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2 years ago

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]2 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 (Here 208 V is three phase voltage and 120 V is single phase voltage)

Since,

$V_{1ph} =\frac{ V_{3ph}}{\sqrt{3} }\\V_{1ph) = \frac{208}{\sqrt{3} }\\$

$V_{1ph} = 120 V$

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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2 years ago

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2 years ago

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3 years ago

Given the vector current density J = 10rho2zarho − 4rho cos2 φ aφ mA/m2:

Xelga [282]

Answer:

(a) Current density at P is $J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\$ .

(b) Total current I is 3.257 A

Explanation:

Because question includes symbols and formulas it can be misunderstood. In the question current density is given as below;

$J=10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}\\$

where $\textbf{a}_{\rho}$ and $\textbf{a}_{\phi}$ unit vectors.

(a) In order to find the current density at a specific point (P), we can simply replace the coordinates in the current density equation. Therefore

$J(P(\rho=3, \phi=30^o,z=2))=10.3^2.2.\textbf{a}_{\rho}-4.3.(\cos(30^o)^2).\textbf{a}_{\phi}\\\\J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\$

(b) Total current flowing outward can be calculated by using the relation,

$I=\int {\textbf{J} \, \textbf{ds}$

where integral is calculated through the circular band given in the question. We can write the integral as below,

$I=\int\{(10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}).(\rho.d\phi.dz.\textbf{a}_{\rho}})\}\\\\I=\int\{(10\rho^2z).(\rho.d\phi.dz)\}\\\\\\$

due to unit vector multiplication. Then,

$I=10\int\(\rho^3z.dz.d\phi$

where $\rho=3,\ 0$ . Therefore

$I=10.3^3\int_2^{2.8}\(zdz.\int_0^{2\pi}d\phi\\I=270(\frac{2.8^2}{2}-\frac{2^2}{2} )(2\pi-0)=3257.2\ mA\\I=3.257\ A$

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3 years ago