ESE 471 β Energy Storage Systems
SECTION 4: ULTRACAPACITORS
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Introduction2
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Ultracapacitors
Capacitors are electrical energy storage devices Energy is stored in an electric field
Advantages of capacitors for energy storage High specific power High efficiency Equal charge and discharge rates Long lifetime
Disadvantages of capacitors for energy storage Low specific energy
Ultracapacitors (or supercapacitors) are variations of traditional capacitors with significantly improved specific energy Useful in high-power energy-storage applications
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Ultracapacitors β Ragone Plot
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Ultracapacitors - Applications
Ultracapacitors are useful in relatively high-power, low-energy applications They occupy a similar region in the Ragone plane as flywheels
Energy recovery and regenerative braking applications Cars
EV, HEV, ICE (e.g. Mazda 6 i-ELOOP) Buses Trains Cranes Elevators
Uninterruptible power supply (UPS) applications Fast-responding, short-term power until generators take over
Wind turbine pitch control Put turbine blades in safe position during loss of power
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Capacitor Fundamentals6
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Fluid Capacitor
Modulus of elasticity, ππ Area, π΄π΄
Incompressible fluid External pumps set
pressure or flow rate at each port
Consider the following device: Two rigid hemispherical shells Separated by an impermeable
elastic membrane
Total volume inside shell is constant
Volume on either side of the membrane may vary
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Fluid Capacitor β Equilibrium
Equal pressuresΞππ = ππ1 β ππ2 = 0
No fluid flowππ1 = ππ2 = 0
Membrane does not deform
Equal volume on each side
ππ1 = ππ2 =ππ2
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Fluid Capacitor β ππ1 > ππ2 Pressure differential
Ξππ = ππ1 β ππ2 > 0
Membrane deforms
Volume differential
Ξππ = ππ1 β ππ2 > 0
Transient flow as membrane stretches, but...
No steady-state flow As π‘π‘ β β
ππ1 = ππ2 = 0
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Fluid Capacitor β ππ1 < ππ2 Pressure differential
Ξππ = ππ1 β ππ2 < 0
Volume differential
Ξππ = ππ1 β ππ2 < 0
Ξππ proportional to: Pressure differential Physical properties, ππ, π΄π΄
Total volume remains constant
ππ1 + ππ2 = ππ
Again, no steady-state flow
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Fluid Capacitor β Constant Flow Rate
Constant flow rate forced into port 1
ππ1 β 0
Incompressible, so flows are equal and opposite
ππ1 = ππ2
Volume on each side proportional to time
ππ1 =ππ2 + ππ1 β π‘π‘
ππ2 =ππ2 β ππ2 β π‘π‘ =
ππ2 β ππ1 β π‘π‘
Volume differential proportional to time
Ξππ = ππ1 β ππ2 = 2ππ1 β π‘π‘
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Fluid Capacitor β Capacitance
Define a relationship between differential volume and pressure
Capacitance
πΆπΆ =ΞππΞππ
Intrinsic device property
Determined by physical parameters:
Membrane area, π΄π΄
Modulus of elasticity, ππ
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Fluid Capacitor β DC vs. AC
In steady-state (DC), no fluid flowsππ1 = ππ2 = 0
Consider sinusoidal Ξππ (AC):Ξππ = ππ sin πππ‘π‘
Resulting flow rate is proportional to: Rate of change of
differential pressure Capacitance
ππ1 = ππ2 = πΆπΆπππππππ‘π‘ = πππΆπΆππ cos πππ‘π‘
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Fluid Capacitor β Time-Varying Ξππ
Equal and opposite flow at both ports
ππ1 = ππ2
Not the same fluid flowing at both ports Fluid cannot permeate the membrane
Fluid appears to flow through the device Due to the displacement of the
membrane A displacement flow
The faster Ξππ changes, the higher the flow rate
ππ β ππ
The larger the capacitance, the higher the flow rate
ππ β πΆπΆ
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Fluid Capacitor β Changing Ξππ
A given Ξππ corresponds to a particular membrane displacement
Step change in displacement/pressure is impossible Would require an infinite
flow rate
Pressure across a fluid capacitor cannot change instantaneously
Forces must balance
Membrane cannot instantaneously jump from one displacement to another
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Fluid Capacitor β Energy Storage
Stretched membrane stores energy Potential energy
Energy released as membrane returns ππ and ππ are supplied
Not a real device, but analogous to other potential energy storage methods PHES CAES Electrical capacitors
Stored energy proportional to: Ξππ Ξππ
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Electrical Capacitor
In the electrical domain, our βworking fluidβ is positive electrical charge
In either domain, we have a potential-driven flow
Fluid Domain Electrical Domain
Pressure β P Voltage β V
Volumetric flow rate β Q Current β I
Volume β V Charge β Q
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Electrical Capacitor
Parallel-plate capacitor Parallel metal plates Separated by an insulator
Applied voltage creates charge differential Equal and opposite charge
ππ1 = βππ2 Zero net charge
Equal currentπΌπΌ1 = πΌπΌ2
What flows in one side flows out the other Schematic symbol:
Units: Farads (F)
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Electrical Capacitor β Electric Field
Charge differential results in an electric field, π¬π¬, in the dielectric Units: ππ/ππ
|π¬π¬| is inversely proportional to dielectric thickness, ππ
Above some maximum electric field strength, dielectric will break down Conducts electrical current Maximum capacitor voltage
rating
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Electrical Capacitor - Capacitance
Capacitance Ratio of charge to
voltage
πΆπΆ =ππππ
Intrinsic device property Proportional to physical
parameters: Dielectric thickness, ππ Dielectric constant, ππ Area of electrodes, π΄π΄
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Parallel-Plate Capacitor
Capacitance
πΆπΆ =πππ΄π΄ππ
ππ: dielectric permittivity π΄π΄: area of the plates ππ: dielectric thickness
Capacitance is maximized by using: High-dielectric-constant
materials Thin dielectric Large-surface-area plates
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Capacitors β Voltage and Current
ππ π‘π‘ = πΆπΆπππππππ‘π‘
Voltage across capacitor results from an accumulation of charge differential Capacitor integrates current
ππ π‘π‘ =1πΆπΆβ« ππ π‘π‘ πππ‘π‘
Current through a capacitor is proportional to Capacitance Rate of change of the voltage
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Voltage Change Across a Capacitor
For a step change in voltage,πππππππ‘π‘
= β
The corresponding current would be infinite
Voltage across a capacitor cannot change instantaneously
Current can change instantaneously, but voltage is the integral of current
limΞtβ0
ΞV = limΞtβ0
οΏ½π‘π‘0
π‘π‘0+Ξπ‘π‘ππ π‘π‘ πππ‘π‘ = 0
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Capacitors β Open Circuits at DC
Current through a capacitor is proportional to the time rate of change of the voltage across the capacitor
ππ π‘π‘ = πΆπΆπππππππ‘π‘
A DC voltage does not change with time, soπππππππ‘π‘
= 0 and ππ π‘π‘ = 0
A capacitor is an open circuit at DC
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Capacitors in Parallel
Total charge on two parallel capacitors is
ππ = ππ1 + ππ2ππ = πΆπΆ1ππ + πΆπΆ2ππππ = πΆπΆ1 + πΆπΆ2 ππ
ππ = πΆπΆππππππ
Capacitances in parallel add
πΆπΆππππ = πΆπΆ1 + πΆπΆ2
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Capacitors in Series
Total voltage across the series combination is
ππ = ππ1 + ππ2
ππ =πππΆπΆ1
+πππΆπΆ2
ππ = ππ1πΆπΆ1
+1πΆπΆ2
=πππΆπΆππππ
The inverses of capacitors in series add
πΆπΆππππ =1πΆπΆ1
+1πΆπΆ2
β1
=πΆπΆ1πΆπΆ2πΆπΆ1 + πΆπΆ2
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Constant Current Onto a Capacitor
Capacitor voltage increases linearly for constant current
ππ π‘π‘ = πΌπΌ π‘π‘βπ‘π‘0πΆπΆ
, π‘π‘ β₯ π‘π‘0
π π π π π π π π π π =πΌπΌπΆπΆ
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Electrical Capacitor β Energy Storage
Capacitors store electrical energy Energy stored in the
electric field
Stored energy is proportional to: Voltage Charge differential
πΈπΈ =12ππππ =
12πΆπΆππ2 =
12ππ2
πΆπΆ
Energy released as E-field collapses ππ and πΌπΌ supplied
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Energy Storage β Example
A capacitor is charged to 100 V The stored energy will be used to lift a 1000 kg elevator car 10 stories (35 m) Determine the required capacitance
The required energy is
πΈπΈ = πππππ = 1000 ππππ β 9.81πππ π 2β 35 ππ
πΈπΈ = 343.4 ππππ
Energy stored on the capacitor is
πΈπΈ =12πΆπΆ 100 ππ 2
The required capacitance is
πΆπΆ =2 β 343.4 ππππ
100 ππ 2 = 68.7 πΉπΉ
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Ultracapacitors30
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Ultracapacitors - Introduction
Energy stored by a capacitor
πΈπΈ =12πΆπΆππ2
Would like to maximize capacitance in order to maximize energy storage Recall the capacitance of a parallel-plate capacitor
πΆπΆ =πππ΄π΄ππ
To increase capacitance: Use a higher-permittivity dielectric Increase surface area of the plates Decrease dielectric thickness
Traditional capacitors do all of these things ππ limited by available materials and dielectric strength π΄π΄ limited by practical overall device size ππ limited by dielectric breakdown field strength
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Traditional Capacitors β Construction
Letβs take a look at the construction of two high-capacitance traditional capacitors Aluminum electrolytic Tantalum electrolytic
Aluminum electrolytic capacitor:
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Traditional Capacitors β Construction
Tantalum electrolytic capacitor:
In both of these types of capacitors, efforts are made to maximize π΄π΄ and minimize ππ
But, a physical dielectric layer of non-zero thickness is used
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Ultracapacitors
In a previous example, we found we needed a capacitance of 68.7 F Impractically large for a traditional capacitor Not so for an ultracapacitor
Ultracapacitors or supercapacitors achieve very high capacitance values by eliminating the solid dielectric layer of traditional capacitors
Energy is stored in an E-field Not in a dielectric layer In an electric double layer (Helmholtz double layer) Electric double-layer capacitors (EDLC)
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Ultracapacitors
Electric double-layer capacitor
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Ultracapacitors
Electrodes are rough and porous Surface area is increased Activated charcoal Aerogel
No charge transfer between the electrolyte and the electrode
Separator is permeable Mechanical separation
preventing contact between electrodes
Source: Yves Brunet, Energy Storage
Thickness of double layers is on the molecular scale
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Ultracapacitors
Two double layers Two capacitors in series
Capacitance values in the range of 1 β¦ 1000s of farads are common
Ultracapacitors are polarized Positive electrode must be kept at a higher potential
Maximum voltage determined by the electrolyte dissociation voltage Typically ~2.5 V For higher-voltage operation, multiple ultracapacitors are connected in series
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Equivalent Circuit Model38
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Equivalent Circuit Model
Ultracapacitor equivalent circuit model
Rs: equivalent series resistance (ESR) Primarily due to ionic conduction in the electrolyte
C0: primary capacitance of the ultracapacitor Cv: voltage-dependent capacitance
Associated with diffusion layers near the double layers Cv = kβ v
Rleak: leakage resistance Typically specified as a leakage current at Vmax
R1, C1, β¦ Rn, Cn: distributed resistance and capacitance of the porous electrodes Models multiple time constants
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Equivalent Circuit Model
We will typically simplify this model significantly Account for only capacitance and ESR Typical ESR values: 0.5 mΞ© β¦ 500 mΞ©
Account for leakage resistance, Rleak, when appropriate Self-discharge Typical leakage resistance: 100 Ξ© β¦ 100 kΞ© Typical leakage currents: 10ππA β¦ 10 mA
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Charging and Discharging41
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Charging and Discharging
The voltage seen across a capacitor is proportional to the stored charge differential
ππ =πππΆπΆ
So, unlike batteries, capacitor voltage does not remain constant as a capacitor discharges
Power electronic circuitry generally required to interface between ultracapacitors and load DC-DC converters Inverters β DC-AC and AC-DC converters
Interface circuitry also provides charge/discharge control Current/power control
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Charging and Discharging
Two primary modes of charging/discharging Constant current Constant power
Unlike batteries, capacitors can be charged and discharged at the same rates
Constant-current charging is simple Both in terms of circuitry and analysis/design
Constant-power charging useful in many applications, such as regenerative braking Charging while drawing constant power from the vehicle Discharging while supplying constant power to the vehicle
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Constant-Current Charging
Voltage drop across Rs during charge/discharge
Constant rate of voltage change
Power varies
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Constant-Power Charging
Varying rate of voltage change
Current varies depending on state of charge Higher current at lower
state of charge Lower current near full
charge
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Cell Balancing46
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Cell Balancing
Typically, ππππππππ = 2.5 ππβ¦ 3.0 ππ Series-connected cells provide higher voltages
Consider a series connection of four cells Equal charge differential, Β±ππ, on each cell The voltage across each capacitor is
ππ1 = πππΆπΆ1
, ππ2 = πππΆπΆ2
, ππ3 = πππΆπΆ3
, ππ4 = πππΆπΆ4
Nominally, all capacitors are equal
πΆπΆ1 = πΆπΆ2 = πΆπΆ3 = πΆπΆ4 = πΆπΆ
Nominally, all voltages are equal
ππ1 = ππ2 = ππ3 = ππ4 =πππΆπΆ = 2.5V
But, capacitances may vary by as much as Β±20%
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Cell Balancing
Consider the following scenario: Total equivalent capacitance
πΆπΆππππ = 0.24πΆπΆ
Stored chargeππ = 10 ππ β 0.24πΆπΆ
Now, cell voltages are imbalanced
ππ1 = ππ2 =ππ
1.2πΆπΆ =10 ππ β 0.24πΆπΆ
1.2πΆπΆ = 2 ππ
ππ3 = ππ4 =ππ
0.8πΆπΆ =10 ππ β 0.24πΆπΆ
0.8πΆπΆ = 3 ππ
If ππππππππ = 2.5 ππ, then πΆπΆ1 and πΆπΆ2 are underutilized πΆπΆ3 and πΆπΆ4 are overstressed
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Cell Balancing
Cell balancing circuitry Safely utilize each cellβs storage
capacity
Two balancing approaches: Resistive balancing Resistors placed in parallel with the cells Slow β not for high-duty-cycle
applications
Active balancing Cell voltages monitored and electronic
switches balance voltages Fast β good for high-duty-cycle
applications
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Efficiency50
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Ultracapacitors β Efficiency
Ultracapacitors have small, but non-zero, ESR They are lossy devices Not all input energy is available for use Efficiency is less than 100%
We will define round-trip efficiency as the efficiency through an entire charge/discharge cycle Ratio of output energy to input energy
πππππ‘π‘ = πΈπΈπππππππΈπΈππππ
β 100% (1)
Efficiency depends on how a capacitor is used Rate of charge/discharge Depth of discharge
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Ultracapacitors β Efficiency
Energy stored by a capacitor is proportional to the capacitor voltage squared
πΈπΈππ = 12πΆπΆππ2 (2)
Capacitorβs effectiveness at storing energy depends on its state of charge (SOC) Energy stored more quickly at high SOC Energy stored more slowly at low SOC
Loss in ESR depends on the current
Therefore, instantaneous efficiency, ππ π‘π‘ , varies with SOC Total round-trip efficiency depends on depth of discharge Ultracapacitors are typically not discharged completely
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Discharge Factor
Discharge factor
ππ = ππππππππππππππππ
(3)
ππππππππ: voltage at the lowest allowable SOC ππππππππ: maximum allowable (fully-charged) capacitor voltage
Weβll now examine the round-trip efficiency for capacitors operated at constant current and at constant power
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Efficiency β Constant Current
For a capacitor operating at a non-zero discharge factor, only some of the stored energy is usable
Usable energy
πΈπΈπ’π’ =12πΆπΆππππππππ2 β
12πΆπΆππππππππ2
πΈπΈπ’π’ =12πΆπΆππππππππ2 β
12πΆπΆ ππππππππππ 2
πΈπΈπ’π’ = 12πΆπΆππππππππ2 1 β ππ2 (4)
Power dissipated in the ESR at constant current, πΌπΌ, isπππ π = πΌπΌ2π π π π
Energy lost in the ESR isπΈπΈπ π = πΌπΌ2π π π π β π‘π‘
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Charging efficiency β Constant Current
During the charging cycle, the efficiency is
ππ1 = πΈπΈπππΈπΈππ+πΈπΈπ π
=12πΆπΆππππππππ
2 1βππ212πΆπΆππππππππ
2 1βππ2 +πΌπΌ2π π π π β π‘π‘ππ
where π‘π‘ππ is the duration of the charging cycle
We can solve for π‘π‘ππ as follows
ππππππππ β ππππππππ = πΌπΌβ π‘π‘πππΆπΆ
β π‘π‘ππ = πΆπΆππππππππ 1βπππΌπΌ
The efficiency then becomes
ππ1 =12πΆπΆππππππππ
2 1βππ212πΆπΆππππππππ
2 1βππ2 +πΌπΌπ π π π πΆπΆππππππππ 1βππ
ππ1 =12ππππππππ 1βππ2
12ππππππππ 1βππ2 +πΌπΌπ π π π 1βππ
(5)
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Round-Trip Efficiency β Constant Current
Similar loss is incurred in the ESR during discharge Energy output is the stored energy minus resistive loss
Round-trip efficiency is
πππππ‘π‘ = πΈπΈπππππππΈπΈππππ
= πΈπΈππβπΈπΈπ π πΈπΈππ+πΈπΈπ π
πππππ‘π‘ =12πΆπΆππππππππ
2 1βππ2 βπΌπΌπ π π π πΆπΆππππππππ 1βππ12πΆπΆππππππππ
2 1βππ2 +πΌπΌπ π π π πΆπΆππππππππ 1βππ
πππππ‘π‘ =12ππππππππ 1βππ2 βπΌπΌπ π π π 1βππ12ππππππππ 1βππ2 +πΌπΌπ π π π 1βππ
(6)
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Instantaneous Efficiency β Constant Current
At lower SOC: Low rate of
energy storage Low efficiency
At higher SOC: Higher rate of
energy storage Higher efficiency
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Efficiency vs. Current β Constant Current
As current increases, loss in ESR increases
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Efficiency vs. Discharge Factor β Constant Current
Greater depth-of-discharge corresponds to lower efficiency
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Efficiency β Constant Power
For constant-power charging/discharging, we start with the same expression for efficiency
πππππ‘π‘ =πΈπΈπ’π’ β πΈπΈπ π πΈπΈπΈπΈ + πΈπΈπ π
Usable energy is the same
πΈπΈπ’π’ =12πΆπΆππππππππ2 1 β ππ2
But, since current is now time varying, the energy lost in the resistance (in one direction) is
πΈπΈπ π = οΏ½0
π‘π‘ππππ2 π‘π‘ π π ππ πππ‘π‘
Things get a bit more complicated, as we now need to solve a differential equation to determine the capacitor voltage
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Efficiency β Constant Power
The input to the capacitor is now constant power Write a power-balance equation
ππ β ππ2 π‘π‘ β π π π π β ππ π‘π‘ ππππ π‘π‘ = 0
Current is given by
ππ π‘π‘ = πΆπΆ πππππππππ‘π‘
The power balance becomes
π π π π πΆπΆ2πππππππππ‘π‘
2+ πΆπΆ ππππππ
πππ‘π‘ππππ π‘π‘ β ππ = 0 (7)
An ordinary differential equation in quadratic form
C
Rs
v
+
β
vc
+
β
i
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Efficiency β Constant Power
Applying the quadratic formula to (7), we get
πππππππππ‘π‘
=βπΆπΆππππ π‘π‘ Β± πΆπΆ2ππππ2 π‘π‘ +4π π π π πΆπΆ2ππ
2π π π π πΆπΆ2
Simplifying, and keeping only the valid β+β solution
πππππππππ‘π‘
=βππππ π‘π‘ + ππππ2 π‘π‘ +4π π π π ππ
2π π π π πΆπΆ(8)
For discharging this becomesπππππππππ‘π‘
=βππππ π‘π‘ + ππππ2 π‘π‘ β4π π π π ππ
2π π π π πΆπΆ(9)
We donβt have a nice closed-form solutions to (8) or (9), but we can solve them numerically
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Instantaneous Efficiency β Constant Power
At lower SOC: High current High πΌπΌ2π π loss Low efficiency
At higher SOC: Lower current Lower πΌπΌ2π π loss Higher efficiency
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Efficiency vs. Current β Constant Power
As power increases, current increases and resistive loss increases
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Efficiency vs. Discharge Factor β Constant Power
Greater depth-of-discharge corresponds to lower efficiency
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Approximate Efficiency β Constant Power
Sometimes we may want a quick way to approximate efficiency for constant-power charging/discharging Calculate an average current Calculate efficiency as you would for constant-current charging/discharging
Approximate average current as the average of the maximum and minimum currents
πΌπΌππππππ βπΌπΌππππππ + πΌπΌππππππ
2where
πΌπΌππππππ = ππππππππππ
and πΌπΌππππππ = ππππππππππ
Then, efficiency is approximately given by
πππππ‘π‘ β12ππππππππ 1βππ2 βπΌπΌπππππππ π π π 1βππ12ππππππππ 1βππ2 +πΌπΌπππππππ π π π 1βππ
(10)
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Series-Connected Capacitor Cells67
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Series-Connected Capacitor Cells
To achieve higher working voltages, multiple capacitor cells are connected in series
How does this effect Energy storage? Efficiency?
Consider a series connection of ππ cells, each with a capacitance of πΆπΆ0 and a maximum voltage of ππππππππππ Equivalent circuit model:
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Series-Connected Capacitor Cells
Energy stored:
πΈπΈ =12πΆπΆ0ππ ππ β ππππππππππ 2
πΈπΈ = ππ β 12πΆπΆ0ππππππππππ
2
As expected, this is ππ times the energy stored in a single cell
For a discharge factor of ππ, the usable stored energy is
πΈπΈπ’π’ = ππ12πΆπΆ0ππππππππππ2 β ππ
12πΆπΆ0 ππππππππππππ 2
πΈπΈπ’π’ = ππ 12πΆπΆ0ππππππππππ2 1 β ππ2 (11)
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Cells in Series β Constant Current Energy lost in the resistance during charging (or discharging):
πΈπΈπ π = πΌπΌ2πππ π 0π‘π‘ππ
For constant-current operation
ππππππππ β ππππππππ = ππππππππ 1 β ππ = ππππππππππππ 1 β ππ
ππππππππππππ 1 β ππ =πΌπΌ π‘π‘πππΆπΆ
=πΌπΌ π‘π‘πππΆπΆ0/ππ
The charging (or discharging) time is
π‘π‘ππ =πΆπΆ0ππππππππππ 1 β ππ
πΌπΌ
So, losses during charging (or discharging) are
πΈπΈπ π = πΌπΌ2πππ π 0πΆπΆ0ππππππππππ 1 β ππ
πΌπΌ
πΈπΈπ π = πππΌπΌπ π 0πΆπΆ0ππππππππππ 1 β ππ (12)
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Cells in Series β Constant Current
Round-trip efficiency is
πππππ‘π‘ =πΈπΈπ’π’ β πΈπΈπ π πΈπΈπ’π’ + πΈπΈπ π
Using (11) and (12), we get
πππππ‘π‘ =ππ 1
2πΆπΆ0ππππππππππ2 1 β ππ2 β πππΌπΌπ π 0πΆπΆ0ππππππππππ 1 β ππ
ππ 12πΆπΆ0ππππππππππ
2 1 β ππ2 + πππΌπΌπ π 0πΆπΆ0ππππππππππ 1 β ππ
πππππ‘π‘ =12ππππππππππ 1 β ππ2 β πΌπΌπ π 0 1 β ππ12ππππππππππ 1 β ππ2 + πΌπΌπ π 0 1 β ππ
This is, of course, the same as for a single cell, but Current can be reduced at a higher maximum voltage
Efficiency will improve
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Cells in Series β Constant Power
For constant-poweroperation, weβll again investigate numerically
Round-trip efficiency vs. # of series-connected cells: Assuming:
πΈπΈ = 100 ππππ ππ = 200 ππ R0 = 3 mΞ© ππ = 0.25 ππππππππππ = 2.5 ππ
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Capacitor Bank Sizing73
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Ultracapacitor Sizing
Sizing a capacitor bank involves determining the following parameters Stored energy: πΈπΈ Available power: ππ Maximum voltage: ππππππππ Minimum voltage: ππππππππ Discharge factor: ππ # of cells in series and/or parallel: πππ π ,ππππ Total capacitance: πΆπΆ Cell capacitance: πΆπΆ0 Efficiency: ππ
Sizing procedure depends on which of these parameters are specified
Weβll outline a procedure assuming the specified requirements are: Energy storage Power Voltage range
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Ultracapacitor Sizing Procedure1. Discharge factor
ππ =ππππππππ
ππππππππ
2. Number of series-connected cells
πππ π =ππππππππ
ππππππππππ3. Total required capacitance
πΈπΈ =12πΆπΆππππππππ
2 1 β ππ2 β πΆπΆ =2πΈπΈ
ππππππππ2 1 β ππ2
4. Cell capacitanceπΆπΆ0 = πππ π β πΆπΆ
5. Determine the resulting efficiency using the required power Evaluate numerically or approximate
6. Iterate if necessary Adjust N as needed
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Ultracapacitor Sizing β Example Size a capacitor bank for an energy recovery system for a tower crane with the
following specifications Height: π = 50 ππ Capacity: ππ = 5,000 ππππ Time to lift max load: π‘π‘ππ = 30 π π
Letβs assume we have a power conversion system that can operate over the range of 60 πππ·π·πΆπΆ β¦ 150 πππ·π·πΆπΆ at an efficiency of πππππππ π = 97%
The required energy to lift the maximum load is
πΈπΈ = πππππ β 1πππππππ π
= 5000 ππππ β 9.81πππ π 2β 50 ππ β
10.97
πΈπΈ = 2.53 ππππ
Performing that lift in 30 π π π π π π corresponds to a required power of
ππ =πΈπΈπ‘π‘ππ
=2.53 ππππ
30 π π = 84.3 ππππ
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Ultracapacitor Sizing β Example
The discharge factor is
ππ =ππππππππππππππππ
=60 ππ
150 ππ= 0.4
The required number of cells in series is
πππ π =ππππππππππππππππππ
=150 ππ2.5 ππ
= 60
The total required capacitance is
πΆπΆ =2πΈπΈ
ππππππππ2 1 β ππ2
πΆπΆ =2 β 2.53 ππππ
150 ππ 2 1 β 0.42
πΆπΆ = 267.6 πΉπΉ
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Ultracapacitor Sizing β Example The capacitance of each individual cell is
πΆπΆ0 = πππ π β πΆπΆ = 60 β 267.6 πΉπΉ
πΆπΆ0 = 16.1 πππΉπΉ
So, the capacitor bank would consist of sixty 16.1 πππΉπΉ capacitors connected in series 16.1 πππΉπΉ is a large capacitance β likely unavailable Connect multiple capacitors in parallel
Letβs say we have access to individual capacitor cells with the following specifications
πΆπΆ0 = 3400 πΉπΉ π π 0 = 0.28 ππΞ©
Five capacitors in parallel will give
πΆπΆ0ππ = πππππΆπΆ0 = 5πΆπΆ0 = 17 πππΉπΉ
π π 0ππ =π π 0ππππ
=π π 05
=0.28 ππΞ©
5= 56 ππΞ©
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Ultracapacitor Sizing β Example
Placing 60 of the 5-capacitor groups in series, the total series resistance is
π π π π = πππ π β π π 0ππ = 60 β 56 ππΞ© = 3.36 ππΞ©
For constant-power charge/discharge, we can approximate efficiency
The maximum current is
πΌπΌππππππ =ππ
ππππππππ=
84.3 ππππ60 ππ = 1.4 πππ΄π΄
The minimum current is
πΌπΌππππππ =ππ
ππππππππ=
84.3 ππππ150 ππ = 562 π΄π΄
The approximate average current is
πΌπΌππππππ βπΌπΌππππππ + πΌπΌππππππ
2 = 981 π΄π΄
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Ultracapacitor Sizing β Example
Using the average current, we can approximate the round trip efficiency for the capacitor bank as
πππππ‘π‘ β12ππππππππππ 1 β ππ2 β πΌπΌπππππππ π 0ππ 1 β ππ12ππππππππππ 1 β ππ2 + πΌπΌπππππππ π 0ππ 1 β ππ
πππππ‘π‘ β1.25 ππ 1 β 0.42 β 981 π΄π΄ β 56 ππΞ© 1 β 0.41.25 ππ 1 β 0.42 + 981 π΄π΄ β 56 ππΞ© 1 β 0.4
πππππ‘π‘ β 0.92 β πππππ‘π‘ β 92%
Note that π π 0ππ is used, because that is the resistance of each of the 60 series-connected parallel combinations
Total round-trip efficiency must include the power conversion system
πππππ‘π‘ = 0.97 β 0.92 β 0.97 = 0.87 β πππππ‘π‘ = 87%
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Ultracapacitor Sizing β Example
Solving numerically, we find that the efficiency of the capacitor bank is a bit higher
πππππ‘π‘ = 94.6%
πΌπΌππππππ overestimates the time-average current
Accounting for conversion losses, we have
πππππ‘π‘ = 89%
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Ultracapacitor Efficiency Summary82
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Efficiency Summary
Configuration Round-trip efficiency
Single capacitor
πππππ‘π‘ =12ππππππππππ 1 β ππ2 β πΌπΌπ π 0 1β ππ12ππππππππππ 1 β ππ2 + πΌπΌπ π 0 1β ππ
πππ π in series
πππππ‘π‘ =12πππ π ππππππππππ 1 β ππ2 β πΌπΌπππ π π π 0 1β ππ12πππ π ππππππππππ 1 β ππ2 + πΌπΌπππ π π π 0 1β ππ
πππππ‘π‘ =12ππππππππ 1 β ππ2 β πΌπΌπ π π π 1β ππ12ππππππππ 1 β ππ2 + πΌπΌπ π π π 1β ππ
πππππ‘π‘ =12ππππππππππ 1 β ππ2 β πΌπΌπ π 0 1β ππ12ππππππππππ 1 β ππ2 + πΌπΌπ π 0 1β ππ
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Efficiency Summary
Configuration Round-trip efficiency
ππππ in parallel
πππππ‘π‘ =
12ππππππππππ 1 β ππ2 β πΌπΌ π π 0ππππ
1 β ππ
12ππππππππππ 1 β ππ2 + πΌπΌ π π 0ππππ
1 β ππ
πππππ‘π‘ =12ππππππππππ 1 β ππ2 β πΌπΌπ π 0ππ 1β ππ12ππππππππππ 1 β ππ2 + πΌπΌπ π 0ππ 1β ππ
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Efficiency Summary
All of the expressions on this and the previous two pages are for constant-current charging/discharging For constant power, use an approximate average current
Configuration Round-trip efficiency
πππ π in series,ππππ in parallel
πππππ‘π‘ =
12πππ π ππππππππππ 1 β ππ2 β πΌπΌ πππ π ππππ
π π 0 1 β ππ
12πππ π ππππππππππ 1 β ππ2 + πΌπΌ πππ π ππππ
π π 0 1 β ππ
πππππ‘π‘ =12ππππππππππ 1 β ππ2 β πΌπΌπ π 0ππ 1β ππ12ππππππππππ 1 β ππ2 + πΌπΌπ π 0ππ 1β ππ
πππππ‘π‘ =12ππππππππ 1 β ππ2 β πΌπΌπ π ππππ 1 β ππ12ππππππππ 1 β ππ2 + πΌπΌπ π ππππ 1 β ππ
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