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algol [13]
2 years ago
7

A wing generates a lift L when moving through sea-level air with a velocity U. How fast must the wing move through the air at an

altitude of 10,000 m with the same lift coefficient if it is to generate the same lift?
Engineering
1 answer:
vredina [299]2 years ago
4 0

Answer:

V1 = 1.721 * V2

Explanation:

To start with, we assume that both lift forces are equal, such that

L2 = L1

1 is that of the level at 10000 m, and 2 is that of the level at sea level.

Next, we try and substitute the general formula for both forces such that

C(l).ρ1/2.V1².A = C(l).ρ2/2.V2².A

On further simplification, we have

ρ1.V1² = ρ2.V2², making V1 subject of formula, we have

V1 = √(ρ2/ρ1). V2²

Using the values of density for air at 10000 m and at sea level(source is US standard atmosphere), we have

V1 = √(1.225/0.4135) * V2

V1 = √2.9625 * V2

V1 = 1.721 * V2

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Answer:

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Explanation:

Given;

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inside  temperature of the heater, T₂ = 50° C

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Rate of heat transfer, P = \frac{7800008400 \ Watt-Hour}{3650 \ Hours}  = 2136988.6 \ W

Rate of heat transfer, P = \frac{K*A *\delta T}{L}

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