Module 01: Thermodynamics of fluid flow Modal 01: Question Number 2 a & 2 b i. Static and Stagnation states ii. Application of first and second law of thermodynamics to Turbo machines iii. Efficiencies of Turbo machines iv. Overall isentropic efficiency, stage efficiency and polytropic efficiency for both compression and expansion processes. v. Preheat and Reheat factor. vi. Previous Year Problems

1. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 1
Turbo Machines
18ME54
Course Coordinator
Mr. THANMAY J. S
Assistant Professor
Department of Mechanical Engineering
VVIET Mysore
Module 01: Thermodynamics of fluid flow
Course Learning Objectives
Understand typical design of Turbo machine, their working principle, application and
thermodynamics process involved.
Course Outcomes
At the end of the course the student will be able to understand Model studies and
thermodynamics analysis of Turbomachines

2. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 2
Contents
Modal 01: Question Number 2 a & 2 b
i. Static and Stagnation states
ii. Application of first and second law of thermodynamics to Turbo machines
iii. Efficiencies of Turbo machines
iv. Overall isentropic efficiency, stage efficiency and polytropic efficiency for both
compression and expansion processes.
v. Preheat and Reheat factor.
vi. Previous Year Problems

3. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 3
Thermodynamics of fluid flow
Static and Stagnation States:
There are two kinds of state for the flowing fluid, namely static state and stagnation state.
(i) Static state: It is the state refers to those properties like pressure, temperature, density
etc. which are measured when the measuring instruments are at rest relative to the flow
of fluid.
Example: 𝑆𝑡𝑎𝑡𝑖𝑐 𝐸𝑛𝑡ℎ𝑎𝑙𝑝𝑦:(ℎ); 𝑆𝑡𝑎𝑡𝑖𝑐 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒: (𝑇); 𝑆𝑡𝑎𝑡𝑖𝑐 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒: (𝑃)
(ii) Stagnation state: It is the final state of a fictitious, isentropic and work free process
during which the final kinetic and potential energies of the fluid reduces to zero in a
steady flow.
Examples:𝑆𝑡𝑎𝑔𝑛𝑎𝑡𝑖𝑜𝑛 𝐸𝑛𝑡ℎ𝑎𝑙𝑝𝑦: (ℎ𝑜); 𝑆𝑡𝑎𝑔𝑛𝑎𝑡𝑖𝑜𝑛 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒: (𝑇𝑜)𝑆𝑡𝑎𝑔𝑛𝑎𝑡𝑖𝑜𝑛 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒: (𝑃𝑜)
First and Second Laws of Thermodynamics Applied to Turbomachines:
Or
Applications of first and second laws of thermodynamics to turbomachines.
Or
Starting from the first law, derive an expression for the work output of a turbomachine
in terms of properties at inlet and outlet.
Or
Deducing an expression, explain the significance of first and second law of
thermodynamics applied to a turbomachine.
Under the assumption, consider single inlet and single output steady state
turbomachine, across the sections of which the velocities, pressures, temperatures and other
relevant properties are uniform.
First law of thermodynamics: The steady flow equation
of the first law of thermodynamics in the unit mass basis
is:
𝒒 + 𝒉𝟏 +
𝒗𝟏
𝟐
𝟐
+ 𝒈 𝒁𝟏 = 𝒘 + 𝒉𝟐 +
𝒗𝟐
𝟐
𝟐
+ 𝒈 𝒁𝟐
Here, (q) and (w) are heat transfer and work transfer per
unit mass flow across the boundary of the control volume respectively.

4. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 4
Since, the stagnation enthalpy: 𝒉𝒐 = 𝒉 +
𝒗𝟐
𝟐
+ 𝒈𝒁
Then equation of the first law of thermodynamics can be derived as:
𝑞 + ℎ1 +
𝑣1
2
2
+ 𝑔 𝑍1 = 𝑤 + ℎ2 +
𝑣2
2
2
+ 𝑔 𝑍2
𝑞 + ℎ𝑜1 = 𝑤 + ℎ𝑜2
𝑞 − 𝑤 = ℎ𝑜2 − ℎ𝑜1 = ∆ℎ𝑜
Generally, all turbomachines are well-insulated devices, therefore 𝒒 = 𝟎.
Then 𝒒 − 𝒘 = ∆𝒉𝒐is derived as −𝒘 = ∆𝒉𝒐
This equation represents that, “the energy transfer as work is numerically equal to the change
in stagnation enthalpy of the fluid between the inlet and outlet of the turbomachine”.
In a power-generating turbomachine, (w) is positive and (Δho) is negative, i.e., the
stagnation enthalpy at the exit of the machine is less than that at the inlet. The machine produces
out work at the shaft.
In a power-absorbing turbomachine, (w) is negative and (Δho) is positive. The
stagnation enthalpy at the outlet will be greater than that at the inlet and work is done on the
flowing fluid due to the rotation of the shaft.
Second law of thermodynamics: The second law equation of states, applied to stagnation
properties is:
𝑻𝒐𝒅𝑺𝒐 = 𝒅𝒉𝒐 − 𝒗𝒐𝒅𝒑𝒐It can be re − written as − 𝒅𝒉𝒐 = −𝒗𝒐𝒅𝒑𝒐−𝑻𝒐𝒅𝑺𝒐
or 𝒅𝒉𝒐 = 𝒗𝒐𝒅𝒑𝒐+𝑻𝒐𝒅𝑺𝒐
but according to First law of Thermodynamics −𝒘 = ∆𝒉𝒐 ≫ 𝒅𝒉 = −𝒅𝒘 therefore
−𝒅𝒘 = 𝒗𝒐𝒅𝒑𝒐+𝑻𝒐𝒅𝑺𝒐
a) In a power-generating machine, 𝒅𝒑𝒐is negative since the flowing fluid undergoes a pressure
drop when mechanical energy output is obtained.
b) In a power-absorbing machine, 𝒅𝒑𝒐 is positive since the flowing fluid undergoes a pressure
increase.
c) In reversible process which has a work output 𝒅𝒘 (𝑹𝒆𝒗) = 𝒗𝒐𝒅𝒑𝒐 and 𝑻𝒐𝒅𝑺𝒐 = 𝟎
d) In a real machine (irreversible machine), ∴ 𝒅𝒘 (𝑰𝒓𝒓𝒆) = 𝒗𝒐𝒅𝒑𝒐 − 𝑻𝒐𝒅𝑺𝒐 and 𝑻𝒐𝒅𝑺𝒐 > 𝟎
e) The power-generating machine gives 𝒅𝒘 (𝑹𝒆𝒗) − 𝒅𝒘 (𝑰𝒓𝒓𝒆) = 𝑻𝒐𝒅𝑺𝒐
f) The power-absorbing machine gives 𝒅𝒘 (𝑰𝒓𝒓𝒆) − 𝒅𝒘 (𝑹𝒆𝒗) = 𝑻𝒐𝒅𝑺𝒐

5. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 5
Efficiency of Turbomachines
Efficiency of power absorbing Turbomachines (Compression Process)
The h-s diagram for the compression process is
shown. The fluid has initially static pressure and
temperature determines by state 1, the state 01 is the
corresponding stagnation state. After passing through the
turbomachine, the final static properties of the fluid are
determined by state 2 and state 02 is corresponding
stagnation state. If the process is reversible, the final fluid
static state would be 2’ while stagnation state would be 02’.
Line 1-2 in static coordinates and line 01-02 in stagnation
coordinates represent the real process.
The actual work input for compression process is,
𝒘 = 𝒉𝟎𝟐 − 𝒉𝟎𝟏
The ideal work input can be calculated by any one of the following four equations:
(i) Total-to-total work input is the ideal work input for the stagnation ends,
𝒘𝑻−𝑻 = 𝒉𝟎𝟐′ − 𝒉𝟎𝟏
(ii) Total-to-static work input is the ideal work input for the stagnation inlet to the static exit,
𝒘𝑻−𝑺 = 𝒉𝟐′ − 𝒉𝟎𝟏
(iii) Static-to-total work input is the ideal work input for the static inlet to the stagnation exit,
𝒘𝑺−𝑻 = 𝒉𝟎𝟐′ − 𝒉𝟏
(iv) Static-to-static work input is the ideal work input for the static inlet to the static exit,
𝒘𝑺−𝑺 = 𝒉𝟐′ − 𝒉𝟏

6. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 6
The efficiency of the compression process can be expressed by any one of the following
equations:
(i) Total-to-total efficiency is defined as the ratio of total-to-total work input to the actual
work input.
𝜼𝑻−𝑻 =
𝒘𝑻−𝑻
𝒘
=
𝒉𝟎𝟐′ − 𝒉𝟎𝟏
𝒉𝟎𝟐 − 𝒉𝟎𝟏
(ii) Total-to-static efficiency is defined as the ratio of total-to-static work input to the actual
work input.
𝜼𝑻−𝑺 =
𝒘𝑻−𝑺
𝒘
=
𝒉𝟐′ − 𝒉𝟎𝟏
𝒉𝟎𝟐 − 𝒉𝟎𝟏
(iii) Static-to-total efficiency is defined as the ratio of static-to-total work input to the actual
work input.
𝜼𝑺−𝑻 =
𝒘𝑺−𝑻
𝒘
=
𝒉𝟎𝟐′ − 𝒉𝟏
𝒉𝟎𝟐 − 𝒉𝟎𝟏
(iv) Static-to-static efficiency is defined as the ratio of static-to-static work input to the actual
work input.
𝜼𝑺−𝑺 =
𝒘𝑺−𝑺
𝒘
=
𝒉𝟐′ − 𝒉𝟏
𝒉𝟎𝟐 − 𝒉𝟎𝟏
Efficiency of power generating turbomachines (Expansion Process)
The h-s diagram for the expansion process is shown
in figure below. The fluid has initially the static pressure
and temperature determined by state 1, the state 01 is the
corresponding stagnation state. After passing through the
turbomachine, the final static properties of the fluid are
determined by state 2 and state 02 is corresponding
stagnation state. If the process is reversible, the final fluid
static state would be 2’ while stagnation state would be 02’.
Line 1-2 in static coordinates and line 01-02 in stagnation
coordinates represent the real process.
The actual work input for compression process is, 𝒘 = 𝒉𝟎𝟏 − 𝒉𝟎𝟐
The ideal work input can be calculated by any one of the following four equations:
(i) Total-to-total work output is the ideal work output for the stagnation ends,
𝒘𝑻−𝑻 = 𝒉𝟎𝟏 − 𝒉𝟎𝟐′
(ii) Total-to-static work output is the ideal work output for the stagnation inlet to the static
exit,
𝒘𝑻−𝑺 = 𝒉𝟎𝟏 − 𝒉𝟐′
(iii) Static-to-total work output is the ideal work output for the static inlet to the stagnation
exit,
𝒘𝑺−𝑻 = 𝒉𝟏 − 𝒉𝟎𝟐′
(iv) Static-to-static work output is the ideal work output for the static inlet to the static exit,
𝒘𝑺−𝑺 = 𝒉𝟏 − 𝒉𝟐′

7. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 7
The efficiency of the Expansion process can be expressed by any one of the following
equations:
(i) Total-to-total efficiency is defined as the ratio of actual work output to the total-to-total
work output.
𝜼𝑻−𝑻 =
𝒘𝑻−𝑻
𝒘
=
𝒉𝟎𝟏 − 𝒉𝟎𝟐′
𝒉𝟎𝟏 − 𝒉𝟎𝟐
(ii) Total-to-static efficiency is defined as the ratio of actual work output to the total-to-static
work output.
𝜼𝑻−𝑺 =
𝒘𝑻−𝑺
𝒘
=
𝒉𝟎𝟏 − 𝒉𝟐′
𝒉𝟎𝟏 − 𝒉𝟎𝟐
(iii)Static-to-total efficiency is defined as the ratio of actual work output to the static-to-total
work output.
𝜼𝑺−𝑻 =
𝒘𝑺−𝑻
𝒘
=
𝒉𝟏 − 𝒉𝟎𝟐′
𝒉𝟎𝟏 − 𝒉𝟎𝟐
(iv)Static-to-static efficiency is defined as the ratio of actual work output to the static-to-static
work output.
𝜼𝑺−𝑺 =
𝒘𝑺−𝑺
𝒘
=
𝒉𝟏 − 𝒉𝟐′
𝒉𝟎𝟏 − 𝒉𝟎𝟐
Overall isentropic efficiency, stage efficiency and polytropic efficiency for both compression
and expansion processes.
Efficiency Compressor Turbines
Infinitesimal Stage
Efficiency or Polytropic
Efficiency Or Or
Stage Efficiency
Overall Efficiency
𝒏 = 𝑷𝒍𝒐𝒚𝒕𝒓𝒐𝒑𝒊𝒄 𝑰𝒏𝒅𝒆𝒙
𝜸 = 𝒓𝒂𝒕𝒊𝒐 𝒐𝒇 𝑺𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝒉𝒆𝒂𝒕 = (
𝑪𝒑
𝑪𝒗
)
𝑷𝒓 = 𝑷𝒓𝒆𝒔𝒔𝒖𝒓𝒆 𝒓𝒂𝒕𝒊𝒐 𝒑𝒆𝒓 𝒔𝒕𝒂𝒈𝒆 = (
𝑷𝟏
𝑷𝟐
)
𝑷𝒓𝟎 = 𝑷𝒓𝒆𝒔𝒔𝒖𝒓𝒆 𝒓𝒂𝒕𝒊𝒐 𝒑𝒆𝒓 𝒔𝒕𝒂𝒈𝒆 = (
𝑷𝟏
𝑷𝑲+𝟏
) = (𝑷𝒓)𝑲

8. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 8
Basic Definitions:
Enthalpy:
Enthalpy is a property of a thermodynamic system, and is defined as the sum of the system's
internal energy and the product of its pressure and volume. H = E + PV or h = E + PV.
Entropy:
a thermodynamic quantity representing the unavailability of a system's thermal energy for
conversion into mechanical work, often interpreted as the degree of disorder or randomness in
the system.
"the second law of thermodynamics says that entropy always increases with time"
Isentropic Process:
In thermodynamics, an isentropic process is an idealized thermodynamic process that is
both adiabatic and reversible. The work transfers of the system are frictionless, and there is no
net transfer of heat or matter. Such an idealized process is useful in engineering as a model of
and basis of comparison for real processes. This is the reason it is called as isentropic (entropy
does not change).
Reversible Process:
In thermodynamics, a reversible process is a process whose direction can be reversed to return
the system to its original state by inducing infinitesimal changes to some property of the
system's surroundings. Throughout the entire reversible process, the system is
in thermodynamic equilibrium with its surroundings
Isothermal Process:
In thermodynamics, an isothermal process is a type of thermodynamic process in which
the temperature of the system remains constant: ΔT = 0.
Isobaric Process:
In thermodynamics, an isobaric process is a type of thermodynamic process in which
the pressure of the system stays constant: ΔP = 0.
Isochoric Process:
In thermodynamics, an isochoric process, also called a constant-volume process,
an isovolumetric process, or an isometric process, i.e., ΔV = 0.
Adiabatic Process:
In thermodynamics, an Adiabatic process is a type of thermodynamic process that occurs
without transferring heat or mass between the thermodynamic system and its environment. An
adiabatic process transfers energy to the surroundings only as work.
Polytropic Process:
A polytropic process is a thermodynamic process that obeys the relation:𝒑𝑽𝒏
= 𝑪 where (p) is
the pressure, (V) is volume, (n) is the polytropic index, and C is a constant. The polytropic
process equation can describe multiple expansion and compression processes which include
heat transfer. Some specific values of n correspond to particular cases:
𝒏 = 𝟎 𝑓𝑜𝑟 𝐼𝑠𝑜𝑏𝑎𝑟𝑖𝑐 𝑃𝑟𝑜𝑐𝑒𝑠𝑠 (CPP) 𝒏 = 𝟏 𝑓𝑜𝑟 𝐼𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑃𝑟𝑜𝑐𝑒𝑠𝑠 (CTP)
𝒏 = ∞ 𝑓𝑜𝑟 𝐼𝑠𝑜𝑐ℎ𝑜𝑟𝑖𝑐 𝑃𝑟𝑜𝑐𝑒𝑠𝑠 (CVP) 𝒏 = 𝜸 𝑓𝑜𝑟 𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝑃𝑟𝑜𝑐𝑒𝑠𝑠 = 𝐶𝑝/𝐶𝑣

9. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 9
Polytropic efficiency for both Compression and Expansion Processes
Polytropic efficiency for Compression (Compressor) Polytropic efficiency for Expansion (Turbines)
𝜼𝑷 𝒄𝒐𝒎𝒑𝒓𝒆𝒔𝒔𝒊𝒐𝒏 =
𝑰𝒔𝒆𝒏𝒕𝒓𝒐𝒑𝒊𝒄 𝑻𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 𝑹𝒂𝒊𝒔𝒆
𝑨𝒄𝒕𝒖𝒂𝒍 𝑻𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 𝑹𝒂𝒊𝒔𝒆
=
𝒅𝑻′
𝒅𝑻
𝜼𝑷 𝑬𝒙𝒑𝒂𝒏𝒔𝒊𝒐𝒏 =
𝑨𝒄𝒕𝒖𝒂𝒍 𝑻𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 𝑫𝒓𝒐𝒑
𝑰𝒔𝒆𝒏𝒕𝒓𝒐𝒑𝒊𝒄 𝑻𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 𝑫𝒓𝒐𝒑
=
𝒅𝑻
𝒅𝑻′
𝒅𝑻 =
𝒅𝑻′
𝜼𝑷
=
𝑻′−𝑻
𝜼𝑷
=
𝑻(
𝑻′
𝑻
−𝟏)
𝜼𝑷
𝒅𝑻 = 𝜼𝑷 𝒅𝑻′
= 𝜼𝑷 (𝑻 − 𝑻′) = 𝜼𝑷 𝑻 (𝟏 −
𝑻′
𝑻
)
𝒃𝒖𝒕 𝒈𝒆𝒏𝒆𝒓𝒂𝒍𝒍𝒚
𝑻′
𝑻
= (
𝑷+𝒅𝑷
𝑷
)
𝜸−𝟏
𝜸
𝒃𝒖𝒕 𝒈𝒆𝒏𝒆𝒓𝒂𝒍𝒍𝒚
𝑻′
𝑻
= (
𝑷+𝒅𝑷
𝑷
)
𝜸−𝟏
𝜸
𝒅𝑻
𝑻
=
(
𝑻′
𝑻
−𝟏)
𝜼𝑷
=
[(
𝑷+𝒅𝑷
𝑷
)
𝜸−𝟏
𝜸
−𝟏]
𝜼𝑷
𝒅𝑻
𝑻
= 𝜼𝑷 (𝟏 −
𝑻′
𝑻
) = 𝜼𝑷 [𝟏 − (
𝑷−𝒅𝑷
𝑷
)
𝜸−𝟏
𝜸
]
𝒅𝑻
𝑻
=
[(𝟏+
𝒅𝑷
𝑷
)
𝜸−𝟏
𝜸
−𝟏]
𝜼𝑷
𝒅𝑻
𝑻
= 𝜼𝑷 [𝟏 − (𝟏 − (
𝒅𝑷
𝑷
))
𝜸−𝟏
𝜸
]
𝑩𝒚 𝒔𝒆𝒓𝒊𝒆𝒔 𝒐𝒇 𝑬𝒙𝒑𝒂𝒏𝒔𝒊𝒐𝒏
(𝟏 + 𝒙)𝒏
= 𝟏 + 𝒏𝒙 +
𝒏(𝒏−𝟏)
𝟐
𝒙𝟐
+ ⋯
(𝟏 + 𝒙)𝒏
= 𝟏 + 𝒏𝒙
(𝟏 +
𝒅𝑷
𝑷
)
𝜸−𝟏
𝜸
= [𝟏 + (
𝜸−𝟏
𝜸
) (
𝒅𝑷
𝑷
)]
𝒅𝑻
𝑻
=
[(𝟏+
𝒅𝑷
𝑷
)
𝜸−𝟏
𝜸
−𝟏]
𝜼𝑷
=
[𝟏+(
𝜸−𝟏
𝜸
)(
𝒅𝑷
𝑷
)−𝟏]
𝜼𝑷
𝑩𝒚 𝒔𝒆𝒓𝒊𝒆𝒔 𝒐𝒇 𝑬𝒙𝒑𝒂𝒏𝒔𝒊𝒐𝒏
(𝟏 − 𝒙)𝒏
= 𝟏 − 𝒏𝒙 +
𝒏(𝒏−𝟏)
𝟐
𝒙𝟐
− ⋯
(𝟏 − 𝒙)𝒏
= 𝟏 − 𝒏𝒙
(𝟏 − (
𝒅𝑷
𝑷
))
𝜸−𝟏
𝜸
= [𝟏 − (
𝜸−𝟏
𝜸
)(
𝒅𝑷
𝑷
)]
𝒅𝑻
𝑻
= 𝜼𝑷 [𝟏 − (𝟏 −
𝒅𝑷
𝑷
)
𝜸−𝟏
𝜸
] = 𝜼𝑷 [𝟏 − [𝟏 − (
𝜸−𝟏
𝜸
) (
𝒅𝑷
𝑷
)]]
∴
𝒅𝑻
𝑻
=
𝟏
𝜼𝑷
(
𝜸−𝟏
𝜸
) (
𝒅𝑷
𝑷
)
𝒅𝑻
𝑻
= 𝜼𝑷 (
𝜸−𝟏
𝜸
)(
𝒅𝑷
𝑷
)
By integration with limits 1 to 2, By integration with limits 1 to 2,
𝒍𝒏 (
𝑻𝟐
𝑻𝟏
) =
𝟏
𝜼𝑷
(
𝜸 − 𝟏
𝜸
) 𝒍𝒏 (
𝑷𝟐
𝑷𝟏
) 𝒍𝒏 (
𝑻𝟐
𝑻𝟏
) = 𝜼𝑷 (
𝜸 − 𝟏
𝜸
)𝒍𝒏 (
𝑷𝟐
𝑷𝟏
)
𝜼𝑷 =
(
𝜸 − 𝟏
𝜸
)𝒍𝒏 (
𝑷𝟐
𝑷𝟏
)
𝒍𝒏 (
𝑻𝟐
𝑻𝟏
)
𝜼𝑷 =
𝒍𝒏 (
𝑻𝟐
𝑻𝟏
)
(
𝜸 − 𝟏
𝜸
) 𝒍𝒏 (
𝑷𝟐
𝑷𝟏
)

10. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 10
Stage efficiency for both Compression and Expansion Processes
Stage efficiency of compressor Stage efficiency of Expansion
𝜂𝑆 =
𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑅𝑎𝑖𝑠𝑒
𝐴𝑐𝑡𝑢𝑎𝑙 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑅𝑎𝑖𝑠𝑒
𝜂𝑆 =
𝐴𝑐𝑡𝑢𝑎𝑙 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝐷𝑟𝑜𝑝
𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝐷𝑟𝑜𝑝
𝜂𝑆 =
𝑇2 ′ − 𝑇1
𝑇2 − 𝑇1
𝜂𝑆 =
𝑇1 − 𝑇2
𝑇1 − 𝑇2′
𝜂𝑆 =
𝑇2 ′ − 𝑇1
𝑇2 − 𝑇1
=
𝑇1[
𝑇2 ′
𝑇1
− 1]
𝑇1 [
𝑇2
𝑇1
− 1]
=
[
𝑇2 ′
𝑇1
− 1]
[
𝑇2
𝑇1
− 1]
𝜂𝑆 =
𝑇1 − 𝑇2
𝑇1 − 𝑇2′
=
𝑇1[1 −
𝑇2
𝑇1
]
𝑇1[1 −
𝑇2′
𝑇1
]
=
[1 −
𝑇2
𝑇1
]
[1 −
𝑇2′
𝑇1
]
𝑏𝑢𝑡
𝑇2′
𝑇1
= (
𝑃2
𝑃1
)
(
𝛾−1
𝛾
)
𝑎𝑛𝑑
𝑇2
𝑇1
= (
𝑃2
𝑃1
)
(
1
𝜂𝑃
)(
𝛾−1
𝛾
)
𝑏𝑢𝑡
𝑇2′
𝑇1
= (
𝑃2
𝑃1
)
(
𝛾−1
𝛾
)
𝑎𝑛𝑑
𝑇2
𝑇1
= (
𝑃2
𝑃1
)
(
1
𝜂𝑃
)(
𝛾−1
𝛾
)
𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑟𝑎𝑡𝑖𝑜 𝑓𝑜𝑟 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑖𝑠
𝑃2
𝑃1
= 𝑃𝑟
𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑟𝑎𝑡𝑖𝑜 𝑓𝑜𝑟 𝐸𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛 𝑖𝑠
𝑃1
𝑃2
= 𝑃𝑟
∴
𝑃1
𝑃2
= Pr 𝑐𝑜𝑛𝑣𝑒𝑟𝑡𝑒𝑑 𝑎𝑠
𝑃2
𝑃1
= 𝑃𝑟−1
𝜂𝑆 =
[
𝑇2 ′
𝑇1
− 1]
[
𝑇2
𝑇1
− 1]
=
[(
𝑃2
𝑃1
)
(
𝛾−1
𝛾
)
− 1]
[(
𝑃2
𝑃1
)
(
1
𝜂𝑃
)(
𝛾−1
𝛾
)
− 1]
𝜂𝑆 =
[1 −
𝑇2
𝑇1
]
[1 −
𝑇2′
𝑇1
]
=
[1 − (
𝑃2
𝑃1
)
−(
1
𝜂𝑃
)(
𝛾−1
𝛾
)
]
[1 − (
𝑃2
𝑃1
)
−(
𝛾−1
𝛾
)
]
∴ 𝜂𝑆 =
[(𝑃𝑟)
(
𝛾−1
𝛾
)
− 1]
[(𝑃𝑟)
(
1
𝜂𝑃
)(
𝛾−1
𝛾
)
− 1]
∴ 𝜂𝑆 =
[1 − (𝑃𝑟)
−(
1
𝜂𝑃
)(
𝛾−1
𝛾
)
]
[1 − (𝑃𝑟)
−(
𝛾−1
𝛾
)
]

11. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 11
Overall isentropic efficiency for both Compression and Expansion Processes
Overall isentropic efficiency for compressor Overall isentropic efficiency for Expansion
𝜂𝑜 =
𝑇𝑜𝑡𝑎𝑙 𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑅𝑎𝑖𝑠𝑒
𝑇𝑜𝑡𝑎𝑙 𝐴𝑐𝑡𝑢𝑎𝑙 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑅𝑎𝑖𝑠𝑒
𝜂𝑜 =
𝑇𝑜𝑡𝑎𝑙 𝐴𝑐𝑡𝑢𝑎𝑙 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝐷𝑟𝑜𝑝
𝑇𝑜𝑡𝑎𝑙 𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝐷𝑟𝑜𝑝
𝜂𝑜 =
𝑇(𝑘+1)′−𝑇1
𝑇(𝑘+1)−𝑇1
𝜂𝑜 =
𝑇1 − 𝑇(𝑘 + 1)
𝑇1 − 𝑇(𝑘 + 1)′
𝜂𝑜 =
𝑇(𝑘+1)′−𝑇1
𝑇(𝑘+1)−𝑇1
=
𝑇1[
𝑇(𝑘+1)′
𝑇1
−1]
𝑇1 [
𝑇(𝑘+1)
𝑇1
−1]
𝜂𝑜 =
[
𝑇(𝑘+1)′
𝑇1
−1]
[
𝑇(𝑘+1)
𝑇1
−1]
𝜂𝑜 =
𝑇1−𝑇(𝑘+1)
𝑇1−𝑇(𝑘+1)′ =
𝑇1[1−
𝑇(𝑘+1)
𝑇1
]
𝑇1[1−
𝑇(𝑘+1)′
𝑇1
]
𝜂𝑜 =
[1−
𝑇(𝑘+1)
𝑇1
]
[1−
𝑇(𝑘+1)′
𝑇1
]
𝑏𝑢𝑡
𝑇(𝑘+1)′
𝑇1
= [(
𝑃2
𝑃1
)
(
𝛾−1
𝛾
)
]
𝐾
𝑎𝑛𝑑
𝑇(𝑘+1)
𝑇1
= [(
𝑃2
𝑃1
)
(
1
𝜂𝑃
)(
𝛾−1
𝛾
)
]
𝐾
𝑏𝑢𝑡
𝑇(𝑘+1)′
𝑇1
= [(
𝑃2
𝑃1
)
(
𝛾−1
𝛾
)
]
𝐾
𝑎𝑛𝑑
𝑇(𝑘+1)
𝑇1
= [(
𝑃2
𝑃1
)
(
1
𝜂𝑃
)(
𝛾−1
𝛾
)
]
𝐾
𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑟𝑎𝑡𝑖𝑜 𝑓𝑜𝑟 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑖𝑠
𝑃2
𝑃1
= 𝑃𝑟
𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑟𝑎𝑡𝑖𝑜 𝑓𝑜𝑟 𝐸𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛 𝑖𝑠
𝑃1
𝑃2
= 𝑃𝑟
∴
𝑃1
𝑃2
= Pr 𝑐𝑜𝑛𝑣𝑒𝑟𝑡𝑒𝑑 𝑎𝑠
𝑃2
𝑃1
=
1
𝑃𝑟
= 𝑃𝑟−1
𝜂𝑜 =
[
𝑇(𝑘+1)′
𝑇1
−1]
[
𝑇(𝑘+1)
𝑇1
−1]
=
[[(
𝑃2
𝑃1
)
(
𝛾−1
𝛾
)
]
𝐾
−1]
[[(
𝑃2
𝑃1
)
(
1
𝜂𝑃
)(
𝛾−1
𝛾
)
]
𝐾
−1]
𝜂𝑜 =
[1−
𝑇(𝑘+1)
𝑇1
]
[1−
𝑇(𝑘+1)′
𝑇1
]
=
[1−[(
𝑃2
𝑃1
)
−(
1
𝜂𝑃
)(
𝛾−1
𝛾
)
]
𝐾
]
[1−(
𝑃2
𝑃1
)
−(
𝛾−1
𝛾
)
]
𝐾
𝜂𝑜 =
[[(
𝑃2
𝑃1
)
(
𝛾−1
𝛾
)
]
𝐾
−1]
[[(
𝑃2
𝑃1
)
(
1
𝜂𝑃
)(
𝛾−1
𝛾
)
]
𝐾
−1]
=
[(𝑃𝑟)
𝐾 (
𝛾−1
𝛾
)
]−1
[(𝑃𝑟)
𝐾 (
1
𝜂𝑃
)(
𝛾−1
𝛾
)
]−1
∴ 𝜂𝑜 =
[1−[(
𝑃2
𝑃1
)
−(
1
𝜂𝑃
)(
𝛾−1
𝛾
)
]
𝐾
]
[1−(
𝑃2
𝑃1
)
−(
𝛾−1
𝛾
)
]
𝐾 =
1−(𝑃𝑟)
−𝐾 (
1
𝜂𝑃
)(
𝛾−1
𝛾
)
1−(𝑃𝑟)
−𝐾 (
𝛾−1
𝛾
)

12. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 12
Comparison between Preheat factor and Reheat factor
Preheat factor Reheat factor
𝑾𝒔 < (∆𝑾𝒔𝟏 + ∆𝑾𝒔𝟐 + ∆𝑾𝒔𝟑) (∆𝑾𝒔𝟏 + ∆𝑾𝒔𝟐 + ∆𝑾𝒔𝟑) > 𝑾𝒔
𝑾𝒔 < ∑ ∆𝑾𝒔 ∑ ∆𝑾𝒔 > 𝑾𝒔
𝟏 <
∑ ∆𝑾𝒔
𝑾𝒔
∑ ∆𝑾𝒔
𝑾𝒔
> 𝟏
Therefore, the Preheat facto(
∑ ∆𝑾𝒔
𝑾𝒔
)is always
less than unity for multistage compressor. This
is due to the preheating of the fluid at the end of
each compression stage and this appears as the
losses in the subsequent stages.
Therefore, the Reheat factor (
∑ ∆𝑾𝒔
𝑾𝒔
) is always
greater than unity for multistage turbine. This is
due to the reheating of the fluid at the end of each
expansion stage and this appears as the losses in
the subsequent stages.

13. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 13
Model Question Papers 01
2
a)
With the application of 1st law of thermodynamics, show the work
transfer is numerically equal to the change in stagnation enthalpy
between the inlet and outlet of a turbomachine.
6
b)
Prove that the overall efficiency is greater than the stage efficiency in
a multistage turbine
6
c)
Air enters a compressor at a static pressure of 1.5 bar, a static
temperature of 150C and a flow velocity of 50 m/s. At the exit the
static pressure is 3 bar, the static temperature is 1000C and the flow
velocity is 100 m/s. The outlet is 1 m above the inlet. Evaluate (a) the
isentropic change in enthalpy, (b) the actual change in enthalpy and
(c) efficiency of the compressor.
8
Model Question Papers 02
2
a)
Obtain an expression for the polytropic efficiency for a compressor
in terms of temperature, pressure and adiabatic index.
6
b)
Explain 1) Static state 2) Stagnation state. With a sketch explain i)
Total to Total efficiency ii) Static to total efficiency iii) Static to
Static efficiency.
6
c)
The flow rate through the compressor is 50 kg/s. The inlet static
conditions are 1 bar and 30°C. Exit temperature from the last stage is
650 K (static). The compressor has five stages of equal pressure
ration of 1.5. Calculate (a) the exit pressure from the last stage (b) the
ideal exit temperature from the last stage, (c) the overall efficiency
(d) the polytropic efficiency and (e) the stage efficiency
8
Model Question Papers 03
2
a)
Show that for expansion process, stage efficiency is higher than
overall efficiency.
8
b)
Find the number of stages of an axial flow compressor with
symmetrical balding in order to produce a total pressure rise from
1bar to 4bar. The blade height is 3cm, the mean diameter is 100cm,
mean speed of the rotor is 2400rpm and the stage efficiency is 82%.
8
Model Question Papers 04
2
a)
Define total to total, total to static, static to static and static to total
efficiencies for power developing and power consuming
turbomachines and write the T-s Diagrams.
8
b)
Total to total efficiency for a power absorbing turbomachine handling
liquid water of standard density is 70%. Suppose the total pressure of
water increased by 4 bar, evaluate (a) the isentropic change in total
enthalpy (b) the actual change in total enthalpy (c) the change in total
temperature of the water and (d) the power input to the water, flow
rate is 30kg/s.
8