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

IceJOKER [234]

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 />

$V_{1ph} = 120 V$

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,

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

4 0

2 years ago

Technician A says test lights are great for performing simple tests. Technician B says you can use a test light to check SRS cir

adoni [48]

The technician that is correct about either testing lights for simple tests or to check SRS Circuits is; Technician A.

<h3>Which Technician is Correct?</h3>

First of all it is pertinent to note that test lights are generally small bulbs that are turned on by the voltage and current flowing through the circuit in analog circuits.

Now, the two values of voltage and current are high and sufficient to light up the bulb. However, in digital circuits, the current is very small in the order of milliamps, and as a result there is not enough power to turn on the lights.

Thus, we can conclude that Technician A is correct.

Read more about Correct Technician at; brainly.com/question/14449935

5 0

2 years ago

For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration will raise the carbon concentratio

diamong [38]

This question is incomplete, the complete question is;

For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration at Temperature T1 will raise the carbon concentration to 0.44 wt% at a point 1.8 mm from the surface. A separate experiment is performed at T2 that doubles the diffusion coefficient for carbon in steel.

Estimate the time necessary to achieve the same concentration at a 4.9 mm position for an identical steel and at the same carburizing temperature T2.

Answer:

the required time to achieve the same concentration at a 4.9 is 83.733 hrs

Explanation:

Given the data in the question;

treatment time t₁ = 11.3 hours

Carbon concentration = 0.444 wt%

thickness at surface x₁ = 1.8 mm = 0.0018 m

thickness at identical steel x₂ = 4.9 mm = 0.0049 m

Now, Using Fick's second law inform of diffusion

$x^2$ / Dt = constant

where D is constant

then

$x^2$ / t = constant

$x^2_1$ / t₁ = $x^2_2$ / t₂

$x^2_1$ t₂ = t₁ $x^2_2$

t₂ = t₁ $x^2_2$ / $x^2_1$

t₂ = ( $x^2_2$ / $x^2_1$ )t₁

t₂ = $($ $x_2$ / $x_1$ $)^2$ × t₁

so we substitute

t₂ = $($ 0.0049 / 0.0018 $)^2$ × 11.3 hrs

t₂ = 7.41 × 11.3 hrs

t₂ = 83.733 hrs

Therefore, the required time to achieve the same concentration at a 4.9 is 83.733 hrs

8 0

2 years ago

Describe three differences between liquids and gases in fluid power systems.<br> Help !!!

scoundrel [369]

Gases, liquids and solids are all made up of atoms, molecules, and/or ions, but the behaviors of these particles differ in the three phases. ... gas are well separated with no regular arrangement. liquid are close together with no regular arrangement. solid are tightly packed, usually in a regular pattern.

3 0

3 years ago

11–17 A long, thin-walled double-pipe heat exchanger with tube and shell diameters of 1.0 cm and 2.5 cm, respectively, is used t

lana [24]

Answer:

the overall heat transfer coefficient of this heat exchanger is 1855.8923 W/m²°C

Explanation:

Given:

d₁ = diameter of the tube = 1 cm = 0.01 m

d₂ = diameter of the shell = 2.5 cm = 0.025 m

Refrigerant-134a

20°C is the temperature of water

h₁ = convection heat transfer coefficient = 4100 W/m² K

Water flows at a rate of 0.3 kg/s

Question: Determine the overall heat transfer coefficient of this heat exchanger, Q = ?

First at all, you need to get the properties of water at 20°C in tables:

k = 0.598 W/m°C

v = 1.004x10⁻⁶m²/s

Pr = 7.01

ρ = 998 kg/m³

Now, you need to calculate the velocity of the water that flows through the shell:

$v_{w} =\frac{m}{\rho \pi (\frac{d_{2}^{2}-d_{1}^{2} }{4} )} =\frac{0.3}{998*\pi (\frac{0.025^{2}-0.01^{2} }{4}) } =0.729m/s$

It is necessary to get the Reynold's number:

$Re=\frac{v_{w}(d_{2}-d_{1}) }{v} =\frac{0.729*(0.025-0.01)}{1.004x10^{-6} } =10891.4343$

Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:

$Nu=0.023Re^{0.8} Pr^{0.4} =0.023*(10891.4343)^{0.8} *(7.01)^{0.4} =85.0517$

The overall heat transfer coefficient:

$Q=\frac{1}{\frac{1}{h_{1} }+\frac{1}{h_{2} } }$

Here

$h_{2} =\frac{kNu}{d_{2}-d_{1}} =\frac{0.598*85.0517}{0.025-0.01} =3390.7278W/m^{2}C$

Substituting values:

$Q=\frac{1}{\frac{1}{4100}+\frac{1}{3390.7278} } =1855.8923W/m^{2} C$

5 0

3 years ago

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