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Participant

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KW Calculation for Small, High-Pressure Centrifugal Fan

03/17/2010 7:32 AM

Hi! Friends,

I am having a problem in calculating Fan power, Plz. help me out,below is the details of problem.

I have to select one High pressure centrifugal fan which will suck air from atmosphere & will deliver in a high pressure fluidisation Air line.The atmospheric temperature varis from 8 degree C(Minm.) to 50 degree C(Maxm.).The capacity of fan stated as 64Nm3/min,with a total pressure of 9678 Pa & Pr.rise of 8500 Pa,Air density at fan inlet is 1.293 Kg/Nm3,Efficiency stated as 70%.

Plz. help in calculating 1.) 'Q' at 50 & 8 degree C &

2.) Power at 8 degree & 50 degree C.

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Guru
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#1

Re: KW Calculation for a Small High Pr. Centrifugal Fan

03/17/2010 8:08 AM

The power required by any fluid mover is the volumetric flowrate mutiplied by the pressure rise, divided by the efficiency.

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#2
In reply to #1

Re: KW Calculation for a Small High Pr. Centrifugal Fan

03/17/2010 9:11 AM

Boss,

As I think first we need to convert flow rate from Nm3/min to m3/min( I need Help) &

secondly there are two terminal temp.(8oC & 50oC) for which we need calculate KW & should take the bigger one.

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#3

Re: KW Calculation for Small, High-Pressure Centrifugal Fan

03/17/2010 1:35 PM

The fan Q will be the same at 8°C and 50°C (64 m3/min), but the mass flowrate will be lower at 50°C due to lower air density. The mass flowrate is the volumetric flowrate (Q) x air density. Use an air properties table to obtain the correct density at temperature – the density you stated appears to be at 0°C.

The power will be higher at lower air density, so the 8°C condition will require more power. Power is calculated as the volumetric flowrate (m3/s) x total static pressure (Pa=N/m2) x 1/efficiency. To calculate power at your design conditions, you need to correct the static pressure based on the air density at each condition. The N in Nm3/min means "Normal" and indicates performance at normal conditions (whatever temperature and pressure your fan performance is given at, maybe 20°C & 1 atm). The correction factor for design conditions will be the density of air at design conditions divided by the density at normal conditions (may be greater than 1 if design temp<normal temp). You will also need to correct the air density for elevation if you are not at sea level (unless fan performance is already given at elevation). Multiply the static pressure at normal conditions by this correction factor and then use this value to calculate power above. Note that this result is shaft power - you will need to size the motor to account for any drive losses (belts) and motor efficiency.

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#5
In reply to #3

Re: KW Calculation for Small, High-Pressure Centrifugal Fan

03/18/2010 8:54 AM

First of all a big thank you for such valuable guideline & secondly Air density can be calculated from = (273+1.293)/(273+t),Air properties chat gives only a approx value ,but how do I calculate(formula?)static pressure based on that density.

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#6
In reply to #5

Re: KW Calculation for Small, High-Pressure Centrifugal Fan

03/18/2010 3:09 PM

I think your formula should be (273 x 1.293) / (273 + t), which does yield the same exact values as the table above. I'd rather interpolate table values than dig up an equation, but that is personal preference.

I am assuming that since you have provided fan data that includes total static pressure (TSP) and efficiency, you have already done a preliminary selection (or is this just catalog data?). If this data is your starting point, then you need to know the temperature for these conditions (0°C based on air density?). Otherwise, you need to start with your actual engineering requirement for external static pressure (ESP or rise) and temperature (i.e. 8500 Pa at 8°C, 20°C, 50°C?). From there, you can correct the ESP to different conditions by taking the ratio of the air densities at the two conditions and multiplying by the initial ESP required (ESP2 = ESP1 x ρ21). You may need to correct your ESP to standard conditions in order to get a fan selection, and thus TSP (includes internal fan SP losses). Then you can correct the TSP (like ESP above) to calculate the shaft power at the desired condition (i.e. 8°C).

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#7
In reply to #3

Re: KW Calculation for Small, High-Pressure Centrifugal Fan

03/23/2010 3:16 AM

Boss,

Coming back to your solution, How the Power (Kw) rating is going to effect if there is an operating dust load of 5 g/m3 to the same system? How mush will be the increment in power(watt)?

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Guru

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#8
In reply to #7

Re: KW Calculation for Small, High-Pressure Centrifugal Fan

03/23/2010 9:32 AM

As noted in prior post #4, the Buffalo Forge "Fan Engineering" manual will give you a very specific answer to this question.

By the way, as you will see in that manual, energy usage by a system moving product AND air will be less than the same system running "EMPTY".

Read the book- all your questions will be answered- Now and in the future.

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Guru

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#4

Re: KW Calculation for Small, High-Pressure Centrifugal Fan

03/18/2010 1:26 AM

Your post states that the high pressure air is going to a "fluidsation" line. I believe that you are talking about a pneumatic conveying system that is receiving the high pressure air from the fan.

If this is correct, get a copy of "Fan Engineering" from Buffalo Forge Fan Company and research the pneumatic conveying section. A major point it that the fan will draw more power when running "empty" than when moving product. It seems irrational, but the manual will explain why this happens AND how to size the fan and motor to be sure that your system does not overload the motor later.

Since you stated a varying temperature, you will like want to use a VFD to maintain proper and consistent air volumes regardless of the inlet air temperature. You can use a pitot tube for the velocity (volume) sensing input because it will always read in the equivalent to "standard" air. Cooler inlet air has a higher density, so the flow will read a higher deltaP in the pitot than the "actual" volumetric flow would do if the flow was "standard" air- in this case, actual flow of, say, 1000 CFM might actually be 1070 SCFM airflow but the pitot would read pressure drop equal to 1070 SCFM of air. n Likewise, if the entering air is warm, and the fan is the same speed as when the air was cool, the actual airflow will still be 1000 CFM (the fan is spinning equal speed so it is displacing equal volume) but the SCFM being moved is likely about 920 SCFM because the air is less dense.

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