Modal 04: Hydraulic Turbines (Question Number 7 a - 7 b & 8a - 8b) i. Definition ii. Classification of Hydraulic Turbines iii. Various efficiencies of Hydraulic Turbines and Various types of Head iv. Pelton Wheel – Principle of working, Velocity triangles, Maximum efficiency Design parameters, Numerical problems. v. Francis turbine – Principle of working Velocity triangles Design parameters Numerical problems vi. Kaplan and Propeller turbines - Principle of working Velocity triangles Design parameters Numerical Problems. vii. Theory and types of Draft tubes.

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 04: Hydraulic Turbines
Course Learning Objectives
Study the various designs of hydraulic turbine based on the working principle.
Course Outcomes
Classify, analyze and understand various type of hydraulic turbine.

2. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 2
Contents
Modal 04: Hydraulic Turbines (Question Number 7 a - 7 b & 8a - 8b)
i. Definition
ii. Classification of Hydraulic Turbines
iii. Various efficiencies of Hydraulic Turbines and Various types of Head
iv. Pelton Wheel – Principle of working,
Velocity triangles,
Maximum efficiency
Design parameters,
Numerical problems.
v. Francis turbine – Principle of working
Velocity triangles
Design parameters
Numerical problems
vi. Kaplan and Propeller turbines - Principle of working
Velocity triangles
Design parameters
Numerical Problems.
vii. Theory and types of Draft tubes.

3. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 3
Hydraulic Turbines
i. Definition
a) Hydraulic turbine is a turbomachine which converts Hydraulic energy into mechanical
energy by dynamic action of water flowing from a high level.
b) Hydraulic turbines are the machines which convert the hydraulic energy in to
mechanical energy.
c) The energy source which does not depend on thermal energy input to produce
mechanical output is hydraulic energy.
d) Hydraulic turbines are the machines which convert the hydraulic energy in to
mechanical energy.
ii. Classification of Hydraulic Turbines
1. Based on the action of water on blades or the energy available at the turbine inlet,
Impulse turbine: In this type of turbine the energy of the fluid entering the rotor is in the form
of kinetic energy of jets. Example: Pelton turbine.
Reaction turbine: In this turbine the energy of the fluid entering the rotor is in the form of
kinetic energy of jets and pressure energy of turbine. Example: Francis turbine and Kaplan
turbine.
2. Based on the direction of fluid flow through the runner
Tangential flow turbine: In this type of turbine water strikes the runner along the tangential
direction, these turbines are also known as peripheral flow turbines. Example: Pelton turbine.
Radial flow turbine: In this type of turbine water flow through the runner along the radial
direction. Example: Francis turbine.
Axial flow turbine: In this type of turbine water flow through the runner along the axial
direction. Example: Kaplan turbine.
Mixed flow turbine: In this type of turbine water enters the runner radially and leaves the
runner axially. Example: Francis turbine.
3. Based on specific speed of runner
Low specific speed turbines: Such turbines have usually high head in the range of 200 m to
1700 m and these machines require low discharge. These turbines have specific speed in the
range of 10 to 30 for single jet and 30 to 50 for double jet. Example: Pelton turbine
Medium specific speed turbines: Such turbines have usually medium head in the range of 50
m to 200 m and these machines require medium discharge. These turbines have specific speed
in the range of 60 to 400. Example: Francis turbine.
High specific speed turbines: Such turbines have usually very low head in the range of 2.5 m
to 50 m and these machines require high discharge. These turbines have specific speed in the
range of 300 to 1000. Example: Kaplan turbine.

4. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 4
iii. Various efficiencies of Hydraulic Turbines
Efficiencies of a turbine
There are following important efficiencies that we will discuss here in this post.
1) Hydraulic Efficiency (𝜼𝒉)
It is defined as the ratio of power developed by the runner to the power supplied by the jet at
entrance to the turbine. (𝑛𝑜𝑡𝑒 𝜌 =
𝑤
𝑔
)
(𝜼𝒉) =
𝑃𝑜𝑤𝑒𝑟 𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑 𝑏𝑦 𝑟𝑢𝑛𝑛𝑒𝑟
𝑃𝑜𝑤𝑒𝑟 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑎𝑡 𝑡ℎ𝑒 𝑖𝑛𝑙𝑒𝑡 𝑜𝑓 𝑡𝑢𝑟𝑏𝑖𝑛𝑒
=
𝜌 𝑄𝑎(𝑉𝑢1±𝑉𝑢2)𝑈
𝑤𝑄𝑎𝐻
=
(𝑉𝑢1±𝑉𝑢2)𝑈
𝑔𝐻
=
𝑯𝒓
𝑯
𝑤ℎ𝑒𝑟𝑒 𝐻𝑟 =
1
𝑔
(𝑉𝑢1 ± 𝑉𝑢2)𝑈 Represents the energy transfer per unit weight of water and is
referred to as the ‘runner head’ or ‘Euler head’
2) Mechanical Efficiency (𝜼𝒎)
It is defined as the ratio of the power obtained from the shaft of the turbine to the power
developed by the runner. These two powers differ by the amount of mechanical losses, viz.,
bearing friction, etc. Mechanical efficiency will be indicated by(𝜼𝒎).
𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 (𝜼𝒎) =
𝑃𝑜𝑤𝑒𝑟 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑠ℎ𝑎𝑓𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑢𝑟𝑏𝑖𝑛𝑒
𝑃𝑜𝑤𝑒𝑟 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑢𝑛𝑛𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑢𝑟𝑏𝑖𝑛𝑒
𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 (𝜼𝒎) =
𝑃
𝜌 𝑄𝑎
(𝑉𝑢1±𝑉𝑢2)𝑈
𝑔
=
𝑷
𝝆 𝑸𝒂𝑯𝒓
3) Volumetric Efficiency (𝜼𝒗)
The volumetric efficiency is the ratio of the volume of water actually striking the runner to the
volume of water supplied by the jet to the turbine.Volumetric efficiency will be indicated
by(𝜼𝒗).
𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 (𝜼𝒗) =
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑤𝑎𝑡𝑒𝑟 𝑎𝑐𝑡𝑢𝑎𝑙𝑙𝑦 𝑠𝑡𝑟𝑖𝑘𝑖𝑛𝑔 𝑡ℎ𝑒 𝑟𝑢𝑛𝑛𝑒𝑟
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑡𝑢𝑟𝑏𝑖𝑛𝑒
=
𝑸𝒂
𝑸
4) Overall Efficiency (𝜼𝒐)
It is defined as the ratio of power available at the turbine shaft to the power supplied by the
water jet. Overall efficiency will be indicated by(𝜼𝒐).
𝑂𝑣𝑒𝑟𝑎𝑙𝑙 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦, (𝜼𝒐) =
𝑃𝑜𝑤𝑒𝑟 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑡𝑢𝑟𝑏𝑖𝑛𝑒 𝑠ℎ𝑎𝑓𝑡
𝑃𝑜𝑤𝑒𝑟 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑤𝑎𝑡𝑒𝑟 𝐽𝑒𝑡
=
𝑷
𝒘 𝑸 𝑯
The values of overall efficiency for a Pelton wheel lie between 0.85 ~ 0.90. The individual
efficiencies may be combined to give
𝑂𝑣𝑒𝑟𝑎𝑙𝑙 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦, (𝜼𝒐) = 𝜼𝒉 × 𝜼𝒎 × 𝜼𝑽 =
𝐻𝑟
𝐻
×
𝑃
𝜌 𝑄𝑎𝐻𝑟
×
𝑄𝑎
𝑄
=
𝑃
𝑤 𝑄 𝐻

5. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 5
Various types of Head
Gross Head
Gross head is basically defined as the difference between
the head race level and tail race level when water is not
flowing. Gross head will be indicated by Hg as displayed
here in following figure.
Net Head
Net head is basically defined as the head available at the inlet of the turbine. Net head is also
simply called as effective head.
When water will flow from head race to the turbine, there will be some losses of head due to
friction between water and penstock. There will also be other losses of head such as loss of
head due to bend, fitting, at entrance of penstock etc. We must note it here that these losses
will be very less and could be neglected when we compare with head loss due to friction.
Net head available at the inlet of turbine could be written as mentioned here.
𝑵𝒆𝒕 𝒉𝒆𝒂𝒅, 𝑯 = 𝑮𝒓𝒐𝒔𝒔 𝒉𝒆𝒂𝒅 (𝑯𝒈)– 𝒉𝒆𝒂𝒅 𝒍𝒐𝒔𝒔 𝒅𝒖𝒆 𝒕𝒐 𝒇𝒓𝒊𝒄𝒕𝒊𝒐𝒏 (𝒉𝒇)
𝒉𝒇 =
𝟒𝒇𝑳𝑽𝒑
𝟐
𝟐𝒈𝑫𝑷
𝒘𝒉𝒆𝒓𝒆 (𝒑) = 𝑷𝒆𝒏𝒔𝒕𝒐𝒄𝒌
iv. Pelton Wheel
Pelton wheel turbine is an impulse turbine, Tangential flow turbine and Low specific speed
Turbine.
Pelton wheel turbine is an impulse turbine working under high head and low discharge. In this
turbine water carried from the penstock enters the nozzle emerging out in the form of high
velocity water jet. The potential energy of water in the penstock is converted in to kinetic
energy by nozzle which is used to run the turbine runner.
Principle of working,
Water flows through these nozzles as a high speed jet striking the vanes or buckets attached to
the periphery of the runner. The runner rotates and supplies mechanical work to the shaft. Water
is discharged at the tail race after doing work on the runner. In a Pelton wheel the jet of water
strikes the bucket and gets deflected by the splitter into two parts, this negates the axial thrust
on the shaft.

6. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 6
 In this type of turbine, the potential and the pressure energy of water is converted to kinetic
energy. A nozzle is used which increases the velocity of water and hence increases the
kinetic energy.
 Pelton wheel is a tangential flow turbine means the water jet will strikes the blade of the
turbine tangentially.
 It is a high head turbine means this turbine is used only in the condition where water is
available at a high head.
 The turbine whose head is very high has very low specific heat. So the turbine is also called
low specific heat turbine. It is also a low discharge turbine.
Construction or Components of Pelton Wheel Turbine:
1. Penstock:
It is a channel or pipeline which controls the flow of water or it also acts as directing medium
for the fluid flow.
2. Nozzle and Spear:
Nozzle: The nozzle is used to increase the kinetic energy of
water which is used to strike the buckets attached to the runner.
Spear: Spear is used to control the quantity of water striking
the buckets. It is a conical needle installed inside the nozzle to
regulate the water flow that is going to strike on the buckets or
vanes of the runner. It is operated by a hand wheel.
The rate of water flow increases and decreases when the spear is moved in a backward direction
and forward direction respectively and that can be handled by means of a hand wheel.
3. Runner and buckets:
The rotating part of the turbine is a runner which is a circular disc and
on the periphery of which a number of buckets are evenly spaced. The
buckets are made of two hemispherical cups joined together. The
splitter acts as a wall joining two hemispherical cups which can splits
the water into two equal parts (i.e. On to the hemispherical cups.) deflected through an angle
of 160 degrees to 170 degrees. The buckets of the Pelton turbine are made up of cast iron, cast
steel bronze or stainless steel.
4. Casing:
The case (outer cover) in which turbine is placed so that water cannot splash outside
(surroundings) called casing.
5. Braking Jet:
To stop the runner in the shortest period of time, a small nozzle is provided which directs a jet
of water at the back of the vanes and that stops the runner of the turbine called as breaking jet.

7. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 7
Velocity triangles of Pelton Wheel
𝐹𝑟𝑜𝑚 𝐼𝑛𝑙𝑒𝑡 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒
𝑽𝟏 = 𝑽𝒖𝟏
𝑉𝑟1 = 𝑉1 − 𝑈
𝐹𝑟𝑜𝑚 𝑜𝑢𝑡𝑙𝑒𝑡 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒
𝑉𝑢2 = 𝑋2 − 𝑈 = 𝑉𝑟2 cos 𝛽2 − 𝑈
𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝐶𝑏 =
𝑉𝑟2
𝑉𝑟1
∴ 𝑉𝑟2 = 𝑉𝑟1. 𝐶𝑏 = (𝑉1 − 𝑈)𝐶𝑏
𝑠𝑜 𝑉𝑢2 = 𝑉𝑟2 cos 𝛽2 − 𝑈 = (𝑉1 − 𝑈)𝐶𝑏 cos 𝛽2 − 𝑈
𝑽𝒖𝟐 = (𝑽𝟏 − 𝑼)𝑪𝒃 𝐜𝐨𝐬 𝜷𝟐 − 𝑼
𝑽𝒖𝟏 + 𝑽𝒖𝟐 = 𝑽𝟏 + (𝑽𝟏 − 𝑼)𝑪𝒃 𝐜𝐨𝐬 𝜷𝟐 − 𝑼
𝑽𝒖𝟏 + 𝑽𝒖𝟐 = (𝑽𝟏 − 𝑼) + (𝑽𝟏 − 𝑼)𝑪𝒃 𝐜𝐨𝐬 𝜷𝟐
𝑽𝒖𝟏 + 𝑽𝒖𝟐 = (𝑽𝟏 − 𝑼)[𝟏 + 𝑪𝒃 𝐜𝐨𝐬 𝜷𝟐]
𝑊𝑒 𝐾𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝑃𝑜𝑤𝑒𝑟 𝑷 = 𝒎𝑼(𝑽𝒖𝟏 + 𝑽𝒖𝟐)
𝑷 = 𝒎𝑼(𝑽𝟏 − 𝑼)[𝟏 + 𝑪𝒃 𝐜𝐨𝐬 𝜷𝟐]
𝜼𝒉 =
𝑷
𝟏
𝟐
𝑽𝟏
𝟐 =
𝟐𝑼(𝑽𝟏−𝑼)[𝟏+𝑪𝒃 𝐜𝐨𝐬 𝜷𝟐]
𝑉1
2 =
𝟐(𝑼𝑽𝟏−𝑼𝟐)
𝑉1
2 [𝟏 + 𝑪𝒃 𝐜𝐨𝐬 𝜷𝟐]
𝜼𝒉 = 𝟐(𝝋 − 𝝋𝟐)[𝟏 + 𝑪𝒃 𝐜𝐨𝐬 𝜷𝟐]
Show that for maximum utilization (maximum efficiency), the speed of the wheel is equal
to half the speed of jet. (VTU, Dec-11)
𝜼𝒉 = 𝟐(𝝋 − 𝝋𝟐)[𝟏 + 𝑪𝒃 𝐜𝐨𝐬 𝜷𝟐]
𝒅𝜼𝒉
𝒅𝝋
= 0
𝒅
𝒅𝝋
[𝟐(𝝋 − 𝝋𝟐)[𝟏 + 𝑪𝒃 𝐜𝐨𝐬 𝜷𝟐]] = 𝟎
𝟐(𝟏 − 𝟐𝝋)[𝟏 + 𝑪𝒃 𝐜𝐨𝐬 𝜷𝟐] = 𝟎
𝑖𝑓 𝝋 =
𝟏
𝟐
𝑡ℎ𝑒𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝟐(𝟏 − 𝟐𝝋)[𝟏 + 𝑪𝒃 𝐜𝐨𝐬 𝜷𝟐] = 𝟎
𝝋 =
𝑼
𝑽𝟏
=
𝟏
𝟐
𝑜𝑟 𝑼 =
𝑽𝟏
𝟐
For maximum utilization (maximum hydraulic efficiency), the speed of the wheel is equal to
half the speed of jet.

8. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 8
Show that the maximum hydraulic efficiency of a Pelton wheel turbine
𝜼𝒉 = 𝟐(𝝋 − 𝝋𝟐)[𝟏 + 𝑪𝒃 𝐜𝐨𝐬 𝜷𝟐]
𝑤ℎ𝑒𝑛 𝝋 =
𝟏
𝟐
𝑡ℎ𝑒 ℎ𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝑜𝑓 𝑃𝑒𝑙𝑡𝑜𝑛 𝑤ℎ𝑒𝑒𝑙 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑚𝑎𝑥𝑖𝑚𝑢𝑚
𝜼𝒉(𝒎𝒂𝒙) = 𝟐 [
𝟏
𝟐
− (
𝟏
𝟐
)
𝟐
] [𝟏 + 𝑪𝒃 𝐜𝐨𝐬 𝜷𝟐]
𝜼𝒉(𝒎𝒂𝒙) = 𝟐 [
𝟏
𝟒
] [𝟏 + 𝑪𝒃 𝐜𝐨𝐬 𝜷𝟐]
𝜼𝒉(𝒎𝒂𝒙) =
[𝟏 + 𝑪𝒃 𝐜𝐨𝐬 𝜷𝟐]
𝟐
𝒊𝒇 𝒕𝒉𝒆 𝒇𝒍𝒐𝒘 𝒊𝒔 𝒇𝒓𝒊𝒄𝒕𝒊𝒐𝒏 𝒍𝒆𝒔𝒔 𝒕𝒉𝒆𝒏 𝑪𝒃 = 𝟏
𝜼𝒉(𝒎𝒂𝒙) =
[𝟏 + 𝐜𝐨𝐬 𝜷𝟐]
𝟐
Design parameters
1. Velocity of jet from the nozzle 𝑽𝟏 = 𝑪𝒗√𝟐𝒈𝑯
Where(𝑪𝒗)is coefficient of velocity for nozzle ranges from 𝟎. 𝟗𝟕 𝒕𝒐 𝟎.𝟗𝟗
2. Tangential velocity of buckets 𝑼 = 𝝋√𝟐𝒈𝑯
Where (𝝋) is speed ratio is varies ranges from 𝟎.𝟒𝟑 𝒕𝒐 𝟎.𝟒𝟖
3. Least diameter of the jet (d):
Total discharge, 𝑸 = 𝒏
𝝅
𝟒
𝒅𝟐
𝑽𝟏 Where (𝒏) is number of jets or nozzles
4. Mean diameter or pitch diameter of buckets (D):
Tangential velocity, 𝑼 =
𝝅𝑫𝑵
𝟔𝟎
5. Angle of deflection usually ranges from 𝟏𝟔𝟓° 𝒕𝒐 𝟏𝟕𝟎° hence vane angle at outlet
𝜷𝟐 = 𝟏𝟖𝟎° − 𝑨𝒏𝒈𝒍𝒆 𝒐𝒇 𝑫𝒊𝒇𝒍𝒆𝒄𝒕𝒊𝒐𝒏
6. Jet ratio 𝒎 =
𝑫
𝒅
it is the ratio of mean diameter of the runner to the minimum diameter of the jet (𝒎)
ranges between 𝟔 𝒕𝒐 𝟑𝟓
7. Minimum number of buckets 𝒁 = 𝟏𝟓 +
𝑫
𝟐𝒅
𝒐𝒓 𝟏𝟓 +
𝒎
𝟐
8. Head loss due to friction in penstock: 𝒉𝒇 =
𝟒𝒇𝑳𝑽𝒑
𝟐
𝟐𝒈𝑫𝒑
Where 𝑫𝒑 is diameter of penstock, 𝑳 is length of the penstock, 𝑽𝒑 is fluid velocity through penstock
and f is friction coefficient for penstock.
9. Width of the bucket𝑩 = 𝟐. 𝟖𝒅 𝒕𝒐 𝟒𝒅; Length of the bucket𝑳 = 𝟐.𝟒𝒅 𝒕𝒐 𝟐.𝟖𝒅 ;
Depth of bucket 𝑻 = 𝟎. 𝟔𝒅 𝒕𝒐 𝟎.𝟗𝟓𝒅
10 Efficiencies
𝛈𝐡 =
𝑼(𝑽𝟏−𝑼)[𝟏+𝑪𝒃 𝐜𝐨𝐬 𝜷𝟐]
𝟏
𝟐
𝐕𝟏𝟐
=
𝟐𝑷
𝐕𝟏𝟐 ;𝛈𝐦 =
𝐏
𝛒 𝑸
(𝐕𝐮𝟏±𝐕𝐮𝟐)𝐔
𝐠
=
𝐏
𝛒 𝑸 𝐅
;𝛈𝐨 =
𝐏
𝛒𝐠𝐐 𝐇
11. Specific Speed of Turbine 𝑵𝒔 =
𝑵√𝑷
𝑯
𝟓
𝟒

9. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 9
Numerical problems.
Modal Question Paper 1
8 b
A single jet Pelton turbine is supplied from a dam 300 m above the center of
nozzle. The diameter of the pipeline supplying the water is 70 cm and 5.6 km
long. The friction coefficient of the pipe is 0.0075. Jet diameter is 10 cm, speed
ratio is 0.47, Cv is 0.97. Outlet vane angle is 15°. The outlet relative velocity of
water is reduced by 15% compared to inlet relative velocity. Assuming
mechanical efficiency as 0.88, find hydraulic power, brake power, hydraulic
and overall efficiency of the system.
10
𝑯𝒈 = 𝟑𝟎𝟎𝒎, 𝑫𝒑 = 𝟎. 𝟕𝒎, 𝑳 = 𝟓𝟔𝟎𝟎𝒎, 𝒇 = 𝟎. 𝟎𝟎𝟕𝟓, 𝒅 = 𝟎. 𝟏𝒎, 𝝋 = 𝟎. 𝟒𝟕,
𝑪𝒗 = 𝟎. 𝟗𝟕, 𝜷𝟐 = 𝟏𝟓°, 𝑽𝒓𝟐 − 𝟏𝟓% , 𝛈𝐦
= 𝟎. 𝟖𝟖, 𝛈𝐡
=? , 𝑷 =? , 𝛈𝐨
=?
𝑸𝒑𝒆𝒏𝒔𝒕𝒐𝒄𝒌 = 𝑸𝒏𝒐𝒛𝒛𝒍𝒆
𝝅
𝟒
𝑫𝒑
𝟐
𝑽𝑷 =
𝝅
𝟒
𝒅𝟐
𝑽𝟏
𝑉𝑃 = [
𝑑
𝐷𝑝
]
2
𝑉1 = [
0.1
0.7
]
2
𝑉1 ∴ 𝑽𝑷 = 𝟎.𝟎𝟐𝟎 × 𝑽𝟏
𝑵𝒆𝒕 𝒉𝒆𝒂𝒅, 𝑯 = 𝑮𝒓𝒐𝒔𝒔 𝒉𝒆𝒂𝒅 (𝑯𝒈)– 𝒉𝒆𝒂𝒅 𝒍𝒐𝒔𝒔 𝒅𝒖𝒆 𝒕𝒐 𝒇𝒓𝒊𝒄𝒕𝒊𝒐𝒏 (𝒉𝒇)
(𝑯𝒈) = (𝑯) + (𝒉𝒇)
∴ 𝑯𝒈 =
𝑽𝟏𝟐
𝟐𝒈
+
𝟒𝒇𝑳𝑽𝒑
𝟐
𝟐𝒈𝑫𝒑
=
𝑉12
2𝑔
+
4𝑓𝐿(0.020 × 𝑉1)2
2𝑔𝐷𝑝
=
𝑉12
2𝑔
+
0.0672(𝑉1)2
2𝑔(0.7)
𝐻𝑔 =
𝑉12
2𝑔
[1 + 0.096] =
𝑉12
2𝑔
[1.096] = 300
∴ 𝑽𝟏 = √
(𝟑𝟎𝟎)𝟐𝒈
𝟏. 𝟎𝟗𝟔
= 𝟕𝟑.𝟐𝟖 𝒎/𝒔 = 𝑽𝒖𝟏
𝑸 =
𝝅
𝟒
𝒅𝟐
𝑽𝟏 =
𝝅
𝟒
(𝟎. 𝟏)𝟐
× 𝟕𝟎. 𝟒𝟖; 𝑸 = 𝟎. 𝟓𝟕𝟓𝟓 𝒎𝟑
/𝒔
𝑽𝟏 = 𝑪𝒗√𝟐𝒈𝑯 ∴ 𝐻 =
(
𝑉1
𝐶𝑣
)
2
2𝑔
∴ 𝑯 = 𝟐𝟗𝟎.𝟖𝟖 𝒎
𝑼 = 𝝋√𝟐𝒈𝑯 = 𝟎.𝟒𝟕√𝟐 𝒈 × 𝟐𝟔𝟗.𝟎𝟖 = 𝟑𝟓.𝟓𝟎 𝒎/𝒔
𝑽𝒓𝟏 = 𝑽𝟏 − 𝑼 = 73.28 − 35.50 = 𝟑𝟕. 𝟕𝟕 𝒎/𝒔
𝑽𝒓𝟐 = 𝟎. 𝟖𝟓 × 𝑽𝒓𝟏 = 𝟎. 𝟖𝟓 × 𝟑𝟕. 𝟕𝟕 = 𝟑𝟐. 𝟏𝟎 𝒎/𝒔
𝑽𝒖𝟐 = 𝑼 − 𝑽𝒓𝟐 𝒄𝒐𝒔𝜷𝟐 = 𝟑𝟓. 𝟓𝟎 − 𝟑𝟐. 𝟏𝟎 × 𝒄𝒐𝒔𝟏𝟓
𝑽𝒖𝟐 = 𝟒. 𝟒𝟖𝟔 𝒎/𝒔
(𝛈𝐦) =
𝐏
𝛒 𝑸
(𝐕𝐮𝟏 ± 𝐕𝐮𝟐)𝐔
𝐠
≫ 𝐏 = (𝛈𝐦)𝛒 𝑸
(𝐕𝐮𝟏 ± 𝐕𝐮𝟐)𝐔
𝐠
𝐏 = (𝟎.𝟖𝟖)𝟗.𝟖𝟏(𝟎.𝟓𝟕𝟓𝟓)
(𝟕𝟑.𝟐𝟖 + 𝟒.𝟒𝟖𝟔)𝟑𝟓.𝟓𝟎
𝟗. 𝟖𝟏
= 𝟏𝟑𝟗𝟖.𝟏𝟑 𝒌𝑾
(𝛈𝐡) =
𝑼(𝑽𝟏 − 𝑼)[𝟏 + 𝑪𝒃 𝐜𝐨𝐬𝜷𝟐]
𝟏
𝟐 𝐕𝟏𝟐
=
𝑷
𝟏
𝟐 𝐕𝟏𝟐
=
𝟏𝟑𝟗𝟖.𝟏𝟑
𝟏
𝟐 𝟕𝟑.𝟐𝟖𝟐
= 𝟎.𝟓𝟐𝟎 ≈ 𝟓𝟐%
(𝛈𝐨) =
𝐏
𝛒𝐠𝐐 𝐇
=
𝟏𝟑𝟗𝟖.𝟏𝟑 × 𝟏𝟎𝟎𝟎
𝟏𝟎𝟎𝟎 × 𝟗.𝟖𝟏(𝟎.𝟓𝟕𝟓𝟓)(𝟐𝟗𝟎. 𝟖𝟖)
= 𝟎. 𝟖𝟓 ≈ 𝟖𝟓%

10. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 10
Modal Question Paper 2
7 c
Pelton wheel has to be designed for the following data: Power to be developed
= 5880 kW, Net head available = 300m, Speed = 550 RPM, ratio of jet diameter
to wheel diameter = 1/10 and overall efficiency = 85%. Find the number of jets,
Diameter of jet, diameter of the wheel and the quantity of water required.
Assume , 𝑪𝒗 = 0.98, , 𝝋 = 0.46.
8
𝑷 = 𝟓𝟖𝟖𝟎 𝒌𝑾, 𝑯 = 𝟑𝟎𝟎𝒎, 𝑵 = 𝟓𝟓𝟎 𝒓𝒑𝒎,
𝟏
𝒎
=
𝒅
𝑫
=
𝟏
𝟏𝟎
, 𝛈𝐨
= 𝟎. 𝟖𝟓, 𝒏 =? , 𝑫 =? , 𝑸 =?
𝑪𝒗 = 𝟎. 𝟗𝟖, 𝝋 = 𝟎. 𝟒𝟔,
(𝛈𝐨) =
𝐏
𝛒𝐠𝐐 𝐇
≫ 𝐐 =
𝐏
𝛒 𝐠 (𝛈𝐨) 𝐇
=
𝟓𝟖𝟖𝟎 × 𝟏𝟎𝟎𝟎
𝟏𝟎𝟎𝟎 × 𝟗.𝟖𝟏 × 𝟎.𝟖𝟓 × 𝟑𝟎𝟎
= 𝟐. 𝟑𝟓
𝒎𝟑
𝒔
𝒘𝒂𝒕𝒆𝒓 𝑸𝒖𝒂𝒏𝒕𝒊𝒕𝒚
𝑽𝟏 = 𝑪𝒗√𝟐𝒈𝑯 = 𝟎.𝟗𝟖√𝟐 × 𝟗.𝟖𝟏 × 𝟑𝟎𝟎 = 𝟕𝟓.𝟏𝟖 𝒎/𝒔
𝑼 = 𝝋√𝟐𝒈𝑯 = 𝟎.𝟒𝟔√𝟐 × 𝟗.𝟖𝟏 × 𝟑𝟎𝟎 = 𝟑𝟓.𝟐𝟗 𝒎/𝒔
𝑼 =
𝝅𝑫𝑵
𝟔𝟎
≫ 𝑫 =
𝑼 × 𝟔𝟎
𝝅 × 𝑵
= 𝟏. 𝟐𝟐𝟓 𝒎
𝒅
𝑫
=
𝟏
𝟏𝟎
≫ 𝒅 =
𝟏.𝟐𝟐𝟓
𝟏𝟎
= 𝟎. 𝟏𝟐𝟐𝟓 𝒎 𝑱𝒆𝒕 𝑫𝒊𝒂𝒎𝒆𝒕𝒆𝒓
𝑸 = 𝒏
𝝅
𝟒
𝒅𝟐
𝑽𝟏 ≫ 𝒏 =
𝑸 × 𝟒
𝝅𝒅𝟐𝑽𝟏
=
𝟐. 𝟑𝟓 × 𝟒
𝝅 (𝟎.𝟏𝟐𝟐𝟓)𝟐 × 𝟕𝟓.𝟏𝟖
= 𝟐.𝟔𝟓𝟐 ≈ 𝟑𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒋𝒆𝒕𝒔
Jan / Feb 2021
7 c
A Pelton wheel is to be designed for the following specifications: Shaft
power =11772kW, Head = 380m. Speed = 750rmp, Overall efficiency =
86%, jet diameter not to exceed 1/6 of wheel diameter, Determine: i)
Wheel diameter ii) jet diameter iii) Number of jets required. Take C v =
0.98, 𝝋 = 0.46.
6
𝑷 = 𝟏𝟏𝟕𝟕𝟐 𝒌𝑾, 𝑯 = 𝟑𝟖𝟎𝒎, 𝑵 = 𝟕𝟓𝟎, 𝛈𝐨 = 𝟎. 𝟖𝟔, 𝒅 =
𝟏
𝟔
𝑫, 𝑪𝒗 = 𝟎. 𝟗𝟖, 𝝋 = 𝟎. 𝟒𝟔,
(𝛈𝐨) =
𝐏
𝛒𝐠𝐐 𝐇
≫ 𝐐 =
𝐏
𝛒 𝐠 (𝛈𝐨) 𝐇
=
𝟏𝟏𝟕𝟕𝟐 × 𝟏𝟎𝟎𝟎
𝟏𝟎𝟎𝟎 × 𝟗. 𝟖𝟏 × 𝟎.𝟖𝟔 × 𝟑𝟖𝟎
= 𝟑. 𝟔𝟕 𝒎𝟑
/𝒔
𝑽𝟏 = 𝑪𝒗√𝟐𝒈𝑯 = 𝟎.𝟗𝟖√𝟐 × 𝟗.𝟖𝟏 × 𝟑𝟖𝟎 = 𝟖𝟒.𝟔𝟏 𝒎/𝒔
𝑼 = 𝝋√𝟐𝒈𝑯 = 𝟎.𝟒𝟔√𝟐 × 𝟗.𝟖𝟏 × 𝟑𝟖𝟎 = 𝟑𝟗.𝟕𝟏 𝒎/𝒔
𝑼 =
𝝅𝑫𝑵
𝟔𝟎
≫ 𝑫 =
𝑼 × 𝟔𝟎
𝝅 × 𝑵
= 𝟏. 𝟎𝟏 𝒎
𝒅
𝑫
=
𝟏
𝟔
≫ 𝒅 =
𝟏.𝟎𝟏
𝟔
= 𝟎.𝟏𝟔𝟖 𝒎 𝑱𝒆𝒕 𝑫𝒊𝒂𝒎𝒆𝒕𝒆𝒓
𝑸 = 𝒏
𝝅
𝟒
𝒅𝟐
𝑽𝟏 ≫ 𝒏 =
𝑸 × 𝟒
𝝅𝒅𝟐𝑽𝟏
=
𝟑.𝟔𝟕 × 𝟒
𝝅 (𝟎.𝟏𝟔𝟖)𝟐 × 𝟖𝟒.𝟔𝟏
= 𝟏. 𝟗𝟓𝟔 ≈ 𝟐 𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒋𝒆𝒕𝒔

11. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 11
Example Problem
A single jet impulse turbine of 10 MW capacity is to work under a head of 500m. If the specific
speed = 10, overall efficiency = 0.8 and the coefficient of velocity = 0.98, find the diameter of the jet
and bucket wheel. Assume 𝝋 = 0.46.
𝑷 = 𝟏𝟎, 𝟎𝟎𝟎 𝒌𝑾, 𝑯 = 𝟓𝟎𝟎𝒎, 𝑵𝑺 = 𝟏𝟎, 𝛈𝐨 = 𝟎. 𝟖, 𝑪𝒗 = 𝟎. 𝟗𝟖, 𝒅 = ? , 𝑫 = ? , 𝝋 = 𝟎. 𝟒𝟔,
𝑵𝒔 =
𝑵√𝑷
𝑯
𝟓
𝟒
≫ 𝑵 =
𝑵𝒔 × 𝑯
𝟓
𝟒
√𝑷
=
𝟏𝟎 × 𝟓𝟎𝟎
𝟓
𝟒
√𝟏𝟎𝟎𝟎𝟎
= 𝟐𝟑𝟔.𝟒𝟑 𝒓𝒑𝒎
𝑽𝟏 = 𝑪𝒗√𝟐𝒈𝑯 = 𝟎. 𝟗𝟖√𝟐 × 𝟗. 𝟖𝟏 × 𝟓𝟎𝟎 = 𝟗𝟕.𝟎𝟔 𝒎/𝒔
𝑼 = 𝝋√𝟐𝒈𝑯 = 𝟎. 𝟒𝟔√𝟐 × 𝟗. 𝟖𝟏 × 𝟓𝟎𝟎 = 𝟒𝟓. 𝟓𝟔 𝒎/𝒔
𝑼 =
𝝅𝑫𝑵
𝟔𝟎
≫ 𝑫 =
𝑼 × 𝟔𝟎
𝝅 × 𝑵
=
𝟒𝟓.𝟓𝟔 × 𝟔𝟎
𝝅 × 𝟐𝟑𝟔.𝟒𝟑
= 𝟑.𝟔𝟖 𝒎
(𝛈𝐨) =
𝐏
𝛒𝐠𝐐 𝐇
≫ 𝑸 =
𝑷
(𝛈𝐨)𝛒𝐠𝐇
=
𝟏𝟎𝟎𝟎𝟎 × 𝟏𝟎𝟎𝟎
(𝟎. 𝟖𝟎)𝟏𝟎𝟎𝟎 × 𝟗. 𝟖𝟏 × 𝟓𝟎𝟎
= 𝟐.𝟓𝟒 𝒎𝟑
/𝒔
𝑸 =
𝝅
𝟒
𝒅𝟐
𝑽𝟏 ≫ 𝒅 = √
𝑸 × 𝟒
𝝅 × 𝑽𝟏
= √
𝟐. 𝟓𝟒 × 𝟒
𝝅 × 𝟗𝟕. 𝟎𝟔
= 𝟎. 𝟏𝟖𝟐𝟖 ≈ 𝟎. 𝟏𝟖𝟑 𝒎 𝑱𝒆𝒕 𝑫𝒊𝒂𝒎𝒆𝒕𝒆𝒓
Dec18/Jan19
6 b A Pelton wheel produces 15500kW under a head of 350m at 500rpm. If the overall
efficiency of the wheel is 84%. Find i) Required number of jets and diameter of
each jet ii) Number of buckets iii) Tangential force exerted Assume : jet ratio as
𝒎 = 𝑫/𝒅 = 𝟗. 𝟓, 𝜽 = 𝟏𝟔𝟎º 𝝋 = 𝟎. 𝟒𝟔
10
𝑷 = 𝟏𝟓𝟓𝟎𝟎 𝒌𝑾,
𝑯 = 𝟑𝟓𝟎 𝒎, 𝑵 = 𝟓𝟎𝟎𝒓𝒑𝒎, 𝛈𝐨 = 𝟎. 𝟖𝟒,
𝒇𝒊𝒏𝒅 𝒏 =? , 𝒅 = ? , 𝒁 =? , 𝑭𝒙 = ?,
𝜷𝟐 = 𝟏𝟖𝟎 − 𝜽 = 𝟏𝟖𝟎 − 𝟏𝟔𝟎 = 𝟐𝟎°
(𝛈𝐨) =
𝐏
𝛒𝐠𝐐 𝐇
≫ 𝑸 =
𝑷
(𝛈𝐨)𝛒𝐠𝐇
=
𝟏𝟓𝟓𝟎𝟎 × 𝟏𝟎𝟎𝟎
(𝟎.𝟖𝟒)𝟏𝟎𝟎𝟎 × 𝟗.𝟖𝟏 × 𝟑𝟓𝟎
= 𝟓. 𝟑𝟕 𝒎𝟑
/𝒔
𝑨𝒔𝒔𝒖𝒎𝒆 𝑪𝒗 𝒂𝒔 𝟎.𝟗𝟓 ~𝟏.𝟎
𝑽𝟏 = 𝑪𝒗√𝟐𝒈𝑯 = 𝟏√𝟐 × 𝟗. 𝟖𝟏 × 𝟑𝟓𝟎 = 𝟖𝟐.𝟖𝟔 𝒎/𝒔 𝑼 = 𝝋√𝟐𝒈𝑯 = 𝟎.𝟒𝟔√𝟐 × 𝟗.𝟖𝟏 × 𝟑𝟓𝟎
= 𝟑𝟖.𝟏𝟏 𝒎/𝒔
𝑼 =
𝝅𝑫𝑵
𝟔𝟎
≫ 𝑫 =
𝑼 × 𝟔𝟎
𝝅 × 𝑵
=
𝟑𝟖.𝟏𝟏 × 𝟔𝟎
𝝅 × 𝟓𝟎𝟎
= 𝟏. 𝟒𝟓𝟓 𝒎
𝒎 =
𝑫
𝒅
= 𝟗. 𝟓 ≫ 𝒅 =
𝑫
𝟗. 𝟓
=
𝟏. 𝟒𝟓𝟓
𝟗. 𝟓
= 𝟎. 𝟏𝟓𝟑 𝒎 𝑱𝒆𝒕 𝑫𝒊𝒂𝒎𝒆𝒕𝒆𝒓
𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑢𝑐𝑘𝑒𝑡𝑠 𝒁 = 𝟏𝟓 +
𝑫
𝟐𝒅
𝒐𝒓 𝟏𝟓 +
𝒎
𝟐
𝒁 = 𝟏𝟓 +
𝑫
𝟐𝒅
= 𝟏𝟓 +
𝟏. 𝟒𝟓𝟓
𝟐 × 𝟎.𝟏𝟓𝟑
= 𝟏𝟗.𝟕𝟓 ≈ 𝟐𝟎 𝑩𝒖𝒄𝒌𝒆𝒕𝒔 𝒓𝒆𝒒𝒖𝒊𝒓𝒆𝒅
𝑸 = 𝒏
𝝅
𝟒
𝒅𝟐
𝑽𝟏 ≫ 𝒏 =
𝑸 × 𝟒
𝝅𝒅𝟐𝑽𝟏
=
𝟓. 𝟑𝟕 × 𝟒
𝝅 (𝟎.𝟏𝟓𝟑)𝟐 × 𝟖𝟐.𝟖𝟔
= 𝟑.𝟓𝟐 ≈ 𝟒 𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒋𝒆𝒕𝒔
𝑻𝒂𝒏𝒋𝒆𝒏𝒕𝒊𝒂𝒍 𝒇𝒐𝒓𝒄𝒆 𝑭𝒙 = 𝒎(𝐕𝐮𝟏 + 𝐕𝐮𝟐)
𝐕𝐮𝟏 = 𝐕𝟏 = 𝟖𝟐.𝟖𝟔 𝒎/𝒔
𝑽𝒓𝟏 = 𝑽𝟏 − 𝑼 = 𝟖𝟐.𝟖𝟔 − 𝟑𝟖.𝟏𝟏 = 𝟒𝟒.𝟕𝟓
𝐕𝐮𝟐 = 𝐔 − 𝐕𝐫𝟐 𝐜𝐨𝐬 𝛃𝟐 = 𝟑.𝟗𝟒 𝐦/𝐬
𝑨𝒔𝒔𝒖𝒎𝒆 𝑪𝒃 = 𝟏 ∴ 𝑽𝒓𝟐 = 𝑽𝒓𝟏;
𝜷𝟐 = 𝟏𝟖𝟎 − 𝟏𝟔𝟎 = 𝟐𝟎°
𝑻𝒂𝒏𝒋𝒆𝒏𝒕𝒊𝒂𝒍 𝒇𝒐𝒓𝒄𝒆 𝑭𝒙 = 𝒎(𝐕𝐮𝟏 + 𝐕𝐮𝟐) ≫ 𝑭𝒙 = 𝝆 𝑸 (𝐕𝐮𝟏 + 𝐕𝐮𝟐) = 𝟏𝟎𝟎𝟎 × 𝟓.𝟑𝟕(𝟖𝟐.𝟖𝟔 + 𝟑. 𝟗𝟒)
𝑭𝒙 = 𝝆 𝑸 (𝐕𝐮𝟏 + 𝐕𝐮𝟐) = 𝟒𝟔𝟔.𝟏𝟏𝟔 𝒌𝑵

12. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 12
Special Problem
A Pelton wheel is running at a speed of 200 rpm and develops 5200kW of power when
working under a head of 220m with an overall efficiency of 80%. Determine its unit speed,
unit discharge, unit power and specific speed.
𝑷 = 𝟓𝟐𝟎𝟎 𝒌𝑾, 𝑯 = 𝟐𝟐𝟎 𝒎, 𝑵 = 𝟐𝟎𝟎 𝒓𝒑𝒎, 𝛈𝐨 = 𝟎. 𝟖,
(𝛈𝐨) =
𝐏
𝛒𝐠𝐐 𝐇
≫ 𝑸 =
𝑷
(𝛈𝐨)𝛒𝐠𝐇
=
𝟓𝟐𝟎𝟎 × 𝟏𝟎𝟎𝟎
(𝟎.𝟖𝟎)𝟏𝟎𝟎𝟎 × 𝟗.𝟖𝟏 × 𝟐𝟐𝟎
= 𝟑. 𝟎𝟏𝟏 𝒎𝟑
/𝒔
unit speed unit discharge unit power
𝑁𝑢 =
𝑁1
√𝐻1
𝑄𝑢 =
𝑄1
√𝐻1
𝑃𝑢 =
𝑃1
𝐻1
3
2
𝑁𝑢 =
200
√220
= 13.48 𝑄𝑢 =
3.011
√220
= 0.2030 𝑃𝑢 =
5200
220
3
2
= 1.59
𝑵𝒔 =
𝑵√𝑷
𝑯
𝟓
𝟒
=
𝟐𝟎𝟎√𝟓𝟐𝟎𝟎
𝟐𝟐𝟎
𝟓
𝟒
= 𝟏𝟕.𝟎𝟐
Example Problem
A Pelton wheel has a water supply rate of 𝟓 𝒎𝟑
𝒔
⁄ at a head of 256m and runs at 500rpm. Assuming
a turbine efficiency of 0.85, a coefficient of velocity for nozzle as 0.985, speed ratio of 0.46, calculate
(a) the power output, (b) the specific speed.
𝑸 = 𝟓 𝒎𝟑
𝒔
⁄ , 𝑯 = 𝟐𝟓𝟔𝒎, 𝑵 = 𝟓𝟎𝟎𝒓𝒑𝒎, 𝛈𝐨 = 𝟎. 𝟖𝟓, 𝑪𝒗 = 𝟎. 𝟗𝟖𝟓, 𝝋 = 𝟎. 𝟒𝟔, 𝑷 =? , 𝑵𝒔 =?
(𝛈𝐨) =
𝐏
𝛒𝐠𝐐 𝐇
≫ 𝐏 = 𝛒𝐠𝐐 𝐇(𝛈𝐨) = 𝟏𝟎𝟎𝟎 × 𝟗. 𝟖𝟏 × 𝟓 × 𝟐𝟓𝟔 × 𝟎.𝟖𝟓 = 𝟏𝟎𝟔𝟕𝟑𝟐𝟖𝟎 = 𝟏𝟎𝟔𝟕𝟑.𝟐𝟖𝐤𝐖
𝑵𝒔 =
𝑵√𝑷
𝑯
𝟓
𝟒
=
𝟓𝟎𝟎√𝟏𝟎𝟔𝟕𝟑.𝟐𝟖
𝟐𝟓𝟔
𝟓
𝟒
= 𝟓𝟎.𝟒𝟒𝟓
Note: why they have given Cv and 𝝋

13. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 13
v. Francis turbine
Francis turbine is a reaction turbine. These turbines have usually medium head in the range of
50 m to 200 m and these machines require medium discharge, hence the specific speed is
medium in the range of 60 to 400. In this type of turbine water enters radially and leaves axially
or vice versa, these turbines are also known as mixed flow turbines.
Construction
(i) Scroll (spiral) casing: It is also known as spiral casing. The water from penstock enters
the scroll casing which completely surrounds the runner. The main function of spiral
casing is to provide a uniform distribution of water around the runner and hence to provide
constant velocity.
(ii) Guide vanes (blades): After the scroll ring water passes over to the series of guide vanes
or fixed vanes, which surrounds completely around the turbine runner. Guide vanes
regulate the quantity of water entering the runner and direct the water on to the runner.
(iii) Runner (Rotor): The runner of turbine is consists of series of curved blades evenly
arranged around the circumference. The vanes or blades are so shaped that water enters
the runner radially at outer periphery and leaves it axially at its center.
(iv) Draft tube: The water from the runner flows to the tail race through the draft tube. A draft
tube is a pipe or passage of gradually increasing area which connect the exit of the runner
to the tail race. The exit end of the draft tube is always submerged below the level of water
in the tail race and must be airtight.
Principle of working
Francis Turbines are generally installed with their axis vertical. Water with high head
(pressure) enters the turbine through the spiral casing surrounding the guide vanes. The water
loses a part of its pressure in the volute (spiral casing) to maintain its speed. Then water passes
through guide vanes where it is directed to strike the blades on the runner at optimum angles.
As the water flows through the runner its pressure and angular momentum reduces. This
reduction imparts reaction on the runner and power is transferred to the turbine shaft.
If the turbine is operating at the design conditions the water leaves the runner in axial direction.
Water exits the turbine through the draft tube, which acts as a diffuser and reduces the exit
velocity of the flow to recover maximum energy from the flowing water In Francis turbine the
pressure and velocity of the fluid decreases as it flows through the moving blades. Hence it
converts both the kinetic energy and pressure energy is converted into work

14. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 14
Velocity triangles
Velocity triangles for Francis turbine: In the slow, medium and fast runners of a Francis turbine
the inlet blade angle (β1) is less than, equal to and greater than 90º respectively. The whirl
component of velocity at the outlet is zero (i.e., Vu2=0).
β1 Inclination Inlet Velocity Triangle Outlet Velocity Triangle
Design Parameters of Francis Turbine:
1. Flow velocity or radial velocity at the turbine inlet is given by, 𝑽𝒇𝟏 = 𝝍√𝟐𝒈𝑯
Where 𝝍 is flow ratio ranging from; 𝟎.𝟏𝟓 𝒕𝒐 𝟎.𝟑𝟎
2. Tangential velocity of the runner or wheel at the inlet is given by, 𝑼𝟏 = 𝝋√𝟐𝒈𝑯
Where 𝝋 is speed ratio ranging from; 𝟎.𝟔 𝒕𝒐 𝟎.𝟗
3. Diameter of runner:
Inlet diameter (𝑫𝟏) of the runner, 𝑼𝟏 =
𝝅𝑫𝟏 𝑵
𝟔𝟎
Outlet diameter (𝑫𝟐) of the runner, 𝑼𝟐 =
𝝅𝑫𝟐 𝑵
𝟔𝟎
Where 𝑼𝟏 and 𝑼𝟐 are inlet and outlet runner velocity respectively
4. Discharge at the outlet is radial then the guide blade angle at the outlet is 𝟗𝟎°.
i.e. 𝜶𝟐 = 𝟗𝟎° 𝒂𝒏𝒅 𝑽𝒖𝟐 = 𝟎
5. Head at the turbine inlet assuming no energy loss is given by,
𝒈𝑯 = (𝑼𝟏 𝑽𝒖𝟏 + 𝑼𝟐 𝑽𝒖𝟐)𝑼 +
𝑽𝟐
𝟐
𝟐
𝒐𝒓 𝑯 =
𝟏
𝒈
[(𝑼𝟏 𝑽𝒖𝟏 + 𝑼𝟐 𝑽𝒖𝟐)𝑼 +
𝑽𝟐
𝟐
𝟐
]
6. Discharge through the turbine is given by, 𝑸 = 𝑨𝒇. 𝑽𝒇 = 𝝅𝑫𝟏𝑩𝟏𝑽𝒇𝟏 = 𝝅𝑫𝟐𝑩𝟐𝑽𝒇𝟐
Where: 𝑨𝒇, is area of flow through the runner, 𝑫 is diameter of the runner, 𝑩 is width of the
runner and 𝑽𝒇 is flow velocity.
If (𝒏) is the number of vanes in the runner and (𝒕) is the thickness of the vane, then
𝑸 = (𝝅𝑫𝟏 − 𝒏𝒕𝟏)𝑩𝟏𝑽𝒇𝟏 = (𝝅𝑫𝟐 − 𝒏𝒕𝟐)𝑩𝟐𝑽𝒇𝟐
Normally it is assumed that, 𝑫𝟏 = 𝟐𝑫𝟐,𝑽𝒇𝟏 = 𝑽𝒇𝟐,𝑩𝟐 = 𝟐𝑩𝟏
7. Ratio of width to diameter is given by, 𝒓 =
𝑩𝟏
𝑫𝟏
ranging from; 𝟎.𝟏𝟎 𝒕𝒐 𝟎.𝟑𝟖.
𝜼𝒉 =
𝒎(𝑽𝒖𝟏 𝑼𝟏)
𝝎𝑸𝑯
=
(𝑽𝒖𝟏 𝑼𝟏)
𝑯
, 𝜼𝒎 =
𝑷
𝝎.𝑸.∆𝑽𝒖.𝑼
, 𝜼𝒐 =
𝑷(𝟏𝟎𝟎𝟎)
𝝆.𝒈.𝑸.𝑯
,

15. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 15
Numerical problems
Modal Question Paper 1
In a radial inward flow turbine, the runner outer diameter is 1.05 m and the inner diameter is 0.5 m. The
runner speed is 400 rpm. Liquid water enters the wheel at a speed of 25 m/s at an angle of 15° to the wheel
tangent at the entry. The discharge at the outlet is radial and absolute velocity is 5 m/s. Find the runner
blade angles at the inlet. Draw the velocity triangles. What is the power output per unit mass flow of water
through the blade? Find also the degree of reaction, utilization factor and the static pressure at the turbine
inlet, if the static pressure at the exit is 1 bar.
(𝑫𝟐) = 𝟏.𝟎𝟓𝒎,
(𝑫𝟏) = 𝟎. 𝟓𝒎,
𝑵 = 𝟒𝟎𝟎𝒓𝒑𝒎, 𝑽𝟏 = 𝟐𝟓𝒎 𝒔
⁄ , 𝜶𝟏 = 𝟏𝟓°,
𝒓𝒂𝒅𝒊𝒂𝒍 𝒊𝒏𝒘𝒂𝒓𝒅 𝒇𝒍𝒐𝒘 𝒕𝒖𝒓𝒃𝒊𝒏𝒆, 𝜷𝟏 = 𝟗𝟎°
𝑫𝒊𝒔𝒄𝒉𝒂𝒓𝒈𝒆 𝒂𝒕 𝒐𝒖𝒕𝒍𝒆𝒕 = 𝑹𝒂𝒅𝒊𝒂𝒍
∴ 𝜶𝟐 = 𝟗𝟎°,𝑽𝒖𝟐 = 𝟎 𝒂𝒏𝒅 𝑽𝒇𝟐 = 𝑽𝟐 = 𝟓 𝒎 𝒔
⁄
Inlet diameter (𝑫𝟏) of the runner,
𝑼𝟏 =
𝝅𝑫𝟏 𝑵
𝟔𝟎
=
𝝅 ×𝟎.𝟓×𝟒𝟎𝟎
𝟔𝟎
= 𝟏𝟎.𝟒𝟕𝒎/𝒔
Outlet diameter (𝑫𝟐) of the runner,
𝑼𝟐 =
𝝅𝑫𝟐 𝑵
𝟔𝟎
=
𝝅×𝟏.𝟎𝟓×𝟒𝟎𝟎
𝟔𝟎
= 𝟐𝟏.𝟗𝟗𝒎/𝒔
𝑐𝑜𝑠𝛼1 =
𝑉𝑢1
𝑉1
≫ 𝑉1 =
25
𝑐𝑜𝑠15
∴ 𝑽𝟏 = 𝟏𝟎. 𝟖𝟑 𝒎/𝒔
𝑝𝑜𝑤𝑒𝑟 𝑜𝑢𝑡𝑝𝑢𝑡 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑚𝑎𝑠𝑠
𝑃 = 𝑚(𝑉𝑢1𝑈1)
𝑃 = 1(10.47 × 10.47)
𝑃 = 𝟏𝟎𝟗. 𝟔𝟐 𝒌𝑾
𝑑𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑹 =
𝑷 −
𝟏
𝟐
(𝑽𝟏
𝟐
− 𝑽𝟐
𝟐
)
𝑷
𝑹 =
𝟏𝟎𝟗.𝟔𝟐−
𝟏
𝟐
(𝟏𝟎.𝟖𝟑𝟐
−𝟓𝟐
)
𝟏𝟎𝟗.𝟔𝟐
𝑹 = 𝟎. 𝟓𝟕
𝑼𝒕𝒊𝒍𝒊𝒛𝒂𝒕𝒊𝒐𝒏 𝑭𝒂𝒄𝒕𝒐𝒓 (𝝐) =
𝑷
𝑷 +
𝟏
𝟐
(𝑽𝟐
𝟐)
=
𝟏𝟎𝟗.𝟔𝟐
𝟏𝟎𝟗. 𝟔𝟐 +
𝟏
𝟐
(𝟓𝟐
)
𝝐 = 𝟎. 𝟐𝟓
𝑡𝑎𝑛𝛽2 =
𝑉2
𝑈2
≫ 𝛽2 = 𝑡𝑎𝑛−1
(
𝑉2
𝑈2
) = 𝑡𝑎𝑛−1
(
5
21.99
) ∴ 𝜷𝟐 = 𝟏𝟐. 𝟖𝟎°

16. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 16
Modal Question Paper 1
In a vertical shaft inward flow reaction turbine, the sum of the pressure and kinetic head at the entrance to
the spiral casing is 120 m and the vertical distance between this section and the tail race level is 3 m. The
peripheral velocity of the runner at entry is 30 m/s, and the radial velocity of flow of water is constant at 9
m/s and the discharge from the runner is without swirl. The estimated hydraulic losses are (a) between
turbine entrance and exit from the guide vanes 4.8 m (b) in the runner 8.8 m (c) in the draft tube 0.79 m
(d) kinetic head rejected to the tail race 0.46 m. Calculate the guide vane angle and the runner blade angle
at the inlet and the pressure heads at entry to and exit from the runner.
𝑼𝟏 = 𝟑𝟎 𝒎/𝒔,
𝑽𝒇𝟏 = 𝟗 𝒎/𝒔,
𝒁𝟐 = 𝟑𝒎
𝒁𝒅 = 𝟎. 𝟕𝟗𝒎
𝑽𝒅𝟐
𝟐𝒈
= 𝟎. 𝟒𝟔𝒎
𝑻𝒐𝒕𝒂𝒍 𝒉𝒆𝒂𝒅 𝒂𝒕 𝒆𝒏𝒕𝒓𝒚
𝑯𝑻 = 𝟏𝟐𝟎 + 𝟑 = 𝟏𝟐𝟑 𝒎
𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒉𝒆𝒂𝒅
𝑯𝑳 = 𝟒. 𝟖 + 𝟖. 𝟖 + 𝟎. 𝟕𝟗 + 𝟎. 𝟒𝟔
𝑯𝑳 = 𝟏𝟒. 𝟖𝟓 𝒎
𝑾𝒐𝒓𝒌 𝒆𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕 𝒉𝒆𝒂𝒅 𝑯
= 𝑯𝑻 − 𝑯𝑳
𝑯 = 𝟏𝟎𝟖. 𝟏𝟓 𝒎
𝑾𝒐𝒓𝒌 𝑫𝒐𝒏𝒆 𝒐𝒓 𝑷𝒐𝒘𝒆𝒓 =
(𝑽𝒖𝟏𝑼𝟏)
𝒈
= 𝟏𝟎𝟖. 𝟏𝟓
∴ 𝑽𝒖𝟏 =
𝟏𝟎𝟖. 𝟏𝟓 × 𝒈
𝑼𝟏
= 𝟑𝟓. 𝟑𝟔 𝒎/𝒔
𝑡𝑎𝑛𝛼1 =
𝑉𝑓1
𝑉𝑢1
=
9
35.36
∴ 𝜶𝟏 = 𝟏𝟒. 𝟐𝟕°
𝑡𝑎𝑛𝛽1 =
𝑉𝑓1
𝑉𝑢1 − 𝑈1
=
9
35.36 − 30
∴ 𝜷𝟏 = 𝟓𝟗. 𝟐𝟐°
𝑽𝟏𝟐
= 𝑽𝒖𝟏𝟐
+ 𝑽𝒇𝟏𝟐
≫ 𝑽𝟏𝟐
= 𝟑𝟓. 𝟑𝟔𝟐
+ 𝟗𝟐
= 𝟏𝟑𝟑𝟏. 𝟑𝟑 𝒎𝟐
/𝒔𝟐
𝒑𝒓𝒆𝒔𝒔𝒖𝒓𝒆 𝒉𝒆𝒂𝒅 𝒂𝒕 𝒆𝒏𝒕𝒓𝒚 = 𝒌𝒊𝒏𝒆𝒕𝒊𝒄 𝒉𝒆𝒂𝒅 𝒂𝒕 𝒕𝒉𝒆 𝒆𝒏𝒕𝒓𝒂𝒏𝒄𝒆 − 𝐤𝐢𝐧𝐞𝐭𝐢𝐜 𝐄𝐧𝐞𝐫𝐠𝐲 𝐚𝐭 𝐭𝐡𝐞 𝐞𝐧𝐭𝐫𝐚𝐧𝐜𝐞 − 𝐡𝐲𝐝𝐫𝐚𝐮𝐥𝐢𝐜 𝐥𝐨𝐬𝐬𝐞𝐬
𝒑𝒓𝒆𝒔𝒔𝒖𝒓𝒆 𝒉𝒆𝒂𝒅 𝒂𝒕 𝒆𝒏𝒕𝒓𝒚 = 𝟏𝟐𝟎 −
𝟏
𝟐𝒈
(𝑽𝟏𝟐) − 𝟒. 𝟖 = 𝟒𝟕. 𝟑𝟒 𝒎
𝑭𝒓𝒐𝒎 𝒐𝒖𝒕𝒍𝒆𝒕 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒕𝒓𝒊𝒂𝒏𝒈𝒍𝒆 𝑽𝟐 = 𝑽𝒇 𝟐 = 𝑽𝒇 𝟏 = 𝟗 𝒎/𝒔
(𝒔𝒊𝒏𝒄𝒆 𝒕𝒉𝒆 𝒇𝒍𝒐𝒘 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒊𝒔 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕)
𝑳𝒆𝒕 𝒑𝟐 𝒃𝒆 𝒕𝒉𝒆 𝒑𝒓𝒆𝒔𝒔𝒖𝒓𝒆 (𝒂𝒃𝒐𝒗𝒆 𝒂𝒕𝒎𝒐𝒔𝒑𝒉𝒆𝒓𝒊𝒄) 𝒂𝒕 𝒆𝒙𝒊𝒕 𝒇𝒓𝒐𝒎 𝒕𝒉𝒆 𝒓𝒖𝒏𝒏𝒆𝒓.𝑨𝒑𝒑𝒍𝒚𝒊𝒏𝒈 𝒆𝒏𝒆𝒓𝒈𝒚
𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝒕𝒉𝒆 𝒊𝒏𝒍𝒆𝒕 𝒂𝒏𝒅 𝒐𝒖𝒕𝒍𝒆𝒕 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒓𝒂𝒇𝒕 𝒕𝒖𝒃𝒆, 𝒘𝒆 𝒉𝒂𝒗𝒆
𝑷𝟐
𝝆𝒈
+
𝑽𝟐𝟐
𝟐𝒈
+ 𝟑 = 𝑷𝒅 +
𝑽𝒅𝟐
𝟐𝒈
+ 𝒁𝒅 ≫
𝑃2
𝜌𝑔
+
92
2 × 9.81
+ 3 = 0 + 0.46 + 0.79 ≫
𝑷𝟐
𝝆𝒈
= −𝟓. 𝟖𝟖𝒎

17. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 17
Modal Question Paper 2
An inward flow reaction turbine with radial discharge with an overall efficiency of 80% is required to
develop 147 kW. The head is 8m. Peripheral velocity of the wheel is0.96√2gH, the radial velocity of the
flow is0.36√2gH. The wheel is to make 150RPM and the hydraulic losses in the turbine are 22% of the
available energy. Determine: (i) The angle of the guide blade at inlet, (ii) The wheel vane angle at inlet (iii)
The diameter of the wheel and (iv) The width of the wheel at inlet
𝑷 = 𝟏𝟒𝟕 × 𝟏𝟎𝟑
, 𝜼𝒐 = 𝟎.𝟖𝟎,𝑽𝒇𝟏 = 𝝍√𝟐𝒈𝑯 = 𝟎.𝟑𝟔√𝟐𝒈𝑯,𝑼𝟏 = 𝝋√𝟐𝒈𝑯 = 𝟎. 𝟗𝟔√𝟐𝒈𝑯 ,
𝑵 = 𝟏𝟓𝟎𝟎,
𝑯𝒍𝒐𝒔𝒔 = 𝟎.𝟐𝟐 × 𝑯, 𝑫𝒆𝒕𝒆𝒓𝒎𝒊𝒏𝒆: 𝜶𝟏,𝜷𝟏, 𝜷𝟐,𝑫𝟐 & 𝑩𝟏
𝜼𝒉 =
𝑯 − 𝑯𝑳𝒐𝒔𝒔
𝑯
=
𝟖 − (𝟎. 𝟐𝟐 × 𝟖)
𝟖
= 𝟎.𝟕𝟖
𝑼𝟏 = 𝝋√𝟐𝒈𝑯 = 𝟎. 𝟗𝟔√𝟐 × 𝟗. 𝟖𝟏 × 𝟖 = 𝟏𝟐.𝟎𝟑 𝒎/𝒔
𝑽𝒇𝟏 = 𝝍√𝟐𝒈𝑯 = 𝟎.𝟑𝟔√𝟐 × 𝟗.𝟖𝟏 × 𝟖 = 𝟒. 𝟓𝟏 𝒎/𝒔
𝜼𝒉 =
𝑽𝒖𝟏 𝑼𝟏
𝒈𝑯
≫ 𝑽𝒖𝟏 =
𝜼𝒉 𝒈 𝑯
𝑼𝟏
=
𝟎.𝟕𝟖 × 𝟗. 𝟖𝟏 × 𝟖
𝟏𝟐. 𝟎𝟑
∴ 𝑽𝒖𝟏 = 𝟓. 𝟎𝟗𝒎 𝒔
⁄ ≫ 𝑵𝒐𝒕𝒆: 𝑼𝟏 > 𝑽𝒖𝟏
𝒕𝒂𝒏 𝜶𝟏 =
𝑽𝒇𝟏
𝑽𝒖𝟏
=
4.51
5.09
≫
𝛼1 = 𝑡𝑎𝑛−1
(
4.51
5.09
) = 41.54°
𝒕𝒂𝒏 𝜷𝟏 =
𝑽𝒇𝟏
𝑼𝟏−𝑽𝒖𝟏
≫
𝛽1 = 𝑡𝑎𝑛−1
(
4.51
12.03 − 5.09
) = 33.01°
𝑼𝟏 =
𝝅𝑫𝟏 𝑵
𝟔𝟎
≫ 𝑫𝟏 =
𝑼𝟏 × 𝟔𝟎
𝝅 × 𝑵
𝑫𝟏 = 𝟎. 𝟏𝟓𝟑𝟏 𝒎
𝜼𝒐 =
𝑷(𝟏𝟎𝟎𝟎)
𝝆.𝒈.𝑸.𝑯
≫ 𝑸 =
𝑷(𝟏𝟎𝟎𝟎)
𝝆.𝒈. 𝜼𝒐.𝑯
𝑸 =
𝟏𝟒𝟕(𝟏𝟎𝟎𝟎)
𝟏𝟎𝟎𝟎.𝟗.𝟖𝟏.𝟎.𝟖𝟎.𝟖
𝑸 = 𝟐. 𝟑𝟒 𝒎/𝒔
𝑸 = 𝝅𝑫𝟏𝑩𝟏𝑽𝒇𝟏 ≫ 𝑩𝟏 =
𝑸
𝝅𝑫𝟏𝑽𝒇𝟏
=
𝟐. 𝟑𝟒
𝝅 × 𝟎. 𝟏𝟓𝟑𝟏 × 𝟒. 𝟓𝟏
= 𝟏. 𝟎𝟕𝟖 𝒎

18. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 18
Modal Question Paper 2015-2016
A medium Francis runner has a diameter of 75cm and with of 10cm. Water leaves the guide vanes at a
velocity of 16m/s inclined at 25° with the runner periphery. The net head is 20m. The overall and hydraulic
efficiencies are 80% and 90% respectively. Assuming that 8% of the flow area is lost due to the runner
vanes thickness. Calculate the runner vane angle at inlet, power output by the runner and speed of the
machine.
(𝑫𝟏) = 𝟎. 𝟕𝟓𝒎,(𝑩𝟏) = 𝟎.𝟏 𝒎,𝑽𝟏 = 𝟏𝟔 𝒎 𝒔
⁄ , 𝜶𝟏 = 𝟐𝟓°,𝑯 = 𝟐𝟎 𝒎,𝜼𝒐 = 𝟎.𝟖, 𝜼𝒉 = 𝟎. 𝟗𝟎
𝑨𝒔𝒔𝒖𝒎𝒆 𝝋 = 𝟎.𝟕𝟓
𝑼𝟏 = 𝝋√𝟐𝒈𝑯 = 𝟎.𝟕𝟓√𝟐 × 𝟗. 𝟏𝟖 × 𝟐𝟎
𝑼𝟏 = 𝟏𝟒.𝟖𝟓 𝒎/𝒔
𝜼𝒉 =
𝑽𝒖𝟏 𝑼𝟏
𝒈𝑯
≫ 𝑽𝒖𝟏 =
𝜼𝒉 × 𝒈 × 𝑯
𝑼𝟏
𝑽𝒖𝟏 = 𝟏𝟏.𝟖𝟗
𝑵𝒐𝒕𝒆: 𝑼𝟏 > 𝑽𝒖𝟏
𝒕𝒂𝒏 𝜶𝟏 =
𝑽𝒇𝟏
𝑽𝒖𝟏
≫ 𝑉𝑓1 = 𝑡𝑎𝑛𝛼1 × 𝑉𝑢1
𝑽𝒇𝟏 = 𝟓. 𝟓𝟒 𝒎/𝒔
𝒕𝒂𝒏 𝜷𝟏 =
𝑽𝒇𝟏
𝑼𝟏 − 𝑽𝒖𝟏
𝛽1 = 𝑡𝑎𝑛−1
(
5.54
14.85 − 11.89
)
𝜷𝟏 = 𝟔𝟏. 𝟖𝟖°
𝑼𝟏 =
𝝅𝑫𝟏 𝑵
𝟔𝟎
= 𝟏𝟒. 𝟖𝟓
𝑵 =
𝑼𝟏 × 𝟔𝟎
𝝅 𝑫𝟏
=
𝟏𝟒.𝟖𝟓 × 𝟔𝟎
𝝅 × 𝟎.𝟕𝟓
= 𝟑𝟕𝟖.𝟏𝟓 𝒓𝒑𝒎
𝑸 = 𝑨𝒇. 𝑽𝒇 = (𝟏 − %𝒐𝒇 𝑩𝒍𝒐𝒂𝒄𝒌𝒂𝒈𝒆)𝝅𝑫𝟏𝑩𝟏𝑽𝒇𝟏 = (𝟏 − %𝒐𝒇 𝑩𝒍𝒐𝒂𝒄𝒌𝒂𝒈𝒆)𝝅𝑫𝟐𝑩𝟐𝑽𝒇𝟐
𝝅𝑫𝟏𝑩𝟏 = 𝝅 × 𝟎. 𝟕𝟓 × 𝟎. 𝟏 = 𝟎.𝟐𝟑 𝒎𝟐 𝑨𝒇 = (𝟏 − 𝟎.𝟎𝟖)𝟎.𝟐𝟑 = 𝟎. 𝟐𝟏𝟏𝟔 𝒎𝟐
𝑸 = 𝑨𝒇𝟏. 𝑽𝒇𝟏 = 𝟎. 𝟐𝟏𝟏𝟔 × 𝟓.𝟓𝟒 = 𝟏.𝟏𝟕𝟐𝒎𝟑
/𝒔 𝜼𝒐 =
𝑷(𝟏𝟎𝟎𝟎)
𝝆. 𝒈. 𝑸. 𝑯
≫ 𝑷 =
𝜼𝒐. 𝝆. 𝒈. 𝑸. 𝑯
𝟏𝟎𝟎𝟎
𝑷 =
𝜼𝒐. 𝝆. 𝒈. 𝑸. 𝑯
𝟏𝟎𝟎𝟎
=
𝟎. 𝟖 × 𝟏𝟎𝟎𝟎 × 𝟏. 𝟏𝟕𝟐 × 𝟐𝟎
𝟏𝟎𝟎𝟎
= 𝟏𝟖. 𝟕𝟓 𝒌𝑾

19. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 19
Modal Question Paper 2015-2016
An inward flow reaction turbine with a supply of 0.6m3/s under a head of 15m develops 75kw at 400 rpm.
The inner and outer diameter of the runner are 40cm and 65cm respectively. Water leaves the exit of the
turbine at 3m/s calculate the hydraulic efficiency and the inlet blade angles. Assume radial discharge and
width to be constant.
𝑸 = 𝟎.𝟔 𝒎𝟑
𝒔
⁄ , 𝑯 = 𝟏𝟓 𝒎,𝑷 = 𝟕𝟓 𝒌𝑾,𝑵 = 𝟒𝟎𝟎𝒓𝒑𝒎,𝑫𝟏 = 𝟎.𝟒 𝒎, 𝑫𝟐 = 𝟎.𝟔𝟓 𝒎,𝑽𝟐 = 𝟑
𝒎
𝒔
𝜼𝒉 =? , 𝜷𝟏 =?
𝑼𝟏 =
𝝅𝑫𝟏 𝑵
𝟔𝟎
; 𝑼𝟏 =
𝝅 × 𝟎.𝟒 × 𝟒𝟎𝟎
𝟔𝟎
= 𝟖.𝟑𝟏 𝒎/𝒔 𝑼𝟐 =
𝝅𝑫𝟐 𝑵
𝟔𝟎
; 𝑼𝟐 =
𝝅 × 𝟎. 𝟔𝟓 × 𝟒𝟎𝟎
𝟔𝟎
= 𝟏𝟑.𝟔𝟏𝒎/𝒔
𝑷𝒐𝒘𝒆𝒓 = (𝑽𝒖𝟏𝑼𝟏) = 𝟕𝟓𝒌𝑾 ∴ 𝑽𝒖𝟏 = 𝟗. 𝟎𝟐𝟓𝒎 𝒔
⁄
𝑵𝒐𝒕𝒆 𝑽𝒖𝟏 > 𝑼𝟏
𝑽𝒇𝟏 = 𝝍√𝟐𝒈𝑯
Where 𝝍 is flow ratio ranging from; 𝟎.𝟏𝟓 𝒕𝒐 𝟎.𝟑𝟎
𝑽𝒇𝟏 = 𝟎. 𝟐𝟓√𝟐𝒈 × 𝟏𝟓 = 𝟒.𝟐𝟖
𝒎
𝒔
𝒕𝒂𝒏 𝜷𝟏 =
𝑽𝒇𝟏
𝑽𝒖𝟏 − 𝑼𝟏
≫
𝛽1 = 𝑡𝑎𝑛−1
(
𝑽𝒇𝟏
𝑽𝒖𝟏 − 𝑼𝟏
)
𝛽1 = 𝑡𝑎𝑛−1
(
𝟒. 𝟐𝟖
𝟗. 𝟎𝟐𝟓 − 𝟖. 𝟑𝟏
) = 80.51°
𝜼𝒉 =
(𝑽𝒖𝟏 𝑼𝟏)
𝒈𝑯
=
(𝟗. 𝟎𝟐𝟓 × 𝟖. 𝟑𝟏)
𝟗. 𝟖𝟏 × 𝟏𝟓
= 𝟎. 𝟓𝟎𝟗𝟔
𝜼𝒉 = 𝟓𝟏%

20. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 20
vi. Kaplan and Propeller turbines
Principle of working
The Kaplan turbine is an axial flow turbine. In the Kaplan turbine, the water enters and exits
the turbine through the runner’s axis of rotation (axial flow). In simple words, the water
enters and exits the turbine in an axial direction but this water flows in a direction parallel to
the runner’s axis of rotation.
Kaplan turbine works on the principle of the axial flow reaction. In an axial flow turbine,
the fluid moves by the impeller in a direction parallel to the impeller’s axis of rotation.
A Kaplan turbine works in the following way:
 First of all, the water introduces into the volute/scroll casing from the pen-stock.
 As water flows inside the volute casing, guide blades direct the water from the casing
toward the impeller blades. These blades are flexible and may change their position
based on flow requirements.
 As the water enters into the impeller area, it takes a turn of 90o
so that it can strike the
impeller blades in an axial direction.
 When the water strikes the impeller blades, these blades start revolving because of the
water reaction force.
 These blades converts K.E of the water into speed and increase the speed of the water.
 After passing through the impeller blades, the water reaches the draft tube, where the
kinetic and pressure energies of the water reduce.
 This draft tube converts the kinetic energy or speed into pressure energy and increases
the pressure of water.
 When the water pressure increases according to the requirements, the water delivers
into the tailrace.
 The increased pressure of the water rotates the turbine. A generator is coupled with the
turbine shaft.
The components of the Kaplan turbine are given below in detail.
1. Runner or Impeller: The runner has a very vital role in the Kaplan turbine working. The
runner or impeller is a rotating component of the turbine. It provides help for electricity
production. The axial water flow acting on the blades causes the rotation of the impeller,
which further rotates the shaft.
2. Hub: Hub includes in the essential components of the Kaplan turbine. The blades mountain
on the hub of the turbine. It controls the rotation of blades. And blades follow it for their
movement. It connects with the central turbine shaft.

21. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 21
3. Draft Tube: In the case of a Kaplan turbine, the atmospheric pressure is higher than the
pressure at the runner outlet area. Therefore, the fluid from the turbine outlet can’t discharge
directly into the tailrace. Due to this reason, a tube having a progressively rising area uses
to discharge the fluid from the outlet into the tailrace. This increasing area tube is known
as a Draft Tube.
4. Runner Blades: The blades are the key components of the turbine. The Kaplan turbine
blade looks like a propeller. Other axial flow turbines have plane blades, while Kaplan
blades have not plane blades but are of twist shape lengthways so that the water swirls at
the inlet-outlet. When the water strikes these blades, they start rotatory motion, which
further rotates the shaft.
5. Shaft: The one end of the turbine shaft is linked with the turbine runner, while the other
end is linked with the generator coil. As the runner rotates due to the rotation of the blades,
the shaft also rotates, which further transmits its rotation to the generator coil. As the
generator coil rotates, it produces electricity.
6. Guide Blades: The guide vane is a regulating component of the entire turbine. It switches
on and off according to the requirements of power. Guide vanes rotate at a specific angle
to regulate the water flow.
7. Scroll or Volute casing: The entire turbine mechanism is surrounded by a housing called
a scroll casing. The scroll casing reduces the cross-sectional area. First of all, the water
flows from the penstock into the volute casing; after that, it flows into the guide vane area.
Kaplan Turbine Velocity triangles
Design parameters
1. Tangential Speed is constant 𝑼𝟏 = 𝑼𝟐 = 𝑼 = 𝝋 √𝟐𝒈𝑯
2. Flow velocity or radial velocity at the turbine inlet is given by, 𝑽𝒇 = 𝝍√𝟐𝒈𝑯
Where 𝝍 is flow ratio ranging from 0.35 to 0.75
3. Flow velocity is remains constant throughout the runner, 𝑽𝒇𝟏 = 𝑽𝒇𝟐 = 𝑽𝒇
4. Discharge through the runner is given by, 𝑸 =
𝝅
𝟒
(𝑫𝟐
− 𝒅𝟐
)𝑽𝒇
Where (𝑫) is tip diameter or outer diameter of the runner and (𝒅) is hub diameter or boss
diameter of the runner.
5. Discharge at the outlet is axial then the guide blade angle at the outlet is 90o.
i.e.𝜶𝟐 = 𝟗𝟎° 𝒂𝒏𝒅 𝑽𝒖𝟐 = 𝟎
6. Head at the turbine inlet assuming no energy loss is given by, 𝑯 =
𝟏
𝒈
[𝑼(𝑽𝒖𝟏 ± 𝑽𝒖𝟐) +
𝑽𝟐
𝟐
𝟐
]
𝜼𝒉 =
𝒎(𝑽𝒖𝟏 𝑼)
𝝎𝑸𝑯
=
(𝑽𝒖𝟏 𝑼)
𝑯
, 𝜼𝒎 =
𝑷
𝝎. 𝑸. ∆𝑽𝒖. 𝑼
, 𝜼𝒐 =
𝑷(𝟏𝟎𝟎𝟎)
𝝆. 𝒈. 𝑸. 𝑯
,

22. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 22
Numerical Problems.
Model Question Paper 02
Determine the efficiency of a Kaplan turbine developing 2940 kW under a head of 5m. It is provided
with a draft tube with its inlet diameter 3m set at 1.6m above the tail race level. A vacuum pressure
gauge connected to the draft tube inlet indicates a reading of 5m of water. Assume that draft tube
efficiency is 78%.
𝑷 = 𝟐𝟗𝟒𝟎 𝒌𝑾, 𝑯 = 𝟓 𝒎, 𝑯 𝑺 = 𝟏. 𝟔 𝒎,
𝑷𝟐
𝝆𝒈
= − 𝟓 𝒎, 𝜼𝒅𝒓𝒂𝒇𝒕 = 𝟎. 𝟕𝟖, 𝜼𝒐 = ?
𝑷𝟐
𝝆𝒈
+
𝑽𝟐𝟐
𝟐𝒈
+ 𝑯 𝑺 =
𝑷𝟑
𝝆𝒈
+
𝑽𝟑𝟐
𝟐𝒈
+ 𝒉𝒇
𝟑 𝒓𝒆𝒑𝒓𝒆𝒔𝒆𝒏𝒕𝒔 𝑫𝒓𝒂𝒇𝒕 𝒐𝒖𝒕𝒍𝒆𝒕
𝑷𝟐
𝝆𝒈
+
𝑽𝟐𝟐
𝟐𝒈
+ 𝑯 𝑺 = 𝟎 +
𝑽𝟑𝟐
𝟐𝒈
+ 𝟎
𝑷𝟐
𝝆𝒈
+ 𝑯 𝑺 =
𝑽𝟑𝟐
𝟐𝒈
−
𝑽𝟐𝟐
𝟐𝒈
= −
(𝑽𝟐𝟐
− 𝑽𝟑𝟐)
𝟐𝒈
𝑷𝟐
𝝆𝒈
+ 𝑯 𝑺 = −
(𝑽𝟐𝟐
− 𝑽𝟑𝟐)
𝟐𝒈
−𝟓 + 𝟏.𝟔 = −
(𝑽𝟐𝟐
− 𝑽𝟑𝟐)
𝟐𝒈
𝒐𝒓
(𝑽𝟐𝟐
− 𝑽𝟑𝟐)
𝟐𝒈
= 𝟑. 𝟒 𝒎 ∴ (𝑽𝟐𝟐
− 𝑽𝟑𝟐) = 𝟔𝟔. 𝟕𝟎𝟖
, 𝜼𝒅𝒓𝒂𝒇𝒕 =
𝑨𝒄𝒕𝒖𝒂𝒍 𝑷𝒓𝒆𝒔𝒔𝒖𝒓𝒆 𝒈𝒂𝒊𝒏 𝒊𝒏 𝒅𝒓𝒂𝒇𝒕 𝒕𝒖𝒃𝒆
𝑷𝒓𝒆𝒔𝒔𝒖𝒓𝒆 𝒂𝒕 𝑰𝒏𝒍𝒆𝒕 𝒕𝒐 𝒕𝒉𝒆 𝒕𝒖𝒃𝒆
=
(𝑽𝟐𝟐
− 𝑽𝟑𝟐
)
𝟐𝒈
(𝑽𝟐𝟐)
𝟐𝒈
=
𝑽𝟐𝟐
− 𝑽𝟑𝟐
𝑽𝟐𝟐
= 𝟎. 𝟕𝟖
𝑉22
− 𝑉32
𝑉22
= 0.78 ≫ 1 −
𝑉32
𝑉22
= 0.78 ∴
𝑉32
𝑉22
= 0.22 𝑜𝑟 𝑽𝟑𝟐
= 𝟎. 𝟐𝟐 𝑽𝟐𝟐
(𝑉22
− 𝑉32) = 66.708 ∴ 𝑉22(1 − 0.22) = 66.708 ≫ 𝑽𝟐 = 𝟗. 𝟐𝟓 = 𝑽𝒇
𝑸 = 𝑨𝒇𝑽𝒇 =
𝝅
𝟒
(𝒅𝟐)𝑽𝒇 =
𝝅
𝟒
(𝟑𝟐) 𝟗.𝟐𝟓
𝑸 = 𝟔𝟓.𝟑𝟖 𝒎𝟑
/𝒔
𝜼𝒐 =
𝑷(𝟏𝟎𝟎𝟎)
𝝆. 𝒈. 𝑸. 𝑯
=
𝟐𝟗𝟒𝟎(𝟏𝟎𝟎𝟎)
𝟏𝟎𝟎𝟎 × 𝟗. 𝟖𝟏 × 𝟔𝟓. 𝟑𝟖 × 𝟓
𝜼𝒐 = 𝟎. 𝟗𝟏𝟔𝟕 = 𝟗𝟏. 𝟔𝟖%

23. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 23
Model Question Bank 1
A Kaplan turbine produces 80,000 HP (58,800 kW) under a head of 25m which has an overall
efficiency of 90%. Taking the value of speed ratio = 1.6, flow ratio = 0.5 and the hub diameter = 0.35
times the outer diameter. Find the diameter and the speed of the turbine.
𝑷 = 𝟖𝟎, 𝟎𝟎𝟎 𝑯𝑷, 𝑯 = 𝟐𝟓𝒎, 𝜼𝒐 = 𝟎. 𝟗, 𝝋 = 𝟏. 𝟔, 𝝍 = 𝟎. 𝟓,
𝒅
𝑫
= 𝟎. 𝟑𝟓, 𝑭𝒊𝒏𝒅: 𝑫, 𝑵.
𝑘𝑖𝑙𝑜𝑤𝑎𝑡𝑡𝑠 = ℎ𝑝 × 0.7457 = 80000 × 0.7457 = 59.656 𝑘𝑊
𝜼𝒐 =
𝑷(𝟏𝟎𝟎𝟎)
𝝆. 𝒈. 𝑸. 𝑯
≫ 𝑸 =
𝑷(𝟏𝟎𝟎𝟎)
𝝆. 𝒈. 𝜼𝒐. 𝑯
= 𝟐𝟕𝟎. 𝟐𝟕𝒎𝟑
/𝒔
𝑽𝒇 = 𝝍√𝟐𝒈𝑯 = 𝟎.𝟓√𝟐 × 𝟗. 𝟖𝟏 × 𝟐𝟓
𝑽𝒇 = 𝟏𝟏.𝟎𝟕 𝒎/𝒔
𝑸 =
𝝅
𝟒
(𝑫𝟐
− 𝒅𝟐
)𝑽𝒇
𝑸 × 𝟒
𝝅 × 𝑽𝒇
= (𝑫𝟐
− 𝒅𝟐) = 𝟑𝟏.𝟎𝟖
(𝑫𝟐
− 𝒅𝟐) = 𝟑𝟏.𝟎𝟖 ≫ 𝑫𝟐
(𝟏 − (
𝒅
𝑫
)
𝟐
) = 𝟑𝟏.𝟎𝟖 ≫ 𝑫𝟐(𝟏 − (𝟎.𝟑𝟓)𝟐) = 𝟑𝟏.𝟎𝟖
𝑫𝟐(𝟎.𝟖𝟕𝟕𝟓) = 𝟑𝟏.𝟎𝟖 ≫ 𝑫𝟐
=
𝟑𝟏.𝟎𝟖
𝟎.𝟖𝟕𝟕𝟓
≫ 𝑫 = √
𝟑𝟏. 𝟎𝟖
𝟎. 𝟖𝟕𝟕𝟓
= 𝟓. 𝟗𝟓𝟏𝟑𝟔 𝒎
𝑼 = 𝝋 √𝟐𝒈𝑯 = 𝟏. 𝟔√𝟐 × 𝟗.𝟖𝟏 × 𝟐𝟓 = 𝟑𝟓.𝟒𝟑 𝒎/𝒔
𝑼 =
𝝅𝑫 𝑵
𝟔𝟎
≫ 𝑵 =
𝑼 × 𝟔𝟎
𝝅 × 𝑫
=
𝟑𝟓.𝟒𝟑 × 𝟔𝟎
𝝅 × 𝟓. 𝟗𝟓𝟏𝟑𝟔
= 𝟏𝟏𝟑.𝟔𝟗 𝒓𝒑𝒎

24. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 24
Model Question bank C-15
A Kaplan turbine produces 10Mw at a head of 25m. The runner and the hub diameters are 3m and
1.2m respectively. The inlet and outlet velocity triangles are right angles triangles. Calculate the
speed and outlet angles of the guide and runner blades if the hydraulic and overall efficiencies are
96% and 85% respectively.
𝑷 = 𝟏𝟎𝟎𝟎𝟎 𝒌𝑾, 𝑯 = 𝟐𝟓𝒎, 𝑫 = 𝟑 𝒎, 𝒅 = 𝟏. 𝟐 𝒎 , 𝜼𝒐 = 𝟎. 𝟖𝟓, 𝜼𝒉 = 𝟎. 𝟗𝟔
𝜼𝒉 =
(𝑽𝒖𝟏 𝑼)
𝑯
= 𝟎. 𝟗𝟔 (𝑽𝒖𝟏 𝑼) = 𝟎. 𝟗𝟔 × 𝟐𝟓 = 𝟐𝟒
𝒃𝒖𝒕 𝑽𝒖𝟏 = 𝑼 ∴ 𝑽𝒖𝟏 𝑼 = 𝑼𝟐
𝑼𝟐
= 𝟐𝟒; 𝑼 = √𝟐𝟒 = 𝟒. 𝟖𝟗 𝒎/𝒔
𝑼 =
𝝅𝑫 𝑵
𝟔𝟎
≫ 𝑵 =
𝑼 × 𝟔𝟎
𝝅 × 𝑫
= 𝟑𝟏. 𝟏𝟖 𝒓𝒑𝒎
𝜼𝒐 =
𝑷(𝟏𝟎𝟎𝟎)
𝝆. 𝒈. 𝑸. 𝑯
≫ 𝑸 =
𝑷(𝟏𝟎𝟎𝟎)
𝝆. 𝒈. 𝜼𝒐. 𝑯
𝑸 = 𝟒𝟕. 𝟗𝟕 𝒎𝟑
/𝒔
𝑸 =
𝝅
𝟒
(𝑫𝟐
− 𝒅𝟐)𝑽𝒇 ≫ 𝑽𝒇 =
𝑸
𝝅
𝟒
(𝑫𝟐 − 𝒅𝟐)
𝑽𝒇 = 𝟖. 𝟎𝟕𝟗 𝒎/𝒔
𝒕𝒂𝒏 𝜶𝟏 =
𝑽𝒇
𝑽𝒖𝟏
≫ 𝜶𝟏 = tan−1
(
𝑽𝒇
𝑽𝒖𝟏
)
𝜶𝟏 = 𝟓𝟖. 𝟖𝟏°
𝒕𝒂𝒏 𝜷𝟏 =
𝑽𝒇
𝑼
≫ 𝜷𝟏 = tan−1
(
𝑽𝒇
𝑼
)
𝜷𝟏 = 𝟓𝟖. 𝟖𝟏°

25. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 25
July 2018
𝑷 = 𝟑𝟎𝟎𝟎𝟎 𝒌𝑾, 𝑯 = 𝟗. 𝟔 𝒎, 𝑵 = 𝟔𝟓. 𝟐 𝒓𝒑𝒎, 𝑸 = 𝟑𝟓𝟎 𝒎𝟑
𝒔
⁄ , 𝑫 = 𝟕. 𝟒 𝒎,
𝒅 = 𝟕. 𝟒 × 𝟎. 𝟒𝟑𝟐 = 𝟑. 𝟏𝟗𝟔𝟖 𝒎 , 𝜼𝒐 = ? , 𝑵𝑺 =? , 𝝋 =? , 𝝍 =?
𝑼 =
𝝅𝑫 𝑵
𝟔𝟎
=
𝝅 × 𝟕.𝟒 × 𝟔𝟓.𝟐
𝟔𝟎
= 𝟐𝟓.𝟐𝟔 𝒎/𝒔 𝑼 = 𝝋 √𝟐𝒈𝑯 ≫ 𝝋 =
𝑼
√𝟐𝒈𝑯
= 𝟏. 𝟖𝟒
𝑸 =
𝝅
𝟒
(𝑫𝟐
− 𝒅𝟐)𝑽𝒇 ≫
𝑽𝒇 =
𝑸
𝝅
𝟒
(𝑫𝟐 − 𝒅𝟐)
=
𝟑𝟓𝟎
𝝅
𝟒
(𝟕. 𝟒𝟐 − 𝟑. 𝟏𝟗𝟐)
𝑽𝒇 = 𝟗. 𝟗𝟗𝟓𝟑 𝒎/𝒔
𝑽𝒇 = 𝝍√𝟐𝒈𝑯 ≫ 𝝍 =
𝑽𝒇
√𝟐𝒈𝑯
= 𝟎.𝟕𝟐
𝜼𝒐 =
𝑷(𝟏𝟎𝟎𝟎)
𝝆.𝒈.𝑸.𝑯
=
𝟑𝟎𝟎𝟎𝟎(𝟏𝟎𝟎𝟎)
𝟏𝟎𝟎𝟎×𝟗.𝟖𝟏×𝟑𝟓𝟎×𝟗.𝟔
𝜼𝒐 = 𝟎. 𝟗𝟏 = 𝟗𝟏%
𝑵𝒔 =
𝑵√𝑷
𝑯
𝟓
𝟒
=
𝟔𝟓.𝟐√𝟑𝟎𝟎𝟎𝟎
𝟗.𝟔
𝟓
𝟒
= 𝟔𝟔𝟖.𝟐𝟗
Jan 2019 and Jan 2020 (Activity Problem)
A Kaplan turbine working under a head of 20 m develops 11772 KW shaft power. The
outer diameter of the runner is 3.5 m and hub dice is 1.75 m. 1 he guide blade angle at the
extreme edge of the runner is 35°. The hydraulic and overall efficiency of the turbine are
88% and 84% respectively. If the velocity of whirl is zero at outlet, determine:
i) Runner vane angles at inlet and outlet at the extreme edge of the runner,
ii) Speed of the turbine.
Jan / Feb 2021(Activity Problem)

26. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 26
vii. Theory and types of Draft tubes.
This is used to increase the pressure from low turbine exit pressure to the pressure of
surrounding to which the fluid is rejected.
It reduces the velocity of the fluid and hence increases the pressure of the fluids to the
atmospheric level at the tailrace.
a) Draft Tube is an important component of a reaction turbine.
b) The component is like a pipe in which area increasing gradually that connects the outlet of
the runner to the tail-race.
c) There are two ends in which one end is connected to the runner outlet and the other end is
submerged below the level of water in the tail-race.
d) It converts excess of kinetic energy into static pressure.
Efficiency of the draft tube
The efficiency of the draft tube can be said as it is the ratio of
actual conversion of kinetic energy into the pressure
energy in the draft tube to the kinetic energy available at the
inlet to the draft tube.
This means, Actual conversion of kinetic energy into pressure energy / kinetic energy
available at the inlet of the draft tube.
𝑽𝟐 = Fluids velocity at the inlet of the draft tube or at the outlet of the turbine,
𝑽𝟑 = Fluids velocity at the outlet of the draft tube,
𝒈 = gravitational acceleration,
𝒉𝒅 = head losses in the draft tube
Types of Draft Tubes
Draft tubes are mainly classified as:
 Simple Elbow draft tube
 Elbow with a varying cross-section area
 Moody Spreading tube
 Conical Diffuser or Straight divergent
1. Simple Elbow draft tube:
 This type is used for the low head.
 The Efficiency of this tube is about 60 percent which is moderate.
 The area of inlet and outlet are the same. A little bit the outlet section is changed.

27. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 27
2. Elbow with varying cross-section:
 It is used in the Kaplan Turbine which I have covered in detail in my previous article,
check out here.
 In this tube the upper section is circular and the outlet is a rectangular section.
3. Moody Spreading Tube:
 This reducing the whirling speed of the water.
 The efficiency around 88 percent which is good.
 It has two passage one inlet and 2 outlets
 It is having central solid core where it distributes two parts of an outlet.
4. Conical Diffuser or Straight Divergent:
 The cone angle less than 10 degrees if the angle is high cavitation will come. The cone
angle is in diagram 4 when you draw a vertical line the angle made will be cone angle.
 Having efficiency is about 90 percent.
Functions of Draft Tube:
1. A reaction turbine is required to be installed above the tail race level for easy maintenance
work, hence some head is lost. The draft tube recovers this head by reducing the pressure head
at the outlet to below the atmospheric level. It increases the working head of the turbine by an
amount equal to the height of the runner outlet above the tail race. This creates a negative head
or suction head.
2. Exit kinetic energy of water is a necessary loss in the case of turbine. A draft tube recovers
part of this exit kinetic energy.
3. The turbine can be installed at the tail race level, above the tail race level or below the tail
race level.