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neonofarm [45]
2 years ago
9

A single-phase transformer circuit feeds a motor and lighting load of 50 kilowatts. At a power factor of .8, the KVA rating of t

he step down transformer would be
Engineering
1 answer:
AveGali [126]2 years ago
3 0

The KVA rating of the step down transformer at the given power factor would be 62.5 kVA.

<h3>What is power factor of a transformer?</h3>

Power factor (PF) is the ratio of working power, measured in kilowatts (kW), to apparent power, measured in kilovolt amperes (kVA).

PF = working power / apparent power

PF =  kW/kVA

kVA = kW/PF

kVA = 50 kW/0.8

kVA = 62.5 kVA

Thus, the KVA rating of the step down transformer at the given power factor would be 62.5 kVA.

Learn more about power factor here: brainly.com/question/7956945

#SPJ1

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Can a 1½ " conduit, with a total area of 2.04 square inches, be filled with wires that total 0.93 square inches if the maximum f
Papessa [141]

Answer:

it is not possible to place the wires in the condui

Explanation:

given data

total area = 2.04 square inches

wires total area = 0.93 square inches

maximum fill conduit =  40%

to find out

Can it is possible place wire in conduit conduit

solution

we know maximum fill is 40%

so here first we get total area of conduit that will be

total area of conduit = 40% × 2.04

total area of conduit = 0.816 square inches

but this area is less than required area of wire that is 0.93 square inches

so we can say it is not possible to place the wires in the conduit

4 0
3 years ago
Example – a 100 kW, 60 Hz, 1175 rpm motor is coupled to a flywheel through a gearbox • the kinetic energy of the revolving compo
rjkz [21]

Answer:

1200KJ

Explanation:

The heat dissipated in the rotor while coming down from its running speed to zero, is equal to three times its running kinetic energy.

P (rotor-loss) = 3 x K.E

P = 3 x 300 = 900 KJ

After coming to zero, the motor again goes back to running speed of 1175 rpm but in opposite direction. The KE in this case would be;

KE = 300 KJ

Since it is in opposite direction, it will also add up to rotor loss

P ( rotor loss ) = 900 + 300 = 1200 KJ

7 0
2 years ago
A gas in a piston–cylinder assembly undergoes a compression process for which the relation between pressure and volume is given
viktelen [127]

Answer:

A.) P = 2bar, W = - 12kJ

B.) P = 0.8 bar, W = - 7.3 kJ

C.) P = 0.608 bar, W = - 6.4kJ

Explanation: Given that the relation between pressure and volume is

PV^n = constant.

That is, P1V1^n = P2V2^n

P1 = P2 × ( V2/V1 )^n

If the initial volume V1 = 0.1 m3,

the final volume V2 = 0.04 m3, and

the final pressure P2 = 2 bar. 

A.) When n = 0

Substitute all the parameters into the formula

(V2/V1)^0 = 1

Therefore, P2 = P1 = 2 bar

Work = ∫ PdV = constant × dV

Work = 2 × 10^5 × [ 0.04 - 0.1 ]

Work = 200000 × - 0.06

Work = - 12000J

Work = - 12 kJ

B.) When n = 1

P1 = 2 × (0.04/0.1)^1

P1 = 2 × 0.4 = 0.8 bar

Work = ∫ PdV = constant × ∫dV/V

Work = P1V1 × ln ( V2/V1 )

Work = 0.8 ×10^5 × 0.1 × ln 0.4

Work = - 7330.3J

Work = -7.33 kJ

C.) When n = 1.3

P1 = 2 × (0.04/0.1)^1.3

P1 = 0.6077 bar

Work = ∫ PdV

Work = (P2V2 - P1V1)/ ( 1 - 1.3 )

Work = (2×10^5×0.04) - (0.608 10^5×0.1)/ ( 1 - 1.3 )

Work = (8000 - 6080)/ -0.3

Work = -1920/0.3

Work = -6400 J

Work = -6.4 kJ

5 0
3 years ago
Given: A graphite-moderated nuclear reactor. Heat is generated uniformly in uranium rods of 0.05m diameter at the rate of 7.5 x
sineoko [7]

Answer:

The maximum temperature at the center of the rod is found to be 517.24 °C

Explanation:

Assumptions:

1- Heat transfer is steady.

2- Heat transfer is in one dimension, due to axial symmetry.

3- Heat generation is uniform.

Now, we consider an inner imaginary cylinder of radius R inside the actual uranium rod of radius Ro. So, from steady state conditions, we know that, heat generated within the rod will be equal to the heat conducted at any point of the rod. So, from Fourier's Law, we write:

Heat Conduction Through Rod = Heat Generation

-kAdT/dr = qV

where,

k = thermal conductivity = 29.5 W/m.K

q = heat generation per unit volume = 7.5 x 10^7 W/m³

V = volume of rod = π r² l

A = area of rod = 2π r l

using these values, we get:

dT = - (q/2k)(r dr)

integrating from r = 0, where T(0) = To = Maximum center temperature, to r = Ro, where, T(Ro) = Ts = surface temperature = 120°C.

To -Ts = qr²/4k

To = Ts + qr²/4k

To = 120°C + (7.5 x 10^7 W/m³)(0.025 m)²/(4)(29.5 W/m.°C)

To = 120° C + 397.24° C

<u>To = 517.24° C</u>

5 0
3 years ago
If a weld is laying into a joint with a concave weld contour with undercutting issues, what might be the cause?
Katarina [22]

Answer:bbbb

Explanation:

4 0
3 years ago
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