Question

\(W_A\) and \(W_B\) are the respective maximum take-off weights of an aircraft for two ambient air conditions given below.
Condition A: \(p = 1\) bar, \(T = 50^{\circ}\)C; Condition B: \(p = 0.66\) bar, \(T = -30^{\circ}\)C
If all other parameters relevant for take-off are kept the same in these two conditions, the ratio \(W_B/W_A\) is ________ (rounded off to 3 decimal places).

Hint:

Since lift equals weight at rotation and S, cl_max, and V are unchanged, weight is proportional to air density; find density in each condition from p = rho R T and take the ratio.

Answer 1

Correct Answer: 0.877

Step 1: Identify the physical link between takeoff weight and air density.
At the moment of lift-off, the wing must generate lift equal to the aircraft weight: \(L = W = \tfrac{1}{2}\rho V^2 S\,c_{l,max}\). The problem states that all other parameters relevant to take-off (wing area \(S\), maximum lift coefficient \(c_{l,max}\), and the take-off/rotation speed \(V\)) are kept the same between the two conditions. That leaves air density \(\rho\) as the only variable, so
\[ W \propto \rho \]
\[ \frac{W_B}{W_A} = \frac{\rho_B}{\rho_A} \]

Step 2: Find the air density in each condition using the ideal gas law.
\[ \rho = \frac{p}{RT}, \qquad R = 287\ \text{J/(kg K)} \]
Condition A: \(p_A = 1\times10^5\) Pa, \(T_A = 50+273.15 = 323.15\) K:
\[ \rho_A = \frac{1\times10^5}{287\times323.15} = \frac{100000}{92744.05} = 1.0782\ \text{kg/m}^3 \]
Condition B: \(p_B = 0.66\times10^5\) Pa, \(T_B = -30+273.15 = 243.15\) K:
\[ \rho_B = \frac{0.66\times10^5}{287\times243.15} = \frac{66000}{69784.05} = 0.9458\ \text{kg/m}^3 \]

Step 3: Take the ratio.
\[ \frac{W_B}{W_A} = \frac{\rho_B}{\rho_A} = \frac{0.9458}{1.0782} = 0.8772 \]

Final Answer:
Rounded to 3 decimal places, \(W_B/W_A \approx 0.877\). The lower pressure at condition B reduces air density more than the lower temperature raises it, so a lighter maximum take-off weight is possible there for the same lift-off requirements.
\[ \boxed{W_B/W_A \approx 0.877} \]