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Gennadij [26K]
3 years ago
14

At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power ge

neration potential of a wind turbine with 90-m-diameter (D) blades at that location. Take the air density to be 1.25 kg/m3.
Engineering
1 answer:
tangare [24]3 years ago
5 0

Answer:

e= 50 J/kg

Explanation:

Given that

Speed ,v= 10 m/s

Diameter of the turbine = 90 m

Density of the air ,ρ = 1.25 kg/m³

We know that mechanical energy given as

E=\dfrac{1}{2}mv^2\ J

That is why mechanical energy per unit mass will be

e=\dfrac{1}{2}v^2\ J/kg

Now by putting the values in the above equation we get

e=\dfrac{1}{2}\times 10^2\ J/kg

e= 50 J/kg

That why the mechanical energy unit mass will be 50 J/kg.

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It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d
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The exit temperature is 586.18K and  compressor input power is 14973.53kW

Data;

  • Mass = 50kg/s
  • T = 288.2K
  • P1 = 1atm
  • P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\

Let's substitute the values into the formula

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K

The exit temperature is 586.18K

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The compressor input power is calculated as

P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW

The compressor input power is 14973.53kW

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