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denpristay [2]
3 years ago
7

Consider a 5 m long, air-filled section of a coaxial transmission line, given that the radius of the inner conductor is 10 cm an

d the inner radius of the outer conductor is 20 cm? If the current is 100 mA, how much magnetic energy is strored in the air medium between the conductors. This problem is similar to Example 2 in the Time Varying Fields folder in the Example Bank.
a) 200.89 nj
b) 3.47 nj
c) 10.45nj
d) 80.9nj
Engineering
1 answer:
Helga [31]3 years ago
4 0

Answer: b) 3.47 nj

Explanation:

Given that;

length l = 5m

radius of inner conductor r = 10cm = 0.1m

radius of outer conductor D = 20cm = 0.2m

current I = 100A = 100×10⁻³ = 0.1

medium between conductor in air u₀ = 4π × 10⁻⁷

Energy in a coaxial cable transmission line is

w = u₀ /2π I² en(b/a)

we substitute

L = 4π × 10⁻⁷ /4π ×10⁻²× 5 en (20/10)

L =3.4657 × 10⁻⁹ J

L = 3.4657 nJ ≈ 3.47 nJ

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A 10-mm steel drill rod was heat-treated and ground. The measured hardness was found to be 290 Brinell. Estimate the endurance s
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the endurance strength  S_e = 421.24  MPa

Explanation:

From the given information; The objective is to estimate the endurance strength, Se, in MPa .

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It is expressed as:

200 \leq H_B \leq 450

S_{ut} = 3.41 H_B

where;

H_B = Brinell hardness number

S_{ut} =  Ultimate tensile strength

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S_{ut} = 3.41 H_B; replace 290 for H_B ; we have

S_{ut} = 3.41 (290)

S_{ut} = 988.9 MPa

We can see that the derived value for the ultimate tensile strength when the Brinell harness number = 290 is less than 1400 MPa ( i.e it is 988.9 MPa)

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The Endurance limit can be represented by the formula:

S_e ' = 0.5 S_{ut}

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S_e ' = 494.45 MPa

Using Table 6.2 for parameter for Marin Surface modification factor. The value for a and b are derived; which are :

a = 1.58

b =  -0.085

The value of the surface factor can be calculate by using the equation

k_a = aS^b_{ut}

K_a = 1.58 (988.9)^{-0.085

K_a = 0.8792

The formula that is used to determine the value of  k_b for the rotating shaft of size factor d = 10 mm is as follows:

k_b = 1.24d^{-0.107}

k_b = 1.24(10)^{-0.107}

k_b = 0.969

Finally; the the endurance strength, Se, in MPa if the rod is used in rotating bending is determined by using the expression;

S_e =k_ak_b S' _e

S_e= 0.8792×0.969×494.45

S_e = 421.24  MPa

Thus; the endurance strength  S_e = 421.24  MPa

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