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2 years ago

14

At a certain location, wind is blowing steadily at 7 m/s. Determine the mechanical energy of air per unit mass and the power gen

eration potential of a wind turbine with 80-m-diameter blades at that location. Also determine the actual electric power generation assuming an overall efficiency of 30 percent. Take the air density to be 1.25 kg/m3.

1 answer:

Kaylis [27]2 years ago

8 0

Answer:

Explanation:

From the information given;

The velocity of the wind blow V = 7 m/s

The diameter of the blades (d) = 80 m

Percentage of the overall efficiency $\eta_{overall} = 30\%$

The density of air $\rho = 1.25 kg/km^3$

Then, we can use the concept of the kinetic energy of the wind blowing to estimate the mechanic energy of air per unit mass by using the formula:

$e_{mechanic} = \dfrac{mV^2}{2}$

here;

m = $\rho AV$

= $1.25 \times \dfrac{\pi}{4}(80)^2 \times 7$

= 43982.29 kg/s

∴

$W = e_{mechanic} = \dfrac{mV^2}{2}$

$= \dfrac{43982.29 \times 7^2}{2}$

$= 1077566.105 \ W$

$\mathbf{ =1077.566 \ kW}$

The actual electric power is:

$W_{electric} = \eta_{overall} \times W$

$W_{electric} = 0.3 \times 1077.566$

$\mathbf{W_{electric} =323.26 \ kW}$

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In a tensile test on a steel specimen, true strain = 0.12 at a stress of 250 MPa. When true stress = 350 MPa, true strain = 0.26

scZoUnD [109]

Answer:

The strength coefficient is $625$ and the strain-hardening exponent is $0.435$

Explanation:

Given the true strain is 0.12 at 250 MPa stress.

Also, at 350 MPa the strain is 0.26.

We need to find $(K)$ and the $(n)$ .

$\sigma =K\epsilon^n$

We will plug the values in the formula.

$250=K\times (0.12)^n\\350=K\times (0.26)^n$

We will solve these equation.

$K=\frac{250}{(0.12)^n}$ plug this value in $350=K\times (0.26)^n$

$350=\frac{250}{(0.12)^n}\times (0.26)^n\\ \\\frac{350}{250}=\frac{(0.26)^n}{(0.12)^n}\\ \\1.4=(2.17)^n$

Taking a natural log both sides we get.

$ln(1.4)=ln(2.17)^n\\ln(1.4)=n\times ln(2.17)\\n=\frac{ln(1.4)}{ln(2.17)}\\ n=0.435$

Now, we will find value of $K$

$K=\frac{250}{(0.12)^n}$

$K=\frac{250}{(0.12)^{0.435}}\\ \\K=\frac{250}{0.40}\\\\K=625$

So, the strength coefficient is $625$ and the strain-hardening exponent is $0.435$ .

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3 years ago

Velocity components in an incompressible flow are: v = 3xy + x^2 y: w = 0. Determine the velocity component in the x-direction.

cupoosta [38]

Answer:

Velocity component in x-direction $u=-\frac{3}{2}x^2-\frac{1}{3}x^3$ .

Explanation:

v=3xy+ $x^{2}$ y

We know that for incompressible flow

$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$

$\frac{\partial v}{\partial y}=3x+x^{2}$

So $\frac{\partial u}{\partial x}+3x+x^{2}=0$

$\frac{\partial u}{\partial x}= -3x-x^{2}$

By integrate with respect to x,we will find

$u=-\frac{3}{2}x^2-\frac{1}{3}x^3$ +C

So the velocity component in x-direction $u=-\frac{3}{2}x^2-\frac{1}{3}x^3$ .

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3 years ago

1. Sewage-treatment plant, a large concrete tank initially contains 440,000 liters liquid and 10,000 kg fine suspended solids. T

Elenna [48]

Answer:

Concentration = 10.33 kg/m³

Explanation:

We are given;

Mass of solids; 10,000 kg

Volume; V = 440,000 L = 440 m³

Rate at which water is pumped out = 40,000 liter/h

Thus, at the end of 5 hours we amount of water that has been replaced with fresh water is = 40,000 liter/h x 5 hours = 200,000 L = 200 m³

Now, since the tank is perfectly mixed, therefore we can calculate a ratio of fresh water to sewage water as;

200m³/440m³ = 5/11

Thus, the amount left will be calculated by multiplying that ratio by the amount of solids;

Thus,

Amount left; = 10000 x (5/11) = 4545 kg

The concentration would be calculated by:

Concentration = amount left/initial volume

Thus,

Concentration = 4545/440 = 10.3 kg/m³

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3 years ago

Lets assume, a represents the edge length (lattice constant) of a BCC unit cell and R represents the radius of the atom in the u

uranmaximum [27]

Answer:

$4\ R=\sqrt 3\ a$

Explanation:

Given that

Lattice constant = a

Radius of unit cell cell =R

Atom is in BCC structure.

In BCC unit cell (Body centered cube)

1.Eight atoms at eight corner of cube which have 1/8 part in each cube.

2.One complete atom at the body center of the cube

So the total number of atoms in the BCC

Z= 1/8 x 8 + 1 x 1

Z=2

In triangle ABD

$AB^2=AD^2+BD^2$

$AB^2=a^2+a^2$

$AB=\sqrt 2\ a$

In triangle ABC

$AC^2=AB^2+BC^2$

AC=4R

BC=a

$AB=\sqrt 2\ a$

So

$16R^2=2a^2+a^2$

$4\ R=\sqrt 3\ a$

So the relationship between lattice constant and radius of unit cell

$4\ R=\sqrt 3\ a$

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3 years ago

Atmospheric air at a pressure of 1 atm and dry bulb temperature of 28 C has a wet-bulb temperature of 20 C.

IrinaK [193]

Answer:

a) $\phi = 48\%$ , b) $\omega = 0.012\,\frac{kg\,H_{2}O}{kg\,Air}$ , c) $h = 58\,\frac{kJ}{kg\,Air}$ , d) $T = 17^{\textdegree}C$ , e) $P_{v} = 1.831\,kPa$

Explanation:

a) The relative humidity is given by the intersection of the dry bulb and wet bulb temperatures:

$\phi = 48\%$

b) The humidity ratio is:

$\omega = 0.012\,\frac{kg\,H_{2}O}{kg\,Air}$

c) The enthalpy is:

$h = 58\,\frac{kJ}{kg\,Air}$

d) The dew-point temperature is:

$T = 17^{\textdegree}C$

e) The water vapor pressure is the product of the relative humidity and the saturation pressure evaluated at dry bulb temperature:

$P_{v} = \phi \cdot P_{sat}$

$P_{v} = 0.48\cdot (3.816\,kPa)$

$P_{v} = 1.831\,kPa$

7 0

3 years ago