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# AHU heat gain due to supply fan motor2

## AHU heat gain due to supply fan motor

(OP)
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.

### RE: AHU heat gain due to supply fan motor

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?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

### RE: AHU heat gain due to supply fan motor

(OP)
It is on the attachment. Q[BTUH] = (2,545 [BTU/HP-Hour])(fan Brake HP)/(motor eff).

### RE: AHU heat gain due to supply fan motor

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....

### RE: AHU heat gain due to supply fan motor

PV=NRT will also cause some temp increase from high air pressures.

### RE: AHU heat gain due to supply fan motor

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.

### RE: AHU heat gain due to supply fan motor

(OP)
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.

### RE: AHU heat gain due to supply fan motor

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.

### RE: AHU heat gain due to supply fan motor

(OP)
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.

### RE: AHU heat gain due to supply fan motor

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.

### RE: AHU heat gain due to supply fan motor

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.

### RE: AHU heat gain due to supply fan motor

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.

### RE: AHU heat gain due to supply fan motor

(OP)
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

### RE: AHU heat gain due to supply fan motor

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.

### RE: AHU heat gain due to supply fan motor

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?

### RE: AHU heat gain due to supply fan motor

You are ok there mate.I have seen in clean room applications fan work contributing to 60% of cooling load

### RE: AHU heat gain due to supply fan motor

(OP)
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?

### RE: AHU heat gain due to supply fan motor

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

### RE: AHU heat gain due to supply fan motor

If the motor's in the airstream, then it seems like all of the energy added to system counts and the efficiency doesn't matter. The energy imparted to the airstream to induce flow will be given up as heat when it slows back down, right?

### RE: AHU heat gain due to supply fan motor

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.

### RE: AHU heat gain due to supply fan motor

(OP)
The exact equation I used is equation 4-7 on this link: http://www.globalspec.com/reference/56319/203279/4...

It is similar to the equations in the ASHRAE Fundamentals Handbook. The 2013 IP edition has them on page 18.6.

In the case of my situation, both motor and equipment operated by the motor are in the conditioned space - direct drive fans in the air handler. The ASHRAE formula uses the motor horsepower rating, but then corrects it with a Motor Load Factor. It also has a motor use factor, but in this case, I am assuming continuous operation. I took the Motor Load Factor as the correction to go from motor horsepower to brake horsepower. The equation in the original reference uses brake horsepower and does not have the Motor Load Factor.

So the brake horsepower is the actual power being used by the fan. My assumed 92% motor efficiency then gives the actual amount of electric power. Since everything is in the air stream, it all gets transferred to the air ..... as energy but not necessarily heat

LittleInch, I see your point about the the potential and kinetic energy. Energy and Heat aren't the same here, but the equations don't seem to draw a distinction. I wonder why? So to reiterate what you said, the assumed 92% motor efficiency gives around 111 BHP of motor power being consumed. So 111 BHP - 102 BHP = 9 BHP is turned into heat. Then of the 102 BHP that the motor is actually delivering, approx 40% is lost due to fan efficiency. So about 40 BHP is lost as heat also.

So 49 BHP is turned into heat, which is about 125,000 BTUH. at 45,000 CFM, that equates to roughly a 2.6 degrees F temp rise.

It all seems logical. Now I just need to wrap my mind around the power use to actually move the air. ASHRAE says it should be included in the calculation, and they call the whole thing, "Instantaneous sensible heat gain from equipment operated by electric motors in a conditioned space." If the power is being used to raise the pressure of air and move it, how is that an instantaneous sensible heat gain?

### RE: AHU heat gain due to supply fan motor

If you are recirculating air in a closed space, all of the electrical energy becomes heat within that space. That is pretty basic conservation of energy.

### RE: AHU heat gain due to supply fan motor

Yes the heat gain is about 6F (5.8F per calc)with the motor & driven equip (fan) in the airstream. At the branch duct to the OR (Operating Room)you would probably need to put a booster in line AHU with DX Coil reheat coil, booster fan and HEPA filter. The booster in line AHU can have recirculated air from the OR to supply up to 60 ACPH air through HEPA laminar flow hood/light assembly over the OR table. Use draw through fan arrangement so the fan heat will heat up the air before the required final HEPA filter downstream of the fan otherwise the saturated air will encourage mold growth on the HEPA filter. Use Sterile Air UV lights downstream of the cooling coil to sterilize the cooling coil downstream side to prevent mold growth.
Try to minimize pressure loss to reduce fan energy. Avoid miter elbows. Use long radius elbows or single vane (not double) miter elbows. Use conical or shoe type branch connection. Be generous in sizing risers. provide volume damper at each branch off the riser.

### RE: AHU heat gain due to supply fan motor

(OP)
Yes but LittleInch is starting to convince me that the energy imparted to the air in order to pressurize and move it does not manifest itself as heat gain .... at least not instantaneously. Friction loss in the ductwork is a source of heat gain, but ultimately the air is discharged into the space with a velocity and hence kinetic energy. Are we saying that all of the energy eventually turns into heat gain? And even is that is true, when is eventually?

My original question is what temperature rise should I expect to be immediately caused by heat rejection of the fan motors? what temperature difference should I expect from the leaving side of the draw-through cooling coil to the leaving side of the fans? The equations were telling me about 6 deg F, but I am starting to think that a value less than 3 deg F is what I should actually see as sensible heat gain.

### RE: AHU heat gain due to supply fan motor

~sorry for confusing the issue and I'm no HVAC expert (at all), but looking at the equations provided, then like a few people have said follow the energy. Ultimately all the energy provided by the fan and motor will turn into heat. This will manifest itself in increased air temp which may not be seen immediately after the fan (my post above), but will be eventually.

Therefore the end consumer of the air will feel hotter than he or she wants to be.

The key issue is whether the fans you have are the most efficient at the duty you're asking them to do. I don't know whether 60% efficient is good or not, but if you could get it to say 85, then you would need less bhp for the same pressure and flow and hence lower heat gain.

Or indeed if you need the high pressure or not. Is that pressure actually needed or not?

Solve the underlying issue first.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

### RE: AHU heat gain due to supply fan motor

(OP)
I know this is going around and around in a couple circles but I am trying to determine what I should see immediately Downstream of the fan in terms of Heat gain. I guess I can safely say somewhere between 3 and 6 degrees but I was hoping to quantify it a little better

### RE: AHU heat gain due to supply fan motor

For motors in air stream, use a 3v-3a power analyzer to read total fan power, install a grid a temp sensors in mixed air entering fans, install grid of temp sensors down stream of fans (allow for mixing), shutoff all cooling & heating and traverse supply duct. Numbers should be within 5% to 10%. Avoiding poor mixing is the key. All power (KW) will show as heat gain. If you have room, Two Onicon thermal grids should give you the information. The power analyzer will verify sensor accuracy

### RE: AHU heat gain due to supply fan motor

(OP)
I just re-read this thread and am more confused now than I was when I first asked the question. I have always struggled with motor heat gain. I think the problem is that "Heat Gain" is not the same as "Temperature Rise" - at least not immediate temperature rise. The equations can be found in the Ashrae handbook or any other number of sources. I used this link for easy access: http://www.globalspec.com/reference/56319/203279/4...

my question is how to determine the immediate downstream temperature rise caused by the fan.

With the motor out of the air stream, the heat gain of the fan is reported to be the (brake HP) * (2545 BTU/HP-Hr). That will give you BTU/Hr of "Heat Gain". But what is this heat gain??? It is an increase in pressure of the air, correct? It is not really heat in the form of a temperature rise, is it? Perhaps if you accounted for the friction heat all the way along the ductwork, it would add up, but I am specifically interested in the temperature rise immediately downstream of the fans. (after a nominal distance for mixing).

With both motor and fan in the airstream, the equation changes to (brake HP)*(2545 BTU/HP-Hr) / (motor efficiency). Since the motor is obviously not 100% efficient, this number will be larger than the situation where the motor is out of the air stream. But again, is "Heat Gain" really heat in the form of a temperature increase? I would think that ONLY the heat of motor inefficiency would immediately contribute to actual temperature increase.

So like I said, I am more confused now. My specific situation has a total brake HP of 102 BHP in a fan system delivering 45,000 CFM. Assuming a 92% eff motor, the numbers are:

Motor out of air stream: (102)(2545) = 259,600 BTU/Hr
Motor in air stream: (102)(2545)/(.92) = 282,200 BTU/Hr

Considering only the difference between these two : 42,600 BTU/Hr. That equates to only about a 1 deg F temp rise by having the motors in the air stream. In my situation, the motors are in the airstream, so I am confident that at least 1 deg F temperature rise should be expected through the fans.

But I am not at all confident about the "heat gain" from anything other than the motor inefficiency. How much (if any) of that original 259,600 BTH/Hr (without the motor) contributes to a temperature rise that I can measure immediately downstream of the fans?

### RE: AHU heat gain due to supply fan motor

If you have a fan that is 60% efficient, 40% of the power is put into the air as sensible heat by increasing the temperature. Is this correct?

What happens to the other 60% and is this power completly accounted for by the bernoulli equation? What are the 3 components of the bernoulli equation and do any of the components of the bernoulli equation involve temperature?

### RE: AHU heat gain due to supply fan motor

2
The conservation of energy principle is very simple and clear. What you are getting confused by is in establishing the boundaries of the system you are analyzing. Your last statement made this pretty clear. The fact that air is being recirculated is irrelevant to your question. Eventually all the energy will show up as only heat. The fan creates kinetic energy and heat through turbulence and friction. This is "fan efficiency". You know how much energy is put into the motor. Subtract out the kinetic energy of the air after the fan and you have the amount of heat energy going into the air. You can also subtract out the potential energy of any pressure rise. You know the mass flow rate of the air so you can calculate the temperature rise of the air passing though the fan. That is what you are seeking. However, the air entering the fan will be slowly increasing in temperature because the air is being recirculated, and the kinetic energy that had been put into the air on previous passes has all been converted to heat.

### RE: AHU heat gain due to supply fan motor

I thought you had this sorted.

I agree 100% with compositepro.

If you ignore the enclosed system issue, then your immediate temperature rise will be the sum of the inefficiencies, which is both motor inefficiency (~8% of electrical input power) PLUS fan inefficiency (a much larger figure of around 40% of shaft power looking back through this thread). This is a real temperature rise as otherwise where does the energy go?? The other energy is the higher energy of the air d/s the fans (mixture of potential energy and kinetic energy). Eventually inside a closed system all this energy will be converted to heat, but immediately D/S the fans this energy hasn't been converted.

Of course if you don't have enough cooling capacity to take out the ambient heat loads PLUS the entire electrical fan load then the temperature will rise over time for a closed loop / re-circulation system.

Any clearer?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

### RE: AHU heat gain due to supply fan motor

(OP)
Ok, I am getting closer. The "Fan Heat Gain" in my original post is perhaps better described as "Fan Energy Gain". The fan in imparting energy to the air, only some of which is immediately seen as heat. The equation:

Fan Heat Gain = (Brake HP) (2545) / (Motor Eff)

Using my numbers. Fan Heat Gain = (102)(2545)/(0.92) = 282,200 BTUH

So 282,200 BTUH of energy is delivered to the air, which is producing 45,000 CFM @ 8.3" TSP. That energy consists of potential energy of the air, kinetic energy of the air, and heat gain in the form of temperature rise. Correct?

I'm still not sure how to subtract away the kinetic and potential energy so that I can quantity just the temperature rise, but at least I think I am understanding the equation better. The description of "Fan Heat Gain" is misleading if you ask me, even though I understand that eventually all the energy is converted to heat.

When I want to know fan heat gain, I want to know the specific air temperature increase that I will see across the fan.

### RE: AHU heat gain due to supply fan motor

#### Quote (BronYrAur)

When I want to know fan heat gain, I want to know the specific air temperature increase that I will see across the fan.

http://pontyak.com/fans/fanpowerandenergy.html

Volume of air is function of the pressure differential that the impeller creates right downstream and upstream the fan via centrifugal effect.

When the unit is in fan only mode, the temperature along the duct system should be slightly different, due to heat transfer and energy losses in form of air turbulence and friction.
Right downstream of the fan, the enthalpy of the air will come from heat losses of the electric motor and the work of the impeller over that mass of air.

### RE: AHU heat gain due to supply fan motor

This does not seem like a plug and chug type problem. Start of with a basic diagram. If you draw a fan and motor inside the air stream, then a box around that to represent the system boundary, you can draw arrows indicating energy entering or leaving the system. Equate the arrows going in tothe arrows going out. Solve for delta T across fan or for leaving air temperature. If you only consider heat energy, drawing the diagram is a little trickier I think.

### RE: AHU heat gain due to supply fan motor

In a Mining Environnement, All work (excluding work done against gravity) translates to Heat Gain.

And for fans, yes, the mesured electric power at fan input will ALL be translated to heat.

### RE: AHU heat gain due to supply fan motor

#### Quote:

Ok, I am getting closer. The "Fan Heat Gain" in my original post is perhaps better described as "Fan Energy Gain". The fan in imparting energy to the air, only some of which is immediately seen as heat. The equation:

Fan Heat Gain = (Brake HP) (2545) / (Motor Eff)

Yes you are getting closer.

Let me try to help even if you have achieved your required results but I will share some knowledge for others.

We send our units for testing all the time, and we obtain our Blower Power Input by testing the unit in the test reports. From there we calculate our gross, net capacity. And I believe that knowledge can help you answer your question but its important to establish some things about motor, blower, BHP, name plate hp etc.

Motor Rating: usually in hp which can be converted to kW. It is written on nameplate of the motor.
Fan/shaft Input: it is the brake horsepower of the blower
Power Input: it is (shaft power)/ motor efficiency .... motor efficiency usually taken 60-80%.

shaft power is BHP (brake horsepower) in kW or HP as you like. Some engineers mistake BHP with the motor hp on the nameplate and they end up getting inaccurate results. There are 2 ways to determine shaft power:

1. If you are from system or client side:

you can obtain shaft power by calculating total duct work static pressure, cfm required, and efficiency and some conversion factors.

2. If you are from equipment side:

and you have the specs of the blower like total SP and cfm. you said you have 2.4 inches of water guage but is it the total SP? Total SP is = ESP+ ISP (evap coil, filter etc), if you have the cfm, plug these values in your respective blower fan software and get the shaft power. Use that shaft power and get the power input as indicated above. Use that power input for your fan heat gain problem.

If power input is in Watts, power input * 3.412 + Net capacity = gross capacity. Please note that this power input is totally sensible heat gain, no latent heat gain is given off to the air. Rule of Thumb is that the gross capacity value will be around 4-6% more than the net capacity value. Rule of thumb is that if the temperature right before the blower is 56 degree fahrenheit, the supply temperature after the blower will be 60 degree fahrenheit at standard temperature ratings (95 F ambient also known as T1) required by client. If your unit is giving way way more than 60 F, then check your unit components and work from there.

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