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4 years ago

15

Three single-phase, 10 kVA, 2400/280 V, 60-Hz transformers are connected to form a three-phase, 2400/480 V transformer The equiv

alent impedance of each single-phase transformer referred to the primary side is 10 + j30 Î©. The transformer bank delivers 30 kVA at 480 V and 0.9 power factor lagging.
a. Draw the three-phase schematic diagram showing the transformer connection. Draw a per-phase equivalent circuit.
b. Determine the primary current and power factor.
c. Determine the primary voltage.
d. Determine the voltage regulation.

1 answer:

Dominik [7]4 years ago

7 0

The answer to this question is letter A

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1 year 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

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5 0

4 years ago

An industrial plant consists of several 60 Hz single-phase motors with low power factor. The plant absorbs 600 kW with a power f

Gelneren [198K]

Answer:

(a) $Q=332$ $kvar$ and $C=5.66$ $uF$

(b) $pf=0.90$ lagging

Explanation:

Given Data:

$P=600kW$

$V=12.47kV$

$f=60Hz$

$pf_{old} =0.75$

$pf_{new} =0.95$

(a) Find the required kVAR rating of a capacitor

$\alpha _{old}=cos^{-1}(0.75) =41.41$ °

$\alpha _{new}=cos^{-1}(0.95) =18.19$ °

The required compensation reactive power can be found by

$Q=P(tan(\alpha_{old}) - tan(\alpha_{new}))$

$Q=600(tan(41.41) - tan(18.19))$

$Q=332$ $kvar$

The corresponding capacitor value can be found by

$C=Q/2\pi fV^{2}$

$C=332/2*\pi *60*12.47^{2}$

$C=5.66$ $uF$

(b) calculate the resultant supply power factor

First convert the hp into kW

$P_{mech} =250*746=186.5$ $kW$

Find the electrical power (real power) of the motor

$P_{elec} =P_{mech}/n$

where $n$ is the efficiency of the motor

$P_{elec} =186.5/0.80=233.125$ $kW$

The current in the motor is

$I_{m} =(P/\*V*pf)$

The pf of motor is 0.85 Leading

Note that represents the angle in complex notation (polar form)

$I_{m} =(233.125/12.47*0.85)$

$I_{m}=18.694+11.586$ $j$ $A$

Now find the Load current

pf of load is 0.75 lagging (notice the minus sign)

$I_{load} =(600/12.47*0.75)$

$I_{load} =48.115-42.433$ $j$ $A$

Now the supply current is the current flowing in the load plus the current flowing in the motor

$I_{supply} =I_{m} + I_{load}$

$I_{supply}= (18.694+11.586)+(48.115-42.433)$

$I_{supply} =66.809-30.847j$ $A$

or in polar form

$I_{supply} =73.58$ °

Which means that the supply current lags the supply voltage by 24.78

therefore, the supply power factor is

$pf=cos(24.78)=0.90$ lagging

Which makes sense because original power factor was 0.75 then we installed synchronous motor which resulted in improved power factor of 0.90

8 0

4 years ago

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MakcuM [25]

Answer: For the center plate to remain stationed in one position without rotating, the bottom plate has to move to the left at a speed of 2m/s, so as to cancel the force acting on it from the top.

The center plate will not move when the bottom plate is moving left in a speed of 2m/s to counter the speed of the top plate, because a body will continue to be at rest if all the forces acting towards the body are equal. The center plate will be at rest because we have directed equal force from the top and bottom of the plate.

3 0

3 years ago

lesya [120]

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5 0

3 years ago

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