Annual energy savings from replacing motors with 93% efficiency models

[Ang1987]

(b) An energy audit was conducted on a building to demonstrate energy savings opportunities by replacing the air handling unit or AHU fans motors, condensing and chilled water pump motors of the water cooled chilled water air-conditioning installation with newer, more efficient motors of the same horse-power ratings.

In the building, there are five units of AHU fans, two sets of condensing water pumps, two sets of chilled water pumps and two sets of Cooling Towers (CT). All four pumps operate continuously for twenty four hours per day.

The existing AHU fans are rated at 10hp with efficiency of 85%, existing condensing water pumps are rated at 60 hp with efficiency of 83%, existing chilled water pumps are rated at 75 hp with efficiency of 83% and existing CT fans are rated at 70 hp with efficiency of 80%. Assume 1 hp = 0.7457 kW.
If the new motors have an efficiency of 93%, determine the annual savings in the electricity consumed given the electricity tariff is S$0.26/kWh.

 

[phyzguy]

You need to make an attempt to solve the problem before we can help. Try calculating the amount of energy used per day with the current motors, and again with the more efficient motors.

 

[Ang1987]

10hp x 0.7457=7.457kw
60hp x 0.7457=44.74kw
75hp x 0.7457=55.93kw
70hpx0.7457=52.2kw

 

[phyzguy]

That's the power output for each motor in kW. The power input (i.e. the power used) is the power output divided by the efficiency. And the energy used is the power used times the number of hours the motor is on. So what is the total energy used by all 4 motors in one day? Your answer should be in kWh.

 

[Ang1987]

10hp x 0.7457=7.457kw *0.24*24hr= 42.95kwh
60hp x 0.7457=44.74kw *0.24*24hr=257.7kwh
75hp x 0.7457=55.93kw *0.24*24hr=322.15kwh
70hp x 0.7457=52.2kw *0.24*24hr=300.67kwh

 

[phyzguy]

Why did you multiply them all by 0.24? Each one has a different efficiency. Why didn't you divide by the efficiency like I told you?

 

[Ang1987]

10hp x 0.7457=7.457kw /0.85*24hr= 210.55
60hp x 0.7457=44.74kw /0.83*24hr=1293.69
75hp x 0.7457=55.93kw /0.83*24hr=1617.25
70hp x 0.7457=52.2kw /0.80*24hr=1566

 

[phyzguy]

OK, good, you're making progress. You have calculated the energy used by each motor in one day, but I think you made a mistake in line 2. Also, there is more than one motor of each type. So try to answer the following questions:

(1) What is the total energy used in one day?
(2) What is the total energy used in one year?
(3) What is the total energy used in one year with the new, higher efficiency motors?
(4) How much money will the higher efficiency motors save in one year?

 

[Ang1987]

One day
AHU fan-10hp x 0.7457=7.457kw /0.85*24hr= 210.55
Condensing pump-60hp x 0.7457=44.74kw /0.83*24hr=1293.69
Chilled pump-75hp x 0.7457=55.93kw /0.83*24hr=1617.25
Cooling tower-70hp x 0.7457=52.2kw /0.80*24hr=1566

One Year
10hp x 0.7457=7.457kw /0.85*24hr= 210.55*365=75798 *5=378990
60hp x 0.7457=44.74kw /0.83*24hr=1293.69*365=472196*2=944392
75hp x 0.7457=55.93kw /0.83*24hr=1617.25*365=590296*2=1180592
70hp x 0.7457=52.2kw /0.80*24hr=1566*365=571590*2=11471180
Total: 3645154x$0.24/kwh= $947,740

One Year ( new motor)
10hp x 0.7457=7.457kw /0.93*24hr= 192.43*365=69278*5=346390
60hp x 0.7457=44.74kw /0.93*24hr=1154.6*365=415649*2=831298
75hp x 0.7457=55.93kw /0.93*24hr=1443.35*365=526824*2=1053648
70hp x 0.7457=52.2kw /0.93*24hr=1347.1*365= 491690*2=983380
Total: 3214716x$0.24/kwh= $771,531

With the new motors, the annual savings is $176,209

 

[phyzguy]

Well, I didn't check all the numbers, but the methodology looks right.

 

[Ang1987]

Thanks . Appreciate your help