*4.3. Determining the Number of Extra Buses to Provide Time to Charge*

We now know the number of base buses needed to drive the traffic, but there is also a need to provide time for the buses to charge, and that may require extra buses. Thus, in this section, we determine the number of extra buses needed for charging. Note that the biogas buses and HVO buses do not require any extra buses beyond the base buses. Besides night charging, which is assumed to allow the buses to start each day fully charged, we divide the charging in three categories to make it easier to build and understand the model. The categories of daily charging are:


**Figure 8.** How the number of buses during peak traffic can be reduced by moving charging to other times.

**Figure 9.** How the number of buses during peak traffic can be reduced by moving charging to other times, when the off-peak traffic is almost as high as the peak traffic.

### 4.3.1. Extra Buses for End-Stop Charging for a Whole Day (EndStop1)

With this charging strategy, the buses are charged after each trip, and it is called **EndStop1** in the calculations. The energy charged equals the energy used during the last trip, which means that the buses always starts each trip with the same battery state-of-charge. The number of extra buses required is determined by calculating how much time is required to charge the bus after each trip. The calculation is made for the peak periods, as that is when the greatest number of buses will be charging simultaneously. The amount of energy that the bus must charge at each end stop is:

$$W\_{\text{TripPeak}} = T\_{\text{TripNetPeak}} \times P\_{\text{BusAv}} \tag{7}$$

**Figure 10.** How the number of buses during peak traffic cannot be reduced if the off-peak traffic is the same as the peak traffic.

The charge time is:

$$T\_{\text{ChgPeakEndStop}} = \frac{W\_{\text{TripPeak}}}{P\_{\text{ChgEndStop}}} \tag{8}$$

This charge time can be used to calculate a factor for how many buses need to be charging per bus in traffic:

$$k\_{\text{busChgPeakEndStop}} = \frac{T\_{\text{ChgPeakEndStop}}}{T\_{\text{TripGrossPenak}}} \tag{9}$$

Now the number of extra buses required to allow for charging at end stops during peak times can be determined:

$$N\_{\text{BusEntraEndstop}} = N\_{\text{BusPeak}} \times k\_{\text{BusChgPeakEndStop}} \tag{10}$$

The number of buses charging off-peak and in the evening can be calculated with Equations (7)–(10) using the net trip time off-peak and in the evening. The extra buses required for charging at the end stops are illustrated by the red area in Figure 7, where the grey area is the buses in traffic from Figure 6. In the diagram, the time to drive from, or back to the depot is also shown as light blue segments. The longer the distance between depot and bus route, the longer the light blue segments will be.

From the diagram in Figure 7, we can see that the maximum number of buses used during the day will be during peak traffic:

$$N\_{\text{BusTotalEndStop}1} = N\_{\text{BusPeak}} + N\_{\text{BusExtraEndspot1}} \tag{11}$$

Since the diagram also shows the highest number of buses which are simultaneously charging, it can also be used to determine the number of end-stop chargers required, which will be completed in a later section.

### 4.3.2. Extra Buses for End-Stop Charging during Off-Peak Time Only (EndStop2)

Since the highest number of buses during the day will determine how many buses must be bought, there is a possibility to save on the bus investment if it is possible to change when the buses charge so that fewer buses are needed during the peak times. This will occupy more buses off-peak, but that does not influence the investment in buses if it does not exceed the bus number in the peak times. The lowest number of buses required is the number of buses in traffic during the peak times, so the best we can do, in terms of reducing the number of buses, is to limit them to the highest number of buses in traffic during the peak times. If we do that, it means that no buses can charge during the peak periods.

Figure 8 shows a charging strategy which adds to the charging off-peak in order to compensate for the elimination of the charging in the peak periods, and it is called EndStop2 in the calculations below. The yellow colour shows the charging which cannot be done during the peak since the number of buses have been reduced. This will be compensated for in two ways. Some of the buses which needs extra charging will be driven to the depot and can charge there, illustrated by the green colour in Figure 8. The other buses, which remains in traffic in the midday period, will need to further charge at the end-stop chargers. That extra charging at the end stops is illustrated by the purple colour in Figure 8, and it allows the buses to charge the battery so that it is full before the afternoon peak starts. The charging shown in red is the charging which is needed between the trips just to keep battery state-of-charge the same from trip to trip, and it is the same as the charging shown in Figure 7.

There is a charge deficit also from the afternoon peak, but it will not be necessary to charge the batteries to full capacity again after that peak. The reason why this deficit does not need to be restored is that there will be enough time during the rest of the day to charge at the end stops so that the battery state-of-charge can remain constant from the start of one trip to the start of the next. This deficit can be compensated for before the next day during the night charging in the depot. Thus, we do not need to analyse any extra charging after the second peak. This does not mean that the buses cannot charge a little extra after that peak; it simply means that such charging is not necessary to consider when sizing the number of buses and chargers. In general, there will be possibilities for additional charging also after the afternoon peak, but it will not be necessary to do that.

The buses which are in traffic between the peak times, besides needing to charge in order to keep the batteries from draining, will need to stay for additional time at the end stops to further charge the battery after each trip to ensure that it is full before the second peak. How much extra time is required for the charging depends on the ratio between how long the first peak was, and how long the period between the peak times are. A conservative (high) estimate of how long the buses drained their batteries for during the first peak can be seen in Figure 8:

$$T\_{\text{ChgDeficit1}} = T\_{\text{stopAMPeak}} - T\_{\text{startAMPeak}} \cdot \tag{12}$$

Based on that, a worst-case charging energy deficit can be determined from the number of departures per hour, the trip time, and the average power consumption:

$$W\_{\text{ChgDeficit1}} = T\_{\text{ChgDefici1}} \times n\_{\text{DeptPellourPeak}} \times T\_{\text{TripNetPeak}} \times P\_{\text{BusAverage}} \tag{13}$$

A conservative (low) estimate of the time they can charge up again can also be seen in Figure 8:

$$T\_{\text{ChgExta1}} = T\_{\text{startPMPeak}} - \left(T\_{\text{stopAMPeak}} + T\_{\text{TripGrossOfPeak}}\right) \tag{14}$$

If we assume that all the charging can be completed at midday, without having to add any extra buses, we can determine the number of buses needed to charge as the sum of the buses needed for normal end-stop charging, plus the extra buses needed to be charged:

$$N\_{\text{busMidday2a}} = \frac{N\_{\text{ChgDeficit1}}}{T\_{\text{ExtraChg1}} \times P\_{\text{ChgEndStop}}} + N\_{\text{busMidday1}}.\tag{15}$$

Note that this number in some extreme cases can become higher than the number of buses in traffic during the peak times. That is taken care of in Equation (22) when this number of buses is compared with a calculation of *N*busMidday2b made for the case when we need to add buses to allow some charging also during the peak times.

If the timetable is such that the number of buses in traffic between the peak times is not significantly lower than in the peak times, we obtain the second case where there may not be enough buses available to charge off-peak in order to compensate for the charging deficit from the peak. It will still be possible to move charging from the peak times to off-peak time, but not fully, so some extra buses will be needed to allow for additional charging. Those extra buses do not only increase the possibility of charging between the peak times but will also allow charging during the peak times. Such a case is illustrated in Figure 9, and it can be seen that the number of buses is higher than the highest number of

buses in traffic during the peak times, but it is still lower than what would have been required if the buses were charged to full capacity after each trip as well as also during the peak times.

Thus, if the number of buses required midday, *N*busMidday2a, is higher than the number of buses in traffic during the peak times, the charge balance equation must be altered to include the need for some buses that can charge both during the two peak times as well as in the midday period. In this case we also know that no buses will drive to the depot in the midday period.

Thus, it can be calculated how much extra charging is possible during the midday period for all the buses which are not needed in traffic during the peak times, *N*busPeak

$$\mathcal{W}\_{\text{ExtraMiddayMax}} = \left(\mathcal{N}\_{\text{busPeak}} - \mathcal{N}\_{\text{busMidday1}}\right) \times P\_{\text{ChgEndStop}} \times T\_{\text{ChgExtra1}}.\tag{16}$$

There will be a remining energy deficit if this maximum midday charging is not sufficient:

$$\mathcal{W}\_{\text{Rernain}} = \mathcal{W}\_{\text{ChgDefici}1} - \mathcal{W}\_{\text{ExtraMiddayMax}} \tag{17}$$

The extra buses that need to charge this energy have to do it during the morning peak, the midday period and during the afternoon peak. It may seem strange that the charging can be done also during the afternoon peak, since we earlier stated that the buses should be fully charged before the second peak starts, but that statement was based on the assumption that no charging could take place during the afternoon peak. It will not be important that all buses are fully charged at the beginning of the second peak if some of them avoid draining too much by charging a little also during the peak. The critical factor is that they shall not be at their minimum charge level before the end of the afternoon peak, not that they are fully charged at the beginning of that peak.

Note that it is not the added buses themselves which charge all this energy, as they will have a full battery when the morning peak starts. Instead, the added buses take over the task of driving the trips in order to relieve the other buses so that they can have more time to charge up. The time which one extra bus can be used to relieve other buses is:

$$T\_{\text{ChgExta2}} = T\_{\text{stopPMPeak}} - T\_{\text{startAMPeak}} \cdot \tag{18}$$

It will take some time for the extra buses to relieve the other buses, and there will always be some waiting time for a bus which has been charging before it can start driving trips again. Therefore, it is not realistic to assume that all of the time added by the extra buses can be used for the charging of buses. The fraction of the added time which can be used for charging is:

$$k\_{\text{UtillExraBuysChg}} = 50\%.\tag{19}$$

The charge balance for the added buses lets us calculate the number of extra buses required:

$$N\_{\text{BusEtraEndStop2}} = \frac{W\_{\text{Remain}}}{T\_{\text{ChgExtra2}} \times P\_{\text{ChgEndStop}} \times k\_{\text{UtilExtraBusChg}}} \,\tag{20}$$

Since the equations do not check that the remaining energy is a positive value, this number of extra buses can become negative. This never happens in reality, but according to Equation (22), it will not be a problem since it is compared with the number for the first case to find out what the right number of buses is. The total number of buses needed for the route at midday, for this second case, is:

$$N\_{\text{BusMidday2b}} = N\_{\text{BusExtra}} + N\_{\text{BusPeak}} \cdot \tag{21}$$

We can now use the number of buses determined for the two cases to decide what the actual need of the buses will be in the midday period:

$$N\_{\text{BusMidday2}} = \min \Big[ N\_{\text{BusMidday2a}} \cdot \max \Big( N\_{\text{BusPeak}} \cdot N\_{\text{BusMidday2b}} \Big) \Big]. \tag{22}$$

A third and even more extreme case is if the timetable has the same number of departures per hour during the whole day. This case is illustrated in Figure 10, and we can see that there is no possibility to move any charging to the midday period, so the number of extra buses will be higher. In this case, the strategy to charge off-peak will no longer add any benefit, and the resulting number of buses becomes the same as the number of buses needed for the normal end-stop-charging strategy.

The TCO variable "Number of buses", can now be determined for the charging strategy EndStop2:

$$N\_{\text{busTotal}} = \max\{N\_{\text{busPeak}} \mid N\_{\text{BusMidday2}}\}.\tag{23}$$
