Rolling Up the Numbers

One way to look at it is in terms of the engine output power based on expected overall efficiency (engine plus drive train) and the energy content of the fuel.

Taking the overall (engine to wheel) efficiency as 0.25, and the fuel energy content at 33 MJ/L, then the rate of energy (power) delivered to the wheels would be,

P = effic * Energy/L * L/km * hrs/sec * km/hr
P = 0.25 * 33000000 * 1/23.3 * 1/3600 * v
P = 98 v

That is, if you are achieving 23.3 km/L then the power delivered to the wheels (in Watts) is approximately 98 times the velocity in km/hr. So for example at 75 km/hr @ 23.3 km/L the power would be approximately 7.4 kW

Of course this relation is not strictly accurate because the fuel efficiency (km/L) would itself be a function of velocity (as would, to some extent, be the engine efficiency). However it does give a nice simple estimate of the power at the wheels.

Another way to approach the problem is from the other end, calculate the losses dues to rolling resistance and air resistance and equate this to the energy required (ignoring elevation changes for now).

The energy losses per unit time (Watts) are :
P = m*g*Cr* (v/3.6) + 1/2 rho * Cd * A (v/3.6)^3

Putting in some numbers, m=1700 kg (includes 150kg load), g=9.8, Cr=0.01, Cd=0.26, rho=1.3, A=2.5 square meters, we get:

P = 0.0091 v^3 + 47 v
Again where the power is in Watts and the velocity in km/h

So for example at 75 km/h the estimated power required at the wheels is 0.0091*75^3 + 47*75 = 7.4 kW, which amazingly is the same result I obtained by coming at it from the other end. Of course the two methods are only going to agree at one velocity because the earlier method inherently ignores the velocity dependance of some important parameters.

 

Honestly that figure was only a guesstimate. I was assuming about 30% thermodynamic efficiency, 5% friction loss in the engine and a further 10% in the transmission (0.3 * 0.95 * 0.9). It's only a ballpark figure.