An industrial customer with a three-phase, 480V service entrance has the followi
ID: 2079008 • Letter: A
Question
An industrial customer with a three-phase, 480V service entrance has the following set of loads: Two 40 HP, 90% efficient lathes, PF = 0.80 lagging One 55 HP, 85% efficient mill, PF = 0.78 lagging One 7 ton air conditioner' with a COP of 1.4 and a 0.85 lagging power factor One 25 kW high intensity discharge (HID) lighting system, unity PF If the lighting system is replaced with a T8 fluorescent system with magnetic ballast that consumes 15.2% of the previous system, but introduces a 0.86 leading power factor, by how much docs the service entrance current change? Consider the case when all systems are fully loaded. Consider the AC load under the new lighting regime. Use the NFPA 70 to determine the minimum allowed gauge of the service conductors. Feeder lines arc copper, with a 60 degree C temperature raring, contained within a raceway with an ambient temperature of 40 degree C. Start with article 310.15; state the specific article(s)/tables used to determine your answer.Explanation / Answer
Total load of given industry is
1) motor load Pm1 = 40*0.7457*0.9 = 26.8485 kW , pf = 0.8 lagging , phi = 36.86 , Qm1 = 20.129 kVar
2) motor load Pm2 = 55*0.7457*0.85 = 34.861475 kW , pf = 0.78 lagging , phi = 38.739, Qm2 = 27.968 kVar
3) air conditioner load Pac = 12*ton/(COP*3.412) = 12*7/(1.4*3.412) = 17.585 kW , pf = 0.85 lagging , phi = 31.788, Qac = 10.898 kVar.
4) lighting load PL = 25 kW, pf = 1
Total load is PT = 26.8485 + 34.861475 + 17.585 + 25 = 104.294 kW
QT = 20.129 + 27.968 + 10.898 + 0 = 59.086 kVar
Complex power ST = PT + j QT = 104.294 + j 59.086 KVA , pf = 0.87 lagging , Magnitude of complex power
S = 119.868 kVA
Magnitude of Current drawn from mains is
I = S/1.732*V = 119.868*1000/(1.732*480) = 144.183 at an angle of -29.540
Now lighting load was replaced with fluorescent light load which consumes Pf = 0.152*25 = 3.8 kW , pf = 0.86 lead
phi = -30.680 and Qf = -2.254 kVar
New complex power Sn = Pn + jQn = 83.094 + j 56.832 KVA , pf = 0.82 , phi = 34.91 lag
Magnitude of complex power Sn = 100.67 KVA ,
New magnitude of current drawn from source is
In = Sn/1.732*V = 100.67*1000/(1.732*480) = 121. 090 A at angle of -34.910
Current change by 144.183 - 121.090 = 23. 093 A (decreased compare to first condition)
from article 310.15(B)(16) the size of copper conductor is
for case 1 : -
Copper conductor size is 3/0 AWG and current rating is 165 A
for case 2: -
Copper conductor size is 2/0 AWG and current rating is 145 A