Distributed Energy Resources - Lecture 5: Batteries and Electric Vehicles

Other notes in this series from Kevin Kircher’s Distributed Energy Resources class are here.

Summary

Lecture 5 covers modelling of batteries and EVs which are just a type of battery that come with some extra constraints.

This was a fun one; some simple models with real world applications and use of bits of theory from lectures 3&4.

Notes

Homework

1. Show that with uniform time step $∆t$ and piecewise constant $p^{chem}(t)$, the continuous-time battery model $\frac{dx(t)}{dt}=-\frac{x(t)}{\tau }+p^{chem}(t)$ can be written in discrete time as $x(k+1)=ax(k)+(1-a)\tau p^{chem}(k)$where $a=e^{-\Delta t/\tau }$

• From last week, if we have the continuous-time vector LDS: $\frac{dx(t)}{dt}=\tilde{A}(t)x(t)+\tilde{B}(t)u(t)+\tilde{w}(t)$
  then the general discrete-time LDS equation is
  $x(k+1)=A(k)x(k)+B(k)u(k)+w(k)$
  • where $A(k)=e^{(t_{k+1}-t_k)\tilde{A}(t_k)}$ and if $\tilde{A}(t_k)$ is invertible:
  • $B(k)=(A(k)-I)\tilde{A}(t_k{)}^{-1}\tilde{B}(t_k)$
  • $w(k)=(A(k)-I)\tilde{A}(t_k{)}^{-1}\tilde{w}(t_k)$
• And the scalar equivalent
  • ${\delta }_x(t)=\tilde{a}(t){\delta }_x(t)+\tilde{b}(t){\delta }_u(t)+\delta \tilde{w}(t)$
  • ${\delta }_x(k+1)=a(k){\delta }_x(k)+b(k)\delta u(k)+{\delta }_w(k)$
  • where
    • $a(k)=e^{\Delta t\tilde{a}(t_k)}$
    • $b(k) = (a(k) − 1)\tilde{b}(t_k)/\tilde{a}(t_k)$$
    • ${\delta }_w(k)=(a(k)-1){\delta }_{\tilde{w}}(t_k)/\tilde{a}(t_k)$
• In this case
  • $\tilde{a}(t)=\frac{-1}{\tau }\Rightarrow a(k)=e^{-\frac{\Delta t}{\tau }}$
  • $\tilde{b}(t)=1\Rightarrow b(k)=-\tau (e^{-\frac{\Delta t}{\tau }}-1)$
  • ${\delta }_{\tilde{w}}(k)=0\Rightarrow {\delta }_w(k)=0$
  • ${\delta }_u(t)=p^{chem}(t)$
• So

  • $x(k+1)=e^{-\frac{\Delta t}{\tau}}x(k) -\tau(e^{-\frac{\Delta t}{\tau}} - 1)p^{chem}(t)$$

  • if we let $a=e^{-\Delta t/\tau }$, then that gives us
    $x(k+1)=ax(k)-\tau (a-I)p^{chem}(t)=\mathbf{ax}(\mathbf{k})+(1-\mathbf{a})\tau {\mathbf{p}}^{\mathbf{chem}}(\mathbf{t})$

2. In the special case of a battery with no self-dissipation, the continuous-time model simplifies to $\frac{dx(t)}{dt}=p^{chem}(t)$ Show that with uniform time step $∆t$ and piecewise constant $p^{chem}(t)$, a discrete-time version of this model is $x(k+1)=x(k)+∆t{p}^{chem}(k)$

• ${\int }_{t_k}^{t_k+1}\frac{dx(t)}{dt}={\int }_{t_k}^{t_k+1}{p}^{chem}(t)$

• $x(t_k+1)-x(t_k)={\int }_{t_k}^{t_k+1}{p}^{chem}(t)dt$

• $x(t_k+1)-x(t_k)=p^{chem}(t){\int }_{t_k}^{t_k+1}dt$ Since $p^{chem}$ is piecewise constant.

• $x(k+1)=p^{chem}\Delta t+x(k)=\mathbf{x}(\mathbf{k})+\mathbf{\Delta t}{\mathbf{p}}^{\mathbf{chem}}(\mathbf{k})$

3. The charge state of a battery, initially at 80% of its energy capacity, drops to 50% of its energy capacity after 30 days unplugged and unused. What is the battery’s self-dissipation time constant?

• $x(t_0)=0.8;x(t_1)=0.5;p^{chem}=0;\tau =?$
• $\frac{dx(t)}{dt}=\frac{-x(t)}{\tau }+p^{chem}(t)=\frac{-x(t)}{\tau }$
• $\int \frac{dx(t)}{dt}=-\frac{1}{\tau }\int x(t)dt$
• $\int \frac{1}{x}dx(t)=-\frac{1}{\tau }\int dt$
• $\ln |x(t)|-\ln |x_0|=-\frac{\Delta t}{\tau }$
• $\ln |\frac{x(t)}{x_0}|=-\frac{\Delta t}{\tau }$
• $\tau =-\frac{\Delta t}{\ln |\frac{x(t)}{x_0}|}=-\frac{(30\ast 24\ast 3600)}{\ln (\frac{0.5}{0.8})}\approx 5514851$

4. Suppose an electric vehicle has an energy intensity of $\alpha =0.3kWh/km$ and a comparable gasoline vehicle gets $\beta =25$ miles per gallon. If burning one gallon of gasoline causes $\gamma =26$ pounds of CO2 emissions (including upstream emissions associated with oil extraction and processing), what is the break-even CO2 intensity of electricity $\mu$ (in units of g/kWh) at which the two vehicles cause the same CO2 emissions per unit distance driven? By what percent would the EV reduce CO2 emissions from driving with the US-average CO2 intensity of electricity, 345 g/kWh? With the average CO2 intensity of electricity in your home state or country? What factors not considered here might complicate this analysis?

• ICE CO2: $26/25=1.04\text{ lbs/mile}=0.47\text{ kg/mile}=0.47/1.61\text{ kg/km}=0.292\text{ kg/km}$
• EV CO2 for breakeven: $292\text{ g/km}=0.3\text{ kWh/km}\ast \mu$
  • $\mu =(292/0.3)\text{ g/kWh}=973.3\text{g/kWh}$
• By what percent would the EV reduce CO2 emissions from driving with the US-average CO2 intensity of electricity, 345 g/kWh?
  • EV emissions/km = $345\text{ g/kWh}\ast 0.3\text{ kWh/km}=103.5\text{g/km}$
  • $\frac{(292-103.5)}{292}\times 100=64.6\%$
• With the average CO2 intensity of electricity in your home state or country?
  • currently 406g/kWh per https://app.electricitymaps.com/zone/IE/72h/hourly
• What factors not considered here might complicate this analysis? Regenerative braking, temperature variation of energy intensity of EV, varying $\mu$ over time.

5. Suppose someone commutes two miles each way, five days per week, riding an electric bike with an energy intensity of $5Wh/km$. If they work 50 weeks per year and electricity costs 0.15 $/kWh:

• How much do they spend on bike electricity per year?
  • $50\times 5\times 2\times (2\times 1.61)\times (5/1000)\times 0.15=1.21$ dollars
• Compare this to the annual fuel cost from the same commute in an automobile that gets 30 miles per gallon with a fuel price of 3 $/gallon
  • $50\times 5\times 2\times 2\times (1/30)\times 3=100$ dollars

6. Like lecture four, there was some not very interesting python scripting.

Batteries

• A simple battery model
  $\frac{dx(t)}{dt}=\frac{-x(t)}{\tau }+p^{chem}(t)$
  • $x(t)\in R(kWh)$ is the stored chemical potential energy
  • $\tau >0(h)$ is the self-dissipation time constant. This $\infty$ for an ideal battery
  • $p^{chem}(t)\in R(kW)$ is the chemical charging power, or discharging if $p^{chem}(t)<0$ - ie: electrical power input/output.
• Electrical charging/discharging power is
  $max\{p^{chem}(t)/{\eta }_c,{\eta }_d{p}^{chem}(t)\}$
  • ${\eta }_c,{\eta }_d\in (0,1]$ are the charging and discharging efficiencies.
• Discrete-time battery model
  $x(k+1)=ax(k)+a(1-a)\tau p^{chem}(k)$
  • $a=e^{\Delta t/\tau }$
• Battery constraints
  • $0\le x(k)\le \bar{x}$
  • $-{\bar{p}}_d\le p(k)\le \overline{p_c}$
    • $\bar{x}\ge 0(kWh)$ is the chemical energy capacity
    • $\overline{p_c}\ge 0(kW)$ is the electrical charging power capacity
    • $\overline{p_d}\ge 0(kW)$ is the electrical discharging power capacity
    • So then power constraints are
      $-\frac{{\bar{p}}_d}{{\eta }_d}\le p^{chem}(k)\le {\eta }_c{\bar{p}}_c$
• unused, a typical battery might lose ∼1 to 3% energy per day
• ${\eta }_c{\eta }_d$ is called the round-trip efficiency, typically ~0.9, and ${\eta }_c$ and ${\eta }_d$ are both ~0.95

EVs

• Just a battery with wheels (and thus special charging/discharging conditions)

• $p^{chem}(k)=-\frac{\alpha (k)d(k)}{\Delta t}$

  • $\alpha (k)(kWh/km)$ is energy intensity of driving (like MPG for ICE cars). Typically 0.15 - 0.4 kWh/km. Ebikes ~0.005kWh/km

• Temperate profile

• plugged in indicator $z(k)=\{\begin{array}{l}1\text{ if the EV is plugged in over time step K}\\ 0\text{ otherwise}\end{array}$

• $\{\begin{array}{ll}-{\bar{p}}_d\le p(k)\le {\bar{p}}_c& \text{ if }z(k)=1\\ p(k)\le 0& \text{ if }z(k)=0\end{array}$
  with ${\bar{p}}_d=0$ for EVs without bidirectional charging

• if $z(k)=1$ (meaning the EV is plugged in) and we want to charge to full:

  • $p^{chem}(k)=min\left\{{\eta }_c{\bar{p}}_c,\frac{\bar{x}-ax(k)}{(1-a)\tau }\right\}$
  • This places $x(k+1)$ at $\bar{x}$ when the battery is nearly full

• if z(k) = 1 and we want to charge steadily to meet a deadline:

  • $p^{chem}(k)=min\left\{{\eta }_c{\bar{p}}_c,\frac{x^{\ast }-a^{k^{\ast }-k}x(k)}{(1+a+\cdots +a^{k^{\ast }-k-1})(1-a)\tau }\right\}$