed water exiting the pump.
Ideally, the mixture leaves the heater as a saturated liquid at the heater pressure.

The advantages of open heater are simplicity, lower cost, and high heat transfer
capacity. The disadvantage is the necessity of a pump at each heater to handle
the large feed water stream.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 74
THE IDEAL REGENERATIVE RANKINE CYCLE WITH AN OPEN
FEEDWATER HEATER.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 75
Closed Feed water Heaters

In closed feed water heater, the heat is transferred from the extracted
steam to the feed water without mixing taking place. The feed water
flows through the tubes in the heater and extracted steam condenses
on the outside of the tubes in the shell. The heat released from the
condensation is transferred to the feed water through the walls of the
tubes.
The condensate (saturated water at the
steam extraction pressure), some times
called the heater-drip, then passes
through a trap into the next lower
pressure heater.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 76
• Closed Feed water Heaters

• This, to some extent, reduces

the steam required by that
heater. The trap passes only
liquid and no vapour.
• The drip from the lowest
pressure heater could
similarly be trapped to the
condenser, but this would be
throwing away energy to the
condenser cooling water.
• To avoid this waste, the drip
pump feed the drip directly
into the feed water stream.
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 77
• A closed heaters system requires only a single pump for the main
feed water stream regardless of the number of heaters. The drip
pump, if used is relatively small. Closed heaters are costly and
may not give as high a feed water temperature as do open heaters.
• In most steam power plants, closed heaters are favored, but at
least one open heater is used, primarily for the purpose of feed
water deaeration. The open heater in such a system is called
deaerator.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 78
Note:
• The higher the number of heater used, the higher will be the
cycle efficiency.
• The number of heater is fixed up by the energy balance of the
whole plant when it is found that the cost of adding another
does not justify the saving in QH or the marginal increase in
cycle efficiency.
• An increase in feed water temperature may, in some cases,
cause a reduction in boiler efficiency.
• So the number of heaters get optimized.
• Five feed water heaters are often used in practice.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 79
THE IDEAL REGENERATIVE RANKINE CYCLE WITH A CLOSED
FEEDWATER HEATER.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 80
 4) In a single-heater regenerative cycle the steam enters the turbine at 30 bar, 400
0C and the exhaust pressure is 0.1 bar. The feed water is a direct-contact type

which operates at 5 bar. Find a) the efficiency and steam rate of the cycle, and b)
the increase in mean temperature of heat addition, efficiency and steam rate, as
compared to the Rankine cycle (without regeneration). Neglect pump work.
Ans :
T5 = 400 0C, P5 = 30 bar
P7 = 0.1 bar, P6 = 5 bar
From super heater tables,
Enthalpy at 30bar, 400 0C,
h5 = 3231 kJ/kg, S5 = 6.921 kJ/kg.K

At 0.1 bar, from saturated steam tables,

hf = 191.8 kJ/kg, hg= 2585 kJ/kg
sf = 0.649 kJ/kgK, sg = 8.15 kJ/kgK

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 81
Steam quality at station 7
6.921 = 0.649 + x(8.15-0.649)
x = 0.8361

Enthalpy at 7, h7 = hf+x(hg-hf)
h7 = 191.8 + 0.8361(2585 – 191.8) = 2192.7 kJ/kg
h1 = 191.8 kJ/kg
s7 = s5 =6.921 kJ/kg.K
s1 = 0.649 kJ/kgK

Since the pump work is negligible, h2 = h1 = 191.8 kJ/kg

h3 is liquid enthalpy at 5 bar = 640.2 kJ/kg
s3 is liquid entropy at 5 bar = 1.861 kJ/kg
Since the pump work is negligible, h4 = h3 = 640.2 kJ/kg

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 82
At 5 bar, from super heated tables

6.921  6.865
Temperature corresponding to entropy of 6.921  200  160  160
7.059  6.865
 171.5 0C
Enthalpy at 5 bar, 171.5 0C , h6 =2790 kJ/kg
m(h6-h3)= (1-m)(h3-h2)
m(2790-640.2) = (1-m)(640.2-191.8)
2149.8m = (1-m)448.4 Thus m=0.1724 kg

Heat input = h5 –h4 = 3231 – 640.2 = 2590.8 kJ/kg

Work output = (h5-h6) + (1-m)( h6-h7)
= (3231-2790)+(1-0.1724)(2790-2192.7) = 935.3 kJ/kg
Efficiency = [935.3/2590.8]×100 = 36.1 %
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 83
 Steam rate = 3600/WT
= 3600 /935.3 = 3.849 kg/kW.hr
Tmean (s7-s4) = Heat input
Tmean (6.921 – 1.861)=2590.8
Tmean = 512.01 K = 239.01 0C
Without regeneration
h5 = 3231 kJ/kg, h7 = 2192.7 kJ/kg, h1 = 191.8 kJ/kg

Work output WT= h5-h7 = 3231 – 2192.7 = 1038.3 kJ/kg

Heat input QS= h5 –h2 = 3231 – 191.8 = 3039.2
Efficiency η = 1038.3 / 3039.2 ×100 = 34.16% With regeneration,
Steam rate = 3600/WT = 3600 /1038.3 = 3.467 kg/kWh • Efficiency increased
• Steam rate increased
Tmean (s7-s1) = Heat input • Mean temperature
Tmean (6.921 – 0.649)=3039.2 increased
Tmean = 484.5 K = 211.6 0C
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 84
5) (GATE 2021)
Ans :
h2 = 2624 kJ/kg
h5 = 226.7 kJ/kg
h6 = 708.6 kJ/kg
m5 = 100 kg/s
m2 == ?

m2h2 + m5h5 = (m2+m5)h6

M2*2624 + 100*226.7 = (m2+100)*708.6
1915.4 m2 = 481.9 *100
m2 = 481.9 * 100/1915.4 = 25.16 kg/s

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 85
 (GATE 2021)

6)

Ans : Wp =h2-h1= 20 kJ/kg W/Q1 = 0.5

WT =h3-h4= 750 kJ/kg Q1 = 2250/0.5 = 4500 kJ/kg
Wl =h5-h6= 1500 kJ/kg Q2=2250 kJ/kg
Eff = 50% h6 – h1 = 2250
h1 =200 kJ/kg h6 = 2250 +200 = 2450 kJ/kg
Hg = 2600 kJ/kg
Steam fraction at station 6
2450 = h1 +x (hg – h1) = 200 + x(2600 – 200)
1-Q2/Q1 = 0.5 x = 0.9375
Q2/Q1 = 0.5

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 86
 7) A certain chemical plant requires heat from process steam at 120 0C at the rate of 5.83 MJ/s and power at
the rate of 1000 kW from the generator terminals. Both the heat and power requirements are met by a back
pressure turbine of 80% brake and 85% internal efficiency which exhausts steam at 120 0C dry saturated. All
the latent heat released during condensation is utilized in the process heater. Find the pressure and
temperature of steam at the inlet to the turbine. Assume 90% efficiency for the generator.
Ans :
T
80% brake and 85%
Turbine internal efficiency
120 0C dry sat

0C
120 0C
120 5.83 MJ/s
Generator 1000 kW
Chemical Plant
90% efficiency
Turbine exhausts steam 120 0C dry saturated and has efficiency of 85%
h4 = hg at 120 0C = 2706 kJ/kg
h1 = 503.7 kJ/kg
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 87
Heat input to process plant per kg of steam = h4 – h1 =2706 – 503.7 = 2202.3 kJ/kg
Process plant requires 5.83 MJ/sec heat input
Mass of the steam from the turbine = 5.83 X 1000 / 2202.3 = 2.647 kg/sec

Generator output = 1000 kW

Generator efficiency = 90%
Input to Generator or Turbine work output or Brake output = 1000/0.9 = 1111.1 kW

Brake output 1111.1

brake   0.8 
Ideal output mh3  h 4 s 

h3-h4s = 524.7 kJ/kg

Turbine efficiency = 85% = 0.85

h3  h4 h3  2706
turbine   0.85 
h3  h4s 524.7
h3 = 3152 kJ/kg h4s = h3 – 524.7 = 3152 – 524.7 = 2627.3 kJ/kg

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 88
Steam quality at the turbine exhaust (Station 4s)
h4s = h4f + x(h4g –h4f)
2627.3 = 503.7 + x(2706 – 503.7)
x = 0.9642
Entropy at 4s = 1.528 + 0.9642(7.13 -1.528) = 6.9294 kJ/kg K

At the turbine entry (Station 3), enthalpy = 3152 kJ/kg and Entropy = 6.924 kJ/kgK

From the Mollier chart /

Super heated steam tables

Pressure at the entry of

turbine = 22.5 bar
Temperature = 360 0C

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 89
 Pressure at the entry of
turbine = 22.5 bar
Temperature = 360 0C

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 90
8) A certain factory has an average electrical load of 1500 kW and requires 3.5 MJ/s for heating purposes. It
is proposed to install a single extraction pass out steam turbine to operate under the following conditions.
Initial pressure 15 bar, Initial temperature 300 0C, Condenser pressure 0.1 bar.
Steam is extracted between the two turbine sections at 3 bar, 0.96 dry, and it isobarically cooled without
sub-cooling in heaters to supply the heating load. The internal efficiency of the turbine ( in the LP section ) is
0.8 and the efficiency of the boiler is 0.85 when using oil of calorific value 44 MJ/kg.
If 10% of boiler steam is used for auxiliaries calculate the oil consumption per day. Assume that the
condensate from the heaters(at 3 bar) and that before being pumped to the boiler. Neglect extraneous
losses.

Ans : 3 HPT m LPT

T G
3 bar,
Boil 0.96 dry
m1
er (m-m1)
3 bar Process
2’ 4l 4 heater
4s Condenser
m1
2
0.1 bar Hot well
1 5s 5
S
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 91
 P3 = 15 bar, T3 = 300 0C (super heated steam) Super heated Steam
h3 = (2993 + 3082)/2 = 3037.5 kJ/kg
s3 = (6.838+6.994)/2 = 6.916 kJ/kg
Steam extracted at 3 bar, 0.96 dry (Station 4)
Enthalpy, h4 = 561.5 + 0.96(2725-561.5) = Saturation Steam
2638.5 kJ/kg
Entropy, s4 = 1.672 + 0.96(6.992 – 1.672) =
6.774 kJ/kg K

Liquid enthalpy , h4l= 561.5 kJ/kg (at 3 bar) Condenser pressure, P5 =0.1 bar
Enthalpy available for heating = h4 –h4l = sf =0.649 kJ/kgK , sg = 8.15 kJ/kgK
2638.5 -561.5 = 2077 kJ/kg Entropy at 4 = Entropy at 5s
6.774 = 0.649 + x(8.15-0.649)
Heating load required by the plant = 3.5 MJ/sec x = 0.81655
Mass of the steam required to provide the Steam quality after expansion in the turbine = 0.816
heating load, m1 = 3.5 *1000 / 2077 = 1.685
h5s = 191.8 + 0.816(2585 – 191.8) = 2144.6 kJ/kg
kg/sec
LP Turbine efficiency = 0.8
h4-h5 = 0.8(h4-h5s) = 0.8(2638.5 – 2144.6) = 395.6
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 92
h5 = h4 - 395.6 = 2638.5 – 395.6 = 2242.9 kJ/kg
h1 = 191.8 kJ/kg (liquid enthalpy at 0.1 bar)

Turbine work output , WT = m (h3 –h4) + (m – m1)(h4 – h5)

= m(3037.5 – 2638.5) + (m – 1.685)(2638.5 – 2242.9)
= 399m + 395.6m – 666.6 = 794.6m – 666.6

Work output required for the plant = 1500 kW

794.6m – 666.6 = 1500 => m = 2.7267 kg/sec = 9815.9 kg/hr of steam

Hot well Heat balance
m1*h4l + (m-m1)h1 = m*h2
1.685 * 561.5 + (2.73 – 1.685)*191.8 = 2.73* h2
h2 = 419.75 kJ/kg
10% of the steam is used for auxiliaries
Steam required = 1.1 m
Heat input required to produce steam= 1.1 m (h3 – h2) = 1.1*2.7267(3037.5 – 419.75) =7851.6 kJ
Boiler efficiency =0.85

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 93
 Oil energy to be burnt =7851.6 / 0.85 = 9237.2 kJ
Calorific value of oil = 44 MJ /kg
Oil consumption = 9237.2 / 44000 = 0.2099 kg/sec = 755.77 kg/hr = 18.138 tons /day
9) A steam turbine gets its supply of steam at 70 bar and 450 0C. After expanding to 25 bar in high pressure
stages, it is reheated to 420 0C at the constant pressure. Next, it is expanded in intermediate pressure stages
to an approximate minimum pressure such that part of the steam bled at this pressure heats the feed water
to a temperature of 180 0C. The remaining steam expands from this pressure to a condenser pressure of 0.07
bar in the flow pressure stage. The isentropic efficiency of the hp. Stage is 78.5%, while that of the
intermediate and lp stages is 83% each. From the above data a) determine the pressure at which bleeding is
necessary and sketch a line diagram of the arrangement of the plant b) sketch on T-s diagram all the
processes c) determine the quantity of steam bled per kg of flow at the turbine inlet and d) calculate the
cycle efficiency. Neglect pump work. u h s
v
60 bar
Ans :
P3 =70 bar, T3 = 450 0C
80 bar
h3 = (3302+3272)/2 = 3287 kJ/kg
S3 = (6.719+6.555)/2 = 6.637 kJ/kgK
From super heated tables

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 94
 m

HPT LPT G

Boil m
er (m-m1) 4
m
m1 4s 6
Condenser 6s
8
FWH 0.07 bar
7s 7

3 bar,
0.96 dry
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 95
P4 =25 bar, T5 = 420 0C

4s 6
6s
8
s3 =s4 =6.637 kJ/kgK
0.07 bar
7s 7
Temperature corresponding to 6.637 kJ/kgK entropy at 25 bar is
6.637  6.5645 320  280  280  297.1 0C
6.7345  6.5645
297.1  280 3056  2958.5  2958.5  3000 kJ / kg
Enthalpy at station 4s, h4s = 320  280

At station 5, 25 bar, 420 0C h5 = 420  400 3350.5  3239.5  3239.5  3283.9 kJ / kg

450  400
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 96
 Entropy at station5, s5 =
420  400 7.1835  7.024  7.024  7.0878 kJ / kg
450  400
HP Stage efficiency = 78.5%
h3  h4
 0.785  h 4  3287  0.7853287  3000   3061.7 kJ / kg
h3  h 4s
Station 6 is intermediate pressure where the steam is bled to heat the feed water heater

Total steam flow = m kg/sec

Steam bled for FWH = m1 kg/sec
Steam expanded in the LP turbine stages till end = m – m1

m1h6 + (m-m1)h1= m h8 1

h1 is liquid enthalpy at condenser pressure = 162.6 kJ/kg

h8 is liquid enthalpy at intermediate pressure = 763.2 kJ/kg

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 97
After heating in FWH, temperature is 180 0C
Saturation pressure corresponding to 180 0C is 10.02 bar
Entropy at 5 = Entropy at 6s =7.0875 kJ/kgK

Temperature corresponding to Entropy at 6s

7.0875  7.046 320  280  280  291.1 0C
7.196  7.046

291.1  280 3094  3008  3008  3031.8 kJ / kg
Enthalpy at 6s, h6s
320  280
LP Turbine efficiency = 83 %
h5  h6
 0.83  h6  3283.9  0.833283.9  3031.8  3074.66 kJ / kg
h5  h6s
From equation 1
m1*3074.67+(m-m1)162.6=m763.2
2912 m1=600.6 m => m1/m = 0.20625

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 98
For every kg of steam inlet to turbine, 0.2 kg is bled after HPT to heat the feed water heater
Heat input = (h3 – h8) + (h5-h4) = (3287 – 763.2) + (3283.9 – 3061.7) = 2746 kJ/kg

Entropy at station 6
 3074.66  3008 
Temperature corresponding to Station 6 =  320  280   280  311 C
0

 3094  3008 
Entropy , s6 =
 311  280 
 7.196  7.046  7.046  7.16225 kJ / kgK
 320  280 

Entropy at 6 = Entropy at 7s
Since station 7s is at condenser pressure of 0.07 bar
7.16225 = 0.557 + x(8.28 – 0.557)
x = 0.85526

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 99
 Enthalpy at 7s, h7s = 162.65 + 0.85526(2572-162.65) =2223.3 kJ/kgK

LP Turbine efficiency = 83%

h6  h7
 0.83  h7  3074.6  0.83(3074.6  2223.3)  2368 kJ / kg
h6  h7 s
Turbine work output = m((h3-h4) +(h5-h6)) + (m-m1)( h6-h7)
=1(3287 – 3061.7 + 3283.9 - 3074.67) + (1-0.20625)*(3074.67-2368) = 995.45 kJ/kg

Cycle Efficiency = Work output / Heat Input = 995.45 / 2746 = 0.3625 = 36.25%

From saturated steam tables

hf hg sf sg

0.07 162.65 2572 0.557 8.28

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 10
 COMBINED VAPOUR POWER CYCLE
It is a combination of Gas power cycle (ex Brayton cycle) and steam power
cycle (Rankine cycle)
• The maximum fluid temperature at the
turbine inlet is about 620°C for steam
turbine cycle and 1425°C for gas-turbine
power plants.
• The use of higher temperatures in gas
turbines is made possible by recent
developments in cooling the turbine
blades and coating the blades with high-
temperature-resistant materials such as
ceramics

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 10
• Gas leaves the gas turbine at very
high temperatures (usually above
500°C)
• In combined vapour power cycle,
high temperature exhaust gases
from the gas turbine are used to
heat the feed water in the steam
power cycle (bottoming cycle).
• The result is a combined gas–
steam cycle.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 10
• The combined cycle increases the
efficiency without increasing the initial T
cost greatly.
• Many new power plants operate on
combined cycles, and many more existing
steam- or gas-turbine plants are being
converted to combined-cycle power
plants.
• Thermal efficiencies well over 40 percent
are achieved as a result of combined cycle.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 10
 CHARACTERISTICS OF AN IDEAL WORKING FLUID
• The maximum temperature that can be used in steam cycles
consistent with the best available material is about 6000C, while
the critical temperature of steam is 3750C, which necessitates large
superheating and permits the addition of only an infinitesimal
amount of heat at the highest temperature.

• The desirable characteristics of the

working fluid in a vapour power cycle
to obtain best thermal efficiency are as
follows:

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 10
1. The fluid should have a high critical temperature so that the
saturation pressure at the maximum permissible temperature
(metallurgical limit) is relatively low. It should have a large enthalpy
of evaporation at that pressure.
2) The saturation pressure at the
temperature of heat rejection
should be above atmosphere
pressure so as to avoid the
necessity of maintaining vacuum in
the condenser.
3) The specific heat of liquid should be
small so that little heat transfer is
required to raise the liquid to the
boiling point.
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 10
4) The saturation vapour line of the T-S diagram should be steep, very
close to the turbine expansion process so that excessive moisture
does not appear during expansion.
5) The freezing point of the fluid should be below room temperature,
so that it does not get solidified while flowing through the pipe
lines.
6) The fluid should be chemically stable
and should not contaminate the
materials of construction at any
temperature.
7) The fluid should be nontoxic, non
corrosive, not excessively viscous, and
low in cost.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 10
 BINARY VAPOUR CYCLES (BVC)

• As the name indicates (BVC) has two cycles operating in tandem.

• Though water is better than any working fluid, at higher temperature range of
applications, there are better fluids like Diphenyl ether (C6H5)2O, Al2Br6, Hg, Na, K,
etc.
• Diphenyl ether not used because it decomposes gradually at higher temperature.
 Al2Br6 is a possibility and is on verge of research
 At p=12 bar, saturation temperature of water, Al2Br6 and Hg are 1870C,
482.50C and 5600C respectively.
• Therefore Hg is actually used in practice, especially at high temperature range.
• But at low temperature, the saturation pressure of Hg is too low and it is
impractical to maintain such a vacuum.
• For this reason, Hg is combined with steam to form a BVC.
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 10
BVC Flow diagram and Ts diagram Topping cycle

Bottoming
cycle

6-1 : super heat created in Super heater

4 -5 : Sensible heating of water in
Economizer
5-6 : Latent heat provided by
condensation of mercury
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 10
 Efficiency of BV Cycle
If m kg/sec of mercury required to generate 1 kg/sec of steam in the
steam cycle
Heat input, Q1 = m(ha – hd) + (h5- h4) + (h1 – h6)
Heat input Heat input in Heat input in
to mercury Economizer Super heater

Heat rejected, Q2 = (h2 – h3)

WT = m(ha – hb) + (h1 – h2)

WP = m( hd –hc) + (h4 – h3)
Q1  Q 2 WT  Wp
cycle  
Q1 Q1
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 10
 EXAMPLE: AMMONIA CYCLE
• A binary vapor power cycle consists of two Rankine cycles with steam
and ammonia as the working fluids
• In the steam cycle, superheated vapor enters the turbine at 8 MPa,
580°C, and saturated liquid exits the condenser at 75°C.
• The heat rejected from the steam
cycle is provided to the ammonia
cycle, producing saturated vapor at
65°C, which enters the ammonia
turbine.
• Saturated liquid leaves the ammonia
condenser at 1.5 MPa.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 11
8) A binary-vapour cycle operates on mercury and steam. Saturated mercury vapour
at 4.5 bar is supplied to the mercury turbine, from which it exhausts at 0.04 bar. The
mercury condenser generates saturated steam at 15 bar which is expanded in a steam
turbine to 0.04 bar. a) Find the overall efficiency of the cycle, b) If 50,000 kg/hr of
steam flows through the steam turbine, what is the flow through the mercury turbine ?
c) Assuming that all processes are reversible, what is the useful work done in the
binary vapour cycle for the specified steam flow ? d) If the steam leaving the mercury
condenser is superheated to a temperature of 300 0C in a super heater located in the
mercury boiler, and if the internal efficiencies of the mercury and steam turbines are
0.85 and 0.87 respectively, calculate the overall efficiency of the cycle. The saturated
properties of the mercury are given below

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 11
 Ans :
Mercury Working fluid turbine :
Turbine entry = 4.5 bar saturated
ie ha =355.98 kJ/kg
Exhaust pressure Pb = 0.04 bar
hc = 29.98 kJ/kg

Vapour quality of mercury after expansion

Sa = Sb = Sf + x(Sg - Sf)
0.5397 = 0.0808 + x(0.6925-0.0808) ∴ x = 0.75
hb = 29.98 + 0.75(329.85-29.98) = 254.89 kJ/kg

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 11
 Steam turbine :
Steam generated at 15 bar ie P5 = 15 bar
h5 = (762.8+908.8)/2 = 835.8 kJ/kg
h6 = (2778 + 2800)/2 = 2789 kJ/kg

Mass flow rate of mercury to produce 1 kg of steam

0.04 bar
𝑚ሶ (hb-hc) = (h6 –h3) 2

𝑚ሶ = (2789 – 121.4)/(254.89 -29.98) = 11.86 kg/sec

Mercury condensate is
Entropy at Station 6, used to heat the water to
s6 = (6.586+6.341)/2 = 6.4635 kJ/kgK saturated vapor condition.
h3 = 121.4 kJ/kg There is no economizer or
super heater

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 11
Entropy at station 6 = Entropy at station 2
6.4635 = 0.423 + x(8.475 – 0.423)  x = 0.7501
h2 = 121.4 + 0.75(2554 – 121.4) = 1945.8 kJ/kg

0.04 bar
2

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 11
a)Turbine output,
WT = m((ha – hb)) + (h6 –h2)
=11.86(355.98 – 254.89) + (2789 - 1945.8) = 2042.2 kJ/sec
Heat input
QS = m(ha – hc)= 11.86(355.98 -29.98) = 3886.36 kJ/kg

Cycle Efficiency η = 2042.2 / 3886.4 = 0.4854 = 52.54%

b) Steam flow rate ms= 50000 kg /hr = 13.89 kg/sec
Mercury flow rate mm= 11.86 × 13.89 = 164.72 kg/sec = 593 tons /hour

c) Work output
For one kg of steam = 2042.2 kJ/sec, ∴ For 13.89 kg/sec of steam,
Work output = 2042.2 × 13.89 = 28363.9 kW = 28.36 MW

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 11
d) Steam super heated to 300 0C at 15 bar in a super heater
h1 = (2993 + 3082)/2 = 3037.5 kJ/kg
S1 = (6.838 + 6.994) /2 = 6.916 kJ/kgK

Steam quality after expansion

6.916 = 0.423 + x( 8.475 – 0.423)
x = 0.80638

h2 = 121.4 + 0.806(2554 – 121.4) = 2083 kJ/kg

Steam turbine efficiency = 0.87

(h1-h2s)/(h1-h2) =0.87
(3037.5 – h2s)/(3037.5- 2083) = 0.87
h2s = 2207 kJ/kg

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 11
Mercury turbine efficiency = 0.85
(ha-hbs)/(ha-hb) =0.85
(355.98 – hbs)/(355.98- 254.89) = 0.85
∴ hbs = 270.05 kJ/kg bs

Mercury required for 1 kg of steam generation

m(hbs – hc) = (h6-h3)
m(270.05 – 29.98) = (2789 – 121.4) 2s
∴ m = 11.11 kg/sec

Heat input = m(ha-hc) + (h1 – h6)

= 11.11(355.98 -29.98) + (3037.5 – 2789) = 3870.9 kJ
Turbine work output = m(ha – hbs) + (h1 – h2s)
= 11.11( 355.98 – 270.05) + ( 3037.5 – 2207) =1785.2 kJ
Cycle efficiency = 1785.2 / 3870.9 = 0.4611 = 46.11 %

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 11
 MCQs

1) Which of the process in the Rankine cycle operating fluid is pure water
A) Isentropic compression in the pump
B) Constant pressure heat addition in boiler Ans
C) Isentropic expansion in turbine
D)Constant pressure heat rejection in condenser

2) Processes in Rankine cycle are similar to

A) Otto Cycle
B) Diesel cycle Ans
C) Dual cycle
D)Closed Brayton cycle

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 11
3) Superheating of steam in Rankine cycle results in
A) Increase in efficiency
B) Increase in work output Ans
C) Increase in work output and efficiency
D)Decrease in heat rejection

4) Most common fluid in the topping cycle of Binary vapour power cycles is
A) Water
B) Mercury
C) Ammonia Ans
D)Sulphur dioxide

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 11
Questions from Previous Year Papers

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 12
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 12
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 12
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 12
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 12
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 12
EXTRA SLIDES

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 12
EFFECT OF LOWERING THE CONDENSER PRESSURE

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 12
THE EFFECT OF SUPERHEATING THE STEAM TO HIGHER
TEMPERATURES ON THE IDEAL RANKINE CYCLE.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 12
THE EFFECT OF INCREASING THE BOILER PRESSURE ON THE
IDEAL RANKINE CYCLE.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 12
 The ideal reheat Rankine cycle.

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 13
THE VARIOUS STEPS TO INCREASE THE EFFICIENCY OF RANKINE
CYCLE ARE AS FOLLOWS
Lowering the condenser pressure

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 13
THE VARIOUS STEPS TO INCREASE THE EFFICIENCY OF RANKINE
CYCLE ARE AS FOLLOWS
Superheating the Steam to High Temperatures

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 13
THE VARIOUS STEPS TO INCREASE THE EFFICIENCY OF RANKINE
CYCLE ARE AS FOLLOWS
Increasing the Boiler Pressure

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 13
Regenerative Rankine Cycle

• Process 1-2: Expansion of steam through the turbine with constant entropy
• Process 2-3: Condensation process through the condenser at constant pressure
• Process 3-4: Pressurization of the working fluid by the feed pump I up to the pressure
at which working fluid will enter in to the feedwater heater for mixing with the
extracted steam
• Process 4-6: Working fluid at state 4 and extracted steam at state 5 will be entered in
to the feedwater heater and hence process 4-6 will be termed as heat energy
addition to the working fluid internally in the heat exchanger or feedwater heater.
• Process 6-7: Pressurization of the working fluid by the feed pump II up to the boiler
pressure at which working fluid will enter in to the boiler for heat energy addition
• Process 7-1: Heat energy addition to the working fluid at constant pressure during
the process 7-1

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 13
Efficiency of regenerative Rankine Cycle
Work done by the turbine
WT = (h1-h5) + (1-Y) (h5-h2)
Net work from the cycle
W Net = WT – (WP1 + WP2)
Let us ignore the pump works as it will be quite small as compared to the work
done by the turbine and therefore we can say that net work from the cycle will be
equivalent to the work from the turbine
W Net = (h1-h5) + (1-Y) (h5-h2)
Net input heat energy
Q =h1-h7
Efficiency of the cycle with regeneration with single feedwater heater will be
calculated as mentioned here
η= [(h1-h5) + (1-Y) (h5-h2)]/ (h1-h7)

Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 13
Approved by AICTE |Affiliated to VTU | Recognized by UGC with 2(f) & 12(B) status |Accredited by NBA and NAAC 13