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luissandoval
29-08-2008, 12:14 AM
HI guys:

I have a big problem, i'm calculating a refrigeration load in a blast freezer but i have a question how can i compute the refrigeration load for the fan? i'm confused. i hope that you help me.

Sorry i don't know to write very well in english.

NH3LVR
29-08-2008, 01:58 AM
Assuming the fan motor is in the space, which is almost always the case, all the electrical imput of the motor turns to heat. I forget home many BTUs per actual Horsepower used, but that is the method.

ntfreezer
29-08-2008, 09:43 AM
dear lussandoval,
please sent emil to me. Maybe i can help you. liweijian@worldbase.cn

smpsmp45
29-08-2008, 12:16 PM
You need to know the Evaporator fan HP or you have to assume it to start with.

HP X 2546.699= btu/hr.

With this you have to see what apacity & what fans are getting selected. If your assumption is not matching, you have to recalculate using the right fan HP.

luissandoval
29-08-2008, 02:53 PM
Guys:

thanks a lot, the blast freezer's parameters:

* Five fans with fan motor inside of the blast freezer with a power of 12.7 KW c/u.

*air temperature inside blast freezer is -35ºC.

So

Q= 5*17.27(HP)*2546.699= 219907.5 btu/h

is correct this Compute???

Dacosta
02-09-2008, 11:24 PM
Following the units I get

5*12.7KW *1000 W/KW * 3.4121 BTU/hr/W=190500 BTU/hr

Dan

US Iceman
03-09-2008, 12:21 AM
You cannot use the simple fan motor input power. It will be too low. The actual power input is increased due to the heavier air density. As the air gets colder it becomes more dense. This requires more input power to compensate for the increased mass flow of air.

Fans are constant volume devices.

The fan performance data is usually based on standard conditions which are sea level and about 70°F.

Don Davis
03-09-2008, 12:33 AM
You cannot use the simple fan motor input power. It will be too low. The actual power input is increased due to the heavier air density. As the air gets colder it becomes more dense. This requires more input power to compensate for the increased mass flow of air.

Fans are constant volume devices.

The fan performance data is usually based on standard conditions which are sea level and about 70°F.

I think the fact that more HP would be required would show itself by a larger motor being employed and that converting HP to BTU would still be the way to go and would in fact be generous if the motor isn't fully loaded.

US Iceman
03-09-2008, 04:45 AM
While the colder air temperatures do overload the motor (higher amps than nameplate) the motor itself does not overheat due to the cold air itself.

I've heard too many stories where electrician's wired up the motor to nameplate amps and found the wiring and starters were too small for the amp draw.

The best way is to ask the manufacturers what the amp draw is at the operating temperature the unit is designed for.

If the motor was operating at ambient temperatures (or elevated temperatures) then oversizing the motor is a good idea for two reasons:


the motor will run cooler
the motor is more efficient around 75-85% of it's full load nameplate rating.

Peter_1
03-09-2008, 06:51 PM
What about apparent power ;) ?

Don Davis
03-09-2008, 08:37 PM
While the colder air temperatures do overload the motor (higher amps than nameplate) the motor itself does not overheat due to the cold air itself.

I've heard too many stories where electrician's wired up the motor to nameplate amps and found the wiring and starters were too small for the amp draw.

The best way is to ask the manufacturers what the amp draw is at the operating temperature the unit is designed for.

If the motor was operating at ambient temperatures (or elevated temperatures) then oversizing the motor is a good idea for two reasons:


the motor will run cooler
the motor is more efficient around 75-85% of it's full load nameplate rating.


US Iceman is right. I also thought of this after making my previous reply. Manufacturers recognize the benefit of low ambient temp and take advantage effectively 'overloading' the motors. So, using the actual operating amps x volts and converting to BTU would be the way to go.

Same with "air-over" motors on cooling towers. Always use nameplate amps to size starters.

US Iceman
03-09-2008, 10:13 PM
The problem is you won't know the motor amps until you are down to temperature. The method I use to help get the estimate closer is to use:

Nameplate FLA X (air density cold / air density ambient) = corrected amp load

Or use this for the motor heat:

Motor Input Power (kW or HP) X (air density cold / air density ambient) = estimated power input. Convert this to heat energy and you have the motor heat load.

Peter_1
04-09-2008, 06:35 AM
Altough this is interesting, we're drifting a litlle bit away from the initial question: determining the heat load from a fan.

Don Davis
04-09-2008, 11:30 PM
Altough this is interesting, we're drifting a litlle bit away from the initial question: determining the heat load from a fan.

No, I don't think we're drifting at all. I look at it like this...
All of the energy supplied to the motor shows up as heat energy to the enclosed space. The electrical energy could just as well operated a resistance grid heater instead of a motor/fan. It's all the same.

smpsmp45
05-09-2008, 06:45 AM
Dear Us Iceman,

Thanks for the nice inputs.

dglaab
01-11-2008, 10:22 PM
Hi all:

I have these design issues regularly in my blast freezers. All energy put into motors is added as heat in the box (either as motor heat or fan friction heat)and must be removed. If the evap manufacturer matched the fan blades for the working temp, your nameplate amps will be a good number, if not, assume you will be running a higher amperage (as much as 20% over nameplate,not a problem usually with a -40 ambient).Both are out there commonly, especially if evaps are used in other than designed conditions (used evaps installed). If you have an operating system, amp clamping will give you a good number for the calculations.
I can't locate the reference, but somewhere I recall a 1.1 factoring for load as well, probably to account for motor inefficiency (usually 93% ish on a high efficiency). I like to err on the high side for heat load, no customer complains if your system overperforms.
I once worked on a blast tunnel that was running 500 hp of fans (spec'd by an airflow guy with no consideration for refer load), they were running 700hp of compressor just to chill the box !. I removed 360 hp of fans and reconfigured evaps for a 300% increase in production.

Hope this helps

Dan

Andy
02-11-2008, 11:30 AM
Hi Guys:D

another way to look at the fan load is start with a good design, fan power input wise.

Blast freezers, 15 - 20% of required refrigeration will be the fans.

Now this is dependant on good airflow design, we should be looking to use a good airflow design.


Another point is smaller motors will be less efficent, it would require less actual power if you use larger motors and less of them.

An 11kW motor times 1.1 for freezer use will use 12.1kW mechanical power or an electrical load of 15.125kW.

Correct me if Im wrong this 15hp motor will load the refrigeration plant by over 15kW.

Kind Regards Andy:)