Answer:
A. To compute the Mach number for drag divergence, we need to use the formula:
Mdd = sqrt(CD0/K)
where CD0 is the zero-lift drag coefficient and K is the lift-induced drag factor.
We can find CD0 and K using the following equations:
CD = CD0 + K(CL^2)
L = W = 11,000 lb
rho = 0.000886 # slugs/ft^3 at 30,000 ft
S = 327.5 # wing area in ft^2
V = M * sqrt(1.4 * 1716 * 30,000) # velocity in ft/s
CL = 2 * W / (rho * S * V**2)
CD = 0.025 + (CL**2) / (pi * 8.8 * 0.9)
CD = CD0 + K(CL^2)
CD0 = CD - K(CL^2)
Now we need to find K. We can use the equation:
K = 1 / (pi * 8.8 * AR)
where AR is the aspect ratio of the wing.
AR = (b^2) / S
where b is the wingspan.
Assuming the wingspan is 35 feet, we get:
AR = (35^2) / 327.5 = 3.745
K = 1 / (pi * 8.8 * 3.745) = 0.00305
CD0 = 0.025 - 0.00305(CL^2) = 0.0056
Now we can compute Mdd:
Mdd = sqrt(CD0/K) = sqrt(0.0056/0.00305) = 1.63
Therefore, the Mach number for drag divergence is 1.63.
B. We have already computed CD0 and K in part A, so we can just use those values.
CD0 = 0.0056
K = 0.00305
Note that there are two nacelles and two tip tanks, so the total wetted area is increased by 25%.
CD0 = CD0 * 1.25 = 0.007
Therefore, CD = 0.007 + 0.00305(CL^2)
At level flight, L = W, so CL = W / (0.5 * rho * V^2 * S) = 2W / (rho * V^2 * S)
Substituting this into the above equation, we get:
CD = 0.007 + 0.00305(4W^2 / (rho^2 * V^4 * S^2))
CD = 0.007 + 0.00305(4W^2 / (0.000886^2 * (M*sqrt(1.4*1716*30000))^4 * 327.5^2))
CD = 0.007 + 0.00835/M^4
Finally, we can solve for CD at M = 0.7:
CD = 0.007 + 0.00835/0.7^4 = 0.0097
Therefore, CD = 0.0097 and K = 0.00305 for this flight condition.
Step-by-step explanation: