Smart Grids
EE 772
Department of Electrical Engineering Indian Institute of Technology Bombay, India
October 5, 2018
Smart Grids EE 772 Department of Electrical Engineering Indian - - PowerPoint PPT Presentation
Smart Grids EE 772 Department of Electrical Engineering Indian Institute of Technology Bombay, India October 5, 2018 Sizing and operation of Battery Storage Devices Smart Grids 2 / 31 Background With storage devices, renewable energy
EE 772
Department of Electrical Engineering Indian Institute of Technology Bombay, India
October 5, 2018
Smart Grids 2 / 31
stant generation base-load plant, or, renewable energy forecast uncer- tainty can be mitigated
possible choice can be the use of historical dataset
is well established:
low-frequency component to be associated with finite cycle batteries Large life-cycle battery to be used in conjunction with high-frequency component Battery mix will be cost-effective
frequency components within a dataset
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PR (t + (n − 1) · ND) + PS (t + (n − 1) · ND) = PB(n, t); 1 t ND, 1 n NY
Injection from the batteries should cancel out the variability within the stor- age device. Therefore, injection from the batteries should be 180◦ phase apart with zero mean.
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Bitaraf et al., Sizing Energy Storage to Mitigate Wind Power Forecast Error Impacts by Signal Processing Techniques
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Objective: We want to contain all the daily variabilities within the batteries for both high and low frequency components for a given cut-off frequency. Challenge: (i) If the sizing of storage devices is given injection from RE-BSD can not be constant. (ii) With asymmetric charging and discharging characteristics, the average injection won’t simply be the daily sample average. Solution: The objective is to ‘minimize the squared sum of injection of variability into the grid’.
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The Objective Function: min
PK
d (t) Total RE-BSD dispatch
− {PK
d (t)}
2
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The constraints:
GK(t) − PK
g (t) − PK d (t) = 0
Λ(t) = PK
g (t)
|(PK
g (t))| + ǫ
b (t) − ηch · PK g (t)
b (t) − PK g (t)
ηdch
kh
PK
b (t) = 0 Smart Grids 9 / 31
Total energy contained: QK(t) = C0 + kh T
t=1 PK b (t)
where, 1 T ND C0 is the initial charge stored within the battery
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To ensure complete utilization of the batteries for the complete historical data set, while maintaining desired depth of discharge difference. This ensures preservation of the life of the battery. Capacity of battery ‘+’: CK
+ = max{QK} − C0
∆SOC Capacity of battery ‘-’: CK
− = C0 − min{QK}
∆SOC
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Two methods to determine the life are compared: (i) Depth of Discharge based method, (ii) Throughput method
(DOD) may not remain constant
calculation will be tedious
be calculated as, Lifebatt = 1 m
i=1 Ni/CFi
as, Υ = kh
|Pb(t)|
Y = F · C
number of cycles at standard operating condition.
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Throughput of the battery: TK = kh
b (t)
PK = max
g (t)
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Challenge: We can calculate capacity ratings, throughput of batteries and power rating of the converter
Benefit of random sampling:
moderate accuracy.
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If the capacity ratings, throughput of batteries and power rating of the converter calculated based on randomly sampled days follows normal distribution 1 C+ = µ(C+) + 3 · σ(C+) C− = µ(C−) + 3 · σ(C−) P = µ(P) + 3 · σ(P) T = µ(T) + 3 · σ(T) P{µ − 3σ < X < µ + 3σ} = 0.99730 = ⇒ events with |X − µ| > 3σ are virtually impossible Selection of σ in sizing is upto the discretion of the planner.
1What if the samples does not follow normal distribution? Smart Grids 15 / 31
C = C+ + C− SOC
′
avg =
C− C+ + C− + ǫ Average SOC may not reside at 50% capacity.
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‘1C-rate’ of the battery is the required constant current output from batteries to discharge it within one hour Crate = |Pb(t)| C ; ∀t Increasing capacity rating decreases C-rate Increasing current-drawn has a negative effect on life Statistically calculated C-rate (R): R =
b (t)
C
R = µ(RK) + 3 · σ(RK) If, R > Rlim, to improve the life of batteries, modify calculation of batteries to: C′ = C · R Rlim
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The prime goal: Find the optimal cut-off frequency, that minimizes the annualized sum of the cost of battery and PE-converter. Challenge: Calculation of sizing of batteries and PE-Converters at each cut-off frequency is computationally intensive. Solution: Statistically select finite number of cut-off frequencies to evaluate the total cost. Based on these ‘costly’ samples generate a probability distribution of the cut-off frequencies so as to enable random selection of new cut-off frequencies where probability of selection of cut-off frequency near the existing optima is the highest.
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