Heat load due to conveyor motors

[MAragorn]

Ok, I feel really dumb about this. It seems like it should be straight forward, but I can't wrap my head around it.

I have a project with about 250 hp of conveyor motors. These conveyors move product both horizontally and vertically. Eventually, one of the conveyors moves the product out of the warehouse at a higher elevation than it comes in.

My question is: Does all the heat from the conveyor motors become part of the A/C load? I understand that if it was a chilled water pump, a fan or an air compressor I would include it all in the A/C load.

But I am getting confused in my head because of the change in potential energy of the product (due to elevation change) leaving the warehouse.

 

[RossABQ]

The motors are doing work, which generates heat. The motor doesn't know whether the energy is "stored" in product as a higher potential energy or not.

 

[MAragorn]

Don't get me wrong, I am not arguing, I just want to make sure I understand this.

If a fan is blowing air, the heat from the motor and the velocity of the air all ends up being converted to heat. This I understand.

However, if in the conveyor, all the energy is converted to heat in the motor...WHERE does the energy to change the elevation of the product come from?

The energy of the moving air turns to heat, but the potential energy of the elevated product does not. This seems to me to not be an equivalent treatment of energy.

 

[AbbyNormal]

You have to cool off elevator machine rooms, however the intermittent loperation helps you out. Myself I have never done a building busy enough to need the cooling that an elevator submittal lists.

What goes in must come out. More power goes in than what is required to do the lifting. So friction and heat from the ineffciency of the motor will heat the space.

Take the "V" out of HVAC and you are left with a HAC(k) job.

 

[BronYrAur]

So to MAragorn's point, how much of the motor heat is rejected to the room, versus the energy "consumed" by moving the conveyor? Simply taking 250HP of motors and converting it to heat is (250HP)*(2,542BTUH/HP) = 635,500 BTUH or 53 tons.

But is all of this heat really rejected to the space? And now that I think about it, I don't fully understand my conversion number of 2,542BTUH/HP. I took it from a book, but how does it account for motor efficiency? A less efficient motor will consume more energy and give off more heat, right?

I'm confused too.

 

[AbbyNormal]

Take a look at table 3A from chapter 30 of the current ASHRAE HOF, your 635,500 is pretty good lol, for a single big motor.

That table shows for 'motor & driven equipment' in the conditioned space the gain on a 250 hp motor (91%) efficient is 699,000 so take your number divided by 0.91 and you are right there.

AHSRAE is basically using the motor power divided by efficiency times a fudge factor for how frequently it runs, where 1 would be continuous and as well as a factor for how loaded that motor is and the load factor would get into the service factor at times THEREFORE GREATER THAN 1.

Table 3A shows them fully loaded with both the motor and the driven equipment in the space, with the driven equipment in the space and motor not in the space, and with the motor in the space and the driven equipment out of the space.

The motor in the space only, the gain on the 250 hp motor is 62,900 Btu/hr got to be the efficiency here, 699,000x.09=62910.

With a horizontal conveyor rolling along loaded, what is the motor power doing after the belts are started and inertia is over come? It is over coming the friction of the conveyor (like rolling losses of car, the bearings, drive losses), the friction of the loaded conveyor on the rollers, plus air drag on the load. So when all friction losses are in the space there is heat to the space.

Loaded belt more rolling losses, more weight on the (for lack of a better word) rollers, plus air drag. Belt is running empty, less weight down on those rollers so less friction, less air drag. Less power the motor draws.

With the inclined belts and vertical 'legs' lifting the load there is also the friction mentioned above. I do not see the energy of the lift itself creating heat outside of the heat shed at the motor.

Smaller motors are less efficient than the big motors.

I spent a few years in grain elevators and at times the equipment ran steady such as when loading 30,000 tons on a ship or other times it was infrequent and very intermittent. Some belts ran steady but were unloaded the majority of the time.


Take the "V" out of HVAC and you are left with a HAC(k) job.

 

[AbbyNormal]

746w X 3.412=2545

Take the "V" out of HVAC and you are left with a HAC(k) job.

 

[BronYrAur]

I have that table and see what you mean. I also see equation 2, which takes efficiency, motor use factor, and motor load factor into account. My simple conversion of 2,545 BTUH/HP was missing a few things.

I have strayed from MAragorn's original question of potentieal energy. The motor is working harder to raise the product than to simple convey it horizontally, but both motor and driven equipment are in the conditioned space, so it would seem that all of the heat is rejectrd into the space.

Now one more question. Let's take 250HP for example from table 3A. With both motor and driven equipment in the conditioned space, the heat is 699,000BTUH. With the motor out and the driven eqipment in, the heat is 636,000 BTUH. Now where is this heat? Is it the motor heat that is not in the conditioned space, or is it the heat from the driven equipment that is in the the conditioned space?

 

[Golestan]

Hi Abbynormal: are you reffering to ahrae handbook of fundamentals? I have been looking for the same type info on the motors runing in a space and how mauch heat do they add to the heat load.
I have the 2001 ashrae handbook, but I can't find the chapter you are reffering to. Am I in the right book?
thanks,

 

[BronYrAur]

In the 2001 book, see table 3A on page 29.7

 

[AbbyNormal]

I think the only extra heat involved with the lift would be the increase you would see similar to just the motor in the space. The energy of the lift is accounted for in the potential energy of the product and not heat is my thinking.

The 636,000 is the work done by the motor on the driven equipment in the space, what you would allow for in a cooling load. The 62,900 is outside of the conditioned space, similar to a big monster air handler with the shaft extending out of the casing and the motor out of the airstream.



Take the "V" out of HVAC and you are left with a HAC(k) job.

 

[AbbyNormal]

Sorry I was referring to the 2005 HOF.

Take the "V" out of HVAC and you are left with a HAC(k) job.

 

[BronYrAur]

AbbyNormal,

So to be clear, if the 250 HP motor is OUT of the conditioned space and the driven equipment in IN the space, I would add 636,000 to the air conditioning load of the space. Correct?

Conversly, if the motor is IN the conditioned space and the driven equipment is OUT, I would only add 62,900 to the air conditioning load. Correct?

This is counter-intuitive to me because I would think that the motor would always be a larger heat source than whatever equipment it may be driving.

 

[MAragorn]

AbbyNormal, what you are saying is exactly what I would expect, and what I was thinking.

In this case, both the motor and the driven equipment are in the space, so the complete load is added to the A/C load.

I may have accidentally misled, in that there are a number of smaller convetors in series, each with its own motor, all totalling 250 HP.

I believe that ONLY the portion of the work of the last conveyor that elevates the product prior to leaving the building would NOT be part of the HVAC load. Except for that all motor loads are included in the A/C load.

 

[AbbyNormal]

So to be clear, if the 250 HP motor is OUT of the conditioned space and the driven equipment in IN the space, I would add 636,000 to the air conditioning load of the space. Correct?

That is my take on it

Conversly, if the motor is IN the conditioned space and the driven equipment is OUT, I would only add 62,900 to the air conditioning load. Correct?

Correctamundo again fonzie or is it Mr. Page?

This is counter-intuitive to me because I would think that the motor would always be a larger heat source than whatever equipment it may be driving.

I think that heat coming off of the motor casing is the inefficiency. That 250 hp example said it was a 91% efficient motor, 699x.91=636, what's left over?


Take the "V" out of HVAC and you are left with a HAC(k) job.

 

[AbbyNormal]

I may have accidentally misled, in that there are a number of smaller convetors in series, each with its own motor, all totalling 250 HP.

Smaller motors tend to be less efficient than the larger ones. They all run steady at the same time?

I believe that ONLY the portion of the work of the last conveyor that elevates the product prior to leaving the building would NOT be part of the HVAC load. Except for that all motor loads are included in the A/C load.

My gut feeling is it will radiate a little more heat off of the casing than the horizontal ones, I don't think it really adds any more heat from the work than the horizontal ones, I guess if you knew how many pounds per hour it was lifting you could correct.



Take the "V" out of HVAC and you are left with a HAC(k) job.

 

[BronYrAur]

Thanks AbbyNormal, and it is Mr. Page

 

[AbbyNormal]

You're welcome Jimmy

Take the "V" out of HVAC and you are left with a HAC(k) job.

 

[AbbyNormal]

Back to the counter intuitive concept.

It is sort of like in the days where they were boring cannon barrels under water to come up with work makes heat

http://www.scienceandsociety.co.uk/results.asp?image=10318052&

http://wwwflag=2&imagepos=8

Take the "V" out of HVAC and you are left with a HAC(k) job.

 

[quark]

I started a similar discussion based upon some prior discussions in thread403-158338 sometime back. The link given in the thread by imok2 gives some good leads for discounting the heat load.

ASHRAE 2005 HOF says, as the no load power of motor is generally high, we have to follow the equations without any reduction (or atleast that is what I understood from the book). Check the same page as that of motor heat loads.