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3 years ago

Comparison of copper and aluminium conductors looking at their properties

Engineering

1 answer:

Alexeev081 [22]3 years ago

7 0

Answer:

The density of the copper is higher than aluminium. Hence it is heavier compared to aluminium conductors it requires strong structures and hardware to bear the weight. More ductile and has high tensile strength.

...

Aluminium & Copper properties.

Property Copper (Cu) Aluminium (Al)

Density (g/cm3) 8.96 2.70

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A mass of 0.3 kg is suspended from a spring of stiffness 0.4 N/mm. The damping is 3.286335345 kg/s. What is the undamped natural

mina [271]

Answer:

f=5.81 Hz

Explanation:

Given that

m= 0.3 kg

K= 0.4 N/mm

K= 400 N/m

C= 3.28

We know that undamped natural frequency given as

$\omega= \sqrt{\dfrac{K}{m}}$

Now by putting the values

$\omega= \sqrt{\dfrac{K}{m}}$

$\omega= \sqrt{\dfrac{400}{0.3}}$

ω = 36.51 rad/s

We know that

ω = 2 π f

36.51 = 2 x π x f

f=5.81 Hz

So the undamped natural frequency is 5.81 Hz.

6 0

3 years ago

A type 3 wind turbine has rated wind speed of 13 m/s. Coefficient of performance of this turbine is 0.3. Calculate the rated pow

Anna [14]

Answer:

Rated power = 1345.66 W/m²

Mechanical power developed = 3169035.1875 W

Explanation:

Wind speed, V = 13 m/s

Coefficient of performance of turbine, $C_p$ = 0.3

Rotor diameter, d = 100 m

Radius = 50 m

Air density, ρ = 1.225 kg/m³

Now,

Rated power = $\frac{1}{2}\rho V^3$

Rated power = $\frac{1}{2}\times1.225\times13^3$

Rated power = 1345.66 W/m²

b) Mechanical power developed = $\frac{1}{2}\rho AV^3C_p$

Here, A is the area of the rotor

A = π × 50²

thus,

Mechanical power developed = $\frac{1}{2}\times1.225\times\pi\times50^2\times13^3\times0.3$

Mechanical power developed = 3169035.1875 W

8 0

3 years ago

A fatigue test is performed on 69 rotating specimens made of 5160H steel. The measured number of cycles to failure (L in kcycles

tensa zangetsu [6.8K]

Answer:

(a) Mean = 122.9, σ = 30.071

(b) No. of failed specimens at less than 115k cycles are 27.

Explanation:

We are given:

L 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210

f 2 1 3 5 8 12 6 10 8 5 2 3 2 1 0 1

(a) First we need to calculate the mean and standard deviation. The formula for calculating mean is:

Mean = ∑fx/∑f

And for standard deviation we have:

S.D. = √Var

Var = ∑fx²/∑f - (Mean)²

∑fx = (2*60) + (1*70) + (3*80) + (5*90) + (8*100) + (12*110) + (6*120) + (10*130) + (8*140) + (5*150) + (2*160) + (3*170) + (2*180) + (1*190) + (0*200) + (1*210)

= 120 + 70 + 240 + 450 + 800 + 1320 + 720 + 1300 + 1120 + 750 + 320 + 510 + 360 + 190 + 0 + 210

∑fx = 8480

Mean = ∑fx/∑f

= 8480/69

Mean = 122.9

∑fx² = (2*60²) + (1*70²) + (3*80²) + (5*90²) + (8*100²) + (12*110²) + (6*120²) + (10*130²) + (8*140²) + (5*150²) + (2*160²) + (3*170²) + (2*180²) + (1*190²) + (0*200²) + (1*210²)

=7200+4900+19200+40500+80000+145200+86400+169000+156800+112500+51200+86700+64800+36100+0+44100

∑fx² = 1104600

Var = ∑fx²/∑f - (Mean)²

= 1104600/69 - (122.9)²

= 16008.69565 - 15104.41

Var = 904.2856

S.D = √Var

σ = √904.2856

σ = 30.071

(b) Let X be the number of failed specimen.

We will use the z-score to calculate the probability. The formula for z-score is:

z = (X-μ)/σ

P(X<115) = P(z<(115-122.9)/30.071)

= P(z<-0.26)

Using the normal distribution probability table, we can compute the value of P(z<-0.26).

P(X<115) = 0.3974

So, no. of failed specimens at less than 115k cycles are: 0.3974*69 = 27 specimens

(c) σ = 30.071

P(x<115) = 0.99

P(z<(115-μ)/30.071) = 0.99

From the normal distribution table we find that 0.99 lies between the z values 2.52 and 2.33. Hence, we get 2.525 as the z-value at which the probability is 0.99.

z = (x-μ)/σ

2.525 = (115 - μ)/30.071

75.93 = 115 - μ

μ = 115 - 75.93

μ = 39.07

4 0

3 years ago

Select the correct answer.

juin [17]

Orthographic projection, common method of representing three-dimensional objects, usually by three two-dimensional drawings in each of which the object is viewed along parallel lines that are perpendicular to the plane of the drawling.

4 0

3 years ago

One gram of Strontium-90 has an activity of 5.3 terabecquerels (TBq), what will be the activity of 1 microgram?

noname [10]

1 micro gram of Strontium-90 has an activity of

0.0000053 terabecquerels (TBq),

Explanation:

Given information denotes that .,one gram of Strontium-90 has an activity of 5.3 terabecquerels (TBq)

the activity of 1 micro gram is

1 gram = 1,000,000 micro gram has activities of 5.3 terabecquerels

therefore 1 micro gram has the activity of (5.3 ÷ 1,000,000 = 0.0000053 )

= $(5.3 ÷ 1,000,000 = 0.0000053 )$

Hence ., 1 micro gram of Strontium-90 has an activity of

0.0000053 terabecquerels (TBq),

8 0

3 years ago