18 me54 turbo machines module 01 question no 1a &1b

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: Introduction to Turbo Machines & 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

Contents
Modal 01: Question Number 1 a & 1 b
i. Introduction:
ii. Definition of Turbo machine,
iii. Parts of Turbo machines,
iv. Comparison with positive displacement machines,
v. Classification of Turbo machine,
vi. Dimensionless parameters and their significance,
vii. Unit and specific quantities,
viii. Model studies and its numerical.
(Note: Since dimensional analysis is covered in Fluid Mechanics subject, questions on
dimensional analysis may not be given. However, dimensional parameters and model studies
may be given more weightage.)
• Simple Numerical; on Model Analysis.
Modal 01: Question Number 2 a & 2 b
i. Thermodynamics of fluid flow:
ii. Application of first and second law of thermodynamics to Turbo machines
iii. Efficiencies of Turbo machines
iv. Static and Stagnation states
v. Overall isentropic efficiency
vi. stage efficiency (their comparison) and polytropic efficiency for both compression and
expansion processes.
vii. Reheat factor for expansion process.
• Simple Numerical; stage efficiency and polytropic efficiency.
* Solved Previous Year Question Papers

Introduction to Turbo Machines:
Turbo or Turbinis is of Latin origin and it implies that which spins or whirls around. A
Turbomachine is a rotary machine, which always involves an energy transfer between a
continuously flowing fluid and a rotor. It is a power or head generating machine. It uses the
dynamic action of the rotor or impeller or runner which changes the energy level of the
continuously flowing fluid through the rotor.
The turbomachine is used in several applications, the primary ones being electrical power
generation, aircraft propulsion and vehicular propulsion for civilian and military use. The units
used in power generation are steam, gas and hydraulic turbines, ranging in capacity
Definition of Turbo machine:
A turbo machine is a device where mechanical energy in the form of shaft work, is transferred
either to or from a continuously flowing fluid by the dynamic action of rotating blade rows.
The interaction between the fluid and the turbo machine blades also results in fluid dynamic
lift. In simple words: “A turbo machine is a device in which energy transfer occurs between
a flowing fluid and rotating element due to dynamic action. This results in change of
pressure and momentum of the fluid”
Parts of Turbo machines:
Example 01 Example 2
The principal components of a turbo machine are:
1. Rotating element: (vane, impeller or blades)– operating in a stream of fluid.
2. Stationary elements: which usually guide the fluid in proper direction for efficient energy
conversion process.
3. Shaft: This either gives input power or takes output power from fluid under dynamic
conditions and runs at required speed.
4. Housing: to keep various rotating, stationery and other passages safely under dynamic
conditions of the flowing fluid.

Comparison with Positive Displacement Machines:
Parameters Positive-displacement Machine Turbomachine
Simple
Examples
Action
It involves a change in volume or a
displacement of fluid.
It involves change in pressure and momentum
of the fluid.
There is a positive confinement of the fluid
in the system.
There is no positive confinement of the fluid at
any point in the system
Operation
It involves a reciprocating motion of the
mechanical element
It involves a purely rotary motion of
mechanical element
Stability
Features
vibrations are more. Hence low speeds are
adopted.
balanced and vibrations eliminated. Hence high
speeds can be adopted.
Heavy foundations are required. Light foundations sufficient.
Mechanical design is complex Design is simple.
Weight per unit output is more. Weight per unit output is less.
Overall
Efficiency
High efficiency because of static energy
transfer.
Efficiency is low because of dynamic energy
transfer.
Volumetric
efficiency
Low fluid handling capacity per unit
weight of machine.
High fluid handling capacity per unit
weight of machine.
Fluid
Impact
No such serious problems are
encountered.
Causes cavitation, Surging or pulsation: These
factors deteriorate the performance of the
machine.
Examples
Rotary & Reciprocating Pumps, I.C
engines etc.
Hydraulic turbines, Gas turbines, Steam
Turbines etc.

Classification of Turbo machine:
1. Based on energy transfer
a) Energy is given by fluid to the rotor - Power generating turbo machine
Example: Turbines
b) Energy given by the rotor to the fluid – Power absorbing turbo machine
Example: Pumps, blowers and compressors
2. Based on fluid flowing in turbo machine
a) Water
b) Air
c) Steam
d) Hot gases
e) Liquids like petrol etc.
3. Based on direction of flow through the impeller or vanes or blades
(with reference to the axis of shaft rotation)
a) Axial flow – Axial pump, compressor or turbine
b) Mixed flow – Mixed flow pump, Francis turbine
c) Radial flow – Centrifugal pump or compressor
d) Tangential flow – Pelton water turbine
4. Based on condition of fluid in turbo machine
a) Impulse type (constant pressure) Example: Pelton water turbine
b) Reaction type (variable pressure) Example: Francis reaction turbines
5. Based on position of rotating shaft
a) Horizontal shaft Example: Steam turbines
b) Vertical shaft Example: Kaplan water turbines
c) Inclined shaft Example: Modern bulb micro

Dimensionless parameters and their significance,
Dimensional Analysis:
The dimensional analysis is a mathematical technique deals with the dimensions of the
quantities involved in the process. Basically, dimensional analysis is a method for reducing the
number and complexity of experimental variable that affect a given physical phenomenon, by
using a sort of compacting technique. The three primary purposes of dimensional analysis are:
1. To generate non-dimensional parameters that help in the design of experiments and in the
reporting of experimental results.
2. To obtain scaling laws so that prototype performance can be predicted from model
performance.
3. To predict the relationship between the parameters.
Fundamental Quantities of Dimensional Analysis: Mass (M), length (L), time (T) and
temperature (ɵ) are called fundamental quantities since there is no direct relation between these
quantities. There are seven basic quantities in physics namely, mass, length, time, electric
current, temperature, luminous intensity and amount of a substance.
Secondary Quantities or Derived Quantities: The quantities derived from fundamental
quantities are called derived quantities or secondary quantities. Examples: area, volume,
velocity, force, acceleration, etc.

Methods of solving Dimensional Analysis
1) Dimensional Homogeneity: An equation is said to be dimensionally homogeneous if
the fundamental dimensions have identical powers of M, L, T on both sides.
Example for Dimensional Homogeneity:
𝐷𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝐹𝑙𝑜𝑤 × 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑄 = 𝐴 . 𝑉
In dimensional form: 𝐷𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒: 𝑄 =
𝐿3
𝑇
; 𝐴𝑟𝑒𝑎: 𝐴 = 𝐿2
; 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦: 𝑉 =
𝐿
𝑇
when these
values are substituted
(𝐿𝐻𝑆) 𝑸 = 𝑨 . 𝑽 (𝑅𝐻𝑆)
𝐿3
𝑇
= 𝐿2
×
𝐿
𝑇
𝐿3
𝑇
=
𝐿3
𝑇
This proves LHS = RHS; then the Equation is said to be in Dimensional Homogeneity
2) Buckingham’s π-Theorem:
The Buckingham’s π-theorem states that “if there are ‘n’ variables in a dimensionally
homogeneous equation and if these variables contain ‘m’ fundamental dimensions such as
M, L, T then they may be grouped into (n-m), non-dimensional independent π-terms”.
Let a variable X1 depends upon independent variables X2, X3,.....Xn. The functional equation
may be written as:
X1 = f (X2, X3.....Xn)
The above equation can also be written as:
f (X1, X2, X3,.....Xn) = C
Where, C is constant and f is some function.
In the above equation, there are ‘n’ variables. If these variables contain ‘m’ fundamental
dimensions, then according to Buckingham’s π-theorem,

Example for Buckingham’s π-Theorem
Performance of a turbomachine depends on the variables discharge (Q), speed (N), rotor
diameter (D), energy per unit mass flow (gH), power (P), density of fluid (ρ), dynamic viscosity
of fluid (μ). Using the dimensional analysis obtain the π-terms. (VTU, Jul/Aug-02)
Proof: General relationship is given as
Quantities Dimension Values Quantities Dimension Values
discharge (Q) power (P),
speed (N), density of fluid (ρ),
rotor diameter (D),
dynamic viscosity of
fluid (μ).
energy per unit mass
flow (gH),
Number of variables, n = 7
Number of fundamental variables, m = 3
Number of π-terms required, (n-m) = 4
Repeating variables are: D, N, ρ = a, b, c
Step 1 Step 2
which is Flow Coefficient which is Head Coefficient
Step 3 Step 4
which is Power Coefficient

Give the significance of the dimensionless terms (i) Flow coefficient (ii) Head coefficient
(iii) Power coefficient with respect to turbomachines. (VTU, Jan-07)
Or,
Explain capacity coefficient, head coefficient and power coefficient referring to a
turbomachine. (VTU, Feb-02, Feb-03, Feb-04, Jan-16, Jul-17)
a) Flow Coefficient: It is also called as capacity coefficient or specific capacity. The term(
𝑄
𝑁 𝐷3)
is the capacity coefficient, which signifies the volume flow rate of fluid through a
turbomachine of unit diameter of runner operating at unit speed. The specific capacity is
constant for dynamically similar conditions. Hence for a fan or pump of certain diameter
running at various speeds, the discharge is proportional to the speed. This is the First fan law.
Speed ratio: The specific capacity is related to another quantity called speed ratio and is obtained
as follows
Where 𝜑 =
𝑈
𝑉
is called the speed ratio, which is defined as the ratio of tangential velocity of
runner to the theoretical jet velocity of fluid. For the given machine, the speed ratio is fixed.
(ii) Head Coefficient: The term (
𝑔𝐻
𝑁 2
𝐷2) is called the head coefficient or specific head. It is a
measure of the ratio of the fluid potential energy (column height H) and the fluid kinetic energy
while moving at the rotational speed of the wheel U. The term can be interpreted by noting that:
𝑔𝐻
𝑁 2𝐷2 ∝
𝑔𝐻
𝑈2 The head coefficient is constant for dynamically similar machines. For a machine
of specified diameter, the head varies directly as the square of the tangential speed of wheel.
This is the Second fan law.
(iii) Power Coefficient: The term(
𝑃
𝜌 𝑁 3
𝐷5) is called the power coefficient or specific power. It
represents the relation between the power, fluid density, speed and wheel diameter. For a given
machine, the power is directly proportional to the cube of the tangential speed of wheel. This is
the Third fan law.

Unit and specific quantities,
Unit Quantities:
1] Unit speed.
The unit speed is the speed of the turbine operating under one meter head. Mathematically, unit
speed,(𝑵𝑼 =
𝑵
√𝑯
) this can be derived as follows:
2] Unit discharge or Unit flow.
The unit discharge is the discharge through a turbine when the head on the turbine is unity.
Mathematically, unit discharge, (𝑸𝑼 =
𝑸
√𝑯
) this is derived as follow:

3] Unit Power.
The unit power is the power developed by a turbine when the head on the turbine is unity.
Mathematically, unit power, (𝑷𝑼 =
𝑷
𝑯
𝟑
𝟐
) this can be derived as follow:
Specific quantities:
1) Specific speed of a turbine is defined as “a speed of a geometrically similar machine
which produces one kilowatt power under a head of one meter”.
Power coefficient: (
𝑃
𝜌𝑁3𝐷5
) ≫ 𝑃 ∝ 𝜌𝑁3
𝐷5
or 𝑃 = (𝐶)𝜌𝑁3
𝐷5
Head Coefficient: (
𝑔𝐻
𝑁 2
𝐷2) ≫ 𝑁 2
𝐷2
∝ 𝑔𝐻 ≫ 𝐷 ∝
(𝑔 𝐻)
1
2
𝑁
substituting in above equation
𝑃 = (𝐶)𝜌𝑁3
(
(𝑔 𝐻)
1
2
𝑁
)
5
we get 𝑃 = (𝐶) (
𝜌 (𝑔 𝐻)
5
2
𝑁2
)
Where C is proportionality constant, from the definition of specific speed of turbine:
𝑃 = (𝐶) (
𝜌 (𝑔 𝐻)
5
2
𝑁2
) ≫ (1) = (𝐶) (
𝜌 (𝑔 (1))
5
2
𝑁𝑆
2 ) ∴ (𝐶) =
𝑁𝑆
2
𝜌 (𝑔 )
5
2
substituting (𝐶) value in equation 𝑃 = (𝐶) (
𝜌 (𝑔 𝐻)
5
2
𝑁2
) we get 𝑃 = (
𝑁𝑆
2
𝜌 (𝑔 )
5
2
) (
𝜌 (𝑔 𝐻)
5
2
𝑁2
)
solving we get
𝑁𝑆 =
𝑁 𝑃
1
2
𝐻
5
4
=
𝑁 √𝑃
𝐻
5
4
𝒔𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝒔𝒑𝒆𝒆𝒅 𝒐𝒇 𝑻𝒖𝒓𝒃𝒊𝒏𝒆

2) Specific speed of Pump: It can be defined as “a speed of geometrically similar
machines discharging one cubic meter per second of water under head of one meter”
Head Coefficient: (
𝑔𝐻
𝑁 2
𝐷2) ≫ 𝑁 2
𝐷2
∝ 𝑔𝐻 ≫ 𝐷 ∝
(𝑔 𝐻)
1
2
𝑁
Flow Coefficient: (
𝑄
𝑁 𝐷3) ≫ 𝑄 ∝ 𝑁 𝐷3
Substituting 𝐷 ∝
(𝑔 𝐻)
1
2
𝑁
value in 𝑄 ∝ 𝑁 𝐷3
,
we get 𝑄 ∝ 𝑁 (
(𝑔 𝐻)
1
2
𝑁
)
3
≫ 𝑄 ∝
(𝑔 𝐻)
3
2
𝑁2 or 𝑄 = (𝐶)
(𝑔 𝐻)
3
2
𝑁2
Where C is proportionality constant, from the definition of specific speed of pump,
Then 𝑄 = (𝐶)
(𝑔 𝐻)
3
2
𝑁2 ≫ (1) = (𝐶)
(𝑔 (1))
3
2
𝑁𝑆
2 ∴ (𝐶) =
𝑁𝑆
2
(𝑔 )
3
2
substituting (𝐶) value in equation 𝑄 = (𝐶)
(𝑔 𝐻)
3
2
𝑁2 we get 𝑄 = (
𝑁𝑆
2
(𝑔 )
3
2
)
(𝑔 𝐻)
3
2
𝑁2
solving we get
𝑁𝑆 =
𝑁 𝑄
1
2
𝐻
3
4
=
𝑁 √𝑄
𝐻
3
4
𝒔𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝒔𝒑𝒆𝒆𝒅 𝒐𝒇 𝑷𝒖𝒎𝒑
Model studies and its numerical.
Model: is the small-scale replica of the actual structure or machine. It is not necessary that
models should be smaller than the prototypes (although in most of the cases it is), they may be
larger than the prototypes.
Prototype: The actual structure or machine
Model analysis: the study of models of actual machine.
Advantages of Model Analysis:
• The performance of the machine can be easily predicted, in advance.
• A relationship between the variables influencing a flow problem is obtained.
• The merits of alternative designs can be predicted with the help of model testing.
Three type of similarities must exist between the model and prototype.
1) Geometric Similarity: The ratio of all corresponding linear dimension in the model and
prototype are equal.

2) Kinematic Similarity: means the similarity of motion between model and prototype.
Thus, kinematic similarity is said to exist between the model and the prototype if the
ratios of the velocity and acceleration at the corresponding points in the model and
prototype are the same in magnitude; the directions also should be parallel.
3) Dynamic Similarity: means the similarity of forces between model and prototype.
Thus, dynamic similarity is said to exist between the model and the prototype if the
ratios of the forces acting at the corresponding points in the model and prototype are
the same in magnitude; the directions also should be parallel.

List of Formulas
a) Use of Unit Quantities
𝐻1𝑎𝑛𝑑 𝐻2 𝑏𝑒 𝑡ℎ𝑒 ℎ𝑒𝑎𝑑𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑡𝑢𝑟𝑏𝑜𝑚𝑎𝑐ℎ𝑖𝑛𝑒
𝑁1𝑎𝑛𝑑 𝑁2 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑠𝑝𝑒𝑒𝑑𝑠
𝑄1𝑎𝑛𝑑 𝑄2 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒. 𝑎𝑛𝑑
𝑃1𝑎𝑛𝑑 𝑃2 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑝𝑜𝑤𝑒𝑟 𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑇𝑢𝑟𝑏𝑜 𝑚𝑎𝑐ℎ𝑖𝑛𝑒
Using the defining equations of unit quantities
𝑁𝑢 =
𝑁1
√𝐻1
=
𝑁2
√𝐻2
𝑄𝑢 =
𝑄1
√𝐻1
=
𝑄2
√𝐻2
𝑃𝑢 =
𝑃1
𝐻1
3
2
=
𝑃2
𝐻2
3
2
b) Flow Coefficient between Model and
Prototype
(
𝑄
𝑁 𝐷3)
𝑚𝑜𝑑𝑒𝑙
= (
𝑄
𝑁 𝐷3)
𝑃𝑟𝑜𝑡𝑜𝑡𝑦𝑝𝑒
c) Head Coefficient between Model and
Prototype
(
𝑔𝐻
𝑁 2
𝐷2)
= (
𝑔𝐻
𝑁 2
𝐷2)
or (
√𝐻𝑚
𝑁𝑚 𝐷𝑚
) =
(
√𝐻𝑝
𝑁𝑝 𝐷𝑝
)
d) Power Coefficient between Model and
Prototype
(
𝑃
𝜌 𝑁 3
𝐷5)
= (
𝑃
𝜌 𝑁 3
𝐷5)
e) Specific speed of Model and Prototype
(𝑁𝑠)𝑇𝑢𝑟𝑏𝑖𝑛𝑒 = (
𝑁 √𝑃
𝐻
5
4
)
= (
𝑁 √𝑃
𝐻
5
4
)
(𝑁𝑠)𝑃𝑢𝑚𝑝𝑠 = (
𝑁 √𝑄
𝐻
3
4
)
= (
𝑁 √𝑄
𝐻
3
4
)
Note: Answers calculated for Power is in Watts
it can be converted to kilo watts
𝑬𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒄𝒚 𝒐𝒇 𝑻𝒖𝒓𝒃𝒐 𝑴𝒂𝒄𝒉𝒊𝒏𝒆𝒔
𝜼 =
𝑷
𝝆𝒈𝑯𝑸
𝒐𝒓
∴ 𝑷 = 𝜼𝝆𝒈𝑯𝑸
𝑬𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒄𝒚 𝒐𝒇 𝑴𝒐𝒅𝒆𝒍:
𝜼𝒎 =
𝑷𝒎
𝝆𝒈𝐻𝑚 𝑸𝒎
𝑬𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒄𝒚 𝒐𝒇 𝑷𝒓𝒐𝒕𝒐𝒕𝒚𝒑𝒆:
𝜼𝒑 =
𝑷𝒑
𝝆𝒈𝐻𝑝 𝑸𝒑
f) Affinity laws

18ME54
Model Question Paper -2 with effect from 2020-21(CBCS Scheme)
Fifth Semester B.E. Degree Examination
TURBO MACHINES
Note:
01. Answer any FIVE full questions, choosing at least ONE question from each MODULE.
02. Thermodynamics Data Hand Book/Steam Table are permitted
03. Missing data may be assumed suitably by giving proper reason.
Q1
(A)
Define Turbo machine. With neat sketch explain the principal parts of general
turbomachine.
(B)
Explain the following with appropriate equation:
(i). Flow coefficient (ii) Head coefficient
(iii) Power coefficient (iv) Specific speed of a turbine.
(C)
From the performance curves of turbines, it is seen that a turbine of one-meter
diameter acting under a head of one meter, develops a speed of 25 RPM. What
diameter should be the prototype have, if it develops 10,000 kW working under a
head of 200m, with a specific speed of 150.
Model Given Data 𝑫𝒎 = 𝟏𝒎, 𝑯𝟏 = 𝟏𝒎, 𝑵𝒎 = 𝟐𝟓 𝒓𝒑𝒎,
Protype Given Data 𝑫𝑷 = ? , 𝑷𝒑 = 𝟏𝟎𝟎𝟎𝟎 𝒌𝑾, 𝑯𝒑 = 𝟐𝟎𝟎 𝒎, 𝑵𝒔𝒑 = 𝟏𝟓𝟎
By Unit Speed Equation
𝑁𝑢 =
𝑁𝑚
√𝐻𝑚
=
𝑁𝑝
√𝐻𝑝
𝑁𝑢 =
𝑁𝑚
√𝐻𝑚
=
1128.18
√200
= 79.77
𝑁𝑢 =
25
√𝐻𝑚
= 79.77
𝐻𝑚 = (
25
79.77
)
2
= 0.09820
Specific speed of Turbine
(𝑵𝒔)𝑻 = 𝟏𝟓𝟎 = (
𝑵 √𝑷
𝑯
𝟓
𝟒
)
𝑷
𝑵𝑷 = (
𝟏𝟓𝟎 × 𝟐𝟎𝟎
𝟓
𝟒
√𝟏𝟎𝟎𝟎𝟎
) = 𝟏, 𝟏𝟐𝟖. 𝟏𝟖 𝒓𝒑𝒎
Head Coefficient between
Model and Prototype
(
√𝐻𝑚
𝑁𝑚 𝐷𝑚
) = (
√𝐻𝑝
𝑁𝑝 𝐷𝑝
)
𝑁𝑝 𝐷𝑝 = (√
𝐻𝑝
𝐻𝑚
) 𝑁𝑚 𝐷𝑚
𝑁𝑝 𝐷𝑝 = 1,128.2342
𝐷𝑝 =
1,128.2342
1128.18
= 1.0000480 ≈ 1.0
𝑁𝑜𝑡𝑒: 𝑓𝑜𝑟 𝑁𝑢 = 79.77 =
𝑁1
√𝐻1
=
𝑁2
√𝐻2
=
𝑁2
√𝐻2
𝑖𝑛 𝑡ℎ𝑖𝑠 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑓 𝐻 = 1 𝑡ℎ𝑒𝑛 𝑁 = 79.77
∴ 𝑈𝑛𝑖𝑡 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝑁𝑢 = 79.77 =
25
√0.09820
=
79.77
√1
=
1128.18
√200
In this Turbo Machines the Diameter of the Blade is kept constant

Model Question Paper -1 with effect from 2020-21(CBCS Scheme)
TURBO MACHINES
Note:
01. Answer any FIVE full questions, choosing at least ONE question from each MODULE.
02. Thermodynamics Data Hand Book/Steam Table are permitted
03. Missing data may be assumed suitably by giving proper reason.
Q1
(A)
A manufacturer desires to double both the discharge and head on a geometrically
similar pump. Determine the percentage change in the rotational speed and
diameter under the new conditions.
(B)
The quantity of water available for a hydel station is 310 m3/s under a head of 1.8
m. Assuming speed of each turbine is 60 RPM and efficiency of 85%, find the
number of turbines required and power produced by each turbine. Each turbine
has a specific speed
of 685.6 (SI)
(C)
A model of a turbine built to a scale of 1:4 is tested under a head of 10 m. The
prototype has to work under a head of 50 m at 450 RPM. (a) What speed should
the model be run if it develops 60 kW using 0.9 m3/s at this speed. (b) What power
will be obtained from the prototype assuming that its efficiency is 3% better than
that of model.
Q1 (A)
𝐺𝑖𝑣𝑒𝑛 𝐷𝑎𝑡𝑎: 𝑄𝑚, 𝑄𝑃 = 2 𝑄𝑚, 𝐻𝑚, 𝐻𝑃 = 2 𝐻𝑚,
𝑇𝑜 𝑓𝑖𝑛𝑑:
𝑁𝑚
𝑁𝑝
× 100 =? ,
𝐷𝑚
𝐷𝑝
× 100 =?
𝑇𝑎𝑟𝑔𝑒𝑡 𝑡𝑜 𝑓𝑖𝑛𝑑: 𝑁𝑚, 𝑁𝑝, 𝐷𝑚 & 𝐷𝑝
𝑏𝑦 𝐹𝑙𝑜𝑤 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 (
𝑄𝑚
𝑁𝑚 𝐷𝑚
3) = (
𝑄𝑝
𝑁𝑝 𝐷𝑝
3) ≫ (
𝑄𝑚
𝑁𝑚 𝐷𝑚
3) = (
2 𝑄𝑚
𝑁𝑝 𝐷𝑝
3)
𝑜𝑟 𝑏𝑦 𝑐𝑟𝑜𝑠𝑠 𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 (
𝑄𝑚
2 𝑄𝑚
) = (
𝑁𝑚 𝐷𝑚
3
𝑁𝑝 𝐷𝑝
3 ) =
1
2
… … … … 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (𝑎)
(
𝐻𝑚
𝑁𝑚
2𝐷𝑚
2) = (
𝐻𝑝
𝑁𝑝
2𝐷𝑝
2) ≫ (
𝐻𝑚
𝑁𝑚
2𝐷𝑚
2) = (
2 𝐻𝑚
𝑁𝑝
2𝐷𝑝
2)
𝑜𝑟 𝑏𝑦 𝑐𝑟𝑜𝑠𝑠 𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 (
𝐻𝑚
2 𝐻𝑚
) = (
𝑁𝑚
2
𝐷𝑚
2
𝑁𝑝
2𝐷𝑝
2 ) =
1
2
… … … … . 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (𝑏)
𝑁𝑢 =
𝑁𝑚
√𝐻𝑚
=
𝑁𝑝
√𝐻𝑝
(
𝑁𝑚
√𝐻𝑚
=
𝑁𝑝
√𝐻𝑝
) = (
𝑁𝑚
𝑁𝑝
=
√𝐻𝑚
√𝐻𝑝
) 𝑜𝑟 (
𝑁𝑚
2
𝑁𝑝
2
=
𝐻𝑚
2𝐻𝑚
) ∴
𝑁𝑚
2
𝑁𝑝
2
=
1
2
𝑆𝑢𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑖𝑛 𝐸𝑞 (𝑏)
(
𝑁𝑚
2
𝐷𝑚
2
𝑁𝑝
2𝐷𝑝
2 ) =
1
2
≫ (
1 × 𝐷𝑚
2
2 × 𝐷𝑝
2 ) =
1
2
𝑜𝑟 (
𝐷𝑚
2
𝐷𝑝
2 ) = 1 𝑜𝑟 𝐷𝑚
2
= 𝐷𝑝
2
𝑖𝑓 𝐷𝑚 = 𝐷𝑝 𝑡ℎ𝑒𝑛 𝑖𝑛 𝐸𝑞 (𝑎) (
𝑁𝑚 𝐷𝑚
3
𝑁𝑝 𝐷𝑝
3 ) =
1
2
≫ (
𝑁𝑚
𝑁𝑝
) =
1
2
≫ 𝑁𝑚 =
1
2
× 𝑁𝑝
∴ 𝐩𝐞𝐫𝐜𝐞𝐧𝐭𝐚𝐠𝐞 𝐜𝐡𝐚𝐧𝐠𝐞 𝐢𝐧 𝐭𝐡𝐞 𝐫𝐨𝐭𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐬𝐩𝐞𝐞𝐝 𝐢𝐬 𝟓𝟎% 𝐚𝐧𝐝 𝐝𝐢𝐚𝐦𝐞𝐭𝐞𝐫 𝐫𝐞𝐦𝐚𝐢𝐧𝐬 𝐬𝐚𝐦𝐞

Q1 (B)
The quantity of water available for a hydel station is 310 m3/s under a head of 1.8
m. Assuming speed of each turbine is 60 RPM and efficiency of 85%, find the
number of turbines required and power produced by each turbine. Each turbine
has a specific speed
of 685.6 (SI)
𝐺𝑖𝑣𝑒𝑛 𝐷𝑎𝑡𝑎: 𝑄 = 310
𝑚3
𝑠
; 𝐻 = 1.8 𝑚; 𝑁 = 60 𝑟𝑝𝑚, 𝜂 = 85%
𝐴𝑠𝑠𝑢𝑚𝑒(𝜌)𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = 998 ~1000 𝑘𝑔/𝑚3
𝑇𝑜 𝑓𝑖𝑛𝑑: 𝑛 =? , 𝑃 =?
Overall Efficiency equation to calculate Overall Power
𝜼 =
𝑷
𝝆𝒈𝑯𝑸
∴ 𝑷 = 𝜼𝝆𝒈𝑯𝑸
𝑃 = 𝜂 × 𝜌 × 𝑔 × 𝐻 × 𝑄 = 0.85 × 1000 × 9.81 × 1.8 × 310 = 46,52,883 = 4652.883 𝑘𝑊
Specific speed of Turbine
𝑵𝒔 = 𝟔𝟖𝟓. 𝟔 =
𝑵 √𝑷
𝑯
𝟓
𝟒
𝟔𝟖𝟓. 𝟔 × 𝑯
𝟓
𝟒
𝑵
= √𝑷 =
𝟔𝟖𝟓. 𝟔 × 𝟏. 𝟖
𝟓
𝟒
𝟔𝟎
= 𝟐𝟑. 𝟖𝟐𝟑𝟕𝟓 ∴ 𝑷 = 𝟓𝟔𝟕. 𝟓𝟕𝟏𝟐𝟑 𝒌𝑾
𝐓𝐡𝐞 𝐏𝐨𝐰𝐞𝐫 𝐩𝐫𝐨𝐝𝐮𝐜𝐞𝐝 𝐛𝐲 𝐞𝐚𝐜𝐡 𝐭𝐮𝐫𝐛𝐢𝐧𝐞 = 𝟓𝟔𝟕. 𝟓𝟕 𝐤𝐖
𝐓𝐨𝐭𝐚𝐥 𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐭𝐮𝐫𝐛𝐢𝐧𝐞𝐬 𝐫𝐞𝐪𝐮𝐢𝐫𝐞𝐝 =
𝟒𝟔𝟓𝟐. 𝟖𝟖 𝒌𝑾
𝟓𝟔𝟕. 𝟓𝟕 𝐤𝐖
= 𝟖. 𝟏𝟗 ≅ 𝟗 𝐓𝐮𝐫𝐛𝐢𝐧𝐞𝐬 𝐑𝐞𝐪𝐮𝐢𝐫𝐞𝐝
Q1 (C)
A model of a turbine built to a scale of 1:4 is tested under a head of 10 m. The
prototype has to work under a head of 50 m at 450 RPM. (a) What speed should
the model be run if it develops 60 kW using 0.9 m3/s at this speed. (b) What power
will be obtained from the prototype assuming that its efficiency is 3% better than
that of model.
Given
Data
𝑫𝒎
𝑫𝒑
=
𝟏
𝟒
; 𝑯𝒎 = 𝟏𝟎𝒎 ; 𝑯𝒑 = 𝟓𝟎𝒎 ; 𝑵𝒑 = 𝟒𝟓𝟎 𝒓𝒑𝒎 ;
𝑵𝒎 = ? ; 𝑷𝒎 = 𝟔𝟎 𝒌𝑾 𝒐𝒓 𝟔𝟎 × 𝟏𝟎𝟑
𝑾 ; 𝑸𝒎 = 𝟎. 𝟗
𝒎𝟑
𝒔
; 𝑷𝒑 = ? ; 𝜼𝒑 = 𝟑%
Head
Coefficient
equation
(
𝐻𝑚
𝑁𝑚
2𝐷𝑚
2) = (
𝐻𝑝
𝑁𝑝
2𝐷𝑝
2) ≫ (
𝐷𝑝
2
𝐷𝑚
2) (
𝐻𝑝
𝐻𝑚
) = (
𝑁𝑚
2
𝑁𝑝
2 )
(
12
42) (
50
10
) = (
𝑁𝑚
2
𝑁𝑝
2 ) = 0.3125 ≫ 𝑁𝑚
2
= 0.3125 × 𝑁𝑝
2
≫ 𝑁𝑚
2
= 0.3125 × 450 2
𝑁𝑚
2
= 63,281.25 ∴ 𝑁𝑚 = 251.5576 𝑟𝑝𝑚
Efficiency
equation
𝜼𝒎 =
𝑷𝒎
≫ 𝜼𝒎 =
𝟔𝟎 × 𝟏𝟎𝟑
𝟏𝟎𝟎𝟎 × 𝟗. 𝟖𝟏 × 𝟏𝟎 × 𝟎. 𝟗
= 𝟎. 𝟔𝟕𝟗𝟓 = 𝟔𝟕. 𝟗𝟓%
3% more
𝜼𝒑 =
𝑷𝒑
= (𝜼𝒑 = 𝜼𝒎 + 𝟑%) =
𝑷𝒑
𝝆𝒈𝑯𝑸
∴ 𝑷 = (𝟎. 𝟔𝟕𝟗𝟓 + 𝟎. 𝟎𝟑)𝝆𝒈𝑯𝑸 ≫
𝑷 = 𝟎. 𝟕𝟎𝟗𝟓 × 𝟏𝟎𝟎𝟎 × 𝟗. 𝟖𝟏 × 𝟓𝟎 × 𝟎. 𝟗 = 𝟑, 𝟏𝟑, 𝟐𝟒𝟑. 𝟓 𝑾 = 𝟑𝟏𝟑. 𝟐𝟒𝟑𝟓 𝒌𝑾

Model Question Paper -1 with effect from 2015-16 (CBCS Scheme)
TURBO MACHINES
TIME: 03 Hours Max. Marks: 100
Q1
(A)
Define turbomachine. Give a comparison between turbomachines and
positive displacement machines
(B)
A single stage centrifugal pump works against a height of 30m, running at
2000rpm, supplies 3m3/s and has an impeller diameter of 300mm.
Calculate (a) the number of stages and (b) the diameter of each impeller
required to pump 6m3/s of water to a height of 220m when running at
1500rpm.
Given
Data
𝑯𝒎 = 𝟑𝟎 𝒎; 𝑵𝒎 = 𝟐𝟎𝟎𝟎 𝒓𝒑𝒎; 𝑸𝒎 = 𝟑
𝒎𝟑
𝒔
; 𝑫𝒎 = 𝟑𝟎𝟎 𝒎𝒎 = 𝟎. 𝟑𝒎;
𝑸𝑷 = 𝟔
𝒎𝟑
𝒔
; 𝑯𝑻𝒐𝒕𝒂𝒍 = 𝟐𝟐𝟎 𝒎; 𝑵𝒑 = 𝟏𝟓𝟎𝟎 𝒓𝒑𝒎;
Specific
speed of
Pump
(𝑁𝑠)𝑃𝑢𝑚𝑝𝑠 = (
𝑵𝒎 √𝑸𝒎
𝑯𝒎
3
4
) = (
𝑵𝒑√𝑸𝒑
𝑯𝒑
3
4
) ≫ 𝑯𝒑
3
4 = (
𝑵𝒑
𝑵𝒎
) × √
𝑸𝒑
𝑸𝒎
× 𝑯𝒎
3
4
𝑯𝒑
3
4 = (
𝟏𝟓𝟎𝟎
𝟐𝟎𝟎𝟎
) × √
𝟔
𝟑
× 𝟑𝟎
3
4 = 13.596189
𝑯𝒑 = 32.44779 𝑚
Number
of Stages
𝑻𝒐𝒕𝒂𝒍 𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝑺𝒕𝒂𝒈𝒆𝒔 =
𝑯𝑻𝒐𝒕𝒂𝒍
𝑯𝒑
=
𝟐𝟐𝟎
𝟑𝟐. 𝟒𝟒𝟕𝟕𝟗
= 𝟔. 𝟕𝟖𝟎𝟏𝟐 ≈ 𝟕
Head
coefficient
(
𝐻𝑚
𝑁𝑚
2𝐷𝑚
2) = (
𝐻𝑝
𝑁𝑝
2𝐷𝑝
2) ≫ 𝐷𝑝
2
= (
𝑁𝑚
2
𝑁𝑝
2
) × (
𝐻𝑝
𝐻𝑚
) 𝐷𝑚
2
𝐷𝑝
2
= (
𝑁𝑚
2
𝑁𝑝
2
) × (
𝐻𝑝
𝐻𝑚
) 𝐷𝑚
2
≫
𝐷𝑝
2
= (
2000 2
1500 2
) × (
32.44779
30
) × 0.32
= 0.173054
𝐷𝑝 = 0.41599 = 41.59 ≈ 41.6 𝑚𝑚

17ME53
Fifth Semester B.E. Degree Examination, Aug./Sept. 2020
TURBO MACHINES
Q1
(A)
Define a turbo machine. List any six differences between turbo machines
and positive displacement machines.
(B)
Identify the following as power generating or power absorbing turbo
machines
(i) Centrifugal compressor (ii) Steam turbine
(iii) Air blower (iv) Kaplan turbine.
(C)
Tests on a turbine runner 1 .25 m in diameter at 30 m head gave the
following results: Power developed = 736 kW, Speed = 180 rpm. Discharge
= 2.7 m 3 /s. Find the diameter, speed and discharge of a runner to operate
at 45 m head and give 1472 kW at the same efficiency. What is the specific
speed of both turbines?
Given
Data
𝑫𝒎 = 𝟏. 𝟐𝟓 𝒎 ; 𝑯𝒎 = 𝟑𝟎 𝒎 ; 𝑷𝒎 = 𝟕𝟑𝟔 𝒌𝑾 ; 𝑵𝒎 = 𝟏𝟖𝟎 𝒓𝒑𝒎 ;
𝑸𝒎 = 𝟐. 𝟕
𝒎𝟑
𝒔
; 𝑫𝒑 = ? ; 𝑸𝒑 = ? ; 𝑯𝒑 = 𝟒𝟓 𝒎 ; 𝑷𝒑 = 𝟏𝟒𝟕𝟐 𝒌𝑾
Same
efficiency
𝜼𝒎 = 𝜼𝒑
(𝜼𝒎 = 𝜼𝒑) ∴
𝑷𝒎
=
𝑷𝒑
𝑷𝒎
𝐻𝑚 𝑸𝒎
=
𝑷𝒑
𝐻𝑝 𝑸𝒑
≫ 𝑸𝒑 =
𝑷𝒑 × 𝐻𝑚 𝑸𝒎
𝑷𝒎𝐻𝑝
≫ 𝑸𝒑 =
𝟏𝟒𝟕𝟐 × 30 × 𝟐. 𝟕
𝟕𝟑𝟔 × 45
≫
𝑸𝒑 = 𝟑. 𝟔
𝒎𝟑
𝒔
Specific
speed of
Turbine
𝑁𝑠 =
𝑵𝒎 √𝑷𝒎
𝑯𝒎
5
4
=
𝑵𝒑 √𝑷𝒑
𝑯𝒑
5
4
𝑁𝑠 =
𝑵𝒎 √𝑷𝒎
𝑯𝒎
5
4
≫
𝟏𝟖𝟎 √𝟕𝟑𝟔
𝟑𝟎
5
4
= 69.55203
𝑁𝑠 = 69.55203 =
𝑵𝒑 √𝑷𝒑
𝑯𝒑
5
4
≫
𝑯𝒑
5
4 × 69.55203
√𝑷𝒑
= 𝑵𝒑
𝑵𝒑 =
𝑯𝒑
5
4 × 69.55203
√𝑷𝒑
=
𝟒𝟓
5
4 × 69.55203
√𝟏𝟒𝟕𝟓
= 𝟐𝟏𝟏. 𝟎𝟕𝟏𝟒 𝑟𝑝𝑚
By Flow
coefficient
(
𝑄𝑚
𝑁𝑚 𝐷𝑚
3) = (
𝑄𝑝
𝑁𝑝 𝐷𝑝
3) ≫ 𝐷𝑝
3
= (
𝑄𝑝 × 𝑁𝑚 𝐷𝑚
3
𝑁𝑝 × 𝑄𝑚
)
𝐷𝑝
3
= (
3.6 × 180 × 1.253
211.0714 × 2.7
) = 2.22081
𝑫𝒑 = 𝟏. 𝟑𝟎𝟒𝟔𝟕𝟗𝟖

17ME53
Fifth Semester B.E. Degree Examination, Aug./Sept. 2020
TURBO MACHINES
Q1
(A)
Define Turbomachine. With neat sketch, explain the parts of Turbo
machine.
(B)
Define specific speed of pump. Derive an expression for the same in terms
of discharge speed and head.
(C)
A Fransis turbine model is built to scale 1 :5 the data for the model is P =
4kW, N = 3500rpm, H = 2m and prototype H = 6m. Assume that the overall
efficiency of the model as 70%. Calculate: i) Speed of the prototype ii)
Power of the prototype. Use Moody’s equation.
Given
Data
𝑫𝒎
𝑫𝒑
=
𝟏
𝟓
; 𝑷𝒎 = 𝟒 𝒌𝑾 = 𝟒 × 𝟏𝟎𝟑
𝑾; 𝑵𝒎 = 𝟑𝟓𝟎𝟎 𝒓𝒑𝒎; 𝑯𝒎 = 𝟐 𝒎 ;
𝑯𝒑 = 𝟔 𝒎 ; 𝜼𝒐 = 𝟕𝟎% = 𝟎. 𝟕;
𝑵𝒑 = ? ; 𝑷𝒑 =?
Head
Coefficient
(
√𝐻𝑚
𝑁𝑚𝐷𝑚
) = (
√𝐻𝑝
𝑁𝑝𝐷𝑝
) ≫ 𝑁𝑝 = (√
𝐻𝑝
𝐻𝑚
) (
𝑫𝒎
𝑫𝒑
) 𝑁𝑚
𝑁𝑝 = (√
6
2
) (
𝟏
𝟓
) (3500)
𝑁𝑝 = 1,212.4355 𝑟𝑝𝑚
Power
Coefficient
𝑷𝒎
𝜌 𝑁𝑚
3𝑫𝒎
5 =
𝑷𝒑
𝜌 𝑁𝑝
3𝑫𝒑
5 ≫
𝑷𝒎
𝑁𝑚
3𝑫𝒎
5 =
𝑷𝒑
𝑁𝑝
3𝑫𝒑
5
𝑷𝒑 = (
𝑫𝒑
𝑫𝒎
)
5
× (
𝑁𝑝
𝑁𝑚
)
3
× 𝑷𝒎 = (
𝟓
𝟏
)
5
× (
1212.4355
3500
)
3
× 𝟒 × 𝟏𝟎𝟑
= 𝟓, 𝟏𝟗, 𝟔𝟏𝟓. 𝟏𝟓
𝑷𝒑 = 𝟓𝟏𝟗. 𝟔𝟏𝟓 𝒌𝑾

18 me54 turbo machines module 01 question no 1a &1b

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18 me54 turbo machines module 01 question no 1a &1b