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
- Thread starter
- #21
The exact equation I used is equation 4-7 on this link:
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?
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?
Compositepro
Chemical
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.
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.
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.
- Thread starter
- #24
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.
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.
LittleInch
Petroleum
~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.
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.
- Thread starter
- #26
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
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
- Thread starter
- #28
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:
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?
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?
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?
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?
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2
- #30
Compositepro
Chemical
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.
LittleInch
Petroleum
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.
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.
- Thread starter
- #32
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.
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.
BronYrAur said:
These formulas could help you obtain an approximate value of temperature:
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.
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.
Kashif Naseem
Mechanical
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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