Fuel Fraction

---

Fill ‘er up!

Now we estimate fuel fraction using a modified form of the Breuget equation. We’ll go through the variables one-by-one.

$$ \scriptsize \frac{W_f}{W_0} = 1-0.975e^{\frac{\normalsize -R \> c_{bhp}}{\normalsize 550 \> \eta_p \> L/D}} $$

Range

Our desired range is 650 mi, but I’d like to have a 45-min. reserve in addition to that. At 200 mph, that’s another 150 miles. For the Breuget equation, we need the range in feet.

$$ \scriptsize R=(650+150)\text{ mi}\cdot \frac{5,280\text{ ft}}{\text{mi}}=4,224,000\text{ ft}$$

Specific fuel consumption

is a measure of the engine’s efficiency in producing power. How many pounds of fuel does it use every second for every horsepower it makes? The figure below is for the engine we’ve initially selected. We can also translate this into a fuel burn rate if we know the cruise power of the engine we’re going to use (not needed for this calculation, but a pilot will want to know this).

$$ \scriptsize c_{bhp} = 0.425 \text{ lb/hp/hr} \> \cdot \frac{\text{hr}}{3,600 \text{ sec}}= 0.000118 \text{ lb/hp/sec}$$

$$ \scriptsize \text{fuel burn rate (cruise)} = \frac{75 \text{% power} \cdot 200 \text{ hp} \cdot 0.425 \text{ lb/hp/hr}}{6 \text{ lb/gal}} = 10.6 \text{ gal/hr} $$

Propeller efficiency

measures the propeller’s efficiency in turning the engine’s power (brake horsepower) into thrust horsepower. Since we want to use a constant-speed propeller, this is about 85% since the blade pitch will be adjusted to a near-ideal angle for all regimes (both climb and cruise).

$$ \scriptsize \eta_p = 0.85 $$

L/D

This is easy, since we already estimated the cruise L/D ratio.

$$ \scriptsize \frac{L}{D} = 7.93 $$

Fuel fraction

Now we have all the values needed to estimate fuel fraction. ( is Euler’s number, approximately 2.71828.)

$$ \scriptsize \frac{W_f}{W_0} = 1-0.975e^{\frac{\normalsize -R \> c_{bhp}}{\normalsize 550 \> \eta_p \> L/D}} $$

$$ \scriptsize \frac{W_f}{W_0} = 1-0.975 \cdot 2.71828^{\frac{\normalsize -4,224,000\text{ ft} \>\cdot \> 0.000118 \text{ lb/hp/sec}}{\normalsize 550 \> \cdot \> 0.85 \> \cdot 7.93}}= 0.148$$

The 0.975 term has already accounted for fuel used in addition to the cruise phase (taxi, takeoff, climb, descent, and landing). However, we’ll bump up the fuel fraction by 3% because we will have a little bit of unusable fuel in the system.

$$ \scriptsize \frac{W_f}{W_0} = 1.03 \>\cdot\> 0.148 = 0.152$$

All we have to do now before reaching our gross weight estimate is to figure the empty weight fraction, which we’ll do in the next installment.