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The chart below shows an hourly load profile over a $24$ h period. The Y $-$Axis is average power in kW and to find the energy of that hour $($energy $=$ power $x$ time$),$ it'd be the same value since you're multiplying by $1$ hour. For example, hour #$3,$ average power is $1$ kW and energy of hour #$3$ is $1kW1h=1kWh.$ Please notice the difference in units $($ kW vs $.$ kWh $).$ Remember that kW$/$h is not a unit. It's either KW or kWh with the different multiples. BC Hydro typically calculates monthly peak demand based on the average power of the maximum three consecutive $5-$min intervals. Other utilities may calculate peak demand slightly differently. For this exercise, assume that the monthly peak demand is based on the maximum power over a $1-$hour interval.

What is the peak demand $(kW)$ in that day and the energy consumption in kWh $?$ Hint: to calculate daily energy consumption, calculate it for each hour as I showed you above and add them all up$.$