## 211 Battery Modelling

2.11.1 The purpose of battery modelling

Modelling (or simulating) of engineering systems is always important and useful. It is done for different reasons. Sometimes models are constructed to understand the effect of changing the way something is made. For example, we could construct a battery model that would allow us to predict the effect of changing the thickness of the lead oxide layer of the negative electrodes of a sealed lead acid battery. Such models make extensive use of fundamental physics and chemistry, and the power of modern computers allows such models to be made with very good predictive powers.

Other types of model are constructed to accurately predict the behaviour of a particular make and model of battery in different circumstances. This model will then be used to predict the performance of a vehicle fitted with that type of battery. This sort of model relies more on careful analysis of real performance data than fundamental physics and chemistry.

In this book we will concern ourselves only with the latter type of performance modelling. However, all modelling of batteries is notoriously difficult and unreliable. The performance of a battery depends on reasonably easily measurable quantities such as its temperature, and performance characteristics such as voltage. However, it also depends on parameters far harder to specify precisely, such as age, and the way the battery has been used (or misused) in the past. Manufacturing tolerances and variation between the different cells within a battery can also have a big impact on performance.

The result of these problems is that all we can do here is give an introduction to the task of battery simulation and modelling.

### 2.11.2 Battery equivalent circuit

The first task in simulating the performance of a battery is to construct an equivalent circuit. This is a circuit made up of elements, and each element has precisely predictable behaviour.

We introduced such an equivalent circuit at the beginning of this chapter. Figure 2.1 is a very simple (but still highly useful) equivalent circuit for a battery. A limitation of this type of circuit is that it does not explain the dynamic behaviour of the battery at all. For example, if a load is connected to the battery the voltage will immediately change to a new (lower) value. In fact this is not true; rather, the voltage takes time to settle down to a new value.

Figure 2.13 shows a somewhat more refined equivalent circuit that simulates or models these dynamic effects quite well. We could carry on refining our circuit more and more to give an ever-closer prediction of performance. These issues are discussed in the literature, for example by Johnson et al. (2001).

The purpose of our battery simulations is to be able to predict the performance of electric vehicles, in terms of range, acceleration, speed and so on, a topic covered in reasonable depth in Chapter 7. In these simulations the speed of the vehicles changes fairly slowly, and the dynamic behaviour of the battery makes a difference that is small compared to the other approximations we have to make along the way. Therefore, in this introduction to battery simulation we will use the basic equivalent circuit of Figure 2.1.

Although the equivalent circuit of Figure 2.1 is simple, we do need to understand that the values of the circuit parameters (E and R) are not constant. The open circuit voltage of the battery E is the most important to establish first. This changes with the state of charge of the battery.

In the case of the sealed lead acid battery we have already seen that the open circuit voltage E is approximately proportional to the state of charge of the battery, as in Figure 2.6. This shows the voltage of one cell of a battery. If we propose a battery variable DoD, meaning the depth of discharge of a battery, which is zero when fully charged

and 1.0 when empty, then the simple formula for the open circuit voltage is:

where n is the number of cells in the battery. This formula gives reasonably good results for this type of battery, though a first improvement would be to include a term for the temperature, because this has a strong impact.

In the case of nickel-based batteries such a simple formula cannot be constructed.

The voltage/state of charge curve is far from linear. Fortunately it now very easy to use

mathematical software, such as MATLAB , to find polynomial equations that give a very good fit to the results. One such, produced from experimental results from a NiCad traction battery is:

/ -8.2816DoD7 + 23.5749DoD6 - 30DoD5 + 23.7053DoD4 \ E = n x , 0 (2.11)

V -12.5877DoD3 + 4.1315DoD2 - 0.8658DoD + 1.37 J ( )

The purpose of being able to simulate battery behaviour is to use the results to predict vehicle performance. In other words we wish to use the result in a larger simulation. This is best done in software such as MATLAB® or an EXCEL® spreadsheet. Which program is used depends on many factors, including issues such as what the user is used

to. For the purposes of a book like this, MATLABw is the most appropriate, since it is

very widely used, and it is much easier than EXCEL to explain what you have done and how to do it.

A useful feature of MATLAB® is the ability to create functions. Calculating the value of E is a very good example of where such a function should be used. The MATLAB® function for finding E for a lead acid battery is as follows:

function E_oc=open_circuit_voltage_LA(x,N) % Find the open circuit voltage of a lead acid % battery at any value of depth of discharge % The depth of discharge value must be between % 0 (fully charged) and 1.0 (flat).

error('Depth of discharge >1')

The function for a NiCad battery is identical, except that the last line is replaced by a formula corresponding to equation (2.11).

Our very simple battery model of Figure 2.1 now has a means of finding E, at least for some battery types. The internal resistance also needs to be found. The value of R

is approximately constant for a battery, but it is affected by the state of charge and by temperature. It is also increased by misuse, and this is especially true of lead acid batteries. Simple first-order approximations for the internal resistance of lead acid and nickel-based batteries have been given in equations (2.3) and (2.9).

### 2.11.3 Modelling battery capacity

We have seen in Section 2.2.2 that the capacity of a battery is reduced if the current is drawn more quickly. Drawing 1A for 10 hours does not take the same charge from a battery as running it at 10 A for 1 hour.

This phenomenon is particularly important for electric vehicles, as in this application the currents are generally higher, with the result that the capacity might be less than is expected. It is important to be able to predict the effect of current on capacity, both when designing vehicles, and when making instruments that measure the charge left in a battery: battery fuel gauges. Knowing the depth of discharge of a battery is also essential for finding the open circuit voltage using equations such as (2.10) and (2.11).

The best way to do this is using the Peukert model of battery behaviour. Although not very accurate at low currents, for higher currents it models battery behaviour well enough.

The starting point of this model is that there is a capacity, called the Peukert Capacity, which is constant, and is given by the equation:

where k is a constant (typically about 1.2 for a lead acid battery) called the Peukert Coefficient. This equation assumes that the battery is discharged until it is flat, at a constant current I A, and that this will last T h. Note that the Peukert Capacity is equivalent to the normal Amphours capacity for a battery discharged at 1 A. In practice the Peukert Capacity is calculated as in the following example.

Suppose a battery has a nominal capacity of 40 Ah at the 5 h rate. This means that it has a capacity of 40 Ah if discharged at a current of:

If the Peukert Coefficient is 1.2, then the Peukert Capacity is:

We can now use equation (2.12) (rearranged) to find the time that the battery will last at any current I.

The accuracy of this Peukert model can be seen by considering the battery data shown in Figure 2.2. This is for a nominally 42 Ah battery (10 h rate), and shows how the capacity changes with discharge time. This solid line in Figure 2.14 shows the data of Figure 2.2 in

Comparison of measured and "Peukert predicted" capacities at different discharge currents

Comparison of measured and "Peukert predicted" capacities at different discharge currents

10 15 20 25 30 Discharge current/Amps

Figure 2.14 Showing how closely the Peukert model fits real battery data. In this case the data is from a nominally 42 V lead acid battery

10 15 20 25 30 Discharge current/Amps

Figure 2.14 Showing how closely the Peukert model fits real battery data. In this case the data is from a nominally 42 V lead acid battery

a different form, i.e. it shows how the capacity declines with increasing discharge current. Using methods described below, the Peukert Coefficient for this battery has been found to be 1.107. From equation (2.12) we have:

Using this, and equation (2.15), we can calculate the capacity that the Peukert equation would give us for a range of currents. This has been done with the crosses in Figure 2.14. As can be seen, these are quite close to the graph of the measured real values.

The conclusion from equations (2.12) and (2.15) is that if a current I flows from a battery, then, from the point of view of the battery capacity, the current that appears to flow out of the battery is Ik A. Clearly, as long as I and k are greater than 1.0, then Ik will be larger than I.

We can use this in a real battery simulation, and we see how the voltage changes as the battery is discharged. This is done by doing a step-by-step simulation, calculating the charge removed at each step. This can be done quite well in EXCEL®, but for reasons explained earlier MATLAB® will be used here.

The time step between calculations we will call St. If the current flowing is I A, then the apparent or effective charge removed from the battery is:

St x Ik

There is a problem of units here. If St is in seconds, this will be have to be divided by 3600 to bring the units into Amphours. If CRn is the total charge removed from the battery by the nth step of the simulation, then we can say that:

St x Ik

It is very important to keep in mind that this is the charge removed from the plates of the battery. It is not the total charge actually supplied by the battery to the vehicle's electrics. This figure, which we could call CS (charge supplied), is given by the formula:

This formula will normally give a lower figure. As we saw in the earlier sections, this difference is caused by self-discharge reactions taking place within the battery.

The depth of discharge of a battery is the ratio of the charge removed to the original capacity. So, at the nth step of a step-by-step simulation we can say that:

Here Cp is the Peukert Capacity, as from equation (2.12). This value of depth of discharge can be used to find the open circuit voltage, which can then lead to the actual terminal voltage from the simple equation already given as Equation (2.1).

To simulate the discharge of a battery these equations are 'run through', with n going from 1, 2, 3, 4, etc., until the battery is discharged. This is reached when the depth of discharge is equal to 1.0, though it is more common to stop just before this, say when DoD is 0.99.

The script file below runs one such simulation for a NiCad battery.

% Simple battery constant current discharge experiment for % a large 5 cell NiCad battery. The time step is set to 50 % seconds, which is sufficiently small for such a constant % current experiment.

% We need to form some arrays for holding data. The array T % is for time, which will run from 0 to 50000 seconds, in 50 % 50 second steps. T=(0:50:50000);

% This corresponds to 1001 values. We form four more arrays, % each also with 1001 elements, and all with initial values % of zero. Dod(n) is used to store values of the depth of % discharge, V(n) stores voltage values, CR(n)and CS(n) % store values of the charge, in Amphours, removed from the % battery and supplied by the battery. CR=zeros(1,1001); % Charged removed from electrodes,

% corrected using Peukert coefficient. DoD=zeros(1,1001); % Depth of discharge, start off fully % charged.

V=zeros(1,1001); % Battery voltage at each time step CS=zeros(1,1001); % Charge supplied by the battery in Ah

% We now set some constants for the experiment I = 30; % Set discharge current to 3 0 Amps

NoCells=5; % 5 cell battery

Capacity=50; % This is the normal 3 h rated capacity of the % battery k=1.045; % Peukert coefficient, not much greater than 1. deltaT =50; % Take 10 second time steps, OK for con I. % Calculated values

Rin = (0.06/Capacity)*NoCells; % Internal resistance, eq 2.9 PeuCap = ((Capacity/3)^k)*3; % See equation 2.12 % Starting voltage set outside loop

V(1) = open_circuit_voltage_NC(0,NoCells) - I*Rin; % Equ 2.1 for n=2:1001

CR(n) = CR(n-1) + ((I^k * deltaT)/3600); % Equation 2.17 DoD(n) = CR(n)/PeuCap; % Equation 2.19 if DoD(n)>1 DoD(n)=1;

V(n)=open_circuit_voltage_NC(DoD(n),NoCells) - I*Rin; % We will say that the battery is "dead" if the % depth of discharge exceeds 99% if DoD(n)>0.99 V(n)=0;

% We now calculate the real amphours given out by the % battery. This uses the actual current, NOT Peukert % corrected, if V(n)>0

CS(n)=CS(n-1)+ ((I*deltaT)/36 00); % Equation 2.18 else

end end

%The bat. V could be plotted against t, but it is sometimes % more useful to plot against Ah given out. This we do here. plot(CS,V,'b.'); axis([0 55 3.5 7]);

XLABEL('Charge supplied/Amphours'); YLABEL('Battery voltage/Volts');

TITLE('Constant current discharge of a 50 Ah NiCad battery');

This script file runs the simulation at one unchanging current. Figure 2.15 shows the graphs of voltage for three different currents. The voltage is plotted against the actual charge supplied by the battery, as in equation (2.18). The power of this type of simulation can be seen by comparing Figure 2.15 with Figure 2.16, which is a copy of the similar data taken from measurements of the real battery.

Constant current discharge of a 50Ah NiCad battery

Constant current discharge of a 50Ah NiCad battery

nominal | |

capacity | |

5 Amps | |

50 Amps |
——__ |

100 Amps |
1 0.1C |

C | |

2C |

10 15 20 25 30 35 40 Charge supplied/Amphours

45 50

10 15 20 25 30 35 40 Charge supplied/Amphours

45 50

Figure 2.15 Showing the voltage of a 6 V NiCad traction battery as it discharges for three different currents. These are simulated results using the model described in the text

Figure 2.16 Results similar to those of Figure 2.15, but these are measurements from a real battery.

Capacity %C5 (Ah)

Figure 2.16 Results similar to those of Figure 2.15, but these are measurements from a real battery.

### 2.11.4 Simulation a battery at a set power

When making a vehicle go at a certain speed, then it is a certain power that will be required from the motor. This will then require a certain electrical power from the battery. It is thus useful to be able to simulate the operation of a battery at a certain set power, rather than current.

The first step is to find an equation for the current I from a battery when it is operating at a power P W. In general we know that:

If we then combine this with the basic equation for the terminal voltage of a battery, which we have written as equation (2.1), we get:

This is a quadratic equation for I. The normal useful solution6 to this equation is:

This equation allows us to easily use MATLABW or similar mathematical software to simulate the constant power discharge of a battery. The MATLAB® script file below shows this done for a lead acid battery. The graph of voltage against time is shown in Figure 2.17.

% A constant P discharge experiment for a lead acid battery. % The system has 10 batteries, each 12 V lead acid, 50 Ah. % We use 10 s steps, as these are sufficiently small for % a constant power discharge. We set up arrays to store the % data.

T=(0:10:10000); % Time goes up to 10,000 in 10 s steps.

% This is 1001 values. CR=zeros(1,1001); % Charge rem. from bat. Peukert corrected. DoD=zeros(1,1001); % Depth of dis., start fully charged. V=zeros(1,1001); % Battery voltage, initially set to zero. NoCells=60; % 10 of 6 cell (12 Volt) batteries.

Capacity=50; % 50 Ah batteries, 10 h rate capacity k=1.12; % Peukert coefficient deltaT =10; % Take 10s steps, OK for constant power.

% Calculated values

Rin = (0.022/Capacity)*NoCells; % Internal re, Equ. 2.2 PeuCap = ((Capacity/10)^k)*10; % See equation 2.12

% Starting voltage set outside loop E=open_circuit_voltage_LA(0,NoCells);

I = (E - (E*E - (4*Rin*P)) A 0 . 5)/(2*Rin); %Equation 2.20

6 As with all quadratics, there are two solutions. The other corresponds to a 'lunatic' way of operating the battery at a huge current, so large that the internal resistance causes the voltage to drop to a low value, so that the power is achieved with a low voltage and very high current. This is immensely inefficient.

Constant power discharge of a lead acid battery

for n=2:1001

E=open_circuit_voltage_LA(DoD(n-1),NoCells) I = (E - (E*E - (4*Rin*P))^0.5)/(2*Rin); CR(n) = CR(n-1) + ((deltaT * I^k)/3600); DoD(n) = CR(n)/PeuCap; if DoD(n)>1 DoD(n)=1;

% We will say that the battery is "dead" if the % depth of discharge exceeds 99%

V(n)=open_circuit_voltage_LA(DoD(n),NoCells) - I*Rin; %Equ 2.1 if DoD(n)>0.99 V(n)=0; end end plot(T,V,'b.');

YLABEL('Battery voltage/Volts'); XLABEL('Time/Seconds');

TITLE('Constant power discharge of a lead acid battery'); axis([0 4000 100 140]);

When we come to simulate the battery being used in a vehicle, the issue of regenerative braking will arise. Here a certain power is dissipated into the battery. If we look again at Figure 2.1, and consider the situation that the current I is flowing into the battery, then the equation becomes:

If we combine equation (2.21) with the normal equation for power we obtain:

P = V x I = (E + IR) x I = EI + RI2 The 'sensible', normal efficient operation, solution to this quadratic equation is:

The value of R, the internal resistance of the cell, will normally be different when charging as opposed to discharging. To use a value twice the size of the discharge value is a good first approximation.

When running a simulation, we must remember that the power P is positive, and that equation (2.22) gives the current into the battery. So when incorporating regenerative braking into battery simulation, care must be taken to use the right equation for the current, and that equation (2.17) must be modified so that the charge removed from the battery is reduced. Also, it is important to remove the Peukert Correction, as when charging a battery large currents do not have proportionately more effect than small ones. Equation (2.17) thus becomes:

We shall meet this equation again in Chapter 7, Section 7.4.2, where we simulate the range and performance of electric vehicles with and without regenerative braking.

### 2.11.5 Calculating the Peukert Coefficient

These equations and simulations are very important, and will be used again when we model the performance of electric vehicles in Chapter 7. There the powers and currents will not be constant, as they were above, but exactly the same equations are used.

However, all this begs the question 'How do we find out what the Peukert Coefficient is?' It is very rarely given on a battery specification sheet, but fortunately there is nearly always sufficient information to calculate the value. All that is required is the battery capacity at two different discharge times. For example, the nominally 42 Amphours (10hour rating) battery of Figure 2.2 also has a capacity of 33.6 Amphours at the 1 hour rate.

The method of finding the Peukert Coefficient from two amphour ratings is as follows. The two different ratings give two different rated currents:

T1 T2

We then have two equations for the Peukert Capacity, as in equation (2.12):

However, since the Peukert Coefficient is Constant, the right hand sides of both parts of equation (2.25) are equal, and thus:

J2J Ti

Taking logs, and rearranging this gives:

This equation allows us to calculate the Peukert Coefficient k for a battery, provided we have two values for the capacity at two different discharge times T. Taking the example of our 42 Ah nominal battery, equation (2.24) becomes:

Putting these values into equation (2.26) gives:

Such calculations can be done with any battery, provided some quantitative indication is given as to how the capacity changes with rate of discharge. If a large number of measurements of capacity at different discharge times are available, then it is best to plot a graph of log(T) against log(I). Clearly, from equation (2.26), the gradient of the best-fit line of this graph is the Peukert Coefficient.

As a general rule, the lower the Peukert Coefficient, the better the battery. All battery types behave in a similar way, and are quite well modelled using this method. The Peukert Coefficient tends to be rather higher for the lead acid batteries than for other types.

### 2.11.6 Approximate battery sizing

The modelling techniques described above, when used with the models for vehicles described in Chapter 7, should be used to give an indication of the performance that will be obtained from a vehicle with a certain type of battery. However, it is possible, and sometimes useful, to get a very approximate guide to battery range and/or size using the approach outlined below.

A designer may either be creating a new vehicle or alternatively may be adapting an existing vehicle to an electric car. The energy consumption of an existing vehicle will probably be known, in which case the energy used per kilometre can be multiplied by the range and divided by the specific energy of the battery to give an approximate battery mass.

If the vehicle is a new design the energy requirements may be obtained by comparing it with a vehicle of similar design. Should the similar vehicle have an IC engine, the energy consumption can be derived from the fuel consumption and the engine/gear box efficiency.

This method is fairly crude, but none the less may give a reasonable answer which can be analysed later.

For example the vehicle may be compared with a diesel engine car with a fuel consumption of 18km.l-1 (50mpg). The specific energy of diesel fuel is approximately 40kWh.kg-1 and the conversion efficiency of the engine and transmission is approximately 10%, resulting in 4kWh of energy per litre of fuel stored delivered at the wheels.

In order to travel 180 km the vehicle will consume 10 litres of fuel, which weights approximately 11 kg allowing for fuel density. This fuel has an energy value of 440 kWh, and the energy delivered to the wheels will be 44kWh (44000Wh) allowing for the 10% efficiency. This can be divided by the electric motor and transmission efficiency, typically about 0.7 (70%), to give the energy needed from the battery, i.e. 62.8 kWh or 62 800 Wh.

Hence if a lead acid battery is used (specific energy 35 Wh.kg-1) the battery mass will be 1257 kg; if a NiMH battery (specific energy 60 Wh.kg-1) is used the battery mass will be 733 kg; if a sodium nickel chloride battery is used of a specific energy of 86 Wh.kg-1 then the battery mass will be 511 kg; and if a zinc air battery of 230 Wh.kg-1 is used a battery mass of 191kg is needed.

Care must be used when using specific energy figures particularly, for example, when a battery such as a lead acid battery is being discharged rapidly, when the specific energy actually obtained will be considerably lower than the nominal 35 Wh.kg-1. However this technique is useful, and in the case quoted above gives a fairly good indication of which batteries would be ideal, which would suffice and which would be ridiculously heavy. The technique would also give a 'ball park figure' of battery mass for more advanced analysis, using the modelling techniques introduced above, and much further developed in Chapter 7.

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