*4.4. Number of Chargers*

There is a need in the depot for one charger per bus for all types of electric buses, and the power demand will depend on the size of the buses' batteries. Thus, there is always a need to build one new substation and pay for one connection to the grid at the depot. For end-stop-charged buses, there is additionally a need for building chargers at both end-stops of the route, so in total three substations and grid connections are needed. Thus, for end-stop-charged buses, the total number of chargers will equal the total number of buses plus the number of end-stop chargers.

The number of end-stop chargers is typically one per end stop for one route, but if the number of buses driving on that route becomes very high, there will be a need to add more chargers if more than one bus at a time need to charge at each end stop. Therefore, the number of end-stop chargers can be determined from the maximum number of buses simultaneously charging at the end stops. This is represented by the red and purple parts in Figures 7–10.

For the EndStop1 charging strategy, the number of charging buses is always highest during the peak periods, while the number is highest between the peak times for EndStop2 charging strategy. It is not realistic to assume that a charger can be used 100% of the time, so we assume that a charger should on average not be used more than *k*MaxUtilEndStopChg of the time, and the maximum utilization is therefore set to 50% in this paper. This will provide a margin to allow for buses to be delayed without having a big influence on other buses' ability to charge. A system with such a margin will also be able to work even if one charger is out of order for a limited time. Thus, the number of end-stop chargers for a bus route with end-stop charging for the whole day will be:

$$N\_{\text{ChgEndStop1}} = \max\left(2, \text{ceil}\left(\frac{N\_{\text{BusEntraEndstop1}}}{k\_{\text{MaxUtillEndStopC}}}\right)\right) \tag{24}$$

where ceil is a function rounding upwards to the nearest integer number.

Note that this means that we can never have less than two end-stop chargers but can have both odd and even number of chargers from three and up. It is possible to have a different number of chargers at the different end stops since bus schedules can be planned so that the buses charge longer on one side of the route than the other. However, we do not allow a system with only one charger per route, since such a system will not be able to maintain service if that single charger fails for longer than a short time.

The strategy to charge between the peak times rather than during the peak times will require another number of end-stop chargers. Still, the number of chargers can be determined by the highest number of simultaneously charging buses; it is just that the highest number of charging buses will occur in the midday period and may require a different number of chargers.

$$N\_{\text{ChgEndStop2}} = \max\left(2, \text{ceil}\left(\frac{N\_{\text{BusMidday2}} - N\_{\text{BusOfPeak}}}{k\_{\text{MaxUellEndStopChg}}}\right)\right). \tag{25}$$

We can now determine the required number of chargers:

$$N\_{\rm chg} = N\_{\rm BusTotal} + N\_{\rm ChgEndStop} \tag{26}$$

The total power of all the chargers is also needed in order to determine the cost of the chargers:

$$P\_{\rm chgTotal} = N\_{\rm busTotal} \times P\_{\rm ChgNlight} + N\_{\rm ChgEndStop} \times P\_{\rm chgEndstop} \tag{27}$$

The end stop charger power is *P*ChgEndStop = 300 kW and the depot charger power *P*ChgNight is 11 kW for EndStop1 and 22 kW for EndStop2.

## *4.5. Calculating Energy Use and Driving Distance*

Energy use can be determined from all the driving by all the buses. For the cost analysis, we do not need to know which bus is driving where, just the sum of all the driving. This is the sum of driving the trips plus driving to and from the depot. The number of trips during a day can be determined from the timetable parameters. All the trip numbers are multiplied by a factor of two since the route has the same departures from both directions. First, it is determined how long during the day the bus route operates at different numbers of departures per hour:

$$T\_{\rm Penk} = \left(T\_{\rm stopPMPeak} - T\_{\rm startAMPeak}\right) + \left(T\_{\rm stopPMPeak} - T\_{\rm startPMPeak}\right) \tag{28}$$

$$T\_{\text{OffPeak}} = T\_{\text{startEvering}} - T\_{\text{start}} - T\_{\text{Peak}} \tag{29}$$

$$T\_{\text{Evering}} = T\_{\text{stop}} - T\_{\text{startEvering}} \tag{30}$$

The number of trips are:

$$N\_{\text{TripOfPeak}} = 2 \times T\_{\text{OffPeak}} \times \eta\_{\text{BusPeller}\text{HororOfPeak}} \tag{31}$$

$$N\_{\text{TripPeak}} = 2 \times T\_{\text{Peak}} \times n\_{\text{BusPellerHourPeak}} \tag{32}$$

$$N\_{\text{TripEven}} = 2 \times T\_{\text{Evening}} \times n\_{\text{BusPerHourEven}} \tag{33}$$

which results in a total trip number:

$$N\_{\text{TripTotal}} = N\_{\text{TripOfPeak}} + N\_{\text{TripPeak}} + N\_{\text{TripEvening}} \tag{34}$$

From this, the total trip distance can be calculated:

$$d\_{\text{TripTotal}} = N\_{\text{TripTotal}} \times l\_{\text{Trip}}.\tag{35}$$

The total energy used during that distance is:

$$\mathcal{W}\_{\text{TripTotal}} = T\_{\text{TripNetTotal}} \times P\_{\text{Average}} \tag{36}$$

where the total net trip time, excluding the layover time, is:

$$\begin{aligned} T\_{\text{TripNetTotal}} &= N\_{\text{TripOfPeak}} \times T\_{\text{TripNetOfPeak}} \\ + N\_{\text{TripPerak}} &\times T\_{\text{TripNetPeak}} + N\_{\text{TripEventing}} \times T\_{\text{TripNetEvening}} \end{aligned} \tag{37}$$

The distance driven during the trips and the energy consumed for it are the same, irrespective of charging strategy. However, the distance driven to and from the depot will differ between the strategies, since the number of buses and the number of the buses which have to drive to the depot in the midday period vary. The number of times a bus has driven to or from the depot during a day is represented by the light blue areas in Figures 7–9. All the blue areas have the same length, *T*PullInOut so we need to determine the number of times a bus drives from the depot per day for each charging strategy. The number of times a bus drives to the depot will of course always be the same.

$$N\_{\text{PullIrOut}} = 2 \times N\_{\text{BusTotal}} + 2 \times \left( N\_{\text{BusTotal}} - N\_{\text{BusMidday}} \right) \tag{38}$$

Based on the number of times a bus drives to and from the depot, the distance driven to and from the depot can be determined:

$$d\_{\text{PullIrOutTotal}} = N\_{\text{PullIrOut}} \times l\_{\text{PullIrOut}} \tag{39}$$

The energy used when driving to and from the depot:

$$\mathcal{W}\_{\text{PullInOutTotal}} = \mathcal{N}\_{\text{PullInOut}} \times T\_{\text{PullInOut}} \times P\_{\text{Average}}.\tag{40}$$

$$\mathcal{W}\_{\text{Total}} = \mathcal{W}\_{\text{Trip Total}} + \mathcal{W}\_{\text{PullIrOutTotal}} \tag{41}$$

$$d\_{\text{Total}} = d\_{\text{TripTotal}} + d\_{\text{PullInOutTotal}}.\tag{42}$$

We have now determined the TCO variables of "Total Energy", "Trip distance" and "Total distance".
