1. **State the problem:** Calculate the total energy consumed per day in kWh given a piecewise linear power load profile $P(t)$ in watts over 24 hours. 2. **Formula:** Energy consumed over a time interval is the integral of power over time. Since power is in watts and time in hours, energy in watt-hours (Wh) is $$E = \int P(t) \, dt$$ To convert Wh to kWh, divide by 1000. 3. **Approach:** The load profile is piecewise linear with given points. We calculate the area under each segment (trapezoids) and sum them. 4. **Segments and calculations:** - From 0:00 to 2:00: power rises from 0 W to 400 W $$E_1 = \frac{(0 + 400)}{2} \times 2 = 400 \text{ Wh}$$ - From 2:00 to 4:00: power rises from 400 W to 600 W $$E_2 = \frac{(400 + 600)}{2} \times 2 = 1000 \text{ Wh}$$ - From 4:00 to 6:00: power constant at 600 W $$E_3 = 600 \times 2 = 1200 \text{ Wh}$$ - From 6:00 to 8:00: power drops from 600 W to 400 W $$E_4 = \frac{(600 + 400)}{2} \times 2 = 1000 \text{ Wh}$$ - From 8:00 to 9:00: power rises sharply from 400 W to 1200 W $$E_5 = \frac{(400 + 1200)}{2} \times 1 = 800 \text{ Wh}$$ - From 9:00 to 10:00: power drops sharply from 1200 W to 200 W $$E_6 = \frac{(1200 + 200)}{2} \times 1 = 700 \text{ Wh}$$ - From 10:00 to 15:00: power rises linearly from 200 W to 1100 W over 5 hours $$E_7 = \frac{(200 + 1100)}{2} \times 5 = 3250 \text{ Wh}$$ - From 15:00 to 16:00: power drops sharply from 1100 W to 600 W $$E_8 = \frac{(1100 + 600)}{2} \times 1 = 850 \text{ Wh}$$ - From 16:00 to 20:00: power drops linearly from 600 W to 200 W over 4 hours $$E_9 = \frac{(600 + 200)}{2} \times 4 = 1600 \text{ Wh}$$ - From 20:00 to 24:00: power constant at 200 W for 4 hours $$E_{10} = 200 \times 4 = 800 \text{ Wh}$$ 5. **Sum all energy segments:** $$E_{total} = 400 + 1000 + 1200 + 1000 + 800 + 700 + 3250 + 850 + 1600 + 800 = 11800 \text{ Wh}$$ 6. **Convert to kWh:** $$E_{total} = \frac{11800}{1000} = 11.8 \text{ kWh}$$ **Final answer:** The energy consumed per day according to the load profile is **11.8 kWh/day**.