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: Set up the weight ratio as a density ratio.
Since $W = \tfrac{1}{2}\rho V^2 S c_{l,max}$ at lift-off and $V$, $S$, $c_{l,max}$ are unchanged between the two conditions, $W_B/W_A = \rho_B/\rho_A$.

Step 2: Combine the ideal-gas expressions before plugging in numbers.
Instead of computing $\rho_A$ and $\rho_B$ separately, write the ratio directly, so the gas constant $R$ cancels:
\[ \frac{\rho_B}{\rho_A} = \frac{p_B/(RT_B)}{p_A/(RT_A)} = \frac{p_B}{p_A}\times\frac{T_A}{T_B} \]

Step 3: Convert temperatures to kelvin and substitute.
$T_A = 50+273.15 = 323.15$ K, $T_B = -30+273.15 = 243.15$ K.
\[ \frac{W_B}{W_A} = \frac{0.66}{1}\times\frac{323.15}{243.15} = 0.66 \times 1.3290 = 0.8772 \]

Step 4: Interpret the result.
Even though condition B is colder (which alone would increase density), its much lower pressure dominates, so the net density, and hence the allowable take-off weight, is lower than in condition A.

Final Answer:
\[ \boxed{W_B/W_A \approx 0.877} \]