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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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Systematic searching is a skill that takes ________ to master.
bagirrra123 [75]

Answer: B, repetitive practice! hope this helps. :)

Explanation:

7 0
4 years ago
Read 2 more answers
A car travels from A, due north to a town B 4 km away. It then travels due east until it arrives town C 5 km from B. determine t
morpeh [17]

Answer:

A to C = 6.4 km

Explanation:

A to B = 4 km

B to C = 5 km

A to C =  using pythagorean theorem

a² + b² = c²

a = A to B = 4

b = B to C = 5

c = A to C

c² = 4² + 5²

c = 6.4 km (A to C)

8 0
3 years ago
A gas expands in a piston-cylinder assembly from p1 = 8 bar, V1 = 0.02 m3 to p2 = 2 bar. The relation between pressure and volum
Charra [1.4K]

Answer:

The heat transfer is 29.75 kJ

Explanation:

The process is a polytropic expansion process

General polytropic expansion process is given by PV^n = constant

Comparing PV^n = constant with PV^1.2 = constant

n = 1.2

(V2/V1)^n = P1/P2

(V2/0.02)^1.2 = 8/2

V2/0.02 = 4^(1/1.2)

V2 = 0.02 × 3.2 = 0.064 m^3

W = (P2V2 - P1V1)/1-n

P1 = 8 bar = 8×100 = 800 kPa

P2 = 2 bar = 2×100 = 200 kPa

V1 = 0.02 m^3

V2 = 0.064 m^3

1 - n = 1 - 1.2 = -0.2

W = (200×0.064 - 800×0.02)/-0.2 = -3.2/-0.2 = 16 kJ

∆U = 55 kJ/kg × 0.25 kg = 13.75 kJ

Heat transfer (Q) = ∆U + W = 13.75 + 16 = 29.75 kJ

7 0
3 years ago
The average age of engineering students at graduation is a little over 23 years. This means that the working career of most engi
RideAnS [48]

Answer:

$916

Explanation:

To solve this, we use the formula

FV = P/i * [(1+i)^n - 1], where

FV = future value of the all the money invested, $5 million

n = time span, = 500 months

P = payment per month

I = interest rate, 9% by 12 months, = 0.0075

Considering that we have been given all in our question, then we substitute directly and solve. So we have,

5000000 = P/0.0075 * [(1+0.0075)^500 -1]

5000000 * 0.0075 = P * [1.0075^500 - 1]

37500 = P * [41.93 - 1]

37500 = P * 40.93

P = 37500/40.93

P = $916.20

Therefore, the engineer needs to save $916 in a month which is the accrued

6 0
3 years ago
g A part made from annealed AISI 1018 steel undergoes a 20 percent cold-work operation. (a) Obtain the yield strength and ultima
nikdorinn [45]

Answer:

A) - Yield strength before operation = 32 kpsi

- Ultimate Strength before operation = 49.5 kpsi

- Yield strength after operation = 61.854 kpsi

- Ultimate Strength after operation = 61.875 kpsi

- Percentage increase of yield strength = 93.29%

- Percentage increase of ultimate strength = 25%

B) ratio before operation = 1.55

Ratio after operation = 1

Explanation:

From online values of the properties of this material, we have;

Yield strength; S_y = 32 kpsi

Ultimate Strength; S_u = 49.5 kpsi

Modulus; m = 0.25

Percentage of cold work; W_c = 0.2

S_o = 90 kpsi

A) Let's calculate the strain(ε) from the formula;

A_o/A = 1/(1 - W_c)

A_o/A = 1/(1 - 0.2)

A_o/A = 1.25

Thus, strain is;

ε = In(A_o/A)

ε = In(1.25)

ε = 0.2231

Yield strength after the cold work operation is;

S'_y = S_o(ε)^(m)

Plugging in the relevant values;

S'_y = 90(0.2231)^(0.25)

S'_y = 61.854 kpsi

Percentage increase of yield strength = S'_y/(S'_y - S_u) × 100% = (61.854 - 32)/32 × 100% = 93.29%

Ultimate strength after the cold work operation is;

S'_u = S_u/(1 - W_c)

S'_u = 49.5/(1 - 0.2)

S'u = 61.875 kpsi

Percentage increase of ultimate strength = S'_u/(S'_u - S_u) × 100% = (61.875 - 49.5)/49.5 × 100% = 25%

B) Ratio of ultimate strength and yield strength before cold work operations is;

S_u/S_y = 49.5/32

S_u/S_y = 1.547

Ratio of ultimate strength and yield strength after cold work operations is;

S'_u/S'_y = 61.875/61.854 = 1

The ratio after the operation is less than before the operation, thus the ductility reduced.

6 0
3 years ago
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