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Keith_Richards [23]

3 years ago

12

Explain the process of energy conversion by describing how energy was converted from the windmill design brief. Discuss the diff

erent forms of energy and what technology was used to convert the energy from one form to another.

1 answer:

cupoosta [38]3 years ago

6 0

Answer:

Wind energy is converted to Mechanical energy which is then converted in to electrical energy

Explanation:

In a wind mill the following energy conversions take place

a) Wind energy is converted into Mechanical energy (rotation of rotor blades)

b) Mechanical energy is converted into electrical energy (by using electric motor)

This electrical energy is then used for transmission through electric lines.

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A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC. Examine each of three cases with respect to the inequa

taurus [48]

Answer:

Explanation:

Given

$T_h=250^{\circ}C\approx 523\ K$

$T_L=30^{\circ}C\approx 303\ K$

$Q_1=6 kW$

From Clausius inequality

$\oint \frac{dQ}{T}=0$ =Reversible cycle

$\oint \frac{dQ}{T}$ =Irreversible cycle

$\oint \frac{dQ}{T}>0$ =Impossible

(a)For $P_{out}=3 kW$

Rejected heat $Q_2=6-3=3\ kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{3}{303}=1.57\times 10^{-3} kW/K$

thus it is Impossible cycle

(b) $P_{out}=2 kW$

$Q_2=6-2=4 kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{4}{303}=-1.73\times 10^{-3} kW/K$

Possible

(c)Carnot cycle

$\frac{Q_2}{Q_1}=\frac{T_1}{T_2}$

$Q_2=3.47\ kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{3.47}{303}=0$

and maximum Work is obtained for reversible cycle when operate between same temperature limits

$P_{out}=Q_1-Q_2=6-3.47=2.53\ kW$

Thus it is possible

6 0

4 years ago

A light bar AD is suspended from a cable BE and supports a 20-kg block at C. The ends A and D of the bar are in contact with fri

babymother [125]

Answer:

Tension in cable BE= 196.2 N

Reactions A and D both are 73.575 N

Explanation:

The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence

$T_{BE}-W=0$ hence

$T_{BE}=W=20*9.81=196.2 N$

Therefore, tension in the cable, $T_{BE}=196.2 N$

Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then

$196.2\times 0.125- 196.2\times 0.2+ D_x\times 0.2=0$

$24.525-39.24+0.2D_x=0$

$D_x=73.575 N$

Similarly,

$A_x-D_y=0$

$A_x=73.575 N$

Therefore, both reactions at A and D are 73.575 N

7 0

3 years ago

vlabodo [156]

Answer:

A. from zero at the center to maximum at the circumference.

Explanation:

The shear stress describes the process where a body deforms as a result of a part sliding over the other.

The value of shear stress which is induced in the shaft due to the applied couple varies from zero at the center to maximum at the circumference.

The shear stress in a shaft is given by this formula;

τ = Tr/I

where;

τ = shear stress (MPa).

T = applied torque (Nmm).

r = distance from center to stressed surface in the given position (mm).

I = polar moment of Inertia of cross-sectional area (mm).

3 0

3 years ago

A venturi meter is to be installed in a 63 mm bore section of a piping system to measure the flow rate of water in it. From spac

Alex Ar [27]

Answer:

Throat diameter $d_2$ =28.60 mm

Explanation:

Bore diameter $d_1=63mm$ ⇒ $A_1=3.09\times 10^{-3} m^2$

Manometric deflection x=235 mm

Flow rate Q=240 Lt/min⇒ Q=.004 $\frac{m^3}{s}$

Coefficient of discharge $C_d$ =0.8

We know that discharge through venturi meter

$Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}$

$h=x(\dfrac{S_m}{S_w}-1)$

$S_m$ =13.6 for Hg and $S_w$ =1 for water.

$h=0.235(\dfrac{13.6}{1}-1)$

h=2.961 m

Now by putting the all value in

$Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}$

$0.004=0.8\times \dfrac{3.09\times 10^{-3} A_2\sqrt{2\times 9.81\times 2.961}}{\sqrt{(3.09\times 10^{-3})^2-A_2^2}}$

$A_2=6.42\times 10^{-4} m^2$

⇒ $d_2$ =28.60 mm

So throat diameter $d_2$ =28.60 mm

4 0

3 years ago

Find the magnitude of the two pulling forces P and Q when their resultant is 50 N at 20° with Q. P 20°

morpeh [17]

Two forces P and Q whose resultant is 10Newton are at right angles to each other. If P makes 30 degrees with resultant.Show me the workings of the magnitude of Q in Newton and the diagram of the vectors.

7 0

3 years ago