Ans: Calculate Annual Savings with 93% Efficient Motors for AHU Fans, Pumps & CT

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Discussion Overview

The discussion revolves around calculating the annual savings from replacing existing motors in an air-conditioning installation with more efficient motors. Participants explore the energy consumption of various motors, their efficiencies, and the resulting financial implications of the switch. The focus is on technical calculations related to energy usage and cost savings.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • An energy audit was conducted to identify savings by replacing existing motors with more efficient ones.
  • Participants calculate the power output of existing motors in kW based on horsepower ratings.
  • There is a discussion on how to calculate energy consumption, with some participants suggesting to divide power output by efficiency to find power input.
  • One participant calculates daily energy usage for each motor but initially uses an incorrect multiplier.
  • Subsequent calculations attempt to correct earlier mistakes and include the number of motors in the total energy consumption calculations.
  • Annual energy usage is computed for both existing and new motors, leading to a proposed annual savings figure.
  • One participant expresses uncertainty about the accuracy of the calculations but acknowledges the methodology appears correct.

Areas of Agreement / Disagreement

Participants generally agree on the methodology for calculating energy usage but there are discrepancies in the calculations presented, particularly regarding the initial energy usage figures. The discussion does not reach a consensus on the final calculations, as some participants question earlier steps.

Contextual Notes

Some calculations depend on the correct application of efficiency ratings and the number of motors, which may not have been consistently applied. There are unresolved issues regarding the accuracy of individual energy consumption figures.

Who May Find This Useful

This discussion may be useful for professionals involved in energy auditing, HVAC system design, or anyone interested in the financial implications of energy efficiency improvements in mechanical systems.

Ang1987
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(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.
 
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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.
 
10hp x 0.7457=7.457kw
60hp x 0.7457=44.74kw
75hp x 0.7457=55.93kw
70hpx0.7457=52.2kw
 
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.
 
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
 
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?
 
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
 
Last edited:
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?
 
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
 
  • #10
Well, I didn't check all the numbers, but the methodology looks right.
 
  • #11
Thanks . Appreciate your help
 

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