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nalin [4]
3 years ago
7

A three-phase transformer bank consists of 3 single-phase transformers to handle 400 kVA witha 34.5kV/13.8kV voltage ratio. Find

the rating of each individual single-phase transformer in the bank (high voltage, low voltage, turns ratio, and apparent power) for the following transformer connections:
Engineering
1 answer:
Contact [7]3 years ago
6 0

Answer:

(a) high voltage = 19.9 kV, low voltage = 7.97 kV, turns ratio = 2.50:1, apparent power = 133 kVA

(b) high voltage = 19.9 kV, low voltage = 13.8 kV, turns ratio = 1.44:1, apparent power = 133 kVA

(c) high voltage = 34.5 kV, low voltage = 7.97 kV, turns ratio = 4.33:1, apparent power = 133 kVA

(d) high voltage = 34.5 kV, low voltage = 13.8 kV, turns ratio = 2.50:1, apparent power = 133 kVA

Explanation:

The turn ratio can be estimated by taking the ratio of the values of the high voltage and the low voltage. For example, the turn ratio for (a) is 19.9 kV/7.97 kV = 2.50:1. Similarly, the apparent power for each transformer is calculated as 400 kVA/3 = 133 kVA. Furthermore, the high and low voltages for the Δ are the same for the three-phase transformer i.e. 34.5 kv/13.8 kV. The high and high voltages for Y connection is calculated as 34.5 kV/1.73 = 19.9 kV and 13.8 kV/1.73 = 7.97 kV.

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Vector A extends from the origin to a point having polar coordinates (7, 70ᵒ ) and vector B extends from the origin to a point h
yaroslaw [1]

Answer:

13.95

Explanation:

Given :

Vector A polar coordinates = ( 7, 70° )

Vector B polar coordinates = ( 4, 130° )

To find A . B we  will

A ( r , ∅ ) = ( 7, 70 )

A = rcos∅ + rsin∅

therefore ; A  = 2.394i + 6.57j

B ( r , ∅ ) = ( 4, 130° )

B = rcos∅ + rsin∅

therefore ;  B = -2.57i + 3.06j

Hence ; A .B

( 2.394 i + 6.57j ) . ( -2.57 + 3.06j ) = 13.95

8 0
3 years ago
Compared to 15 mph on a dry road, about how much longer will it take for
Marysya12 [62]

Answer:

8 to 10 times

Explanation:

For dry road

u= 15 mph        ( 1 mph = 0.44 m/s)

u= 6.7 m/s

Let take coefficient of friction( μ) of dry road is 0.7

So the de acceleration a = μ g

a= 0.7 x 10  m/s ²                         ( g=10 m/s ²)

a= 7 m/s ²

We know that

v= u - a t

Final speed ,v=0

0 = 6.7 - 7 x t

t= 0.95 s

For snow road

μ = 0.4

de acceleration a = μ g

a = 0.4 x 10 = 4 m/s ²

u= 30 mph= 13.41 m/s

v= u - a t

Final speed ,v=0

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t'=7.5 s

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We can say that it will take 8 to 10 times more time as compare to dry road for stopping the vehicle.

8 to 10 times

7 0
3 years ago
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If the specific surface energy for aluminum oxide is 0.90 J/m2 and its modulus of elasticity is (393 GPa), compute the critical
vampirchik [111]

Answer:

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Explanation:

given data

specific surface energy = 0.90 J/m²

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internal crack length = 0.6 mm

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critical stress required for the propagation

solution

we will apply here critical stress formula for propagation of internal crack

( σc ) = \sqrt{\frac{2E\gamma s}{\pi a}}    .....................1

here E is modulus of elasticity and γs is specific surface energy and a is half length of crack i.e 0.3 mm  = 0.3 ×10^{-3} m

so now put value in equation 1 we get

( σc ) = \sqrt{\frac{2E\gamma s}{\pi a}}

( σc ) = \sqrt{\frac{2*393*10^9*0.90}{\pi 0.3*10^{-3}}}

( σc ) = 27.396615 ×10^{6} N/m²

so critical stress required for the propagation is 27.396615 ×10^{6} N/m²

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4 years ago
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