Converting Tons Per Hour to KWH

[rockman7892]

I have been asked by management to find out how many kWH are used in conveying 1 ton of material across a number of conveyor belts that we have.

My approach is going to be to take the total amount of hp used for the motors on these belts to come up with a max hp number. I'm going to assume that we are running the max designed Tons per hour (800tph) across these belts so are therefore using close to the max hp on each motor and therfore use the total hp calculated from the sum of all motors.

Once I have the max hp for conveying 800tph I will convert this to a kW value. If I know that 800tons takes one hour at max hp for all motors then I can take my kW value calculated and multiply this by 1hr.

I will then have a kWH value for conveying 800tons per hour, so in order to get 1 ton per hour I will need to divide this number by 800?

Does this aproach sound right for coming up with an estimate for what I'm trying to do?

 

[stevenal]

Tons per hour is a time based rate. kWh is energy, and not a rate. kW is a rate.

Your method looked fine until you got to the 4th paragraph. You have a kWh value for moving 800tons (with no time divisor). Divide by 800 to get kWh per ton, the value management asked for.

 

[rockman7892]

Stevenal

Here is a sample of my calculation.

Conveyor 1 = 60hp
Conveyor 2 = 300hp
Conveyor 3 = 25hp
Conveyor 4 = 50hp
-------

Total Hp = 435hp

Convert 435hp to kw --> 435kW * .746 = 324.51kW

Since feed rate on belts is 800tph I know that it takes 1hr to move 800 tons. During this hour we can calculate a kWH rate of 324.5 * 1h = 324.5kWH

800 tons across belts = 324.5kWH

1 ton across belts = 324.5kwH / 800tons = .40kWH per ton

So the answer I will give managment is .40kWh per ton. Does this look correct?

 

[7anoter4]

I don't intend to disappoint you but the relation between Q [tons/hour] and P [KW] is more complicated for a belt conveyor.
A manufacturer uses his calculation mode and special software in order to calculate the required motor power.
I found somewhere a very simple relation for a standard belt conveyor:
P = 0.9445*(0.0002 * Q * LZ + 0.0037 * Q * H + 0.04 * LZ * B * V + Q * KS) [kW]
Where:
Q [ton/h]
LZ=CONVEYOR HORIZONTAL PROJECTION[m]
H =CONVEYOR VERTICAL PROJECTION[m]
B=BELT WIDTH[m]
V= BELT SPEED [m/sec]
KS=0 IF THE UNLOADING IS STREIGHT
KS=0.07 IF THE UNLOADING IS 90 degree.
If you want to see how a manufacturer may conduit his calculation, see:

http://www.forbo-siegling.com/Siegling/Siegling.nsf/pages/calculationprolink

Usually are about 40 factors to take into consideration for a suitable calculation.
Good luck!
Regards

 

[dpc]

You are leaving out the motor losses. You are also assuming that the motors are running at full load amps - they probably are not.

 

[rockman7892]

Fortunately I've only been asked for a Rough estimate. Do you think that leaving out motor losses and assuming motor is running fully loaded the way I did my calculation is a good enough aproximation to get a rough estimate?

 

[dpc]

It is a rough approximation.

 

[itsmoked]

You almost certainly will not be using any of those motors at their full load continuously. So using the motor nameplate HP will be very conservative, and will counter the losses issue.

Sounds viable to me.

Another point. You have a 300hp conveyor. It could seriously skew your estimate. It would be way better to measure its power consumption verses 'name plating' it.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[rockman7892]

The 300hp conveyor is actually a crusher I just listed it as a conveyor which is my mistake.

I have a couple of motors in this overall calculation that are varialbe loaded and variable speed controlled which I realize will lead to errors. For example I have a crusher as one piece of equipment, and a dragline as another piece of equipment that are seeing variable loads as they transition during their operation.

With all that said and understanding that there will be some errors in this calculation will it still be a reasonable rough aproximation considering I was asked for only an estimate?

 

[waross]

Now is a good time to ask again for that clip-on watt meter you have been wanting.

If you can't use this as an excuse for some new instruments try the following method. It will give quite accurate results.

Assume that the motors are running at 90% power factor at full load, unless the actual power factor is marked on the motor. If it is marked, use that figure.
The relationship between load and amps is non linear because of the magnetizing current which is 90deg out of phase with the real current.
The magnetizing current is relatively constant and is quite easy to factor out.
Lets use your 300 hp motor for an example.
I will assume 300 amps FLA at 600 volts.
This is an example. Use the actual full load amps marked on the motor.
First we will find the magnetizing current.
The full load current times the power factor will give the real current of the motor. (Don't confuse real current with the current you read with an ammeter. Real current means the part of the current that is in phase with the voltage. This is the only current that produces watts and this is what we are looking for.
SO:
Nameplate current times power factor equals real current.
Nameplate current = 300 amps (use the actual nameplate value here)
power factor = 0.9
Real power = 300A x 0.9 = 270A.
Now we can find the magnetizing or reactive current.
Use Pythagoras' theorem. Construct a right triangle.The base is the real current, the hypotenuse is the full load current and the altitude is the magnetizing current.
Nameplate amps squared = 300 x 300 = 90000
Real amps squared = 270 x 270 = 72900
90000 - 72900 = 17100
The root of 17100 = 131A magnetizing current.

Now, when you take a current measurement, square it. Then subtract the square of the magnetizing current that you have calculated.
Take the square root of the result and express it as a fraction over your real current (270A in the example.)
Example 2:
Suppose that you read 200 amps on the motor.
The quick answer is 2/3 load or 67%.
But, 200 X 200 = 40000.
40000 - 17100 = 22900
The root of 22900 is 151 amps.
151/270 = 56%
Quite an error from 67%
You don't have to calculate the percentages. This was done to show the different results between the different methods.
You may use the real amps that have been calculated (151A) directly to calculate kW or kWHr.
The accuracy of this method depends heavily on the accuracy of the power factor assumption. However, any reasonable assumption for power factor will result in much more accuracy than using a simple fraction of measured amps to full load amps. Another source of error is the losses. This method does not accurately prorate the loss currents at less than full load, but these errors are much less than the errors caused by ignoring the magnetizing current.
And remember, this is a rough estimate today. In a few years it will be gospel and may be used as the basis of decisions involving considerable sums of money.Hope this helps.
Questions welcome. If you have some real world figures please post them.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[rockman7892]

What is the difference between converting a hp to kW by using a multiplier of .746 or using the FLA of the motor and calculating a kVA and then using the power factor to find the kW.

The only thing that I can see is that by using the .746 multiplier you dont account for the efficency of the motor where as converting to KVA first you do?

 

[waross]

When the motor is running at less that full load, the kW draw will be less. Estimating the load or kW based on current alone is not accurate because of the magnetizing current. For instance if your 300 hp motor is running at much less than full load, this method of factoring out the magnetizing component of the motor current will give a much more accurate estimate.
This method does take efficiency into account. If this method is used to estimate hp, there will be some errors due to the losses. In your case, you don't want to know the hp loading, you want to know the kW or kWHr. Losses are part of the cost of running the motor so you may use the real current calculated to calculate your kWHr.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[jraef]

rockman7892,
Too bad you are doing this now as opposed to a year from now; I might be able to share some real values with you. I am currently helping to implement a program at a large aggregate producer to accurately measure and record power consumption right down to each individual continuous load motor. There are aggressive energy savings programs in our area that require proving the reduction in consumption, which of course implies that you know the use under previous normal circumstances. We are installing a system of communications capable motor protection relays to continuously monitor them, but in the mean time, we have been doing preliminary observations as a way to justify the expenditure. Here are our initial observations, based solely on clamp-on ammeters compared to motor nameplate FLAs.

Crushers (cones, HSIs, Rollers etc.) tend to run at about 85 - 95% FLA, in fact most operators are monitoring crusher current as a way to maximize yield by controlling the speed of the infeed conveyors. I even once implemented a PID feedback loop on one where I looked at crusher motor current as the PV and infeed conveyor speed as the OP so the operator just dialed in 80 - 90% of the crusher FLA as the SP. I really wanted to use a belt weighing system as the PV feedback and use TPH as a setpoint, but it was all "too automated" for the operators (you probably know what I mean) and they went back to a manual system. But it was interesting.

There is a trend now to implement VFDs on VSI crushers as a way to tailor the output sizing, and if turned down, the power can drop significantly. But so does the TPH, so I would leave even those at 80-90% FLA for your purposes.

Stackers and inclines tend to run at or near FLA, in fact some even run into the service factor if they are telescoping and run out to their peak height. I attribute that to the motor selection software that is out there now and the fact that many OEMs are on tight budget constraints so they no longer leave that "fudge factor" in if they don't have to.

Under crusher and infeed conveyors tend to run at 70-80% FLA (less if there is a VFD), probably because NOBODY wants to dig one out if it overloads!

Overland conveyors are a crap shoot. My user stays conservative and wants the motors sized with that 25% fudge factor because unlike the stackers and inclines, changing out the motors is a bigger deal, usually out in a high traffic area where they have to shut down operations to let the riggers and electricians change out a motor. But I have certainly seen my share of those running close to the bone as well.

Screens tend to run around 70 - 75% FLA because most people apply standaqrd Design B motors to them and have to oversize them to get the necessary starting torque.

So all in all, I think for your rough budgetary guesstimate approximation, your process is sound, but not likely as conservative as you might think. The biggest problem with estimating energy usage in crushing operations only on motor loading only is what is called the "Recycle Rate" which does not show up as motor HP or kW, it only shows up as kWH. What I mean by that is, if more of your rock doesn't make it through the screen and goes back through the crusher again, you are expending energy to recycle and re-crush that rock. The crusher and return conveyor HP use remains the same, but the usable TPH is lower! That means then that your system has to run longer to finish the job, increasing your kWH. So in my opinion if you really want a conservative value, add about another 10% to your kWH figures to cover the recycle rate.


"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
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[7anoter4]

As dpc and waross already explained above:
Tons/hour is not energy. T/h=1000 kg/3600 sec=1/3.6 kg/sec.
If you add somewhere the conveyor length you'll get a POWER as 1kgf*m/sec=9.81w
If the conveyor run upright and the height is 100m then the required power[net]=9.81*T/h*100m=9.81/3.6*100/1000=9.81/36 kW=0.2725KW=0.3653 HP
But the conveyor run horizontally on rollers so with a friction factor of 0.07 [rolling] we get 0.2725*.07=0.019 kW.
I think motor losses issue is very important. But very important losses are mechanical losses of the conveyor.
Even if the material flow QTH [tons/h] will be 0 the conveyor requires power in order to move with a certain speed. For a small QTH and large belt the efficiency may be 1-3%.
If one use the above relation
P [KW] = 0.9445*(0.0002 * QTH * Horizontalprojection+ 0.0037 * QTH * Verticalprojection + 0.04 * Horizontalprojection * Beltwidth * Speed + QTH * KS)
If QTH=0 still 0.04 * Horizontalprojection * Beltwidth * Speed >0
Let's take:
QTH=0 ; Horizontalprojection=100 m ; Verticalprojection=0 ;
KS=0
Beltwidth=0.3 m;Speed= 1 m/sec
P= 1.1334 KW [noload!]
If we take QTH=1 T/h and all other parameter the same as above then:
P= 1.15229 KW and the Putil=1.15229-1.1334= 0.01889 kw Efficiency =1.6%!
If we take QTH=100 T/h and all other parameter the same as above then:
P= 3.0224 KW and the Putil= 3.0224-1.1334= 1.889 kw Efficiency = 62.5%
If we take QTH=1000 T/H and all other parameter the same as above then:
P= 20.02 KW and the Putil= 20.02 -1.1334= 18.89 kw Efficiency = 94.34%
So kwh is approximatelly propotional with tons/hour for one conveyor only if the conveyor will be well loaded.May be kwh=tons/h*k and k is different from one conveyer to another and may be proportional with conveyor length.
Regards

 

[rockman7892]

Thanks for the responses guys.

Jraef thanks for some historical data. This will help for we have several cursher conveyors and rollers around the plant that this info will come in handy for.