FLUID MACHINERIES

ENGR. PETER CLYDE B. LAMADRID,M.E.

S.Y. 2022-23, 1ST SEMESTER
 TURBINES

✓ Turbines are defined as the hydraulic machines

which convert hydraulic energy into mechanical
energy.
✓ A turbine is a rotary mechanical device that
extracts energy from a fast moving flow of
water, steam, gas, air, or other fluid and
converts it into useful work.
✓ A turbine is a turbo-machine with at least one
moving part called a rotor assembly, which is a
shaft or drum with blades attached.
 TURBINES

WORKING PRINCIPLE:
1. Fluid strikes the blade to produce
rotational energy
2. The turbine shaft is directly coupled to an
electric generator mechanical energy is
converted into electrical energy called
hydroelectric power
 TURBINES
BASIC TYPES OF TURBINES:

1. WATER TURBINE
2. STEAM TURBINE

3. GAS TURBINE
4. WIND TURBINE

Although the same principles apply to

all turbines, their specific designs
differ sufficiently to merit separate
descriptions.
 WATER TURBINES

Turbines that use water as the working fluid for

the production of power are known as HYDRAULIC
TURBINES.
CLASSIFICATION OF HYDRAULIC TURBINES:
1. IMPULSE TURBINES
2. REACTION TURBINES
This classification is based on the interaction of
fluid with turbine blades. however, the turbines can
further be classified on the basis of head available at
the inlet, specific speed, and according to flow
direction. there are three main turbines named after
their inventors, namely, PELTON, FRANCIS, AND
KAPLAN.
 WATER TURBINES

The turbines can be classified under different

headings:
1. Action of water (Impulse or Reaction)
2. Direction of flow
3. Available Head
a. High Head (H>300m)
b. Medium Head (50m<H<300m)
c. Low Head (H<50m)
4. Specific Speed
 WATER TURBINES
HEADS AND EFFICIENCIES:

In fluid mechanics, the term head indicates the energy in

units of distance. there are two types of heads as far as
turbines are concerned:

1. GROSS HEAD (ℎ𝑔 )

𝒉𝒈 = 𝒉𝒉𝑾 − 𝒉𝒕𝑾 ;
ℎℎ𝑊 = HEADWATER ELEVATION
ℎ𝑡𝑊 = TAILWATER ELEVATION

2. NET HEAD /EFFECTIVE HEAD (h)

h = 𝒉𝒈 - 𝒉𝒇

𝒇𝑳𝑽𝟐
𝒉𝒇 = ; Darcy’s equation
𝟐𝒈𝑫

𝟐𝒇𝑳𝑽𝟐
𝒉𝒇 = ; Morse equation
𝒈𝑫
 WATER TURBINES
HEADS AND EFFICIENCIES:
TURBINE EFFICIENCIES
1. HYDRAULIC EFFICIENCY (𝑒ℎ ) - is defined as the ratio of power developed by the 𝑃𝑟𝑢𝑛𝑛𝑒𝑟 ℎ𝑤
𝑒ℎ = or 𝑒ℎ =
turbine runner to power available at turbine inlet 𝑃𝑖 ℎ

2. TURBINE EFFICIENCY (𝑒𝑡 ) − is defined as the ratio of turbine power output to the 𝑃
𝑒𝑡 = 𝑃 𝑡
water power output 𝑤

3. MECHANICAL EFFICIENCY (𝑒𝑚 ) - is the ratio of power available at the shaft to the 𝑃𝑠
𝑒𝑚 =
power developed by the runner 𝑃𝑟𝑢𝑛𝑛𝑒𝑟
4. GENERATOR EFFICIENCY (𝑒𝑔𝑒𝑛 ) - the ratio of the electrical power produced by the 𝑃𝑔𝑒𝑛
𝑒𝑔𝑒𝑛 =
generator to mechanical power available at turbine–generator shaft 𝑃𝑠
5. OVERALL EFFICIENCY (𝑒𝑜 ) - is the efficiency of whole turbine-generator system 𝑃𝑜
which can be obtained by dividing the power output of generator to the hydraulic 𝑒𝑜 =
𝑃𝑖
power input to turbine
𝑄𝑎
6. VOLUMETRIC EFFICIENCY (𝑒𝑣 ) - is defined as the ratio of actual discharge to the 𝑒𝑣 =
total discharge 𝑄𝑡
WATER TURBINES
PELTON TURBINE

The Pelton turbine or Pelton

wheel is a tangential flow impulse
turbine. The water strikes the bucket
along the tangent of the runner. Pelton
turbine is a high head and low discharge
impulse turbine. Pelton turbine works on
the principle of conversion of available
hydraulic energy first into kinetic energy
of the jet and then into mechanical
energy of the rotating wheel or runner,
also known as Pelton wheel.
 PELTON TURBINE
MAIN PARTS OF PELTON WHEEL
USEFUL FORMULAS: PELTON TURBINE
1. Net head /effective head (h)
𝒇𝑳𝑽𝟐
h = 𝒉𝒈 - 𝒉𝒇 𝒉𝒇 = 𝟐𝒈𝑫 ; D= Diameter of penstock

2. theoretical Jet Velocity(𝑉𝑗 )

𝑽𝒋 = 𝟐𝒈𝒉
𝑽𝒋𝒂 = 𝑪𝒗 𝟐𝒈𝒉 𝑪𝒗= Coefficient of velocity

3. Peripheral Velocity (𝑉𝑝 )

𝑽𝒑 = 𝝅𝑫𝑵 D= Mean Runner Diameter

4. Specific Speed of Hydraulic turbine (𝑁𝑆 )

𝑁 𝑃𝑜 𝑷𝒐 𝐢𝐧 𝐇𝐩
𝑁𝑆 = 5 h in ft
ℎ ൗ4
5. Speed ratio / Peripheral Coefficient (𝑘𝑆 )
𝑽𝒑
𝒌𝑺 = 𝑽𝒋
6. Flow ratio (𝑘𝑓 )
𝑽𝒇
𝒌𝒇 = 𝑽𝒋
7. Jet ratio 𝑚
𝑷𝒊𝒕𝒄𝒉 𝒅𝒊𝒂𝒎𝒆𝒕𝒆𝒓 𝒐𝒇 𝒕𝒉𝒆 𝑷𝒆𝒍𝒕𝒐𝒏 𝑾𝒉𝒆𝒆𝒍
m= 𝑱𝒆𝒕 𝑫𝒊𝒂𝒎𝒂𝒕𝒆𝒓
 PELTON TURBINE
EXAMPLE

A 30MW power plant uses a Given: Solution:

𝑃𝑜 = 30𝑀𝑊 𝐷𝑗 =?
double overhung Pelton turbine h = 400m Q = AV
𝜋
under the net head of 400m at 𝐷𝑗 =? Q = 4 (𝐷𝑗 ) 2 (𝑉𝑗𝑎 );

its inlet. Find the jet diameter, 𝐷𝑟𝑢𝑛𝑛𝑒𝑟 = ? 𝑉𝑗𝑎 = 𝐶𝑣 2𝑔ℎ

𝑁𝑟𝑢𝑛𝑛𝑒𝑟 =? 𝑉𝑗𝑎 = 0.95 2(9.81𝑚/𝑠 2 (400𝑚)
mean runner diameter, runner 𝑁𝑠 =? 𝑉𝑗𝑎 = 84.16 m/s
speed, and specific speed of 𝑒𝑔𝑒𝑛 = 0.95 𝑃𝑜 = 30MW =PE + KE
30000 = (1/2m𝑉𝑗𝑎 2 )(2)(𝑒𝑡𝑜𝑡𝑎𝑙 )
the turbine. Assume generator 𝑒𝑚 = 0.95
𝑒ℎ = 0.90 30000= 𝜌𝑄𝑉𝑗𝑎 2 (𝑒ℎ 𝑒𝑚 𝑒𝑣 )
efficiency 95%, mechanical 𝐶𝑣 = 0.95 30000kJ/s = (1000kg/𝑚3 )(84.16m/s) 2 (Q) (0.90*0.95)
efficiency 95%, hydraulic Q = 4.95 𝑚3 /s;
𝑘𝑠 = 0.46
Therefore;
efficiency 90%, coefficient of m = 10 𝜋
4.95= 4 (𝐷𝑗 ) 2 (84.16);
velocity 0.95, speed ratio 0.46, 𝑫𝒋 = 0.27m
and jet ratio 10.
 PELTON TURBINE
EXAMPLE #1

Given: Solution: Solution: Solution:

𝑃𝑜 = 30𝑀𝑊 𝑁𝑠 =?
h = 400m 𝐷𝑟𝑢𝑛𝑛𝑒𝑟 =? 𝑁𝑟𝑢𝑛𝑛𝑒𝑟 =?
𝐷𝑗 =? 𝑃𝑖𝑡𝑐ℎ 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑃𝑒𝑙𝑡𝑜𝑛 𝑊ℎ𝑒𝑒𝑙 𝑉𝑝 = 𝜋𝐷𝑟𝑢𝑛𝑛𝑒𝑟 𝑁𝑟𝑢𝑛𝑛𝑒𝑟 ; 𝑉𝑝 = ? 𝑁 𝑃𝑜
𝐷𝑟𝑢𝑛𝑛𝑒𝑟 = ? m= 𝑁𝑆 = 5
𝐽𝑒𝑡 𝐷𝑖𝑎𝑚𝑎𝑡𝑒𝑟
ℎ4
𝑁𝑟𝑢𝑛𝑛𝑒𝑟 =? 𝑉𝑝
𝑃𝑜 = 30MW for double overhung PT
𝑁𝑠 =? 𝐷𝑟𝑢𝑛𝑛𝑒𝑟 𝑘𝑆 = 𝑉
m= 𝑗
𝑃𝑜 = 30000 = (2)(𝑒𝑔𝑒𝑛 )(Power/runner)
𝐷𝑗
𝑒𝑔𝑒𝑛 = 0.95
𝑉𝑝 = 0.46(84.16m/s)
𝑒𝑚 = 0.95 𝐷𝑟𝑢𝑛𝑛𝑒𝑟 =10 x 0.27m 30000
Power/runner = (2)(0.95) = 21165.51Hp
𝑒ℎ = 0.90 𝑉𝑝 = 38.71m/s
𝑫𝒓𝒖𝒏𝒏𝒆𝒓 = 2.7m
𝐶𝑣 = 0.95 274 21165.51
𝑘𝑠 = 0.46 38.71𝑚/𝑠 𝑁𝑆 = 5
𝑁𝑟𝑢𝑛𝑛𝑒𝑟 = 𝑥 60 13124
m = 10 𝜋(2.7𝑚)
𝑵𝑺 = 5.05rpm
𝑵𝒓𝒖𝒏𝒏𝒆𝒓 = 274rpm
PELTON TURBINE