You have been hired for the summer by Duke Energy as a technical analyst intern, with the task of
finding a way to reduce costs. Your performance in this summer job will in influence whether Duke
offers you a permanent job when you graduate. You are asked in the first week to analyze what
electrical transmission savings might be realized while supplying one of Dukes largest industrial
customers in NC. The particular plant in question when in full operation uses 2.3 MW of power.
Currently, because of numerous motors, generators, and electromagnets, it represents an overall load
on Dukes transmission line which is inductive with a phase angle of +25 degrees. The power is supplied
to the plant from Dukes substation 4.5 km away from the plant. The 60 Hz line voltage supplied into the
plant is 40 kV, and the resistance of the transmission line from the substation to the plant is 5.2 Ω. The
cost per kW-h is $0.08, but the plant pays only for the actual energy it uses. Duke pays to transmit the
electricity from its substation to the plant, and wants to reduce this cost.
So to understand Dukes cost of local power delivery, first you calculate
a) the current in the power lines and what the rms voltage must be at the substation to maintain
40 kV at the plant; then you find
b) the resistance and inductive reactance of the plants total load and
c) how much power is lost in transmission.
d) Suppose the phase angle of the plants impedance were reduced to 18 degrees by adding a
bank of capacitors in series with the load. How much money would be saved by Duke Energy during one
month of operation, assuming the plant operates 16 h/day?
e) What must be the capacitance of this bank of capacitors?
SOLUTION:
a) To begin with we need to calculate the current in the power line and the Vrms at the substation
required to maintain 40kV.
𝑃𝑎𝑣𝑔 = 𝜀𝑟𝑚𝑠 𝐼𝑟𝑚𝑠 cos 𝜑
We know we have to use the full equation for average power here because the current and
voltage are not in phase. We can now solve for the current with the given parameters from the
problem
𝑃𝑎𝑣𝑔
2.3𝑒6𝑊
𝐼𝑟𝑚𝑠 =
=
= 63.4𝐴
𝜀𝑟𝑚𝑠 cos 𝜑 4𝑒4𝑉 cos 25°
The voltage required for the substation is the sum of the voltage required at the plant and
voltage over the power lines. However, we have to be careful and add these voltages as
phasors. Since the power plant is inductive and the power lines are all resistive, they will not be
in phase with each other. We know that the plant has a phase angle of 25 degrees. So, we have
to add these voltages as vectors. 𝑉𝑠𝑠,"𝑅" = 𝑉𝑙𝑖𝑛𝑒 + 𝑉𝑝𝑙𝑎𝑛𝑡 𝑐𝑜𝑠25 = (63.4𝐴)(5.2Ω) +
40000𝑉𝑐𝑜𝑠25 = 36582𝑉 𝑎𝑛𝑑 𝑉𝑆𝑆,"𝐿" = 𝑉𝑝𝑙𝑎𝑛𝑡 𝑠𝑖𝑛25 = 16904𝑉
When we find the final voltage we have
𝑉𝑠𝑢𝑏𝑠𝑡𝑎𝑡𝑖𝑜𝑛 = √𝑉𝑆𝑆,"𝑅" 2 + 𝑉𝑆𝑆,"𝐿" 2 = √(36582𝑉)2 + (16904𝑉)2 = 40299𝑉 at an angle
16904𝑉
𝑡𝑎𝑛𝜙 ′ = 36582𝑉 𝑤ℎ𝑖𝑐ℎ 𝑚𝑒𝑎𝑛𝑠 𝜙 ′ = 24.8 𝑑𝑒𝑔𝑟𝑒𝑒𝑠
b) The resistance of the plant’s total load can be found from the current we determined from
𝑉𝑟𝑚𝑠
) 𝑐𝑜𝑠𝜙
𝑅𝑝𝑙𝑎𝑛𝑡
power: 𝐼𝑟𝑚𝑠,𝑝𝑙𝑎𝑛𝑡 = (
40000𝑉
) 𝑐𝑜𝑠25
63.4Ω
(
𝑉𝑟𝑚𝑠
) 𝑐𝑜𝑠𝜙
𝐼𝑟𝑚𝑠
which means that 𝑅𝑝𝑙𝑎𝑛𝑡 = (
=
= 571.8Ω
To find the inductive reactance of the plants total load, we begin with the equation for phase
angle and solve for XL
𝜒𝐿 − 𝜒𝐶
tan 𝜑 =
=> 𝜒𝐿 = 𝑅 tan 𝜑 + 𝜒𝐶 = (571.8Ω) tan 25° + 0 = 266.6Ω
𝑅
c) The power lost in transmission is (in the line, the current and voltage are in phase, it is just
resistive, so there is no phase angle).
𝑃𝑙𝑖𝑛𝑒 = 𝐼 2 𝑅𝑙𝑖𝑛𝑒 = (63.4𝐴)2 (5.2Ω) = 20,902𝑊
This is approximately 1% power loss.
d) If the phase angle is reduced to 18⁰, then the current through the line and the resulting power
are (following the same logic/steps as above)
𝑃𝑎𝑣𝑔
2.3𝑒6𝑊
𝐼𝑟𝑚𝑠 =
=
= 60.46𝐴 => 𝑃𝑙𝑖𝑛𝑒 = 𝐼 2 𝑅 = 19,007𝑊
𝜀𝑟𝑚𝑠 cos 𝜑 4𝑒4𝑉 cos 18°
So, the difference in work required to transmit the electricity with this change in phase angle is:
∆𝑊 = 20,902𝑊 − 19,007𝑊 = 1,895𝑊
This difference in work means that Duke will save approximately
16ℎ𝑟𝑠 30 𝑑𝑎𝑦𝑠 $0.08
1.895𝑘𝑊
= $72.77/𝑚𝑜𝑛𝑡ℎ
1𝑑𝑎𝑦 1 𝑚𝑜𝑛𝑡ℎ 𝑘𝑊ℎ𝑟𝑠
e) The capacitance required to cause the phase angle to change from 25⁰ to 18⁰ can be found by
solving the equation for ϕ for XC (which is the inverse of ωC)
𝜒𝐿 − 𝜒𝐶
1
tan 𝜑 =
=> 𝜒𝐶 = 𝜒𝐿 − 𝑅 tan 𝜑 =
𝑅
𝜔𝐶
1
1
𝐶=
=
= 32.7𝜇𝐹
𝜔(𝜒𝐿 − 𝑅 tan 𝜑) 60𝐻𝑧(2𝜋)(267Ω − 572Ω tan 18°)