TURBINE
ME 343
TURBINE
A turbine is a mechanical device
that converts fluid energy (such as
steam, water, gas, or air) into
mechanical energy through the
rotation of blades or vanes. This
mechanical energy is often used to
generate electricity or drive
machinery.
Aa
REACTION HYDRO POWER PLANT IMPULSE HYDRO POWER PLANT
REACTION HYDRO POWER PLANT
A reaction turbine is a power-generating prime mover that
operates on the principle of continuous pressure and velocity
changes across both fixed and moving blades, where the fluid
accelerates through both components to generate torque on
the rotor.
1. High efficiency over a wide range of
flow conditions
2. Suitable for low to medium head
applications
3. Continuous conversion of pressure and
velocity into mechanical energy ensures
smooth operation
REACTION HYDRO POWER PLANT
Aa
IMPULSE HYDRO POWER PLANT
An impulse turbine is a type of turbine that works by
converting the pressure energy of a fluid into kinetic energy
using nozzles, then using that high-speed jet to strike the
turbine blades and cause rotation.
1. Simple construction
2. High efficiency at high heads
Aa
3. Easy maintenance
IMPULSE HYDRO POWER PLANT
Nomenclature
1. Reservoir – stores the water coming from the upper river or waterfalls.
2. Head Water – is the water in the reservoir
3. Spillway – is a weir in the reservoir which discharges excess water so that the head
of the plant will be maintained
4. Dam – is the concrete structure that encloses the reservoir.
5. Silt Sluice – is a chamber which collects the mud and through which the mud is
discharged.
6. Penstock – is the channel that leads the water from the reservoir to the turbine.
7. Surge Chamber – a standpipe that connected to the atmosphere and attached to
the penstock so that the water will be at atmospheric pressure.
Aa
8. Tail Race – is the channel which leads the water from the turbine to the tail water.
9. Tail Water – is the water that discharged from the turbine.
10. Draft Tube – is a device that connects the turbine outlet to the tail water so that
the turbine can be set above the tail water level.
hhw
REACTION HYDRO POWER PLANT
hhw= head water elevation
hhw
htw
REACTION HYDRO POWER PLANT
hhw= head water elevation
htw= tail water elevations
FORMULA
1. Gross Head ,hg - is the difference between head water and
tail water elevations.
hg= hhw-htw
hhw hg
htw
REACTION HYDRO POWER PLANT
hhw= head water elevation
htw= tail water elevations
Aa
hhw
IMPULSE HYDRO POWER PLANT
hhw= head water elevation
Aa
hhw
htw
IMPULSE HYDRO POWER PLANT
hhw= head water elevation
htw= tail water elevations
Aa
hhw
htw
IMPULSE HYDRO POWER PLANT
hhw= head water elevation
htw= tail water elevations
D = wheel diameter
FORMULA
1. Gross Head ,hg
hg= hhw-htw-D
Aa
hhw hg
D
hg= hhw-htw
htw
IMPULSE HYDRO POWER PLANT
hhw= head water elevation
htw= tail water elevations
D = wheel diameter
FORMULA Penstock efficiency, ep
- is the ratio net head to the gross head.
Net head or Effective head, h- is the difference between the ep=hnet /hg
gross head and the friction head loss.
Hydraulic Efficiency, eh`
hnet= hg- hf
Hydraulic efficiency is the ratio of the utilized
Friction Head Loss, hf – is the head lost by the flow in a head to the net head.
stream or conduit due to frictional disturbances set up by eh = hw/hnet
the moving fluid and it’s containing conduit and by
where:
intermolecular friction.
hw = utilized head
A. Using Darcy’s equation B. Using Morse equation
h= net head
hf = fLV2/2gD hf = 2fLV2/gD
Head of Impulse Turbine (Pelton)
𝑷
hnet= + (V2/ 2g)
ɣ
Head of Reaction Turbine (Francis and Kaplan)
hnet= ɣ + Δz + (ΔV2/ 2g)
𝑷
FORMULA
Water Power, Pw
Water power is the power generated from an elevated water
supply by the use of hydraulic turbines.
Pw = ɣ Q hnet
ɣ = specific weight of water
ɣ = 9.81kN/ m3 or 62.4 lb/ft3
Turbine efficiency, et
Turbine efficiency is the ratio of the turbine power output to
the water power output.
et= Turbine Power/ Water Power
et = Pt/Pw or et= Pt /ɣ Q h
FORMULA
Electrical or Generator Efficiency, egen
Electrical or generator efficiency is the ratio of the generator
output to the turbine power output.
egen= Pgen/Pt or Pgen = Pt egen = ɣ Q h et egen
Generator Speed, N
N= 120f/P
Where:
N= angular frequency,rpm
f= frequency(usually 60hertz)
P= no. of poles
FORMULA
Peripheral Coefficient, ɸ
- is the ratio of the peripheral velocity (Vp) to the velocity of
the jet(Vj).
ɸ= Vp / Vj
ɸ = πDN/ 𝟐𝒈𝒉 Turbine type recommendation based on head
where: Net head Type of Turbine
Up to 70ft Propeller type
D= diameter of runner 70ft to 110ft Propeller or Francis Type
N= angular speed 110ft to 800ft Francis type
800ft to 1300ft Francis type or Impulse type
h= net head 1300ft and above Impulse type
example
1. A hydraulic turbine receives water from a reservoir at an elevation of
100 m above it. What is the minimum water flow in kg/s to produce a
steady turbine output of 50 MW.
a. 50 247
b. 50 968
c. 50 672
d. 59 465
example
2.A hydro-electric power plant consumes 60 000 000 Kw-hr per year.
What is the net head if the expected flow is 1 500 m 3 min and over-all
efficiency is 63 %.
a. 34.34 m
b. 44.33 m
c. 43.43 m
d. 33.44 m
example
3. A hydroelectric plant discharges water at the rate of 0.75 m3/s and
enters the turbine at 0.35 m/s with a pressure of 275 kpa. Runner
inside diameter is 550mm, speed is 520 rpm and the turbine efficiency
is 88%. Find the turbine speed factor.
a. 0.314
b. 0.432
c. 0.043
d. 0.638
example
4. A hydroelectric plant has a 20 MW generator with an efficiency of
96%. The generator is directly coupled to a vertical Francis type
hydraulic turbine having an efficiency of 80%. The total gross head on
the turbine is 150m while the loss of head due to friction is 4% of the
gross head. The runaway speed is not to exceed to 750rpm. Determine
the flow of water through the turbine in cfs.
example
5. A pelton type turbine was installed 30m below the head gate of the
penstock. The head loss due to friction is 15% of the given elevation.
The length of the penstock is 80 m and the coefficient of friction is
0.00093. Determine the power output in kW.
STEAM TURBINES
Impulse Blading
In an impulse turbine, a fixed nozzle blows steam into a series of moving blades, also called
buckets. Steam enters the buckets at one side and leaves at the other end of the bucket during which there
is no pressure drop. An impulse turbine commonly consists of multiple nozzles.
𝐕𝐛 = bucket velocity
𝐕𝟏 = absolute velocity of the jet leaving the nozzle
𝛂 = nozzle angle
𝐕𝐫𝟏 = velocity of the jet entering, relative to the bucket
𝛃𝟏 = required angle of the jet entering
the bucket for smooth entrance
𝐕𝐫𝟐 = exit velocity of the jet relative to the bucket
𝛃𝟐 = exit angle of the jet
𝐕𝟐 = absolute velocity of the jet leaving the bucket
Va = axial velocity
Vw = velocity of whirl
for the work done by the turbine: For one unit weight of substance, W=1
𝑊𝑡 = 𝐹𝑑 𝑉 ∆𝑉
𝑊𝑡 =
𝑔
Where:
Where:
F = force applied in the direction of motion
V = velocity of the bucket
D= distance travelled by the blade due to the force
∆V = total change of velocity in the direction of
𝑊 ∆𝑉 the motion relative to the bucket
𝐹 = 𝑚𝑎 =
𝑔 𝑡
𝑉 = 𝑉𝑏
𝑑 = 𝑉𝑡
∆𝑉 = 𝑉𝑟1 𝑐𝑜𝑠𝛽1 + 𝑉𝑟2 𝑐𝑜𝑠𝛽2
The equation of work then becomes:
𝑊 ∆𝑉 𝑊 Then:
𝑊𝑡 = 𝑉𝑡 = 𝑉 ∆𝑉 𝑉𝑏 𝑉𝑟1 𝑐𝑜𝑠𝛽1 +𝑉𝑟2 𝑐𝑜𝑠𝛽2
𝑔 𝑡 𝑔 𝑊𝑡 = 𝑔
𝑉𝑏 𝑉𝑟1 𝑐𝑜𝑠𝛽1 + 𝑉𝑏 𝑉𝑟2 𝑐𝑜𝑠𝛽2
= Eq 1
𝑔
Using the law of cosines on the velocity triangles,
At inlet
𝑉𝑏2 2
+ 𝑉𝑟1 − 2𝑉𝑏 𝑉𝑟1 cos 180 − 𝛽1 = 𝑉12 𝑉𝑏 𝑉𝑟1 cos 𝛽1 =
𝑉12 − 𝑉𝑏2 − 𝑉𝑟1
2
2
Where: cos 180 − 𝛽1 = −𝑐𝑜𝑠𝛽1
𝑉𝑏2 + 𝑉𝑟22 − 𝑉22
At outlet 𝑉𝑏 𝑉𝑟2 cos 𝛽2 =
2
𝑉𝑏2 + 𝑉𝑟2
2
− 2𝑉𝑏 𝑉2 cos 𝛽2 = 𝑉22
This shows that the work of the turbine comes from
𝑉12
Substituting in the work equation, Eq 1 the kinetic energy of the jet. The term
2𝑔
is the total
available energy of the jet entering the bucket and
𝑉22
is the kinetic energy of the jet leaving the
𝑉12 − 𝑉𝑏2 − 𝑉𝑟1
2 + 𝑉2 + 𝑉2 − 𝑉2
𝑏 𝑟2 2
2𝑔
𝑊𝑡 = buckets which is the unused portion of the energy
2𝑔 2 −𝑉 2
𝑉𝑟1
of the jet. While the term 𝑟2
represents the
2𝑔
𝑽12 − 𝑽22 − 𝑽2𝒓1 − 𝑽2𝒓2 losses in the buckets.
𝑾𝒕 =
2𝒈
Bucket Efficiency Bucket velocity coefficient
𝑊𝑡 𝑽𝒓𝟐
𝑛𝑏 = 𝑪𝒃 =
𝐴𝐸 𝑽𝒓𝟏
Where AE is the total available energy to the buckets
Velocity ratio (ratio of blade speed to steam speed)
𝑽𝟐𝟏 − 𝑽𝟐𝟐 − (𝑽𝟐𝒓𝟏 − 𝑽𝟐𝒓𝟐 )
𝑽𝒃
𝟐𝒈 𝝆=
𝒏𝒃 = 𝑽𝟏
𝑽𝟐𝟏
𝟐𝒈
𝑽𝟐𝟏 − 𝑽𝟐𝟐 − (𝑽𝟐𝒓𝟏 − 𝑽𝟐𝒓𝟐 )
𝒏𝒃 =
𝑽𝟐𝟏
example
• In a simple impulse stage, steam leaves the nozzles with a velocity of 1200 ft/s. The nozzle angle is
15°. Assume the bucket entrance and exit angles are the same and that the bucket velocity coefficient
is 0.88. The bucket speed is 580 ft/s and steam is supplied to the turbine at the rate of 6000 lb/hr.
Find:
a. the required bucket entrance angle for the given conditions
b. the bucket work, Btu/lb
c. the power developed in the buckets, hp
d. the available energy to the buckets, Btu/lb
e. the bucket efficiency
