Fan Heat Gain 1

[PS]

Dear Engineers,

I read in a book where FAN HEAT GAIN is given by the formula:

Fan Heat Gain = Nameplate ShaftPower/Overall efficiency of motor+drive+fan.

I believed the heat gain to be a part of energy that is being lost due to inefficiency. In other words, "Power input*(1-overall efficiency)". When this is written in terms of break power "Break Power*{(1-eff)/eff)". But after reading the above formula for heat gain, I am just wondering how come all the power absorbed contribute to heat gain? How come the part of energy that helps air to gain pressure and velocity account for heat gain?

Even I referred ASHRAE Fundamentals 1997, Table 4 on Chapter 28, Pg: 28.10. The values given there are in line with the formula given above. I'm interested to find the heat gain for an AHU motor which falls under the "motor in, drive equipment in" category.

I am struck up here. Kindly do advice or suggest me an article regarding this.

Thanks in advance for the support.

 

[ChasBean1]

What you're talking about is academic, nothing wrong with that. For an HVAC system the fan heat gain is between about 1°F and 3°F when the motor is in the air stream. In the grand scheme of system design it is a small variable.

 

[PS]

Thanks @ChasBean1. But I just wanted to know how it is being calculated. I have also read that it is a better option to assume 5% of Room Sensible Heat as Fan heat Gain.

 

[IRstuff]

You seem to be a bit confused; what do you think the inefficiency represents? Is this for school? Student posting is forbidden.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

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[PS]

@IRstuff I'm NOT a student.

I just need a clarification that how come a 0.75kW 3ph motor will contribute a heat gain of 0.993kW in a motor in, drive in configuration. Please refer ASHRAE 1997 Fundamentals, Chapter 28-Table 4 on Pg:28.10.

I asked this question just to increase my level of understanding. I wish NOT to follow anything only because it is being followed by some.

 

[LittleInch]

See

http://www.eng-tips.com/viewthread.cfm?qid=416011

This should help

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

 

[PS]

Thanks @LittleInch. I have gone through the link. It states the same. Here, to obtain the heat gain, the shaft power is divided by the efficiency. Clearly this valve will be greater than the shaft power. This fact "HEAT GAIN > SHAFT POWER" actually surprises me and this is what made to post a thread.

I also know there might be some reasons for doing so. I'm trying to figuring it out. Any justifications for the above will be highly appreciated.

 

[LittleInch]

That's just not correct. It should be multiply by 1-efficiency for the instant heat gain. Eventually it all turns to heat so the system needs to account for that energy.

The only way you can get more than shaft power going into the air is if the motor is in the air flow and then you need to add electrical motor efficiency.

Maybe that's what it is trying to calculate?

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

 

[Compositepro]

The name plate on the motor is the max continuous rating. It is not the actual shaft power. That is governed entirely by the mechanical load on the motor.

 

[IRstuff]

What does not produce useful work is lost as heat.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

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[lilliput1]

All the input power becomes heat gain to the space if the motor and driven equipment is in the space to be cooled. Input = work/efficiency

 

[capliard]

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.

Ingenieur Minier. QuTbec, Canada.

 

[lilliput1]

capliard I think the portion of the work against gravity that does not translate to heat is only the porti that has counterbalance weight to counteract gravity.

 

[lilliput1]

All input is will be cooling load only if the motor is continuously on. If not a cooling load factor based on the number of operating hours and hours after the equipment turns off, In labs the peak cooling load usually is only about 1/3 the equipment nameplate power input. It is important not to oversize cooling equipment so they would run long enough to properly dehumidify the space.

 

[capliard]

@ Lilliput1.

No. The Energy goes into Potential Energy of Rock. Or Heat.

Explanation :

The energy balance is the same, whatever the counterweigh. Ore is still hoisted to surface. The energy required to do that is not reduced by the counterweigh.

Counterweigh is there to reduce required Power.

Ingenieur Minier. QuTbec, Canada.

 

[LittleInch]

capliard - Are you answering a different post? We're talking about fans here, not lifting things....

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

 

[ChasBean1]

PS (Mechanical)(OP) 30 Jul 17 07:08 post:

How it's being calculated? I didn't. My bad. As I get older (maybe to my detriment) calculations of unnecessary values takes a back seat to things that matter. Like I said, it's honestly 1-3°F repeatedly and routinely. Higher temperature rises might indicate a problem.

 

[BronYrAur]

To chime in here, I always used the 1-3 deg F rule of thumb too, but then I ran across a situation where I was observing a 6 deg delta across the fan. That led me to research the formulas and post the question I did in thread

http://www.eng-tips.com/viewthread.cfm?qid=416011

I am still not completely clear on the actual sensible heat gain that is experienced across the fan. It is a conservation of energy problem, but the pressure imparted to the air does not manifest as temperature gain across the fan. So I have found it to be difficult to quantify the actual sensible temperature gain of the air.

 

[PS]

Thanks all for your comments.

 

[PS]

Thanks @BronYrAur. This comment "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." from @Compositepro justifies the usage of total power.

And I believe this can be a solid reason.