Question

2.14 Repeat Example 2.12, assuming that there are eight houses connected on each DT and that there are a total of 120 DTs and 960 residences supplied by the primary feeder Example 2.15 Assume that a 5 kW air conditioner would run 80 % of the time (80% duty cycle) and, during the peak hour, might be limited by utility remote control to a duty cycle of 65%. Determine the following a. The number of minutes of operation denied at the end of 1 h of control of the unit b. The amount of reduced energy consumption during the peak hour if such control is applied simultaneously to 100,000 air conditioners throughout the system c. The total amount of reduced energy consumption during the peak d. The total amount of additional reduction in energy consumption in part (c) if T&D losses of the T&D system at peak is 8% Solution a. The number of minutes of operation denied is (0.80 0.65) x (60 min/h) 9 min b. The amount of reduced energy consumption during the peak is (0.80 0.65) x (5 kW 0.75 kw c. The total amount of energy reduction for 100,000 units is ASt) S xD -D xN (5 kW)x (0.80-0.65] x 100,000 =75 MW www.Technical Books Pdf.com Electric Power Distribution Engineering 72 d. The total additional amount of energy reduction due to the reduction in the T&D losses is (75 MW) x 0.08 6 MW Thus, the overall total reduction is 75 MW6 MW 81 MW The earlier example shows attractiveness of controlling air conditioners to utility company. 2.14 Repeat Example 2.12, assuming that there are eight houses connected on each DT and that there are a total of 120 DTs and 960 residences supplied by the primary feeder

Answer 1

a. During the peak hour, duty cycle is limited to 65% = 65/100 = 0.65 instead of 80% = 0.80.

Therefore, the % of time during which operation is denied during peak hour = (80-65)% = 15% = 0.15

Now, at the end of 1 hour, number of minutes of operation denied = 0.15x60 minutes = 9 minutes

b. Amount of reduced energy consumption per air conditioner during peak hour

= Reduction in duty cycle x power consumption of air conditioner

= 0.15 x 5 kW

= 0.75 kW

c. Amount of reduced energy consumption throughout the system during peak hour

= Reduction in consumption per air conditioner unit x total no. of air conditioners in the system

= 0.75 kW x 100000

= 750 x 10⁵ W

= 75 MW

d. If transmission and distribution (T & D) losses during peak hour is 8% = 0.08, then to deliver 75 MW power the total T & D loss = 75x0.08 MW = 6 MW

To deliver 75 MW less power, total energy saving = 75 MW + 6 MW = 81 MW