Q_Goest, I would like to go over your idea in post #20.
One assumption you made that is wrong, is the estimation of volumetric flow rate; You estimated it at WOT, which will never happen, as a lot less power is needed to maintain a cruising speed. Here's the appropriate math:
The
BSFC is defined by:
BSFC = \frac{\dot{m}_f}{P} = \frac{\frac{\dot{m}_a}{AFR}}{P} = \frac{\frac{\rho \dot{V}}{AFR}}{P}
or:
\dot{V} = \frac{AFR (BSFC)}{\rho} P
Where \dot{V} is the volumetric air flow rate that enters the engine, AFR is the air-fuel ratio, \rho is the air density, P the engine output power. \dot{m}_f and \dot{m}_a are the mass flow rate for the fuel and air, respectively.
We already said that the exhaust volumetric flow rate \dot{V}_{ex} will be related to the inlet volumetric flow rate by their temperature ratio:
\dot{V}_{ex} = \frac{T_{ex}}{T}\dot{V} = \frac{T_{ex}}{T} \frac{AFR (BSFC)}{\rho} P
The power needed to maintain a constant speed is mostly due to aerodynamic drag and you have to assume some drivetrain losses (\eta):
P = \frac{0.5\rho C_d A v^3}{\eta}
Or:
\dot{V}_{ex} = \frac{T_{ex}}{T} AFR (BSFC) \frac{0.5 C_d A v^3}{\eta}
The velocity over the frontal area - as we calculated in our previous post - is:
\frac{\dot{V}_{ex}}{A} = 0.5 \frac{T_{ex}}{T} \frac{AFR (BSFC) C_d}{\eta} v^3
And comparing that velocity to the car speed:
\frac{^{\dot{V}_{ex}} / _A}{v} = 0.5 \frac{T_{ex}}{T} \frac{AFR (BSFC) C_d}{\eta} v^2
If that ratio is less or equal to 1, then the car is always «catching up» the front of the bubble. The minimum speed to break that pattern, such that the car is inside the bubble, is:
v \geq \sqrt{\frac{\eta}{0.5 \frac{T_{ex}}{T} AFR (BSFC) C_d}}
Typical value for the temperature ratio is 3; a good air-fuel ratio at cruising speed will be at most 15;
Typical drag coefficient is 0.325;
Typical BSFC is about 250 g/kW/h (7 x 10
-8 kg/W/s) and you can put a drivetrain efficiency of 0.85.
This gives a minimum speed of 1289 m/s (over 4600 km/h or 2800 mph). No comments necessary.
But you said «let's mix it with some fresh air». That will change the following:
\dot{V}_{ex} = (X+1)\frac{T_{ex}}{T}\dot{V}
Where X is the ratio of fresh air you will mix with the exhaust (mass-wise). Furthermore, to get that extra flow rate, you will need to take some energy somewhere, which will come from the engine power somehow. The minimum power required will be the pressure differential across the pump (\Delta p) times the flow rate, which is X times larger than \dot{V}. So:
\dot{V} = \frac{AFR (BSFC)}{\rho} P \left( 1 + \frac{\Delta p \left(X \dot{V}\right)}{P}\right)
Or:
\dot{V} = \frac{\frac{AFR (BSFC)}{\rho} P}{1 - X \frac{\Delta p}{\rho} BSFC}
Where we could even replace \frac{\Delta p}{\rho} by R\Delta T, which is the specific gas constant for air and the temperature difference across the pump.
All of this modifies our speed equation to:
v \geq \sqrt{\frac{\eta}{0.5 \frac{T_{ex}}{T} AFR (BSFC) C_d} \left(\frac{1 - X (R\Delta T) BSFC}{X+1}\right)}
Even if we assume the addition of air flow is free (\Delta T = 0), and that X = 100 like in your post #20, then our previously calculated minimum speed drops by a factor of 10, i.e. approx. 460 km/h; Still way too high.
Even if we consider the pumping losses, we can drop it to zero by setting X (R\Delta T) BSFC = 1. But in that case, \dot{V} becomes infinite; Cleary, the fuel consumption will also be infinite, defying our first goal of decreasing drag in the first place.
Conclusion: As others previously mentioned, you will only end up with a thin layer of hot air across the car - no matter what - because the car goes too fast for the quantity of exhaust gas that comes out.
Q_Goest said:
Instead of asking about exhausting gas from our car in front of our vehicle, we might ask how does this increased air temperature around a car change drag?
If there are so many other cars around, turbulences will probably be a bigger factor in drag (or if you
tailgate like in Nascar racing 
).
Setting aside the theoretical exercise for a moment...unless there is some amazing heat exchanger in play here, the vehicle is being wrapped in exhaust gases. At what point would someone like to consider how the engine continues to run without fresh air being available? Or the driver, for that matter! Just a thought. 8) Back to the theory!
