How do I calculate WORK for material I've mixed in a blender? 7

[Cahill14]

I am mixing 500kg of a granola mix in a horizontal blender. I want to know how to calculate WORK in terms of kilowatts/kg or some units very close to that. I've collected kW of energy throughout the mixing cycle. I have the total cycle time and the total amount in kilograms that was mixed. Example: 7.19 total kW used over 14.33 minutes for a 500kg batch. How much WORK did I put into the 500kg batch? Thanks for your help.

 

[cowski]

The metric unit for work is the joule (Newton meter or kg m^2/s^2). The watt is the metric unit for power (watt=joule/sec). If you want to know how much work you put in, convert your time to seconds multiply that number by your watts and you will end up with work in joules. If you want to make up your own unit for purely comparison purposes (batch to batch) kW/kg would be fine (just bear in mind it is not a unit of work).

 

[Maui]

The short answer to your question is that you expended approximately 12.4 kJ/kg.

You can calculate this answer by first determining the amount of electrical energy that was required for your process. You stated that you used 7.19 kW over a time period of 14.33 minutes. Since 1 kW = 1 kJ/s (where the J stands for Joules, an SI unit of energy), and 14.33 minutes = 859.8 seconds, you can determine the energy used by taking the ratyio of these two quantities.

Energy expended = Power*Time

= (7.19 kJ/s)*(859.8 s)

= 6182 kJ

Divide this result by the mass of the batch that was processed to get the final result.

Energy expended per unit mass = 6182 kJ/500 kg

= 12.4 kJ/kg


Maui

 

[sreid]

Kilowatts is a unit of power. Kilowatt-Hours is a unit of energy or work.

 

[CoryPad]

The kilowatt hour unit is unnecessary when the SI energy unit joule is available. Also, neither newton nor joule require capitalization unless they begin a sentence.

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[Barry1961]

This may not be a factor but....

If you try to extrapolate the work per kg of granola very far from the 500kg test point your data may get skewed. This would be due to the efficiency of the mixer drive and work required to spin the mixer empty.

For example if you were to mix 1kg of granola in a mixer capable of mixing 500kg batches it would take a very large amount of work per kg.

Barry1961

 

[MintJulep]

You asked "How much WORK did I put into the 500kg batch?". All of the answers you have gotten are how much work the motor did to produce the batch. As Barry points out, various efficinecies must be considered if the question you asked is really the one you need answered.

However...

If there is no appreciable temperatuare increase of the granola durring the mixing process, I believe that the answer to the question you asked is zero.

Energy is the potential to do work, or conversely, if you do work on somthing, you increase its energy state.

It seems apparent that mixed granola has no more potential to do work than un-mixed granola components.

 

[Dennyd]

A

http://www nuts!!!!

 

[sreid]

No FLAME intended here but;

When I read a sentence "I collected KW of energy through the mixing cycle" I'm not certain what was collected. One can make some assumptions and do some math but it may be "garbage in, garbage out."

I pay for electricity in kilowatt-hours. It may be an unnecessary unit of energy but then so would another dozen or so units of work/energy if that is the rule.

Units base on names are usually capitalized as in Newton and Joule.

 

[Cahill14]

Sreid, the controls technician who collected the data for me (I'm guessing) used a kilowatt meter. I asked him to capture kW readings from both motors on the mixer. The batch cycle has seven steps - each with their own respective step time in seconds. The mass in the mixer stayed constant until discharge. Per Maui, I calculated kJ of energy expended per unit mass for each of the seven steps and cumulatively added them. For the discharge cycle I am struggling to attempt to integrate the area under the curve during step 7's discharge cycle. Here, mass varies with time as does loading on the motors. I need rough numbers - not building a rocket - just a granola bar. I appreciate attention to detail, but am not concerned yet about whether it is kiloJoule or kilojoule - I will use kJ for now. Thanks.

 

[sreid]

As Cowski points out you are deriving a unit of measure for your own purposes, Kj/Kg. While it may be true that both the mass and the power are decreasing durning the dump cycle you may not care since the derived unit is on a per batch basis. The energy during the dump cycle is simply the area under the power-time graph.

 

[CoryPad]

I don't mean to detract from the content of this thread, rather to add to it by mentioning some of the SI rules. It helps everyone understand what is being discussed, and apparently two other members found my previous contribution useful.

[blue]sreid[/blue],

Your last post brings more confusion because K is for kelvin and j is nothing in SI. Kilojoules divided by kilograms is kJ/kg. More information regarding SI and its use can be found in the documents published by the National Institute for Science and Technology (NIST):

http://ts.nist.gov/ts/htdocs/200/202/mpo_pubs.htm

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[friartuck]

When I was a lad (too long ago to recall accurately), we did an experiment about waterfalls where the temperature at the bottom of the waterfall was measured and compared with that at the top. It followed that the potential energy at the top was converted to heat energy after plastering the bottom of the fall.

Consequently, we have a problem to solve on where the energy is used.

The motor rating is no good since we dont know if it is partially or fully loaded. But we could measure the full load current during operation. The energy used would vary depending on the loading..(ie it would be a high initial load which would fall as the granola (whatever that is) was munched into pieces. So a graph could be produced.

However the mixer would generate noise and the granola would also rise in temperature (we are breaking molecular bonds which has an energy implication). Plus there would be friction generated by movement. At the same time, the mixer is not perfectly insulated so heat would be lost simultaneously.

So the answer is, yes you can measure it, but which bit are you interested in..the energy going into the food, the noise energy, the electrical energy used or what??

Friar Tuck of Sherwood

 

[mloew]

Cory,

Thank you for always being there to help elucidate the membership on SI. I for one appreciate it.

Best regards,

Matthew Ian Loew
"I don't grow up. In me is the small child of my early days" -- M.C. Escher

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[SMF1964]

What happens if you run your tests, get your data, and then re-run your tests without any granola in the machine? Could you then determine the energy used by the machine with no load, subtract it from the energy used by the machine on a full load and the difference would the energy put into the granola?

Is the granola the cinnamon brown sugar kind? 'cause I like that stuff.