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2 years ago

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The line voltage of a balanced three-phase transmission line is 4200 V rms. The transmission line has an impedance of 4 6 Z j l

  per phase. Assuming that the load receives a total of 1 MVA at 0.75 power factor lagging, determine (a) the complex power, (b) the power loss in the three-phase line, (c) the line voltage at the sending end of the transmission line.

1 answer:

balu736 [363]2 years ago

4 0

Answer:

1.23MVA, 226.74kW, 5.16kV

Explanation:

Parameters Given

line impedance, Zl = (4 + j6) ohms per phase

load voltage, Vl = 4200V

received complex power, S = 1 × 10⁶VA

power factor, cosФ = 0.75

Ф = 41.41°

sinФ = 0.66

Solution:

S = √3 * Vl * I (that is √3 × line voltage × line current)

1 × 10⁶ = √3 × 4200 × I

I = 137.46A

Vl = 4200∠0

I = 137.46∠- 41.41° lagging

source voltage, Vs = load voltage, Vl + voltage drop along the line, Vd

Vd = Zl * I where ( Zl = 4 + j6, = √(4² + 6²)∠tan⁻¹(6/4), = 7.21∠56.31° )

Vd = 7.21∠56.31° × 137.46∠- 41.41°

= 991.22∠14.9°

Vs = Vl + Vd

= 4200∠0° + 991.22∠14.9°

= 4200(cos 0° + j sin 0°) + 991.22(cos 14.9° + jsin 14.9°)

= 4200 + 957.68 + j254.88

= 5157.68 + j254.88

or

= 5163.97∠2.83° V (line voltage at the sending end of the transmission line)

Sending end current, I = 137.46∠-41.41 A

(a) Complex power = √3 × Vs × I

= √3 × 5163.97∠2.83° ×137.46∠-41.41

= 1229477.76∠-38.58°VA

= 1.23∠-38.58MVA

complex power = 1.23MVA

(b) power loss in the three phase line, Pl = 3 × square of line current, I × line impedance, Rl

Pl =3 × I² × Rl where Zl = R + j X = 4 + j6 hence R = 4

= 3 × 137.46² × 4

= 226743.02W

= 226.74kW

(c) from the above, the line voltage at the sending end of the transmission line is = 5163.97V

= 5.16kW

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Answer:

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:

$n = \frac{S_{uts}}{\tau_{max}}$

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$n$ - Safety factor, dimensionless.

$S_{uts}$ - Ultimate shear strength, measured in pascals.

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

The maximum allowable shear stress is consequently cleared and computed: ( $n = 4.2$ , $S_{uts} = 320\times 10^{6}\,Pa$ )

$\tau_{max} = \frac{S_{uts}}{n}$

$\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}$

$\tau_{max} = 76.190\times 10^{6}\,Pa$

Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:

$\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}$

Where:

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

$V$ - Shear force, measured in kilonewtons.

$A$ - Cross section area, measured in square meters.

As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:

$V = \frac{P}{5}$

$V = \frac{450\,kN}{5}$

$V = 90\,kN$

The minimum allowable cross section area is cleared in the shearing stress equation:

$A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}$

If $V = 90\,kN$ and $\tau_{max} = 76.190\times 10^{3}\,kPa$ , the minimum allowable cross section area is:

$A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}$

$A = 1.640\times 10^{-3}\,m^{2}$

The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:

$A = \frac{\pi}{4}\cdot D^{2}$

The diameter is now cleared and computed:

$D = \sqrt{\frac{4}{\pi}\cdot A}$

$D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})$

$D = 0.0457\,m$

$D = 45.7\,mm$

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