Does a bomb calorimeter give a value of heat energy per gram of material or electricity?

May 3, 2011 4:59 AM

Does a bomb calorimeter give a value of heat energy per gram of material or electricity?

I have been using a bomb calorimeter to measure the energy output of some samples. it has given a mean (gross heat of combustion) value of 21601 joules per gram, which I calculated as 6kWh per kg.

Is the value of energy given in heat form or electricity?

posted by sockpim to science & nature (13 answers total)

Energy is energy.
posted by gjc at 5:05 AM on May 3, 2011 [1 favorite]

An answer that was more constructive would be swell.

I mean what kind of energy is it? If it is heat energy is there a standard calculation I can do to give a figure of how much electricity the samples would produce?
posted by sockpim at 5:10 AM on May 3, 2011

Those are both measures of energy per unit mass. Are you confused because kWh is often used to quantify electrical energy? A bomb calorimeter does, as you said, measures the heat output from your sample, and any units of energy per unit mass is fine to use for that. I'm not sure if that answers your question.
posted by Salvor Hardin at 5:12 AM on May 3, 2011

It's just over 5 kcal/gram, which is more than cheese, less than butter, and about the same place as dark chocolate, based on the contents of my fridge.

What units do you want your results to be in? Use them. If it's for a class and the notes are unhelpful, ask whoever you're handing the assignment in to.
posted by Lebannen at 5:13 AM on May 3, 2011

Essentially I wish to create a model from my results to show how much electricity could be produced from mass burning the samples in an incinerator.

My question put in another way would be to ask if I can just convert this energy into kWh and say that 6kWh of electricity could be produced from a sample if burned. Or, if it was heat energy, would I need to work out how much of this energy would be lost by powering a steam turbine to generate electricity (for example a steam turbine with 90% efficiency would convert 6kWh of heat energy into 5.4kWh of electricity).
posted by sockpim at 5:19 AM on May 3, 2011

Ah, didn't see your clarification. You are measuring thermal energy, and it would be possible to turn it into electrical energy using some kind of heat engine. Heat engines have theoretical efficiency limits, and there will be some losses in electrical generation. So depending on the efficiency of conversion, you'd get something a bit shy of 6 kWh/g.
posted by Salvor Hardin at 5:19 AM on May 3, 2011

To your second clarification: yes.
posted by Salvor Hardin at 5:20 AM on May 3, 2011

What you are measuring is the energy density of the materials: amount of energy available per amount of material. The amount of electricity produced for any given sample is dependent on the efficiency of the method used to convert that energy to electricity (e.g., only about 30% of the energy density of coal can be recovered as electricity in a power plant). See the energy density section for coal from wikipedia for a little more info...
posted by masher at 5:24 AM on May 3, 2011

I will be really surprised, by the way, if the efficiency of any process you're thinking about using is anywhere near 90%. Take a look at the Carnot cycle; there's a fundamental theorem that says that you can't do any better than that.
posted by Johnny Assay at 5:29 AM on May 3, 2011

Bomb calorimeters measure the heat of combustion of their contents. In other words the energy change between the sample substance and whatever is left over after combustion with oxygen.

If you want to know how much electricity you could produce from the complete combustion of your sample you need to use the HHV or LHV. Which one you need depends on fuel characteristics and combustion scenarios. HHV is the same as heat of combustion, LLV can be found from HHV if you know the stoichometry of the combustion reaction (it has to do with how much water vapour is produced).

You now need to adjust your gross energy (HHV or LHV) for thermodynamic efficiency of the engine in which you want to burn it. Assuming a carnot cycle (a perfect engine), your efficiency will be 1 - (T_cold / T_hot).

So if the bomb calorimeter gives you 21601 joules per gram, then to determine an upper limit on the electrical energy you could produce from a gram of the sample fuel, do the following:

(21601) x [1- (300 / T_combustion)] x η_Electrical

Where I've assumed that you can cold-sink the engine to 300K, and that:
T_combustion is the combustion temperature
η_Electrical is the conversion efficiency between mechanical and electrical energy
posted by atrazine at 5:36 AM on May 3, 2011

For a typical steam plant electricity produced is going to be about a third of the thermal energy of the fuel.
posted by ihadapony at 7:15 AM on May 3, 2011

Converting heat into electricity is actually fairly complicated, and the real-world efficiencies for power plants are much lower than I would have expected. For starters, see this wikipedia page which gives some of the thermodynamics, and goes on to mention that many power plants only have about 40% efficiency in converting the heat of combustion to electrical energy.

I suspect that the answer to this in any real case (i.e. if you were actually building an incinerator to produce electricity) is fairly complicated and would definitely require consulting an engineer. If you just want a back of the envelope estimate, this paper says that many waste-to-energy plants have an efficiency of 22%, so you could assume that.
posted by pombe at 10:52 AM on May 3, 2011

You are basically confused about the nature of energy.

"Electricity" is not a form of energy; it is a phrase describing a particular kind of power source. Other such phrases are "steam power", and "manpower".

gjc is correct, whether you like his answer or not: energy is energy. What you wanted to ask is something like "If I could contain this energy, and use it to generate electrical power, how much could I reasonably make?" However, the conversion factor is merely a statement of the conversion efficiency, and has nothing whatsoever to do with the bomb nor the calorimeter.
posted by IAmBroom at 7:12 PM on May 3, 2011