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anygoal [31]
3 years ago
15

A cable is checked for power installation and is found to be too small. The cable has 37 strands of copper wire each having a di

ameter of 82.2 mils. Find the size of the cable in circular mils. Carry the answer to two decimal places?
Engineering
1 answer:
satela [25.4K]3 years ago
7 0

Answer:

Length = 9558.7\ mils

Explanation:

Given

diameter = 82.2\ mils

Strands = 37

Required

The length of the wire

First, we calculate the circumference (C) of 1 strand of the wire

C = \pi d

C = \frac{22}{7} * 82.2

C = \frac{22* 82.2}{7}

C = \frac{1808.4}{7}

The length of the 37 strands is:

Length = 37 * C

Length = 37 *  \frac{1808.4}{7}

Length = \frac{37 * 1808.4}{7}

Length = \frac{66910.8}{7}

Length = 9558.68571429

Length = 9558.7\ mils --- approximated

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A.  implementation

Explanation:

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3 years ago
Ammonia gas is diffusing at a constant rate through a layer of stagnant air 1 mm thick. Conditions are such that the gas contain
fiasKO [112]

Answer:

The solution to this question is 5.153×10⁻⁴(kmol)/(m²·s)

That is the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

Explanation:

The diffusion through a stagnant layer is given by

N_{A}  = \frac{D_{AB} }{RT} \frac{P_{T} }{z_{2} - z_{1}  } ln(\frac{P_{T} -P_{A2}  }{P_{T} -P_{A1} })

Where

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z = Thickness in layer of transfer

R = universal gas constant

P_{A1} = Pressure at first boundary

P_{A2} = Pressure at the destination boundary

T = System temperature

P_{T} = System pressure

Where P_{T} = 101.3 kPa P_{A2} =0, P_{A1} =y_{A}, P_{T} = 0.5×101.3 = 50.65 kPa

Δz = z₂ - z₁ = 1 mm = 1 × 10⁻³ m

R =  \frac{kJ}{(kmol)(K)} ,    T = 298 K   and  D_{AB} = 1.18 \frac{cm^{2} }{s} = 1.8×10⁻⁵\frac{m^{2} }{s}

N_{A} = \frac{1.8*10^{-5} }{8.314*295} *\frac{101.3}{1*10^{-3} }* ln(\frac{101.3-0}{101.3-50.65}) = 5.153×10⁻⁴\frac{kmol}{m^{2}s }

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5.153×10⁻⁴(kmol)/(m²·s)

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3 years ago
An 18-in.-long titanium alloy rod is subjected to a tensile load of 24,000 lb. If the allowable tensile stress is 60 ksi and the
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Answer:

Required Diameter = 302.65 inches

Explanation:

We are given;

Allowable tensile stress = 60 ksi

Weight of tensile load = 24,000 lb

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Original length = 18 in

We'll need to check the diameters under stress and strain.

Now, we know that the formula for stress is;

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Thus,

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So for this stress, area required is;

A_req = 24000/60 = 4000 in²

So let's find the required diameter here.

Area = πd²/4

So, 4000 = πd²/4

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Required diameter here is;

d = √5092.96

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Formula for strain is;

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strain = elongation/original length = 0.05/18 = 0.00278

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Making A the subject to obtain;

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So, 71942 = πd²/4

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Required diameter here is;

d = √91599.4

d = 302.65 in

The larger diameter is 302.65 inchesand it's therefore the required one.

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