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The Solar Resource
The Hydro Resource and Micro Hydroelectricity Systems
Overview
• Review of the Hydrologic Cycle• System components• Measuring head and flow• Generating power from water (examples)
6/22/2009 http://retc.morrisville.edu 2
Hydrologic Cycle
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• Key terminology» Insolation» Evaporation» Transpiration» Evapotranspiration» Sublimation» Condensation» Precipitation» Infiltration
» Sub-surface flow» Ground water
discharge» Overland (surficial)
flow» Freshwater storage» Oceanic storage
Hydrologic Cycle
6/22/2009 http://retc.morrisville.edu 4
condensationsublimationInsolation
Precipitation
Surficialflow
Freshwater storage
Transpiration Evaporation
Infiltration
Subsurface flow
Oceanic storage
Groundwater discharge
Hydro Power
• For most hydro systems, we are interested in only certain processes in this cycle» Oceanic storage (wave, tidal, ocean current)» Freshwater storage (wave, pumped storage, dams)» Overland flow (streams and rivers)
• Though our systems use these processes, we must keep in mind that it is a cycle» Water is replenished in our systems due to
incoming solar energy
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Measuring the hydro resource
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In central New York, when do we get most of our precipitation?
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec0
1
2
3
4
5
MSC mean rainfall (2003-2008)
Inch
es o
f Rai
nfal
l
28 inches per year
Measuring the hydro resource
6/22/2009 http://retc.morrisville.edu 7
Chenango River Discharge
0
500
1000
1500
Janu
ary
Fabru
ary
Mar
chApr
ilM
ayJu
ne July
Augus
t
Septe
mbe
r
Octobe
r
Novem
ber
Decem
ber
Cu
bic
Fee
t p
er S
eco
nd
Mean
2006
http://waterdata.usgs.gov/nwis/
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System components: Intake
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• Water enters penstock through the intake
• Remove debris• High maintenance• Accessible
System components: Penstock
• PVC» Cheap, light, and rigid» Low pressure systems» Easily available at
hardware stores» Low losses (in straight
sections)
» Freezing issues
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System components: Penstock
• Polyethylene tube» Flexible» Longer lengths» Lower losses in sweeping bends» Freeze resistant
» Expensive components» Difficult to purchase
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System components: Turbine
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• High head, low flow• 1, 2, and 4 nozzle designs• 12, 24, 48, VDC options• 120 VAC options• Pelton wheel with bronze
runner
System components: Batteries
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• Lead-acid • Deep cycle• Generally 2 to 6V• Wet cell or sealed (gel)
System components: Charge controller
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• Monitors battery bank voltage• When the battery bank is “full”,
electrons are diverted to a diversion load (a.k.a. dump load)
• Can be jumped from 12,24, and 48 VDC depending upon input and battery bank (they must match!)
System components: Diversion Load
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• Waste electrons as quickly as possible• Resistance heating elements• Protect the battery bank
System components: Inverter
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• Converts direct current (DC) to alternating current (AC)
• Can match the utility signal (voltage, shape and frequency)
Generating power
Now that you understand the system components, how does one actually generate power with a micro hydro system?
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Measuring the hydroelectric resource
• Power generation from water is dependent on five variables:» P=ηρgQH» Power in watts (P)» Turbine efficiency (eta, η)» Water density (rho, ρ; usually 1000 kg/m3)» Acceleration of gravity (g, 9.81 m/s2)» Quantity of water flow (Q, in m3/s)» Vertical distance (head, H, in meters)
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Measuring a stream – flow
Flow rate (Q)• Quantity of water passing a given point over a
given amount of time
» Cubic meters per second» Gallons per minute» 1 GPM = 0.000063 m3/s
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Measuring flow
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Measuring the hydro resource - head
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Head (H)
• Head is the vertical distance of the hydro system (from intake to turbine)
• Relationship of head and pressure
2.31 feet
1 psi
Measuring head
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Stream profile diagram
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1,110 feet of penstock
Hydro power - example
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• Small stream:» 20 GPM flow, 140 feet of head, 85% turbine
efficiency
• Pressure:
• Flow:
• Head:
psi 60.6ft 2.31
psi 1ft 140
/sm 0.00126GPM 1
/sm 0.000063 GPM 20 3
3
m 42.7ft 1
m 0.305ft 140
Hydro power: example
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• P= η ρ g Q H» Power = 0.85*1000 kg/m3*9.81 m/s2*0.00126 m3/s *
42.7 m» Power = 448.6 watts
• Yearly energy in kWh?» 448.6 W *24 hrs/day * 365.25 days/yr = 3,932 kWh/yr
• My house uses about 4,000 kWh/yr
Hydro power: what if?
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• If we go from 20 GPM flow and 140 ft of head to 140 GPM and 20 ft of head?
• What power (watts) should I expect?• P= η ρ g Q H
» Power = 0.85*1000 kg/m3*9.81 m/s2*0.00882 m3/s * 6.1 m
» Power = 448.6 watts
Hydro power
• Head and flow have equal importance in determining power (and energy) in a hydro system
» What we have just calculated does not take penstock losses into account
» This will reduce power output
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Hydro power: a comparison
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• 20 GPM and 140 ft of Head• Yearly energy in kWh?
» 448.6 W *24 hrs/day * 365.25 days/yr = 3,932 kWh/yr• My house uses about 4,000 kWh/yr
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…to wind!
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• Class 3 site (7 m/s average; 15 mph)• Turbine at 30% efficiency• P=0.5* η ρ A V3
• 450 W = 0.5*0.3*1.2 kg/m3*(3.14*r2)*(7 m/s)3
• r = 1.5 meters, diameter = 3 meters
This means to get an equivalent amount of energy, I need a 10’ wind turbine rotor!
So, what bother with micro hydro?
• (Relatively) inexpensive• Constant power production (not intermittent)• Minimal impacts• Turbines have high efficiency (80% to 90+%)
Challenges• Not considered “renewable and sustainable”• Permitting process may be required• Highly selective sites• Currently cannot be net metered• Little knowledge of our resource
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Phil Hofmeyer, Ph.D.Assistant ProfessorPh: 315-684-6515Email: [email protected] Web: http://people.morrisville.edu/~hofmeypv/
Ben Ballard, Ph.D.Director, RETCAssistant ProfessorPh: 315-684-6780 Email: [email protected] Web: http://people.morrisville.edu/~ballarbd/
Contact Information
http://retc.morrisville.edu