Just need a double-check please. I know the formulas are quite clear on this, but I was not expecting the answer I just calculated. I have an AHU with a fan wall, so motors are in the air stream. Please see attached. I am coming up with almost a 6 degree temperature rise from the fans. I just wasn't expecting such a high number. Fans are on VFD's, but "Design Condition" shown is the maximum that I am considering. I had to assume a fan motor efficiency.
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AHU heat gain due to supply fan motor 2
- Thread starter BronYrAur
- Start date
LittleInch
Petroleum
Looks to me like you've put in the entire motor power as heat, but it's a little difficult to tell as you've not included the formula.
I suspect all you need to use is the motor power which isn't converted to shaft power, i.e. 8% of 100bhp?
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I suspect all you need to use is the motor power which isn't converted to shaft power, i.e. 8% of 100bhp?
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
- Thread starter
- #3
It seems reasonable, you have a high static pressure, so your bhp per cfm is a little higher and the associate temperature rise matches that. The entire motor load is included, but the heat may not be realized immediately, as some of the heat gain will be due to friction through the duct. At least that is what's going through my head....
LittleInch
Petroleum
BTU per HOUR divided by Cubic feet per MINUTE??
My point about the formula was that you need to use consistent units and not just put in constants.
So your first equation calculates total amount of energy in one hour. I still don't understand what efficiency is doing in this calculation when you include the total electrical bhp (energy) going into the motor. You can't get more energy out that you've put in... If your motor efficiency was only 50% this number would double....
The second is the amount of heat to raise 1 cubic foot of air one degree F (1.08 BTu/ft3)
So you need to calculate the total amount of air in cubic feet in one hour to get everything in the same units.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
My point about the formula was that you need to use consistent units and not just put in constants.
So your first equation calculates total amount of energy in one hour. I still don't understand what efficiency is doing in this calculation when you include the total electrical bhp (energy) going into the motor. You can't get more energy out that you've put in... If your motor efficiency was only 50% this number would double....
The second is the amount of heat to raise 1 cubic foot of air one degree F (1.08 BTu/ft3)
So you need to calculate the total amount of air in cubic feet in one hour to get everything in the same units.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
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- #7
Sorry for the confusion but the units work out. I have seen a couple different versions of the motor heat equation. The one I have takes the brake horsepower and divides it by the motor efficiency. I have seen others that just use the motor nameplate horsepower and convert it to BTUH, but that gives you the highest possible value. It would seem that brake horsepower would be more accurate.
I have ten 15-HP motors in the air stream, so a straight conversion would be (150 HP)(2,545) = 381,750 BTUH. With both motor and driven equipment in the airstream, motor efficiency should not matter.
My other equation is the sensible heat transfer for air. The 1.08 converts mass to volume, specific heat, etc. The units all work out.
So if I use the number above: BTUH = (CFM)(1.08)(Delta-T). I would get a delta-T of 7.85 degs with this calculation.
The reason I question this is because the "rule of thumb" is usually around 2 degrees of temperature change added by the fan. I think it is the high static pressure that is driving this delta-T way above the "rule of thumb." I just needed a reality check on my calculation.
I have ten 15-HP motors in the air stream, so a straight conversion would be (150 HP)(2,545) = 381,750 BTUH. With both motor and driven equipment in the airstream, motor efficiency should not matter.
My other equation is the sensible heat transfer for air. The 1.08 converts mass to volume, specific heat, etc. The units all work out.
So if I use the number above: BTUH = (CFM)(1.08)(Delta-T). I would get a delta-T of 7.85 degs with this calculation.
The reason I question this is because the "rule of thumb" is usually around 2 degrees of temperature change added by the fan. I think it is the high static pressure that is driving this delta-T way above the "rule of thumb." I just needed a reality check on my calculation.
Compositepro
Chemical
There really is only one way to rigorously check calculations, and that is with dimensional analysis. The units of every number in the equation must be explicitly written down and the operation that is done to the numbers is simultaneously done to the units. If your answer ends up with the correct units then the answer is pretty likely correct.
You still need good engineering judgement, like is motor efficiency important? In this case, with the motor in the air flow you might think not. All of the electrical energy put into the motor will end up as heat in the air. But you aren't starting with electrical input to the motor in your calculation. You are starting with the brake horse power of the fan based off of the fan curve. So, yes the motor efficiency must be factored in.
You still need good engineering judgement, like is motor efficiency important? In this case, with the motor in the air flow you might think not. All of the electrical energy put into the motor will end up as heat in the air. But you aren't starting with electrical input to the motor in your calculation. You are starting with the brake horse power of the fan based off of the fan curve. So, yes the motor efficiency must be factored in.
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- #9
Thanks but that is why I did factor efficiency into my original calculation. I fully understand the importance of dimensional analysis and how if can often lead to inaccurate calculations. However, I fear this thread is getting sidetracked down that line. I don't mean to dismiss it as though it is not important. I just know in this case that the dimensions all work out correctly.
What I was hoping to receive back was validation that a temp rise such as mine at 6 degrees is reasonable for a fan system like the one attached. I don't deal with such high static on a daily basis and was just looking for someone with a similar unit to weigh in. There are no dimensional mistakes in the calculations.
What I was hoping to receive back was validation that a temp rise such as mine at 6 degrees is reasonable for a fan system like the one attached. I don't deal with such high static on a daily basis and was just looking for someone with a similar unit to weigh in. There are no dimensional mistakes in the calculations.
Compositepro
Chemical
But if you had used dimensional analysis then you, and we, would immediately have determine that your calculations are correct and that the answer was reasonable. There are ovens designed for handling flammable solvents that have no heating elements. There is only 2 horsepower blower and the oven can reach 500F. Of course the air is recirculated.
LittleInch
Petroleum
I tried the same calculated in metric and it looks about right. It's much easier to keep.same units..... Your issue is that you're working the low side of the fan. You could easily have three times the airflow for more or less the same.power so your temp.rise would.be back to.your 2F. Of course you wouldn't have the pressure.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
The math is correct, it’s the fan selection. A rise GT 2.5F tells me it’s a selection issue. Your HP/CFM is very high for the application. If this is a conversion to a fan wall, your SP is very high unless the cabinet losses are crazy. In most conversions, you cannot fit that many modules in the tunnel. Use the VFDs to vary flow.
You will not pass IECC 2015 – a FEG67 & 15% fan will be required. I get the 9 unit selection; I look at the base design limited to 60hz. I bet you could run the base 10 unit with a high E 10hp motor. You are only in the service factor under full load. A draw thru will help too. I guess you have a 5x2 wall so the tunnel is rectangular. Have you looked at 4 or 3 wide single height? What are you doing for reverse flow on the failed fan? BDDs eat static if not selected correctly.
If this is a standard VAV application, I start with 3” on the duct and 3” on the central plant. The only time I see 8” would be an induction unit AHU.
You will not pass IECC 2015 – a FEG67 & 15% fan will be required. I get the 9 unit selection; I look at the base design limited to 60hz. I bet you could run the base 10 unit with a high E 10hp motor. You are only in the service factor under full load. A draw thru will help too. I guess you have a 5x2 wall so the tunnel is rectangular. Have you looked at 4 or 3 wide single height? What are you doing for reverse flow on the failed fan? BDDs eat static if not selected correctly.
If this is a standard VAV application, I start with 3” on the duct and 3” on the central plant. The only time I see 8” would be an induction unit AHU.
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- #13
thank you all for your comments. It is a draw-thru unit which is part of my issue. it is an operating room air handler for a large Hospital. they want 50 degree discharge air which is giving me trouble as I am trying to reset chilled water temperatures. there is a central chilled water plant that feeds everything. I am trying to make improvements there resetting chill water temperature as ambient permits.
I was not anticipating such a large heat gain from the fan. this is causing the chill water valve to stay at 100% even on an extremely mild day in the 50's. I am unable to reach that point due to the additional heat of the fans.
so I think that clears it up. Due to the high static application with a very large horsepower, the 6 degree temperature rise seems reasonable.
thank you
I was not anticipating such a large heat gain from the fan. this is causing the chill water valve to stay at 100% even on an extremely mild day in the 50's. I am unable to reach that point due to the additional heat of the fans.
so I think that clears it up. Due to the high static application with a very large horsepower, the 6 degree temperature rise seems reasonable.
thank you
I get what you are trying to do now. I would look at a blow thru design to help out on low LAT and air quality issues of an OR. MTI may not be the best fit. For 45K flow, I would look at 1x4 or 2x3 design. I believe you are in Chicago, Try Midwest Applied Solutions for an alternate application, they are very good at fan walls & custom AHUs. Go Cubs.
What is making the static so high? I worked mostly semiconductor cleanrooms, which frequently have the same kinds of elements -- carbon filters, HEPA/ULPA filters, HW/CHW coils, etc. and I can't remember ever seeing a pressure requirement that high, even for the conditions where dirty filters are assumed. Is this a retrofit on existing ductwork?
![[pipe]](/data/assets/smilies/pipe.gif "[pipe] [pipe]")
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- #17
I can't answer that question because I'm not the designer of the system. I strongly suspect a static set point is way too high. It's currently set at 2.4 inches of discharge off the unit. It feeds multiple operating rooms but I don't know anything about the ductwork configuration.
I have made many significant improvements to this facility in terms of Energy savings on the chilled water plant. However, it is always a battle to get them to implement any recommendations. They want to save money, but they don't want to change anything that's currently working. Even if it is working very inefficiently.
So bearing in mind the high static and the brake horsepower, do my numbers seem reasonable for temperature rise?
I have made many significant improvements to this facility in terms of Energy savings on the chilled water plant. However, it is always a battle to get them to implement any recommendations. They want to save money, but they don't want to change anything that's currently working. Even if it is working very inefficiently.
So bearing in mind the high static and the brake horsepower, do my numbers seem reasonable for temperature rise?
Howdy Bron,
Assuming that 15hp motors are 93% efficient; this allows 7 % of the motor O/P to be rejected (as heat) into the air stream.
Therefore, the heat rejected into the air-stream = 800W (max) or 2700BTU (per motor).
The reamining 93% is work used to move the air. This should not impact the temperature rise (of the air stream) signifcantly, n'est pas?
GG
The pessimist sees difficulty in every opportunity. The optimist sees the opportunity in every difficulty.
Winston Churchill
Assuming that 15hp motors are 93% efficient; this allows 7 % of the motor O/P to be rejected (as heat) into the air stream.
Therefore, the heat rejected into the air-stream = 800W (max) or 2700BTU (per motor).
The reamining 93% is work used to move the air. This should not impact the temperature rise (of the air stream) signifcantly, n'est pas?
GG
The pessimist sees difficulty in every opportunity. The optimist sees the opportunity in every difficulty.
Winston Churchill
LittleInch
Petroleum
I think you need to follow the energy.
The inlet power to the fan is the shaft power which is a max of 102 bhp (76kW)according to your diagram.
The motor only supplies what it needs regardless of motor rating. However it has an efficiency of lets say 92%. Therefore the electrical power going into the motors is 111 bhp (83kW). This loss of efficiency comes out as heat - (7kW).
Then you have the efficiency of the fan - 60%. Most of this loss of efficiency will be expressed as heat (where else does the energy go??), hence 102 or 76kW X 0,4 = 40.8 bhp / 30.5kW.
The remaining energy being input into the fan is converted to a mixture of potential energy (pressure) and kinetic energy (flow). Eventually this energy will be converted to heat, but spread out over the ducting and not instantaneous heating.
So you then end up with total heat being input into the air by the fan and motor of 7kW plus 30.5 kW - total of 37.5kW.
Flow is 45000 cfm = 1250m3/min = 20.9 m3/sec = 25.6 kg/sec
Heat capacity of air is approx. 1kJ/kg/K
So with energy of 37.5kJ per second into 25.6kg air, temp rise is 1.46 degrees C = 2.63F
Make sense?
This means that the extra heat caused by putting the motor in the airflow is about a 25% increase - sounds good to me and corresponds to the ROT of 2 degree F without the motor.
I still don't understand your original equation and I think it is the wrong one for this purpose.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
The inlet power to the fan is the shaft power which is a max of 102 bhp (76kW)according to your diagram.
The motor only supplies what it needs regardless of motor rating. However it has an efficiency of lets say 92%. Therefore the electrical power going into the motors is 111 bhp (83kW). This loss of efficiency comes out as heat - (7kW).
Then you have the efficiency of the fan - 60%. Most of this loss of efficiency will be expressed as heat (where else does the energy go??), hence 102 or 76kW X 0,4 = 40.8 bhp / 30.5kW.
The remaining energy being input into the fan is converted to a mixture of potential energy (pressure) and kinetic energy (flow). Eventually this energy will be converted to heat, but spread out over the ducting and not instantaneous heating.
So you then end up with total heat being input into the air by the fan and motor of 7kW plus 30.5 kW - total of 37.5kW.
Flow is 45000 cfm = 1250m3/min = 20.9 m3/sec = 25.6 kg/sec
Heat capacity of air is approx. 1kJ/kg/K
So with energy of 37.5kJ per second into 25.6kg air, temp rise is 1.46 degrees C = 2.63F
Make sense?
This means that the extra heat caused by putting the motor in the airflow is about a 25% increase - sounds good to me and corresponds to the ROT of 2 degree F without the motor.
I still don't understand your original equation and I think it is the wrong one for this purpose.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
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