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3 years ago

A 240-V rms 60-Hz supply serves a load that is 10 kW (resistive), 15 kVAR (capacitive), and 22 kVAR (inductive). Find:

Physics

1 answer:

Rudiy273 years ago

8 0

This question is incomplete, the complete is;

A 240-V rms 60-Hz supply serves a load that is 10 kW (resistive), 15 kVAR (capacitive), and 22 kVAR (inductive). Find:

a) the apparent power

b) the current drawn from the supply

c) the kVAR rating and capacitance required to improve the power factor to 0.96 lagging

d) the current drawn from the supply under the power-factor conditions.

Answer:

a) value of apparent power 12.21 kVA

b) current drawn from the supply is 50.86∠-35° A

c) value of capacitance is 188.03 uF

d) value of current drawn from supply 43.4∠-16.26° A

Explanation:

to calculate the value of apparent power, we say;

S = (10 × 10³) - (j15 × 10³) +( j22 × 10³ )

S = (10 - j15 + 22j ) × 10³

S = ( 10 + 17j ) KVA

now calculating the apparent power

║S║ = √ (( 10 × 10³) + (17j × 10³ ))

= √ ( 100 + 49 × 10³)

= 12.21 kVA

Therefore the value of apparent power 12.21 kVA

we calculate the current drawn from the supply

S = VI°

I° = S/V

I° = (( 10 + 17j ) ₓ 10³) / 240

I° = ( 0.041467 + j0.029167) × 10³

I°= 41.67 + j29.167 A

Therefore I = 41.67 - j29.167 A

I = 50.86∠-35° A

so current drawn from the supply is 50.86∠-35° A

from S = { 10 + 17j KVA, }

we calculate the power factor

∅₁ = tan⁻¹ ( Q / P )

∅₁ = tan⁺¹ ( 7 ₓ 10³ ) / ( 10 ₓ 10³)

∅₁ = tan⁻¹ ( 0.7 )

∅₁ = 35°

Now consider the new power factor, we know cos∅ form the question is 0.96

calculate the new value of power factor angle

∅₂ = cos⁻¹ ( 0.96 )

∅₂ = 16.26°

Now calculate the reduction in the reactive power caused by the shunt capacitor

Qc = Q₁ - Q₂

= P( tan∅₁ - tan∅₂ )

= ( 10 × 10³) (tan(35°) - tan(16.26°))

= ( 10 × 10³) ( 0.7 - 0.29166)

= (10⁴)(0.4083)

Qc = 4.083 kVAR

Now we Calculate value of capacitance

C = Qc / ωV²rms

C = (4.083 × 10³) / 2π(60) ( 240)²

C = (4.083 × 10³) / 21.715 × 10⁶

C = 1.8803 × 10⁻⁴

C = 188.03 uF

Therefore value of capacitance is 188.03 uF

To calculate the current drawn from the new power factor condition, we say;

S₁ = P₁ + jQ₁

P₁ = P = 10 kW

Q₁ = Q - Qc = ( 7 - 4.083) × 10³ = 2.917 kVAR

S₁ = 10 + j2.917 kVAR

We calculate the value of current drawn from supply

S₁ = Vl₁°

l₁° = S₁ / V

l₁° = (10 + j2.917) × 10³ / 240

l₁° = ( 0.04167 + j0.012154) × 10³

l₁° = 41.67 + j12.15 A

So l₁ = 41.67 - j12.15 A

l₁ = 43.4∠-16.26° A

the value of current drawn from supply 43.4∠-16.26° A

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Answer:

E = 1.45 x 10⁹ J = 1.45 GJ

Explanation:

According to the law of conservation of energy:

Potential Energy Lost by Satellite = Kinetic Energy + Internal Energy

$mgh = \frac{1}{2} mv^2 + E\\\\E = mgh - \frac{1}{2} mv^2$

where,

E = Internal Energy = ?

m = mass = 400 kg

g = acceleration due to gravity = 9.81 m/s²

h = height = 500 km = 500000 m

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Therefore,

$E = (400\ kg)(9.81\ m/s^2)(500000\ m)-\frac{1}{2} (400\ kg)(1600\ m/s)^2\\E = 1.962\ x\ 10^9\ J - 0.512\ x\ 10^9\ J$

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8 0

3 years ago

The electric field on the surface of an irregularly shaped conductor varies from 74.0 kN/C to 14.0 kN/C.

mrs_skeptik [129]

Answer:

(a). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

Explanation:

Given that,

Electric field $E_{1}=74.0\ kN/C$

Electric field $E_{2}=14.0\ kN/C$

When the radius of curvature is greatest, the electric field at the surface will be smaller.

Where the radius of curvature is greatest

(a). We need to calculate the local surface charge density at the point on the surface

Using formula of charge density

$\sigma=\epsilon_{0}E_{2}$

Put the value into the formula

$\sigma=8.85\times10^{-12}\times14\times10^{3}$

$\sigma=1.239\times10^{-7}\ C/m^2$

$\sigma=123.9\times10^{-9}\ C/m^2$

$\sigma=123.9\ nC/m^2$

The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). We need to calculate the local surface charge density at the point on the surface where the radius of curvature of the surface is smallest

Using formula of charge density

$\sigma=\epsilon_{0}E_{1}$

Put the value into the formula

$\sigma=8.85\times10^{-12}\times74\times10^{3}$

$\sigma=6.549\times10^{-7}\ C/m^2$

$\sigma=654.9\times10^{-9}\ C/m^2$

$\sigma=654.9\ nC/m^2$

The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

Hence, (a). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

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3 years ago

Could you guys tell me whether the photo represents a balanced or unbalanced equation

NARA [144]

Answer:

that is preatty balenced m8

Explanation:

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4 years ago

A satellite is in elliptical orbit with a period of 7.20 104 s about a planet of mass 7.00 1024 kg. At aphelion, at radius 5.1 1

soldi70 [24.7K]

Answer:

1.64x $10^{-4}$ rad/s

Explanation:

Given:

satellite is in elliptical orbit with a period T= 7.20 x $10^{4}$ s

Mass of planet m= 7.00x $10^{24}$ kg

Satellite's angular speed ωa- = 4.987 10-5 rad/s

At aphelion, at radius Ra= 5.1 x 10^7 m

Assuming torque is zero for this system, therefore, the conservation of angular momentum can be defines as

Lp = La

Ip ωp = Ia ωa--->eq(1)

Where,

'a' represents aphelion and 'p' represents perihelion

ω represents angular velocity

and I represents the rotational inertia

Since, I= 1/2 mR²

Here R is the radius at aphelion / perihelion

m = mass of planet

eq(1)=> ωp= Ia ωa/ Ip

ωp= (1/2 m Ra² ωa) / 1/2 m Rp²

ωp= (Ra/Rp)² ωa --->eq(2)

In order to find Rp, we use : 2a= Rp + Ra

where a represents semimajor axis

with the help of Kepler's third law for elliptical orbits

a= ∛(GmT²/4π²)

a= ∛ 6.67x $10^{-11}$ x 7.00x $10^{24}$ x (7.20 x $10^{4}$ )² / 4π²

a= 3.97 x $10^{7}$ m

2a= Rp + Ra

Rp= 2a-Ra

Rp= 2 x 3.97 x $10^{7}$ - 5.1x $10^{7}$

Rp= 2.84 x $10^{7}$ m

Substituting all the required values in eq 2, we have

ωp= (Ra/Rp)² ωa

ωp= (5.1x $10^{7}$ /2.84 x $10^{7}$ )² x 4.987 x $10^{-5$

ωp= 3.224 x 4.987 x $10^{-5$

ωp= 1.64x $10^{-4}$ rad/s

Therefore, angular speed at perihelion is 1.64x $10^{-4}$ rad/s

3 0

3 years ago

A transformer has a primary voltage of 115 V and a secondary voltage of 24 V. If the number of turns in the primary is 345, how

Nata [24]

Answer:

C. 72

Explanation:

Transformer: A transformer is an electromagnetic device that uses the property of mutual inductance to change the voltage of alternating supply.

In a ideal transformer,

Vs/Vp = Ns/Np ............................................. Equation 1

Where Vp = primary voltage, Vs = secondary voltage, Ns = Secondary turn, Np = primary turn.

Making Ns the subject of the equation,

Ns =(Vs/Vp)Np .......................................... Equation 2

Given: Vs = 24 V, Vp = 115 V, Np = 345.

Substitute into equation 2

Ns = (24/115)345

Ns = 72 turns.

Thus the number of turns in the secondary = 72 turns.

The right option is C. 72

4 0

3 years ago