# Fuel consumption of a Porsche

1. Apr 9, 2017

### mastermechanic

I was looking at the technical specs of Porsche 911. It has 427 kW power and 7.5 liters/100 km fuel consumption on highway. I made a quick calculation but I got strange result. Please tell me why ;

First I assumed that the car has 90 km/h constant speed on highway so it will take 1.11 hours to travel 100 km.

$$1.11 x 3600sec= 4000 sec$$
$$427kW.4.10^3 = 1708 MJ$$
$$Q= m.H$$ where the m is mass of fuel and H is heating value
Gasoline's heating value = 45 MJ/kg
$$1708 MJ = m. 45\frac{MJ}{kg}$$
$$m= 38 kg$$
and density of gasoline= $$680\frac{kg}{m^3}$$ =>$$\frac{38}{680} = 0.055 m^3$$

$$0.055 m^3 . (\frac{1000L}{1 m^3})= 55 L$$ and it's far away from 7.5 L

Edit: I guess we should take into account 25% efficiency but it implies that tech. specs don't show the real power of the car, it shows theoretical power? My mind has blown :D

Last edited: Apr 9, 2017
2. Apr 9, 2017

### phyzguy

The engine is putting out far less than its maximum rated power of 427 kW when traveling down the highway at 90 km/hr. Ask yourself, if you are driving down the highway at 90 km/hr, do you have the accelerator pressed all the way to the floor?

3. Apr 9, 2017

### NTL2009

I didn't verify your calculations, but assuming they are all correct, the delta is due to an invalid assumption. The engine won't need to produce 427 kW just to maintain 90 km/h. The 427 kw is used to accelerate like a bat-outa-hell. Then the engine 'relaxes' to a fraction of that power level to maintain a highway speed.

If your other calculations are correct, that fraction would be 7.5 ∕ 55 ≈ 0.13636364 or ~ 58 kw.

Some related discussion here:

further edit/add: If you assume ~ 25 % efficiency in the ICE, that 58 kW in the fuel ends up as ~ 14.5 kW or ~ 19 HP, close to some assumptions I made in that other thread.

4. Apr 9, 2017

### mastermechanic

Okay so you're saying that it wont need to use the whole power at 90km/h. Okay but you probable heard that if Bugatti Veyron travels at max speed of 407 km/h it will out of fuel in 8 min. Let's calculate it.

Bugatti has 100 liters fuel tank, 1200 hp = 895 kW power and again gasoline has 45 MJ/kg heating value and 680 kg/m^3 density. Because the fuel tank is 100 liters, the tank will be 68 kg when it's full. And I will choose and X which represents the time. So;

$$895\frac{kJ}{s}.X = 68kg.45\frac{MJ}{kg} =$$
$$X= \frac{ 3060.10^6 J}{895.10^3 J} = 3418 sec$$ which is about 55 min and I know it is impossible again at 407 km/h.

5. Apr 9, 2017

### NTL2009

Are you forgetting ICE efficiency again? 55 minutes times ~ 25% efficiency gets you in the ballpark at ~ 13.75 minutes. But an engine designed for raw power might sacrifice efficiency? Assume ~ 15% and you have your 8 minutes. Again, I didn't check your other calcs, but all this seems to fit.

6. Apr 9, 2017

### mastermechanic

But isn't it weird? These power values are the "output" values so they should have been already reduced by Ice efficiency. If it's vice versa, why componies show the theoretical values? If the total power is 895kw then they should show 220 kw.

7. Apr 9, 2017

### NTL2009

No, I don't think so.

You state the Bugatti has 895 kW of engine output power. If we assume 25% efficiency, it needs 4x that for input power, 3.58 MW, but the Bugatti efficiency maybe is closer to the 15% we mentioned above, so maybe ~ 5.97 MW?

8. Apr 10, 2017

### Staff: Mentor

Right -- the output is after the efficiency, but the input is before the efficiency.
It's not "theoretical", it is peak power, and it really is the output power. So if the efficiency is 25% (about right), the input is 4x higher.

9. Apr 10, 2017

### billy_joule

Another issue is that manufacturers generally state peak engine power as measured at the flywheel. After losses in a complex 4wd drivetrain like in a Veyron power at the wheels can be 30% less.