Transcript pptx Slides

Power vs. energy delivery profile technologies
• Ragone chart:
• More information and charts can be found in Holm et. al., “A Comparison of
Energy Storage Technologies as Energy Buffer in Renewable Energy Sources
with respect to Power Capability.”
© A. Kwasinski, 2016
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Lead-acid batteries calculations
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• Most calculations are based on some specific rate of discharge and then a linear
discharge is assumed.
•The linear assumption is usually not true. The nonlinearity is more evident for faster
discharge rates. For example, in the battery below it takes about 2 hours to discharge
the battery at 44 A but it takes 4 hours to discharge the battery at 26 A. Of course, 26x2
is not 44.
• A better solution is to consider the manufacturer discharge curves and only use a linear
approximation to interpolate the appropriate discharge curve.
• In the example below, the battery can deliver 10 A continuously for about 12 hours.
Since during the discharge the voltage is around 12 V, the power is 120 W and the
energy is about 14.5 kWh
10 A continuous
discharge curve
approximation
Discharge
limit
Nominal curve
© A. Kwasinski, 2016
Lead-acid batteries
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Lead-acid batteries efficiency
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• Consider that during the charge you apply a constant current IC, a voltage VC
during a time ΔTC. In this way the battery goes from a known state of charge to
be fully charged. Then the energy transferred to the battery during this process
is:
Ein = ICVC ΔTC
• Now the battery is discharged with a constant current ID, a voltage VD during a
time ΔTD. The final state of charge coincides with the original state of charge.
Then the energy delivered by the battery during this process is:
Eout = IDVD ΔTD
• So the energy efficiency is  E 
VD I D TD
 VC
VC I C TC
• Hence, the energy efficiency equals the product of the voltage efficiency and
the Coulomb efficiency. Since lead acid batteries are usually charged at the
float voltage of about 2.25 V/cell and the discharge voltage is about 2 V/cell, the
voltage efficiency is about 0.88. In average the coulomb efficiency is about
0.92. Hence, the energy efficiency is around 0.80
© A. Kwasinski, 2016
Wind power Conversion efficiency
• Why is it that output power of real wind turbines do not follow a cubic
relationship?
• Because not all the wind power is transmitted through the blades into the
generator.
• Consider the next figure:
vb
Upwind
vu
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Downwind
vd
Rotor area
A
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Wind power conversion efficiency
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• The wind energy “absorbed” by the wind turbine rotor equals the kinetic
energy lost by the wind as it pass through the blades. Hence, the energy
transmitted by the wind to the rotor blades is the difference between the upwind
and the downwind kinetic energies:
Pb 
d ( Eu  Ed ) 1 dm 2 2

(vu  vd )
dt
2 dt
In the last equation it is assumed that there is no turbulence and the air passes
through the rotor as a steady rate.
• If it is assumed that vb is the average between vu and vd, then the mass flow
rate is
v v
dm
 A u d
dt
2
• If we define the ratio
vd

vu
© A. Kwasinski, 2016
Wind power conversion efficiency
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• Then
1
 vu   vu
Pb   A 
2
2

1
 2
2 2
3 1
2 
(
v


v
)


Av
(1


)(1


)
d
u 
 u
2
2



Power in
the wind
Fraction
extracted
Rotor efficiency
Cp
• The rotor efficiency is maximum when λ is 1/3. For this value, Cp is 0.593.
• Still, we still need to know how much of the “absorbed” power by the blades is
transmitted to the generator. This conversion stage is characterized based on
the tip-speed ration (TSR):
rotor tip speed rpm    D
(TSR) 

wind speed
60vb
© A. Kwasinski, 2016
Wind power conversion efficiency
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Concentrated solar
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Concentrated solar
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