I am a little skeptical that air resistance is a huge factor for an 80,000 pound fully loaded semi. Probably most friction losses are due to rolling resistance. As it’s a sunday and I’m daydreaming anyway, I might as well try some back of the napkin calculations:

http://www.engineeringtoolbox.com/drag-coefficient-d_627.htm... http://www.engineeringtoolbox.com/rolling-friction-resistanc... https://en.wikipedia.org/wiki/Semi-trailer_truck

Before I start, I don’t know if the Engineering Toolbox answer is wrong, because I get 217.5 N for (0.29 * 0.5 * 1.2 * ((90 * 1000)/3600)^2) * 2. At least it’s close to the 181 N they claim. So my results could be off but I’m hopeful they’re within the correct order of magnitude..

If we assume a drag coefficient of 1 for a flat plate representing the front of the truck, with most losses coming from form drag, A semi with a frontal area of 2.5x4 meters at 105 km/hr (65 mph or 29 m/s) would feel a force of:

Fd = 1.0 1/2 (1.2 kg/m3) ((105 km/h) (1000 m/km) / (3600 s/h))2 (10 m2) = 5104 N = 520 kg = 1147 lb

If we assume the high drag coefficient of 0.01 for truck tires, the rolling resistance of a 36287 kg (80000 lb) truck (idealized as independent of velocity) is:

Fr = 0.01 (36287 kg) (9.81 m/s2) = 3560 N = 363 kg = 800 lb

Power P = F * v so these are:

Air resistance power = 520 * 29 = 15080 W = 20.22 hp Rolling resistance power = 363 * 29 = 10527 W = 14.12 hp

So on flat ground, amazingly it only takes about 26.1 kW (35 hp) to keep a semi moving at 105 km/hr (65 mph)! This makes sense to me because a human can pull a semi at about 5 mph. Power goes up roughly by the square of velocity so a human with a power of 300 W would need 13 * 13 = 50.7 kW (68 hp) to go 65 mph. We’ve seen here that roughly half the power goes to rolling resistance which stays roughly constant with velocity, so 35 hp is conceivable.

As a sanity check, climbing a 5% grade at 105 km/h is a climb rate of 5.25 km/h or 1.46 m/s. The power required to lift the mass of the truck is:

Power P = F * v = m * g * v = 36287 * 9.8 * 1.46 = 519194 W = 696 hp

So it takes toughly 20 times more power to go up a 5% grade, and that’s why a typical semi truck engine with 600 hp can’t climb a 5% grade at 65 mph.

Semi trucks get 5.5 miles per gallon, so use about 12 gallons per hour. There are 38 kWh of energy in a gallon of diesel fuel, so that’s 456 kW burned in an hour, or a continuous power used of 456 kW (612 hp). If we compare that to the ideal of 35 hp needed, we can see that:

Engine to road efficiency of a semi truck = 26.1 kW/456 kw (or 35 hp/612 hp) = 5.7%

That’s really quite remarkably low and doesn’t surprise me, since semi truck technology has not changed much in 50 years. I read somewhere that cars are in the 8-15% efficiency range. That’s why it’s so trivial for Tesla to beat an internal combustion engine, because electric motors are 95% efficient vs. about 25% efficient for an internal combustion engine.

So the real increase in efficiency of a hybrid gas turbine/electric semi comes from the turbine, which has an efficiency of 40% vs about 25 or 30% for a reciprocating diesel engine. Improved aerodynamics and decreased rolling resistance of a carbon fiber trailer are mostly for show. Hybrid turbine/electric propulsion has been used on trains for decades and I’m kind of flabbergasted why it never took off in semis, or cars for that matter.