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adelina 88 [10]
3 years ago
10

A single-phase load is located 2800 ft from its source. The load draws a current of 86 A and operates on 480 V. The maximum volt

age drop at the load cannot be greater than 3%. What size aluminum conductors should be installed to operate this load?
Engineering
1 answer:
Akimi4 [234]3 years ago
6 0

check photo for answer

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What is bubble sort?what is bubble sort ​
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A single crystal of a metal that has the BCC crystal structure is oriented such that a tensile stress is applied in the [100] di
VARVARA [1.3K]

Answer:

For [1 1 0] and  [1 0 1] plane, σₓ = 6.05 MPa

For [0 1 1] plane, σ = 0; slip will not occur

Explanation:

compute the resolved shear stress in [111] direction on each of the [110], [011] and on the [101] plane.

Given;

Stress direction: [1 0 0] ⇒ A

Slip direction: [1 1 1]

Normal to slip direction: [1 1 1] ⇒ B

∅ is the angle between A & B

Step 1: cos∅ = A·B/|A| |B| = \frac{[100][111]}{\sqrt{1}.\sqrt{3}  } ⇒ cos∅ = 1/\sqrt{3}

σₓ = τ/cos ∅·cosλ

where τ is the critical resolved shear stress given as 2.47MPa

Step 2: Solve for the slip along each plane

(a) [1 1 0]

cosλ = [1 1 0]·[1 0 0]/(\sqrt{2}·\sqrt{1})        

note: cosλ = slip D·stress D/|slip D||stress D|

cosλ = 1/\sqrt{2}

∵ σₓ = τ/\frac{1}{\sqrt{2} } ·\frac{1}{\sqrt{3} } = \sqrt{6} * 2.47MPa = 6.05MPa

Hence, stress necessary to cause slip on [1 1 0] is 6.05MPa

(b) [0 1 1]

cosλ = [0 1 1]·[1 0 0]/(\sqrt{2}·\sqrt{1}) = 0

∵ σₓ = 2.47MPa/0, which is not defined

Hence, for stress along [1 0 0], slip will not occur along [0 1 1]

(c) [1 0 1]

cosλ = [0 1 1]·[1 0 0]/(\sqrt{2}·\sqrt{1})

cosλ = 1/\sqrt{2}

∵ σₓ = τ/\frac{1}{\sqrt{2} } ·\frac{1}{\sqrt{3} } = \sqrt{6} * 2.47MPa = 6.05MPa

See attachment for the space diagram

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3 years ago
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