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Effectus [21]
3 years ago
15

An airplane flies horizontally at 80 m/s. Its propeller delivers 1300 N of thrust (forward force) to overcome aerodynamic drag (

backward force). Using dimensional reasoning and unity conversion ratios, calculate the useful power delivered by the propeller in units of kW and horsepower.
Engineering
1 answer:
Nataly [62]3 years ago
8 0

Answer:

Power in kW is 104 kW

Power in horsepower is 139.41 hp

Solution:

As per the question:

Velocity of the airplane, v_{a} = 80 m/s

Force exerted by the propeller, F_{p} = 1300 N

Now,

The useful power that the propeller delivered, P_{p}:

P_{p} = \frac{Energy}{time, t}

Here, work done provides the useful energy

Also, Work done is the product of the displacement, 'x' of an object when acted upon by some external force.

Thus

P_{p} = \frac{F_{p}\times x}{time, t}

P_{p} = \frac{F_{p}\times x}{time, t}

P_{p} = F_{p}\times v_{a}

Now, putting given values in it:

P_{p} = 1300\times 80 = 104000 W = 104 kW

In horsepower:

1 hp = 746 W

Thus

P_{p} = \frac{104000}{746} = 139.41 hp

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W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V}  } \, dV

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W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP

a=v_2\\

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W=12.59 kJ/mol

The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series  

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