# Homework Help: Calculating energy savings using a variable speed drive

1. Apr 25, 2017

### nothing909

1. The problem statement, all variables and given/known data
I have a task right now that requires me to calculate the energy savings that would be made by using a variable speed drive.

Question:

The factory process water distribution system consists of a 150mm cast iron ring main at a pressure of 5 bar (50m). The local water is pumped into the factory water ring main from a storage tank.

Water ring main pressure and flow rates produced by this pump are mechanically controlled by a pressure relief valve. This method of control causes the motor to run near the top end of the power range on a continuous basis while the plant is in operation.

The pressure in the water ring main is presently controlled by the pressure relief valve which diverts excess water back into the storage tank. This system causes the water pump to be fully loaded at all times.

The factory works a 46 week year and runs continuously seven days per week as shown on the process bar chart.

Assume the cost per killowatt hour to be £0.08 when calculating the annual savings in energy costs

3. The attempt at a solution

There's 4 different flow rates:

50m3/h
80m3/h
100m3/h
170m3/h

From the pump chacateristics graph, you can see that the power required for each flow rate is (going along from 50m head)

50m3/h = 13kW
80m3/h = 16kW
100m3/h = 18.5kW
170m3/h = 33kW

The motors efficiency is 90% so the power into motor is

50m3/h = 13kW / 0.9 = 14.44kW
80m3/h = 16kW / 0.9 = 17.78kW
100m3/h = 18.5kW / 0.9 = 20.6kW
170m3/h = 33kW / 0.9 = 36.7kW

I'm stuck from here. I need to calculate what the price of energy per year is without the variable speed drive, and then calculate what the price is with the variable speed drive, and then subtract them to find the savings achieved per year.

On the pump characteristics graph, what is the thick black line referring to?

Is the powers I calculated with or without the variable speed drive?

Do I just add up all the powers and then calculate how much it costs to run it for 46 weeks? I don't understand.

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Last edited: Apr 25, 2017
2. Apr 26, 2017

### CWatters

That's pump upper limit curve. eg the pump can only produce 65 bar from 0-80m^3/h. At higher flow rates the pressure falls off so that at about 190m^3/h the pressure is down to 50bar.

Without the variable speed drive the power can be calculated based on the statements in the problem....

That appears to be about 37kW.

With the variable drive.

You will be aware that Electricity is sold in units of kWH. At each flow rate you have calculated the power required in kW so the missing factor here is time. Use the bar graph to give you the time spent at each flow rate during the day in hours.

So for the first 6 hours the flow rate is 80m^3 and the power 17.78kW so the energy used in those 6 hours is 6 * 17.78 = 106.78 kWH

3. Apr 27, 2017

### nothing909

Can you just confirm for me if this is correct? :

To calculate the power used per year without a variable speed drive, I just do 41kW(37kW/0.9 because the motor is 90% efficient) x 24(per day) x 7(per week) x 46 (runs continuously for 46 weeks a year)

and to calculate with a variable speed drive, I just look at the bar graph and since there are different power requirements throughout the day, I just go through it adding up the power required at each part of the day? After I have it for a day, I just x7 x 46, to find it for a year?

Then to calculate the power savings, I just subtract the power without the variable speed from the the power with the variable speed drive and multiply it by £0.08?

4. Apr 27, 2017

### CWatters

That all looks correct to me.

5. Apr 27, 2017

### CWatters

Just to be sure.. You are adding up energy not power....