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forsale [732]
3 years ago
9

A 1000 KVA three phase transformer has a secondary voltage of 208/120. What is the secondary full load amperage?

Engineering
1 answer:
IceJOKER [234]3 years ago
4 0

Answer:

The three phase full load secondary amperage is 2775.7 A

Explanation:

Following data is given,

S = Apparent Power = 1000 kVA

No. of phases = 3

Secondary Voltage: 208 V/120 V <em>(Here 208 V is three phase voltage and 120 V is single phase voltage) </em>

<em>Since,</em>

<em />

<em />V_{1ph} =\frac{ V_{3ph}}{\sqrt{3} }\\V_{1ph) = \frac{208}{\sqrt{3} }\\<em />

V_{1ph} = 120 V

The formula for apparent power in three phase system is given as:

S = \sqrt{3} VI

Where:

S = Apparent Power

V = Line Voltage

I = Line Current

In order to calculate the Current on Secondary Side, substituting values in above formula,

1000 kVA = \sqrt{3} * (208) * (I)\\1000 * 1000 = \sqrt{3} * (208) * (I)\\I = \frac{1000 * 1000}{\sqrt{3} * (208) }\\ I = 2775.7 A

 

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A shunt regulator utilizing a zener diode with an incremental resistance of 8 ohm is fed through an 82-Ohm resistor. If the raw
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Answer:

\triangle V_0=0.08V

Explanation:

From the question we are told that:

Incremental resistance  R=8ohms

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7 0
3 years ago
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa √m (50 ksi √in.). If, during
astra-53 [7]

Answer:

0.024 m = 24.07 mm

Explanation:

1) Notation

\sigma_c = tensile stress = 200 Mpa

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\lambda= length of a surface crack (Variable of interest)

2) Definition and Formulas

The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.

By definition we have the following formula for the tensile stress:

\sigma_c=\frac{K}{Y\sqrt{\pi\lambda}}   (1)

We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for \lambda

Multiplying both sides of equation (1) by Y\sqrt{\pi\lambda}

\sigma_c Y\sqrt{\pi\lambda}=K   (2)

Sequaring both sides of equation (2):

(\sigma_c Y\sqrt{\pi\lambda})^2=(K)^2  

\sigma^2_c Y^2 \pi\lambda=K^2   (3)

Dividing both sides by \sigma^2_c Y^2 \pi we got:

\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2   (4)

Replacing the values into equation (4) we got:

\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m

3) Final solution

So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.

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