ME 314 – Energy
Engineering
Hydro Power
GREG M. CUBIO, DEng
USTP
The Hydrologic Cycle
Worlds Hydro Power Capacity
• Produces 24% of world’s electricity supply
• 675,000 MW combined power output
• 2,000 hydro power plants in US with a total
capacity of 92,000 MW
• Largest hydropower plant in US is the 7,600
megawatt Grand Coulee Power Station,
Columbia River, Washington State.
Water Power
• Theoretical Power (Pt)
QH
Pt (hp) 
33,000
• Where
Pt = power in hp
Q = mass flow rate, lbs/min
Q = ρAv
v = velocity of water stream, ft/min
A = cross sectional area of water stream, ft2
ρ = density of water, 62.4 lbs/ft3
H = head, ft
Water Power
• Theoretical Power (Pt)
QH
Pt (kW ) 
102
• Where
Pt = power in kW
Q = mass flow rate, kg/sec
Q = ρAv
v = velocity of water stream, m/sec
A = cross sectional area of water stream, m2
ρ = density of water, 1000 kg/m3
H = head, m
Example
• Given: Volumetric flow rate = 35 liters/sec and 1
meter head. Determine the theoretical power
possible and actual power if the conversion efficiency
was 60%.
• Solution
1. Q (kg / sec)  35secli x 1likg  35 kg / sec
35 kg 1 m N  sec2 9.8m sec W
kW
Power (kW )t  Q  H 
x
x
x
x
x
 0.343 kW
2.
sec
kg  m
sec2 N  m 1000W
3. Power (kW)a  Q  H  Eff  0.343* 0.6  0.206kW  206W
24hrs 365days
Yearly
Output
(
kWh
/
yr
)

0
.
206
kWx
x
 1805kWhr / yr
4.
day
yr
5. Note that 35 lps = 555 gpm.
Inefficiencies
• Hydraulic loses in conduits and turbines
• Mechanical losses in bearings and power
transmission system
• Electrical losses in generator, station use and
transmission (for hydroelectric power)
• Overall effect is to reduce the theoretical
power by a factor of 0.60 – 0.80
Water Power Generating Devices
• Hydraulic Rotating Prime Movers
– Waterwheels (overshot, undershot, breast)
– Tub wheel/flutter wheel
– Turbines (Reaction: Francis, Nagler, Kaplan,
Impulse: Pelton, Turgo-wheels)
• Hydraulic Rams
• Hydraulic Air Compressor
Waterwheels
Overshot (35 ft x 40” wide)
Undershot (10 ft x 24” wide)
Undershot Waterwheel
Application
Application: Water Powered Sawmill
Tubwheel: A small undershot wheel
mounted horizontally
Water Turbines
Reaction Turbines
● Reaction turbines are acted on by water, which
changes pressure as it moves through the turbine
and gives up its energy. They must be encased to
contain the water pressure (or suction), or they must
be fully submerged in the water flow.
● Newton’s third law (law or reciprocal reactions or
conservation of momentum) describes the transfer
of energy for reaction turbines.
● Most water turbines in use are reaction turbines.
● They are used in low and medium head applications.
Impulse Turbines
• Impulse turbines change the velocity of a water jet. The jet
impinges on the turbine's curved blades which reverse the
flow. The resulting change in momentum (impulse) causes a
force on the turbine blades. Since the turbine is spinning, the
force acts through a distance (work) and the diverted water
flow is left with diminished energy.
• Prior to hitting the turbine blades, the water's pressure
(potential energy) is converted to kinetic energy by a nozzle
and focused on the turbine. No pressure change occurs at the
turbine blades, and the turbine doesn't require a housing for
operation.
• Newton’s second law (F=ma) describes the transfer of energy
for impulse turbines.
• Impulse turbines are most often used in very high head
applications.
Francis Turbine Runner
Propeller Type Kaplan Runner
Impulse: Pelton Runner
Turgo-wheels
Typical Range of Heads (m)
• Kaplan turbine – axial flow reaction turbine
– Head:
2 < H < 70 m
– Specific speeds: 300 - 1100 rpm
• Francis turbine – mixed flow reaction turbine
– Head:
300 < H < 500 m
– Specific speeds: 60 - 400 rpm
• Pelton turbine – an impulse turbine
– Head:
50 < H < 1300 m
– Specific speeds: 4 - 70 rpm
• Turgo wheels
– Head:
0 < H < 250 m
• Deriaz turbine – reaction turbine, mixed flow
– Head:
1 < H < 200 m
Primary Characterization of Hydro
Turbine: Specific Speed, ηs
• ηs = NxP/(h(5/4))
• This number describes the speed of the turbine at its
maximum efficiency with respect to the power and
flow rate. The specific speed is derived to be
independent of turbine size.
• Given the fluid flow conditions and the desired shaft
output speed, the specific speed can be calculated
and an appropriate turbine design selected.
• The specific speed, along with some fundamental
formulas can be used to reliably scale an existing
design of known performance to a new size with
corresponding performance.
Banki Turbine (Mitchel, Crossflow or
Ossberger)
Uses nozzles and
blades instead of
buckets (similar
to overshot
waterwheels)
How a Hydro Power Plant Works
Hydro Power Monograph
Types of Hydro Power Plant
• On the basis of operation
– Base load plant and Peak load plant
• Plant capacity
–
–
–
–
Micro (< 5 MW)
Medium capacity (5 – 100 MW)
High capacity plant (100-1000 MW)
Super power plant (> 1000 MW)
• Based on head
– Low head (<15 m)
– Medium head plant (15-50 m)
– High head plant (>50 m)
Grouping done
arbitrarily and not
sacrosanct
Types of Hydro Power Plant
• Based on hydraulic features
– Conventional
– Pumped storage
– Tidal power type
• Based on Construction Feature
– Run-of-river
– Valley dam type
– Diversion canal
– High head diversion
Diversion Canal: Tazimina Project,
Alaska
Types of Hydro Plant
• Run-of-the-river
• Diversion Canal
• Valley Dam type
• High head diversion
Types of Pumped
Storage Hydro
Electric Power
System
800 kW Hydro Power Facility, King
Cove, Alaska (700 residents)
Hoover Dam (17 generators @ 133
MW each produces >2,000 MW)
The Grand Coulee Power Plant
7,600 MW
Largest Hydroelectric Power Plant in
the World (13,320 MW), Brazil
Sample Calculations
• Given: H = 294 meters, pipe diameter = 6 m; terminal
velocity = 10 m/s; Efficiency 80%. Determine the
power output.
• Solution:
• 1. Theoretical Power
QH 294m  (6m) 2 10m 1000kg N  s 2 9.8m W  s
Pt (W ) 

x
x
x
x
x 2 x
3
102
4
s
m
kg  m s
N m
Pt ( MW ) 
814,642W
MW
x
 815MW
6
1x10 W
• 2. Actual Power Output
Pa ( MW )  815MWx 0.80  650MW
Sample2
Condition:
H = 5 m, Q = 16.5galx3.875(L/gal)/2.5s
= 25 L/s x 1 m3 per 1000L = 0.025 m3 per second
Assumed Efficiency = 70%
Solution:
P = ɣQH (Eff), watts
= 9800N/ m3 (0.025m3 /s)(5 m) (0.7)
= 857 watts
Environmental Issues
• Water quality (losses oxygen and gains
nitrogen and phosphorus)
• Damage to surroundings
• Obstruction of aquatic life
• Siltation
• Affect irrigation scheduling
• Overflowing of dams