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mars1129 [50]
4 years ago
8

A three-phase voltage source with a terminal voltage of 22kV is connected to a three-phase transformer rated 5MVA 22kV/220V. The

transformer has a per-phase series impedance of 0.05 pu and is connected to a distribution line with series impedance of 0.01+0.05Ω per phase. The line supplies a balanced three-phase load rated 1MVA, 220V with 0.8 lagging power factor.
a. Draw the one-line diagram of the system described.
b. Draw the equivalent per phase circuit of the system indicating all impedances in per unit. Use 1MVA, 220V as the base at the load.
c. Calculate the real losses in the transmission line in kW.

Engineering
1 answer:
amid [387]4 years ago
6 0

Answer:

A) See Attachment B) See Attachment C) 186.153

Explanation:

C) Per phase Voltage= 220/∛3

                                  = 6.037

S=V²/Z

  = (6.037)²/(0.01+i0.05)

  = 62051.282-i310256.41

Active power losses per phase= 62.051kW

total Active power losses= 62.051×3

                                         =186.153kW

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A train starts from rest at station A and accelerates at 0.4 m/s^2 for 60 s. Afterwards it travels with a constant velocity for
mash [69]
<h3><u>The distance between the two stations is</u><u> </u><u>3</u><u>7</u><u>.</u><u>0</u><u>8</u><u> km</u></h3>

\\

Explanation:

<h2>Given:</h2>

a_1 \:=\:0.4\:m/s²

t_1 \:=\:60\:s

v_{i1} \:=\:0\:m/s

a_2 \:=\:0\:m/s²

t_2 \:=\:25\:min\:=\:1500\:s

a_3 \:=\:-0.8\:m/s²

v_{f3} \:=\:0\:m/s

\\

<h2>Required:</h2>

Distance from Station A to Station B

\\

<h2>Equation:</h2>

a\:=\:\frac{v_f\:-\:v_i}{t}

v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}

v\:=\:\frac{d}{t}

\\

<h2>Solution:</h2><h3>Distance when a = 0.4 m/s²</h3>

Solve for v_{f1}

a\:=\:\frac{v_f\:-\:v_i}{t}

0.4\:m/s²\:=\:\frac{v_f\:-\:0\:m/s}{60\:s}

24\:m/s\:=\:v_f\:-\:0\:m/s

v_f\:=\:24\:m/s

\\

Solve for v_{ave1}

v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}

v_{ave}\:=\:\frac{0\:m/s\:+\:24\:m/s}{2}

v_{ave}\:=\:12\:m/s

\\

Solve for d_1

v\:=\:\frac{d}{t}

12\:m/s\:=\:\frac{d}{60\:s}

720\:m\:=\:d

d_1\:=\:720\:m

\\

<h3>Distance when a = 0 m/s²</h3>

v_{f1}\:=\:v_{i2}

v_{i2}\:=\:24\:m/s

\\

Solve for v_{f2}

a\:=\:\frac{v_f\:-\:v_i}{t}

0\:m/s²\:=\:\frac{v_f\:-\:24\:m/s}{1500\:s}

0\:=\:v_f\:-\:24\:m/s

v_f\:=\:24\:m/s

\\

Solve for v_{ave2}

v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}

v_{ave}\:=\:\frac{24\:m/s\:+\:24\:m/s}{2}

v_{ave}\:=\:24\:m/s

\\

Solve for d_2

v\:=\:\frac{d}{t}

24\:m/s\:=\:\frac{d}{1500\:s}

36,000\:m\:=\:d

d_2\:=\:36,000\:m

\\

<h3>Distance when a = -0.8 m/s²</h3>

v_{f2}\:=\:v_{i3}

v_{i3}\:=\:24\:m/s

\\

Solve for v_{f3}

a\:=\:\frac{v_f\:-\:v_i}{t}

-0.8\:m/s²\:=\:\frac{0\:-\:24\:m/s}{t}

(t)(-0.8\:m/s²)\:=\:-24\:m/s

t\:=\:\frac{-24\:m/s}{-0.8\:m/s²}

t\:=\:30\:s

\\

Solve for v_{ave3}

v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}

v_{ave}\:=\:\frac{24\:m/s\:+\:0\:m/s}{2}

v_{ave}\:=\:12\:m/s

\\

Solve for d_3

v\:=\:\frac{d}{t}

12\:m/s\:=\:\frac{d}{30\:s}

360\:m\:=\:d

d_3\:=\:360\:m

\\

<h3>Total Distance from Station A to Station B</h3>

d\:= \:d_1\:+\:d_2\:+\:d_3

d\:= \:720\:m\:+\:36,000\:m\:+\:360\:m

d\:= \:37,080\:m

d\:= \:37.08\:km

\\

<h2>Final Answer:</h2><h3><u>The distance between the two stations is </u><u>3</u><u>7</u><u>.</u><u>0</u><u>8</u><u> km</u></h3>
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Answer:There is an idea floating around that being an optimist is good for you. In her book, ‘Smile or Die’, Barbara Ehrenreich traces the origins of this idea, from early religious eccentrics to modern day psychology and motivational gurus. Related notions are that self belief is important to achievement, that confidence is needed, and you get what you expect. I freely admit that there was a time when I also thought that expectations influenced results, and sometimes they do.

However, since then I have examined this much more closely and changed my views. I now realise that there are downsides to optimism too, and that optimism and pessimism are not the only options. If you are interested in exploring this in more detail, especially if you have been uncomfortable with the self-delusion that positive thinking usually involves, then this article may provide a clearer understanding of a reasonable alternative that works and makes sense.

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Answer:

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A steel rotating-beam test specimen has an ultimate strength Sut of 1600 MPa. Estimate the life (N) of the specimen if it is tes
ziro4ka [17]

Answer:

the life (N) of the specimen is 46400 cycles

Explanation:

given data

ultimate strength Su = 1600 MPa

stress amplitude σa = 900 MPa

to find out

life (N) of the specimen

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su   .............1

Se = 0.5 × 1600

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and we know

Se for steel is 700 Mpa for Su ≥ 1400 Mpa

so we take endurance limit Se is = 700 Mpa

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