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

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.

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:

a)

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

b)

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

c)

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

d)

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

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow

an energy transformation flow diagram is shown

where is it ?