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Electric cars Appendix 1

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Interpreting the data.  It is the efficiency ratios and the emission ratios shown in blue at the bottom of the spread sheets A or (B) that compare the vehicles’ energy consumptions and emissions. Sensitivity tests can be carried out rapidly by varying the parameters used.

The data that determines the relative fuel efficiency and carbon emissions of the vehicles are:

  1. The energy lost refining and transporting fuel to filling stations.
  2. The thermal efficiency of the internal combustion (IC) engines.
  3. The proportion of the energy burnt in power stations that reaches the plug.
  4. The efficiency of the charging, battery and motor combination.
  5. The drive chain losses.
  6. The energy lost in braking.
  7. The energy, if any, recovered from braking.
  8. The rolling loss.
  9. The weight of the vehicle.
  10. The idling losses, if any.
  11. The carbon emitted per KWh of primary energy burnt.

The values used are as follows:

Losses power station to plug set to 64.1%, see footnote to spread sheets

Losses in refineries and attributable to transporting the fuel to filling stations set to 10%

Charging losses are set to 8%

Battery discharging losses are set to 10% for current EVs falling to 5% for improved ones

Electric motor losses are set to 10%.

Engine efficiencies for diesels have two sets of five values.  These are linear interpolations covering (a) the current realistic range of 25% to 30% and (b) the improved range of 36% to 40% where the higher value of each pair may be appropriate for constant speed or rural running and the lower value for variable speed and the most congested urban conditions.  The range for the MUSIC is set to 43.2% to 48% following MUSI Engineering’s advice.

Drive chain losses have been set to (a) 25% for existing EV and ICV and (b) 0% for the future or improved EVs on the basis the wheels and motors will then be combined as one so eliminating mechanical transmission losses.

Braking losses will vary greatly according to the driving conditions. We have provided the range 5% (representing uncongested rural conditions) to 40% (representing congested urban conditions).

Regenerative braking for EVs appears to be established and is said to recover 50% of braking energy losses.  For conventional vehicles Flybrid systems have a flywheel based arrangement which they claim may recover up to 70% of braking losses and which may be available in 4 to 6 years on standard vehicles.  Where applicable we have used 50% for both systems.

Rolling loss has been set to 50% of the energy reaching the wheel, leaving the balance for overcoming wind resistance.

Weight: EVs are said to weigh 25% more than conventional ones. Flybrid’s mechanical regenerative braking may add 5% to the vehicle weight. Hence, where applicable, the mechanical transmission, braking and rolling losses are inflated by those amounts.

Idling losses on EVs need not arise.  Idling losses for conventional vehicles have been assigned the range 2.5%, when the braking loss is 5%, through to 20%, when the braking loss is 40%, except that, where regenerative braking is assumed, the idling loss has been set to zero (following the BMW system which switches the engine off when the vehicle is stationary).

Energy used in battery manufacture: The energy used in manufacture has been set (a) equal to that transmitted during the battery’s life and (b) equal to half that transmitted.  The basis for that is in the main text.

Fuel costs: Tables 4(1) to 4(3) provide a range of costs per KWh available to overcome wind resistance for the various scenarios examined.  In all cases it is clear that the price of diesel, currently 60 pence per litre of 6 pence per KWh void of tax, would have to rise a great deal before the cost of the EV’s fuel would be the cheaper


The principles. There are various points in the chain from primary burn to end point at which the fuel efficiency of a vehicle could be calculated.  Our previous calculation did that with respect to the upstream end of the vehicle’s drive chain (immediately downstream from the engine or motor).  The present calculation does that with respect to the downstream side of rolling.  That provides the ratio of (a) the energy available to overcome wind resistance to (b) the energy used in the primary burn. The ratio is subsequently referred to as the vehicles efficiency. The reason is that transmission loses within the vehicles and the rolling resistances are proportional to vehicle weight and an EV is heavier than its ICV competitor on account of the batteries. 

Dividing the efficiency of one vehicle by the same for another vehicle provides the relative energy consumptions needed to power the down steam requirements. I.e. if the ratio of vehicle one’s efficiency to vehicle two’s is 1.5 then vehicle two uses 50% more energy than vehicle one.

In more detail we define the losses across each phase in percentage terms as follows:

Refinery/Generating industry burn to plug losses             L1 leaving residual of (1-L1) =E1        

Battery charging losses  )

Battery discharging losses         )

Electric motor losses                 )                                   L2 leaving residual of (1-L2) =E2        

IC engine losses                        )

Transmission losses                                                       L3 leaving residual of (1-L3) =E3

Braking losses                                                              L4 leaving residual of (1-L4) =E4

Rolling losses                                                                L5 leaving residual of (1-L5) =E5        

Hence, if P is the primary energy burnt and if X is the residual for overcoming wind resistance we have an efficiency, defined as X/P = E1 x E2 x E3 x E4 x E5 .......................(1)

If there is regenerative braking returning R% of the initial braking loss, L4,  then the actual braking loss becomes (L4 - (L4 x R)) and E4 should be replaced by (1 - (L4 - (L4 x R))) which reduces to: 1 + L4(R-1) ...............(1a)

Hence the spread calculations use the formula:  

Efficiency, X/P =  (1-L1) x (1-L2) x (1-L3) x (1+L4(R-1)) x (1-L5)................ (1b)

Where L4 is the braking loss prior to regeneration.

Weight: If energy losses in transmission, braking and rolling are proportional to weight then before comparing the efficiency of a heavy vehicle, such as an electric powered car, with a conventional one the energy at the drive chain of the heavier vehicle should be increased by adding the additional losses attributable to the greater weight.

For example, if the losses in transmission amounted to e.g. L3% of the energy available to the drive chain then, if the downstream energy is to remain the same when the weight is increased by W% the upstream energy should be increased by adding (L3 x W) to the upstream energy’s so providing a multiplier to the light vehicle upstream energy of: 1 + (L3 x W) ... (2)

The overall efficiency of the heavier vehicle for comparison with the lighter vehicle would then be the lighter vehicle’s efficiency divided by (1 + (L3 x W)) ....................................... (3)

Hence the divisor yielding the comparative heavier vehicle efficiency, reference (1) and (1b) above, would be   [(1 + (L3 x W)) x (1 + (L4 x W)) x (1 + (L5 x W))] ................................ (4)

That could be extended to:

[(1 + (L3 x K3 x W)) x (1 + (L4 x (R-1) x K4 x W)) x (1 + (L5 x K3 x W))]  ..................... (5)

Supposing anyone can estimate the K values.

Emissions: Dividing the carbon emission per KWh produced by the primary burn by the fuel efficiency provides the carbon emission per KWh available to overcome the downstream requirements.  The ratios of carbon those carbon emissions then compare the vehicles.

Carbon emissions – electricity generation

Table 5C of the DUKES provides 497 tonnes per GWh supplied. According to Julian Prime of the BERR that value relates to the supply of 373,322 GWh in table 5.6. Instead of using the 497 we note the energy burnt (including nuclear) was 952,722 GWh of which 342,127 reached "the plug".  Hence there were 542 Tonnes of CO2 per GWh at the plug and 195 Tonnes per GWh burnt in the power stations.

The difference between the 497 and the 542 highlights what appears to be a systematic error in the ARUP and other reports as to the emissions associated with the KWh consumed by end users. Those reports overlook either the grid losses or energy industry use or both.

We also note that the values do not take account of the emissions attributable to mining and delivering the fuel to the power stations.

Carbon emissions from diesel

The 3.15 Kg of CO2 are emitted per Kg of diesel burnt. The KWh per Kg diesel is to 13.1. Hence there are 0.24 Kg of CO2 per KWh of diesel burnt.  This value does not take account of the emissions in refineries or attributable to transporting the fuel to filling stations but those are captured by the efficiency calculations.

Fuel costs

Tables 4(1) to 4(3) provide the costs per KWh of the residual energy available to overcome wind resistance for each of the various scenarios using the following.

The overall efficiency between primary burn or input to refinery and overcoming wind resistance, E, is the product of the intermediate efficiencies.  Hence the efficiency of the element between the plug or the filling station and overcoming wind resistance is the overall efficiency divided by the efficiency, E1, of the link between power station and plug or refinery and filling station.  Dividing the price at the plug or filling station by that efficiency yields the cost per kWh to overcome wind resistance.

In summary, if (a) p is the cost per KWh at the plug or at the filling station (b) E is the overall efficiency and (c) E1 is the efficiency, power station burn to plug, or refinery to filling pump then the price per KWh available to overcome the wind is p/E/E1.

It is the comparison of those costs which compares the vehicles.


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