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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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An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at

Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

$L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft$

$G_2$ is given as

$G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%$

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

$A=\frac{L}{K}\\A=\frac{460}{115}\\A=4$

A is given as

$-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%$

With initial grade, the elevation of PVC is

$E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\$

The station is given as

$St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\$

Low point is given as

$x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft$

The station of low point is given as

$St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\$

The elevation is given as

$E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft$

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

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Drupady [299]

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A 1000 kg turbine has a rotating unbalance of 0.1 kg.m. The turbine operates at a speed between 500 to 750 rpm. What is the maxi

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Answer:

maximum isolator stiffness k =1764 kN-m

Explanation:

mean speed of rotation $=\frac{N_1 +N_2}{2}$

$Nm = \frac{500+750}{2} = 625 rpm$

$w =\frac{2\pi Nm}{60}$

=65.44 rad/sec

$F_T = mw^2 e$

$F_T = mew^2$

= 0.1*(65.44)^2

F_T =428.36 N

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transmission ratio $= \frac{1}{[\frac{w}{w_n}]^{2} -1}$

$0.7 =\frac{1}{[\frac{65.44}{w_n}]^2 -1}$

SOLVING FOR Wn

Wn = 42 rad/sec

$Wn = \sqrt {\frac{k}{m}$

k = m*W^2_n

k = 1000*42^2 = 1764 kN-m

k =1764 kN-m

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