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Chapter 9: Problem 108

A design group has two possible wing designs \((A \text { and } B\) ) for an airplane wing. The planform area of either wing is \(130 \mathrm{m}^{2}\) and each must provide a lift of \(1,550,000 \mathrm{N}\)

Short Answer

Expert verified
The coefficients of lift for designs A and B can be calculated using the formula \(Cl = L / (0.5 * p * A * V^2)\) after substituting known values. Detailed calculations will vary and depend on assumed velocities and acceptable Cl limits in practical aerodynamics. The suitable wing design is the one which has a manageable lift coefficient at a lower velocity.

Step by step solution

01

Identify Knowns and Unknowns

First, identify the known values and the variable we are trying to find. We know the lift force is \(1,550,000 N\), the planform area is \(130 m^2\) and we are trying to find the value of Cl, the lift coefficient, for both designs A and B.
02

Rearrange the Lift Equation for Cl

Rearrange the lift formula to solve for Cl, so the formula becomes \(Cl = L / (0.5 * p * A * V^2)\). Assuming standard atmosphere conditions (p = 1.225 kg/m^3) and the same operating conditions for both designs (let's surmise it as Va for design A and Vb for design B).
03

Substitute Known Values into the Cl Formula for Each Design

Substitute the known values into the Cl formula for both design A and B. Cl equation for design A becomes \(Cl_A = 1,550,000 / (0.5 * 1.225 * 130 * Va^2)\) and for design B it's \(Cl_B = 1,550,000 / (0.5 * 1.225 * 130 * Vb^2).\) Here Va and Vb will be the velocities (assumed different) for designs A and B respectively.
04

Calculations

Here is the tricky part. Since both designs A and B have the same lift and same area, we are left with the differences in velocity and lift coefficient. The suitable wing design could be assessed based on these differences. If the design with lower velocity has a manageable lift coefficient, it may be a better design. These calculations are subject to what is actually feasible and acceptable in terms of aerodynamics.

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Most popular questions from this chapter

As is discussed in Section \(9.3,\) the drag on a rough golf ball may be less than that on an equal-sized smooth ball. Does it follow that a 10 -m-diameter spherical water tank resting on a \(20-\mathrm{m}\) -tall support should have a rough surface so as to reduce the moment needed at the base of the support when a wind blows? Explain.

On a day without any wind, your car consumes \(x\) gallons of gasoline when you drive at a constant speed, \(U\), from point \(A\) to point \(B\) and back to point \(A\). Assume that you repeat the journey. driving at the same speed, on another day when there is a steady wind blowing from \(B\) to \(A\). Would you expect your fuel consumption to be less than, equal to, or greater than \(x\) gallons for this windy round-trip? Support your answer with appropriate analysis.

(See The Wide World of Fluids article "At \(12,600 \mathrm{mpg}\) It Doesn't cost Much to 'Fill 'er Up," section 9.3.3.) (a) Determine the power it takes to overcome aerodynamic drag on a small \(\left(6 \mathrm{ft}^{2}\right.\) cross section), streamlined \(\left(C_{D}=0.12\right)\) vehicle traveling \(15 \mathrm{mph}\) (b) Compare the power calculated in part (a) with that for a large \(\left(36 \mathrm{ft}^{2} \text { cross-sectional area }\right),\) nonstreamlined \(\left(C_{D}=0.48\right) \mathrm{SUV}\) traveling 65 mph on the interstate.

A 5 -m-diameter parachute of a new design is to be used to transport a load from flight altitude to the ground with an average vertical speed of $3 \mathrm{m} / \mathrm{s}$. The total weight of the load and parachute is 200 N. Determine the approximate drag coefficient for the parachute.

For small Reynolds number flows, the drag coefficient of an object is given by a constant divided by the Reynolds number (see Table 9.4 . Thus, as the Reynolds number tends to zero, the drag coefficient becomes infinitely large. Does this mean that for small velocities (hence, small Reynolds numbers) the drag is very large? Explain.
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