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algol [13]
3 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]3 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:

Due to differences in the nature of the adhesive force and cohesive forces in the molecules of the individual substances and the glass tube.

Explanation:

To understand why this is so, a deep understanding of adhesive and cohesive force is required.

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Note: Adhesive and cohesive forces are best discussed under macroscopic level.

Adhesive force is the Intermolecular forces that exist between atoms of different molecules e g the forces explain when unlike charges stick together.

Cohesive force on the other hand is the Intermolecular force that exist between atoms of the same molecules e.g the force between hydrogen bonding.

Hence to explain the case scenario above, the adhesive force between the water molecules and the glass molecules is higher than the cohesive force between the water molecules. Hence the high rise.

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Explanation:

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An ideal gas turbine operates using air coming at 355C and 350 kPa at a flow rate of 2.0 kg/s. Find the rate work output
GuDViN [60]

Answer:

The rate of work output = -396.17 kJ/s

Explanation:

Here we have the given parameters

Initial temperature, T₁ = 355°C = 628.15 K

Initial pressure, P₁ = 350 kPa

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Assuming an isentropic system, from tables, we look for the saturation temperature of saturated air at 4.287 kJ/(kg·K) which is approximately

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The saturation temperature at the given

T₂ = 79°C

The rate of work output \dot W = \dot m×c_p×(T₂ - T₁)

Where;

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\dot m =  The mass flow rate = 2.0 kg/s

Substituting the values, we have;

\dot W = 2.0 × 0.7177 × (79 - 355) = -396.17 kJ/s

\dot W = -396.17 kJ/s

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