The aerodynamic behavior of a flying insect is to be investigated in a wind tunnel using a ten-times scale model. It is known that the insect’s velocity depends on its size (characteristic length L), wing flapping frequency ω, surrounding fluid’s density rho and viscosity μ. If the insect flaps its wings 50 times a second when flying at 1.25 m/s, determine the wind tunnel air speed and wing oscillation frequency required for dynamic similarit

Explanation:

Write the equation for Reynolds number as follows:

Re = VL/v

For dynamic similarity,

(VL/v)m + (VL/v)p…… (1)

Since, the model and prototype are in same medium, the kinematic viscosity remains same.

From equation (1), we can write

(VL)m = (VL)p

Here, L represents length, and V is the velocity.

Re-write the equation as follows:

Vm = Lp/Lm x Vp

Substitute 1/8 for Lp/Lm and 1.5m/s for Vp .

Vm = 1/8 x 1.5

Vm = 0.1875m/s

Therefore, the wind tunnel air speed is 0.1875m/s.

this is true

explanation:

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the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation: