How can my CD0 estimation be improved?

Question

In this question (link) @PeterKämpf asked me to give a basis for my results. I will share my method for estimation here to see if you guys agree it is correct and how I can improve on it. Note that I use ISA equations where relevant.

Firstly, I estimate the Breguet constant from the PL-R diagrams of the A319-100 and B737-700 (found in the airport planning manuals). For that I use this equation, which is essentially the Breguet equation applied between points A and B (A being MPL+MTOW, B is MTOW+MFW):

$$ K_{sem,i} = \left( \frac{R_B - R_A}{\ln \left( \frac{OEW + PL_A}{OWE + PL_B} \right)} \right)_{sem,i} $$

From the PL-R diagrams: A319-100: R_A = 1463 km, PL_A = 15825 kg, R_B = 5800 km, PL_B = 3800 kg, OEW = 41100 kg (weight variant 00). B737-700: R_A = 3956 km, PL_A = 16860kg, R_B = 6236 km, PL_B = 10800 km, OEW = 38342 kg. — Results: K_A319 = 18277 km, K_B737 = 19607 km.

Using the definition of the Breguet constant, I can estimate their average aerodynamic efficiency, with the following equation:

$$ \left( \frac{L}{D} \right)_{sem,i} = K_{sem,i} \left( \frac{gc_{j,cr}}{V} \right) $$

For that I need cruising speed (A319: M0.78 at 11900m ISA = 230.15 m/s, B737: M0.785 at 11705m ISA+10ºC = 236.91 m/s) and sfc in cruising conditions. I have sfc at 35000 ft and M0.8 16.98 mg/Ns for the A319 and 17.02 mg/Ns in the same conditions for the B737). I adjust the sfc to cruising conditions using the following proportionality (note that beta = 0.5 for turbofan engines):

$$ c_j = \text{cte}(M)^\beta \sqrt{T} $$

I get adjusted values of 16.68 mg/Ns for the A319, 17.2 mg/Ns for the B737. Average aerodynamic efficiencies come out at 13 and 13.96, respectively. I can calculate the average lift coefficient using L=W:

$$ C_{L_{sem,i}} = \left( \frac{2 \frac{W_{cr}}{S_W}}{\rho V^2} \right)_{sem,i} = \left( \frac{2 \cdot 0.8 \frac{W_{TO}}{S_W}}{\rho V^2} \right)_{sem,i} $$

where I've estimated average cruising wing loading as 80% of take-off wing loading. The take-off wing loading of the A319 is 5129 Pa and for the B77 it's 5485 Pa. Density is in cruising conditions. Average lift coefficients come out as 0.4906 and 0.5023, respectively. Using the definition of the polar curve, I calculate CD0 using the following expression:

$$ C_{D0_{sem,i}} = C_{L_{sem,i}} \left( \frac{L}{D} \right)^{-1}_{sem,i} - \frac{C^2_{L_{sem,i}}}{\pi A_{sem,i} \varphi_{sem,i}} $$

where I've taken $\varphi = 0.9$ for both planes (since they have winglets). The aspect ratio of the A319 is 9.5, 9.4 for the B737. CD0 comes out as 0.0274 for the A319 and 0.0265 for the B737. This is off somewhat from the 0.014-0.02 range given by Torenbeek for high-subsonic jets.

Excuse the length, but hopefully you guys can help me improve this estimation or detect errors.

Answer 1

Thank you for sharing your calculation! Now I can see how you arrived at the results. Never mind the length: All that detail is essential for improving your results.

Regarding the Breguet equation I prefer to use the one here and here. This, using your data for the A319, gives $$\frac{L}{D} = \frac{g\cdot b_f\cdot R}{v\cdot ln\left(\frac{m_2}{m_1}\right)} = 17.63$$ with $v$ = 230 m/s, $R$ = 4,337,000 m, $g$ = 9.81 m/s², $b_f$ = 0.00001668 kg/Ns, $m_1$ = 62,925 kg and $m_2$ = 74,950 kg. You will notice that I changed the masses: You cannot use two different payloads for the Breguet equation; instead, all mass differences must be from fuel consumption. I now used the difference in payloads to compute the mass ratio between range 1 and range 2 and use that for the mass of the fuel consumed to cover $R$.

The result is a bit lower than what is usually used for the A319, which probably results from different payloads used for the two different ranges.

Your lift coefficient looks quite credible, even though you could do better than pick a mass reduction of 80% once you know the mass together with the altitude at one point during cruise. The Breguet equation is only valid under the assumption of a constant lift coefficient which is only possible when the airplane climbs continuously as fuel is consumed. In reality, flights get assigned flight levels and do a step climb if traffic control permits. But using the lift coefficient from one data point should be close enough.

Also your way of calculating $c_{D0}$ is good in principle, although I would prefer to lower the wing efficency to 0.85 – swept wings rarely perform better, winglets notwithstanding. If you allow me to use the more usual $\epsilon$ instead of $\varphi$, and if L/D is indeed 18 for the A319, the zero-lift drag coefficient becomes $$c_{D0} = \frac{c_L}{\frac{L}{D}} - \frac{c_L^2}{\pi\cdot AR\cdot\epsilon} = \frac{0.5}{18} - \frac{0.25}{3.14159\cdot9.5\cdot0.85} = 0.0179$$ which lies within the range given by Torenbeek. I only did the numbers for the A319; the B737 data should produce a very similar result.

Your biggest error was in applying the Breguet equation with the wrong masses. Also the masses themselves are too low: Just OEW and payload will not allow the airplane to fly: You need to add some fuel. And the mass fraction must be from the same flight at different times, with the range the distance covered between those times.

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EDIT: Now that I know where your payload-range numbers are from, I can check your results. Since the inclination of the part of the diagram where payload is traded against fuel mass is the same for all MTOW figures, the resulting L/D numbers should also be the same.

From the brown line (75,500 kg MTOW) I read 7,500 km range with 10,000 kg payload and 5,000 km with 16,000 kg payload. This means the airplane consumes 6,000 kg of fuel for 2,500 km range. Since the range/fuel ratio becomes worse with increasing range, this figure cannot be used in isolation but helps to figure out the mass at the end of a median trip. Let's use a 6,250 km flight which is the middle of that range where payload is traded against fuel mass, so a fuel mass of 15,000 kg is consumed at the end of the flight. That makes the mass ratio 75,500 / 60,500 = 1.248 and the L/D = 17.6 with your SFC number. Close enough to 18, given my rounding of the mass figures.