Michael Ernie F.
Rodriguez
MSEE
EE249 – Homework 2
Part 1 – Electricity Cost
1. It is desired to improve the power factor of a 230 V, 60 Hz, 3 phase system from 0.7 to 0.92
lagging. A wye connected capacitor bank with capacitance of 10 microfarad per phase is to be
used. What will be the kW of the load?
Solution: PL = P N
Solving for QC,
θN
1 1
XC = = =265.258 Ω QN
2 πfC 2 π ( 60 ) ( 10 × 10−6 )
θL
2 QL
E
( )
Q C =3 p =3 ¿
XC QC
Solving for the angle of the power factors,
θ L =(0.7)=45.573 °
θ N =(0.92)=23.074 °
From power triangle,
Q L=P L tan tan θL and Q N =P N tan tan θ N
Since,
Q L=QC +Q N and P L=P N
we have,
QC =Q L −QN
Q C =P L ( tan tan θ L−tan tan θ N )
QC 199.428
P L= = =335.622 W =0.3356 kW
tan tan θL −tan tan θ N tan tan(45.573 °)−tan tan (23.074 °)
2. Using the rate schedule on our lecture/module, for a medium size customer:
Demand meter reading = 120 kW
Billing Demand = 150 kW
Energy Consumed = 36000 kW-h
a. Calculate the monthly power bill under the following conditions above.
b. Referring to the residential rate in lecture, calculate the cost per kWh if only 20 kWh
consumed in a given month.
c. The heating element of an electric stove is rated 1200 W. Using the same rate schedule,
what is the least possible cost in running it for 1 hr?
Solution:
a.
For a medium size customer, the rate schedule are as follows,
Demand Charge: $3.00 per month per kW of billing demand
Energy Charge: 4 cents/kWh for the first 100 hours of billing demand
2 cents/kWh for the next 50,000 kWh per month
1.2 cents/kWh for the remaining energy
Since the billing demand is greater than demand meter reading, use the billing demand.
The demand charge is
150 kW x $3.00/kW = $450.00
Next, the energy charge for the first 100 hours is
150 kW x 100 hrs = 15,000 kWh
15,000 kWh x 4₡/kWh = $600.00
The remaining energy is 36,000 kWh – 15,000 kWh = 21,000 kWh. This is less than 50,000 kWh.
The energy charge for the remaining energy is
21,000 kWh x 2₡/kWh = $420.00
The monthly power bill amounts to
$450.00 + $600.00 + $420.00 = $1,470.00
b.
For residential, the rate schedule are as follows:
Minimum monthly charge: $5.00 plus
First 100 kWh per month at 5 cents/kWh
Next 200 kWh per month at 3 cents/kWh
Excess over 200 kWh per month at 2 cents/kWh
Thus, the cost for 20 kWh consumption is
Minimum = $5.00
20 kWh x 5₡/kWh = $1.00
$5.00 + $1.00 = $6.00
Therefore, the cost per kWh is
$6.00 / 20 kWh = $0.3/kWh
c.
The energy consumption of the stove is
P = 1200 W = 1.2 kW
E = Pt = 1.2 kW x 1 hr = 1.2 kWh
Thus, the cost for the consumption of the stove is
Minimum = $5.00
1.2 kWh x 5₡/kWh = $0.06
$5.00 + $0.06 = $5.06
Part 2 – Metering (2WM and 3WM)
3. A balanced wye connected load with impedance of 10+j30 ohms per phase is supplied by a 230
V, 3 phase source. Two wattmeters are connected to the circuit to measure total power of the
load. Calculate the reading of each wattmeter in watts.
Solution:
Solving for the power factor angle of the balanced load,
θ= ( 3010 )=71.565°
The line current is
230∠ 0°
I=
√❑
The wattmeter readings are
W 1= EL I L cos cos ( 30−θ )=( 230 )( 4.199 ) cos cos(30−71.565)=722.593W
W 2 =EL I L cos cos ( 30+θ )=( 230 ) ( 4.199 ) cos cos (30+71.565)=−193.617 W
4. A 3 phase, 3 wire feeder has the following line currents and voltages:
Vab = 220 cis 0° Ia = 30 – j30
Vbc = 220 cis 240° Ic = 29 + j39.8
Determine the power factor, real and reactive powers supplied by the feeder. Let line b as your
reference.
Solution:
V ab=220 ∠ 0 ° V ; V bc=220 ∠240 ° V
I a=30− j 30=42.426 ∠−45 ° A
I c =29+ j 39.8=49.245 ∠53.921° A
P1=V ab I a cos cos θ1=( 220 ) ( 42.426 ) cos cos(0−(−45 ))=6,599.937 W
P2=V bc I c cos cos θ2 =( 220 )( 49.425 ) cos cos(240−53.921)=−10,772.979 W
PT =P1+ P 2=6,599.937+10,772.979=17,371.916 W
Q 1=V ab I a sin sin θ1= ( 220 )( 42.426 ) sin sin(0−(−45 ))=6,599.937 VAr
Q 2=V bc I c sin sinθ 2=( 220 ) ( 49.425 ) sin sin(240−53.921)=−1,151.5VAr
Q T =Q 1+ Q2=6,599.937+1,151.5=7,751.437VAr
7,751.437
pf =cos cos (( 17,371.916))
=0.9132 lagging
