Kelley Fletcher
Dustin Eseltine
Ryan Sargent
Group 5
• The Need:
– With the constant rise in cost of non-renewable energy
sources, alternative sources of renewable energy are
becoming more important
– Energy produced by ocean waves is constant.
– Constant energy = infinite supply
– Infinite supply = Lower Cost
• Objective:
– Design a device that converts ocean waves into
useable electrical energy
• Heaving Floats
– Rise and fall of
waves causes float
to rise and fall
creating energy.
• Pitching Device
– Rise and fall or waves causes
the float to pitch
Wave Direction
– Pitching motion is then
converted to energy.
• Oscillating Water
Column
– Waves cause a pressure
change inside a chamber.
– Oscillating air or water drives
energy device (eg. turbine)
• Surge Device
-Ocean Waves flow into narrowing
chamber
-Water forced into reservoir
-Energy flows through turbine back
into ocean.
Archimedes Wave Swing
•Passing Waves causes the top
chamber to rise and fall
•Rise and fall of top chamber
cause pressure difference
inside device
•Pressure difference runs
hydraulic motor
Oscillating Water
Column
•Waves coming into coast cause
rise and fall inside chamber
•Rise and fall of water in
chamber pushes air through
turbine.
•Dual cycle, needs bi-directional
turbine
• After researching current designs the
following design goals were created:
–
–
–
–
Not a coastal based system.
Not a hydraulic based system.
Make the system scalable.
Design system to be relatively safe from natural
occurrences such as storms
•D-Rings attach buoy to anchoring
cables.
•The three cables from the buoy
are attached to the single cable
using a turnbuckle.
•This system allows for minor
height adjustments after
installation.
•Single cable attaches to buried
concrete anchor.
• Passing waves cause a
differential pressure change
in a submerged chamber.
• The pressure change causes
an airflow through a nozzle.
• The airflow is used to run a
Pelton Turbine.
• ½-wave = 1stroke
• 2 strokes in a cycle:
– Compression
– Suction
Compression
Suction
L= Wave Length
H= Wave Height
D= Water Depth
Note:
•One wave = Crest to Crest –or- Trough to Trough
•Particle depth is considered a negative value
•Design calculations based upon ½wave
Deep Water
•Circular velocity
profile
Shallow-Transitional
•Elliptical velocity
profile
Particle Velocity Equations
H gT cosh[2 ( z  d ) / L]
 2 t 
cos

2 L
cosh(2 / L)
 T 
H gT sinh[2 ( z  d ) / L]
 2 t 
w
sin 

2 L
cosh(2 / L)
 T 
u
V 
w2  u 2
Underwater particle
velocities are related
to:
•Wave Height(H)
•Wave Length(L)
•Water Depth(d)
•Particle depth(z)
•Wave Period(T)
Equation of State – P1V1 = P2V2
P0  0  PSite Site
P0  0  Constant
PSite Site  C   Site 
C
PSite
Bernoulli’s Equation
1
1
2
VSite
 Z site  Patm  V02  Z 0
2
2
1
1
2
PSite  V02  VSite
 Z site
2
2
PSite 
Substituting for Velocity
Psite 
-Bernoulli’s Eq. Allowed Psite to
be solved.
-Once Psite was known the
particle velocity equations were
substituted for surface and site
velocities.
1 
 2 t 
 2 t   1 
 2 t 
 2 t  
  A sin 
  B cos
     X sin 
  Y cos
   Z site
2 
T
T
2
T
T









Psite _ abs 
1 
 2 t 
 2 t   1 
 2 t 
 2 t  
  A sin 
  B cos
     X sin 
  Y cos
   Z site  Patm  h
2 
 T 
 T  2 
 T 
 T 
Volumetric Airflow
Q
 Site
C PSite
 2
t
PSite  t
-Once Psite was known, volumetric
airflow could be derived.
Volumetric Airflow – Q-Bar
Q
Q
T /2
T /2
0
0
Qt  
T /2
 Site
C PSite
t    2
t
t

t
P
Site
0
-Integrating Q gave us Q-bar.
Allowing t to be replaced by
T/2 (one cycle time.)
C
  2 2  2 t 
 2 t 
2
2  2 t  
2
2  2 t  
 A 2 sin 2 
  B cos 
   Z Site   X sin 
  Y cos 
   Patm  h
2
2
 T 
 T 
 T 
 T 

Q-Bar = Volumetric air flow for one cycle
•Using the National Oceanic and Atmospheric Administration’s
website buoy #46212 was chosen.
•Data was downloaded from the NOAA website for Wave Height,
Wave Period, and Atmospheric Pressure.
•Water data was compiled using excel, equations,
and buoy data.
•Average values were tabulated for each day and
then for each month.
•Volumetric airflow data compiled from buoy data
•Average values were compiled daily, monthly, and
then yearly.
•Yearly values were used for turbine calculations.
•Daily values allowed the group to calculate the
turbines power output for any day of the year.
Double Acting Turbine
•Bi-direction turbine
•Possibly self starting
Wells Turbine
•Bi-directional Turbine
•Not self starting
•Blades symmetric to rotation axis
Paddle Wheel Design
•Simplistic in operation and
construction
•Self starting
•Advanced paddle wheel design
•”Buckets” increase amount of energy
extracted from jet stream
•Scalable design
•Self Starting
•Turbine is up to 91% efficient
Volumetric air flow was used to
calculate the following:
•Pitch Circle Diameter (PCD)
Pitch Circle Diameter
•Jet Diameter
•Jet Area
- PCD
4Q
3
PCD 
•Jet Velocity
.0121   determines
turbine size
Jet Diameter
D jet  0.11PCD
-Jet diameter = Nozzle Diameter
Jet Velocity
V jet 
Q
Q

2
A jet   D Jet

 4





Jet area = cross sectional area of nozzle
-Jet velocity is determined from average
flow rate and jet diameter.
Turbine Power Output
W Shaft 
2
 Q V Jet
4
1  cos  
-Shaft work = Theoretical power Output
•At 100% efficiency and Flow the Turbine Produces 56.85 Watts
•Normal overall system efficiency for Pelton Turbines is 60%
•About 40 Watts would be produced at 60% efficiency
•Generator is only capable of handling 18 Watts of Continuous
power. 12 volts x 1.5 amps
Using COSMOSWorks, material data, and calculated values
a brief analysis was completed.
Turbine
Max. Deflection – 0.003 in.
Max. Stress - 485 PSI
Turbine Blades
Max. Deflection – 4.2e-04 in.
Max. Stress – 56.69PSI
Safety Factor - 111
Infinite Life – SLA Model
•Fully scalable turbine system
•Submerged design protects device
•Power output of one “buoy” = 18 Watts
The prototype design concept is complete.
The next steps in the project are:
1) Build a prototype model
2) Prototype would be tested for:
•Turbine efficiency, Air vs. Water
•Stability
•Actual power output
•Actual volumetric flow rate
3) Safety mechanisms may need to be
designed to prevent water entering system.
Questions?
Thank you for your time.