What's the right way to calculate power consumption over a month given a rate per minute?

Question

When I was searching for power required to lift an object, i found that, for example:

$100\ \mathrm{kg}$ to be lifted 3 metres in 5 seconds. (vertical)

Answer:

$$\begin{align} \text{mass}\times\text{gravity}\times\frac{\text{distance}}{\text{time}} &= 100 \times 9.8 \times \frac{3}{5} \\ &= 588\ \text{watts} \end{align}$$ Assuming an efficiency loss of 22% (588/78%), this gives 750 watts required to lift a object weighing $100\ \mathrm{kg}$ 3 metres high in 5 seconds.

I don't think there is any problem in the above calculation. But then I want to calculate the monthly electricity consumption (kWh) of a 750 watt motor if run for the above said purpose 5 times per minute, i.e run for 25 seconds each minute at full capacity of 750 watt.

I have two ideas:

1. Is it $$\begin{align} \text{Total cycles in a month} &= \frac{\text{5 cycles}}{\mathrm{min}}\times\frac{60\ \mathrm{min}}{\mathrm{hr}}\times\frac{24\ \mathrm{hr}}{\text{day}}\times 30\ \text{day} \\ &= 216000\ \text{cycles} \end{align}$$ With power consumption per cycle of 750 watts, the total power consumption is $$750\ \mathrm{W}\times 216000\ \text{cycles} = 162000000\ \mathrm{W} = 162000\ \mathrm{kWh}$$ I think this answer is absurd

2. Or $$\begin{align} \text{Total hours run in a day} &= \frac{25\ \text{seconds}}{60\ \text{seconds}}\times\frac{24\ \mathrm{hr}}{\text{day}} \\ &= 10\ \mathrm{hr} \\ \text{Total hours in a month} &= 10 \times 30 = 300\ \text{hours in a month} \end{align}$$ Since $750\ \mathrm{W}$ is 75% of $1\ \mathrm{kW} = 300 \times 75\% = 225\ \mathrm{kWh}$ or 225 units (I think this makes more sense)

Answer 1

In your first option, the $750\ \mathrm{W}$ is already a rate. A watt quantifies energy per unit time. It doesn't make sense to multiply that by the number of cycles; you need to multiply it by a time to get an amount of energy. (As in $\mathrm{kWh}$, kilowatt-hour, which is a kilowatt times an hour.)

By analogy, it would be like saying your car goes 60 miles per hour, and you drive it for 8 hours in a day, then at the end of the day your car has traveled 480 mph. Do you see why that doesn't make sense? That's nearly the same mistake you made in your first approach.

There are a couple ways you can calculate the amount of energy needed: either find the total time taken by all those operational cycles and multiply that by the power of $750\ \mathrm{W}$, or as CuriousOne mentioned in a comment, find the amount of energy taken per cycle and multiply that by the number of cycles. I highly encourage you to track the units through your calculation; they can be multiplied and divided just like variables. If you had been paying attention to the units, you would have immediately noticed something was wrong when you came out with a number in watts ($162000000\ \mathrm{W}$, or if you were being more careful it would have been $1.62\times 10^8\ \mathrm{W}\,\text{cycles}$) despite needing a result in kilowatt-hours. You can't arbitrarily change units of power like the watt to units of energy like the kilowatt-hour.

Answer 2

1) As per my question Total energy consumed (Kwh) will be the same IRRESPECTIVE of Power (Kw)

2) What matters as per my question is the "work done" (Energy consumed in joules)

3) Power is derived from the rate of energy consumed.. ( Here, in how much seconds i want a cycle to be completed)

So total energy consumed = [(100 kg * 9.8 * 3 metres) / 78% efficiency] * 216000 cycles = 810000000 Joules= 810000000 watt seconds

Since i want the power consumed in kwh = 810000000 watt seconds /3600 seconds = 225000 Watts Hour= 225 Kwh = 225 Units of ENERGY CONSUMED

Hence to derive the power = Work done per cycle / time required per cycle = 3750 joules / 5 sec = 750 Watts of POWER Required

If i want it to be completed in 4 seconds = Power required will be 937.5 Watts (3750 Joules/ 4 seconds)

But the total ELECTRICITY CONSUMPTION to complete the said work will be 225 units or KWH, irrespective of whether the cycle is 4 or 5 seconds each...