Thank you! An industrial assembly hall Ls continuously lighted by one hundred 40

Question

Thank you!

An industrial assembly hall Ls continuously lighted by one hundred 40-W mercury vapor lamps supplied by a 120-V and 60-Hz source with a power factor of 0.65. Due to the low power factor, a 25 percent penalty is applied at billing. If the average price of 1 kWh Ls $0.01 and the capacitor's average price Ls $50 per millifarad. compute .after how many days of operation the penalty billing covers the price of the power factor correction capacitor. (To avoid penalty, the power factor must be greater than 0.85.)

Explanation / Answer

Active Power = 40*100 =4000W = 4KWH

Power factor = 0.65 => Cos fi =0.65 => sin fi=0.76

Appaarent power = 4/0.65 = 6.153 KVA

Reactive power = 6.153*0.76=4.67KVAr

Now consider pf=0.85 to avoid penalty billing

Now active power remains same but

Apparent power = 4/0.85=4.705 KVA

Reactive Power = sqrt(4.705*4.705 - 4*4) = 2.48 KVAr

Therefore reactive power supplied to avoid penalty = 4.67 - 2.48 = 2.19 KVAr =2190 VAr

Now Capactive reactive power = V^2*w*C where w=2*pi*frequency = 2*3.14*60 = 376.8 rad

Therefore C=2190/(376.8*120*120)=0.403 mF.

For 1mF the cost is $50, the cost for 15mF=0.403*$50 = $20.15

One day billing = 4KW*24 Hrs*$0.01 =$0.96.

Penalty on one day billing = 25% *0.96 = $0.04.

No of days of operation the penalty billing costs capacitor = $20.15/$0.04 = 503.75 days.

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