If the initial interconnection time error is zero seconds and experiences four hours of 60.02 Hz followed by three hours of 59.98 Hz, what is the total accumulated time error?

If the initial interconnection time error is zero seconds and experiences four hours of 60.02 Hz followed by three hours of 59.98 Hz, what is the total accumulated time error?

• A. 1.2 seconds
• B. 4.8 seconds
• C. 1.2 seconds per hour
• D. 4.8 seconds per hour

Answer: (D)

To determine the total accumulated time error given the frequency deviations, first, it's essential to recognize the relationship between frequency and time. In a power system, the standard frequency is typically 60 Hz, which corresponds to a full cycle occurring every 1/60 of a second or 1.6667 milliseconds. The scenario describes an initial frequency of 60.02 Hz for four hours, followed by 59.98 Hz for three hours. 1. **Calculation for Frequency of 60.02 Hz**: - This frequency is above the standard 60 Hz, which means that for every cycle, the system completes more cycles than it should in the same amount of time. To find the time error over the four-hour period at 60.02 Hz: - The cycle length at 60.02 Hz is approximately 0.01665 seconds. - One hour at this frequency would yield 60.02 cycles, where every cycle is equal to 1 second / 60.02 cycles. - Over four hours, the difference in time is 4 hours of operation at this frequency leads to an extra = (60.02 - 60) cycles x 4 hours = 0.02 cycles

To determine the total accumulated time error given the frequency deviations, first, it's essential to recognize the relationship between frequency and time. In a power system, the standard frequency is typically 60 Hz, which corresponds to a full cycle occurring every 1/60 of a second or 1.6667 milliseconds.

The scenario describes an initial frequency of 60.02 Hz for four hours, followed by 59.98 Hz for three hours.

1. Calculation for Frequency of 60.02 Hz:

• This frequency is above the standard 60 Hz, which means that for every cycle, the system completes more cycles than it should in the same amount of time. To find the time error over the four-hour period at 60.02 Hz:

• The cycle length at 60.02 Hz is approximately 0.01665 seconds.

• One hour at this frequency would yield 60.02 cycles, where every cycle is equal to 1 second / 60.02 cycles.

• Over four hours, the difference in time is 4 hours of operation at this frequency leads to an extra = (60.02 - 60) cycles x 4 hours = 0.02 cycles