Hello Everyone,
Ours is a Residential Project with PTAC units and DOAS in the Proposed
design. The DOAS nameplate motor HP is 7.5 where as the fan absorbed power
is 4.73 BHP. If we model 7.5 HP; we are falling short of the minimum LEED
energy performance requirements. Modelling
4.73 BHP just gives us enough savings.
Which one is the correct fan power to model?
I have been through the archives and didn't find appropriate answer. Please
help!
Regards,
Mayank.
If the fan is using 4.73BHP, that's the value to consider for energy cost.
It is very common for a motor to have greater capacity than is actually
used.
Terminology can be a beast with fan/pump motor power calcs.
?Fan absorbed power? isn?t a phrase I?ve encountered frequently but suggests to me work done by the fan on the air (or vise versa), presumably at design conditions for your case. I would intuitively use this in combination with your knowledge of the nameplate HP rating to determine appropriate fan power inputs. Leverage the nameplate HP rating of 7.5 to determine a reasonable motor efficiency first.
So ? quick aside for Recommended further reading:
? The ?I don?t have time!? suggestion: Chapter 10 tables from 90.1
? ?Hit me with the good stuff!? https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0ahUKEwitnt6Y_ZLVAhUCeSYKHcRdCZ0QFggiMAA&url=https%3A%2F%2Fenergy.gov%2Fsites%2Fprod%2Ffiles%2F2014%2F04%2Ff15%2Famo_motors_handbook_web.pdf&usg=AFQjCNE2OJ-TJR5_EFu34EoAiPO1PMP9BA
? Highlights include:
o Table 3-3 detailing solid documentation for full load efficiencies for older large motors (excludes winding degradation losses for motors that haven?t been serviced).
o Table 2-1 should look familiar to those perusing 90.1 over the past decade.
o I recall on skimming today that Chapter 3 was super helpful for me learning to both digest and intelligently request direct measurements of operating motor performance in the field.
So back to the problem. Descriptively: [Power In] = [Work accomplished by the fan: 4.73 bhp] / [Motor efficiency]. For a new motor that should lead you to something in-between 4.73 and 7.5 hp.
Since you CC?ed eQuest-users specifically: For developing eQuest inputs, additionally consider your flow rate associated with that design operating point to determine a kW/CFM input: [kW/CFM] = 4.73* 0.7457 / motor-eff / CFM. Best paired with a corresponding system design airflow rate input that isn?t getting overridden due to site elevation (air density) or zonal flow demands.
A related nugget perhaps helpful to share, as this confused me for some years: motor efficiencies in broad strokes do vary, but not much between 50-100% of nominal horsepower load with all else being equal (distinct from motor speed!). It?s pretty normal to see motors ?oversized/underloaded? motors with high nameplate horsepower relative to their actual full-load operating power requirement, and that isn?t necessarily a bad thing. The above document includes some examples and further explanatory text illustrating where ?right-sizing? a fan motor can result in identical or worse power input, where the points of efficiency gained for operating closer to full capacity are countered by the points of efficiency lost inherent to smaller motor sizes.
~Nick
[cid:image001.png at 01D2FFA2.BC049F10]
Nick Caton, P.E., BEMP
Senior Energy Engineer
Regional Energy Engineering Manager
Energy and Sustainability Services
Schneider Electric
D 913.564.6361
M 785.410.3317
F 913.564.6380
E nicholas.caton at schneider-electric.com
15200 Santa Fe Trail Drive
Suite 204
Lenexa, KS 66219
United States
[cid:image002.png at 01D2FFA2.BC049F10]
Hello Mayank,
The correct fan power to model is the input power to the motor. This can be determined by dividing the brake horsepower by the motor efficiency (i.e. 4.73 BHP/90% = 5.26 hp) and should be slightly larger than the bhp and smaller than the motor hp.
All the best,
Daniel
?
Daniel Knapp PhD, P Phys, LEED AP
danielk at arborus.ca
Energy Efficiency Expert
Arborus Consulting
2-43 Eccles Street
Ottawa ON K1R 6S3
(613) 234-7178
Terminology can be a beast with fan/pump motor power calcs.
?Fan absorbed power? isn?t a phrase I?ve encountered frequently but suggests to me work done by the fan on the air (or vise versa), presumably at design conditions for your case. I would intuitively use this in combination with your knowledge of the nameplate HP rating to determine appropriate fan power inputs. Leverage the nameplate HP rating of 7.5 to determine a reasonable motor efficiency first.
So ? quick aside for Recommended further reading:
? The ?I don?t have time!? suggestion: Chapter 10 tables from 90.1
? ?Hit me with the good stuff!? https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0ahUKEwitnt6Y_ZLVAhUCeSYKHcRdCZ0QFggiMAA&url=https%3A%2F%2Fenergy.gov%2Fsites%2Fprod%2Ffiles%2F2014%2F04%2Ff15%2Famo_motors_handbook_web.pdf&usg=AFQjCNE2OJ-TJR5_EFu34EoAiPO1PMP9BA
? Highlights include:
o Table 3-3 detailing solid documentation for full load efficiencies for older large motors (excludes winding degradation losses for motors that haven?t been serviced).
o Table 2-1 should look familiar to those perusing 90.1 over the past decade.
o I recall on skimming today that Chapter 3 was super helpful for me learning to both digest and intelligently request direct measurements of operating motor performance in the field.
So back to the problem. Descriptively: [Power In] = [Work accomplished by the fan: 4.73 bhp] / [Motor efficiency]. For a new motor that should lead you to something in-between 4.73 and 7.5 hp.
Since you CC?ed eQuest-users specifically: For developing eQuest inputs, additionally consider your flow rate associated with that design operating point to determine a kW/CFM input: [kW/CFM] = 4.73* 0.7457 / motor-eff / CFM. Best paired with a corresponding system design airflow rate input that isn?t getting overridden due to site elevation (air density) or zonal flow demands.
A related nugget perhaps helpful to share, as this confused me for some years: motor efficiencies in broad strokes do vary, but not much between 50-100% of nominal horsepower load with all else being equal (distinct from motor speed!). It?s pretty normal to see motors ?oversized/underloaded? motors with high nameplate horsepower relative to their actual full-load operating power requirement, and that isn?t necessarily a bad thing. The above document includes some examples and further explanatory text illustrating where ?right-sizing? a fan motor can result in identical or worse power input, where the points of efficiency gained for operating closer to full capacity are countered by the points of efficiency lost inherent to smaller motor sizes.
~Nick
[cid:image001.png at 01D2FFA2.BC049F10]
Nick Caton, P.E., BEMP
Senior Energy Engineer
Regional Energy Engineering Manager
Energy and Sustainability Services
Schneider Electric
D 913.564.6361
M 785.410.3317
F 913.564.6380
E nicholas.caton at schneider-electric.com
15200 Santa Fe Trail Drive
Suite 204
Lenexa, KS 66219
United States
[cid:image002.png at 01D2FFA2.BC049F10]
Thanks to Jim, Christopher, Patrick, Daniel and Nick for giving insight of
fan power modelling. What I understood from the conversation is: The fan
power consumption is the function of BHP and motor efficiency NOT the rated
motor HP.
Thanks again!
Regards,
Mayank
Yes, with the caveat/understanding that the nominal/nameplate HP rating directly informs the motor efficiency.
Regards,
~Nick
[cid:image001.png at 01D3014B.D70C2A30]
Nick Caton, P.E., BEMP
Senior Energy Engineer
Regional Energy Engineering Manager
Energy and Sustainability Services
Schneider Electric
D 913.564.6361
M 785.410.3317
F 913.564.6380
E nicholas.caton at schneider-electric.com
15200 Santa Fe Trail Drive
Suite 204
Lenexa, KS 66219
United States
[cid:image002.png at 01D3014B.D70C2A30]
Thanks to Jim, Christopher, Patrick, Daniel and Nick for giving insight of
fan power modelling. What I understood from the conversation is: The fan
power consumption is the function of BHP and motor efficiency NOT the rated
motor HP.
Thanks again!
Regards,
Mayank
Yes, with the caveat/understanding that the nominal/nameplate HP rating directly informs the motor efficiency.
Regards,
~Nick
[cid:image001.png at 01D3014B.D70C2A30]
Nick Caton, P.E., BEMP
Senior Energy Engineer
Regional Energy Engineering Manager
Energy and Sustainability Services
Schneider Electric
D 913.564.6361
M 785.410.3317
F 913.564.6380
E nicholas.caton at schneider-electric.com
15200 Santa Fe Trail Drive
Suite 204
Lenexa, KS 66219
United States
[cid:image002.png at 01D3014B.D70C2A30]
Thanks Nick for the correction :)
Thanks Nick for the correction :)
Hi Mayank,
I think that the 7.5 HP is the rated horse power of the motor.
Whereas the 4.73 BHP is the break horse power which is the actual power
available at the shaft of the engine.
The rated power is the maximum power which a motor can output from the
engine.
HP is the nameplate power used in the industry.
In this way the 4.73 BHP is the power with which the fan rotates so you can
have the fan power as 4.73 BHP.
*Thanks,*
Sharad.Kumar
India.
............................................................
............................................................................
Hello Everyone,
Ours is a Residential Project with PTAC units and DOAS in the Proposed
design. The DOAS nameplate motor HP is 7.5 where as the fan absorbed power
is 4.73 BHP. If we model 7.5 HP; we are falling short of the minimum LEED
energy performance requirements. Modelling
4.73 BHP just gives us enough savings.
Which one is the correct fan power to model?
I have been through the archives and didn't find appropriate answer. Please
help!
Regards,
Mayank.
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