dr2much Posted March 28, 2014 Share Posted March 28, 2014 Let's assume a machine with substance A as the fuel. The output of the machine is electricity plus a product B. 1 kg of product A yields 1 kg of product B and C kWh of electricity. If the calorific value of product A is for example 20 MJ/kg and of product B it is 15 MJ/kg The electric output C of the machine from 1 kg of fuel A is 2,777 KWh. 1 kg A ---> 1 kg B + 2,777 kWh electricity A simple attempt to do an energy balance based on the calorific values gives the following: 20 MJ ----> 15 MJ + 2,7777 kWh 20 MJ ----> 15 MJ + 10 MJ Is this a possible situation? If so, how can it be accounted for that the energy content of the reaction product (based on its calorific value) plus the electric power produced is higher than the energy content of the fuel? Can this be explained by enthalpy changes for example vaporization and expansion of the fuel during combustion? Can someone please explain this in laymen's terms? (assuming that a machine with the input and output as described above is theoretically possible) Link to comment Share on other sites More sharing options...
Sensei Posted March 28, 2014 Share Posted March 28, 2014 Is product A burned? If so, oxygen is taken from air. Link to comment Share on other sites More sharing options...
dr2much Posted March 28, 2014 Author Share Posted March 28, 2014 Yes, fuel A is burned. However, I always understood that oxygen as such has no energy content (calorific value). Link to comment Share on other sites More sharing options...
studiot Posted March 28, 2014 Share Posted March 28, 2014 (edited) Let's assume a machine with substance A as the fuel. The output of the machine is electricity plus a product B. 1 kg of product A yields 1 kg of product B and C kWh of electricity. I can't comment on the numbers in your subsequent example as there is not enough information. But the above quote is a sound basis to start. Both A and as Sensei pointed out atmospheric oxygen, have chemical bonds which are broken and release energy. The combustion results in B, and some of this released energy goes into forming the new bonds in B. Some of this energy may go into a change of state of, if for instance the fuel is coal or oil and the result is carbon dioxide and water. What is left over appears as heat and is converted to electricity, ignoring the inefficiency of the generation process (this is far from negligable). Further, what Sensei was probably trying to say is that 1kg of A plus some kg of oxygen makes more than 1kg of B. That is you need to take the mass of oxygen into account in your reckoning. The relative masses of hydrogen, carbon and oxygen are 1, 12 and 16 So for every 1kg of hydrogen completely burned 8kg of oxygen are required. For every 1k of carbon, 16*2/12 = 2.7 kg of oxygen are required. So you see the oxygen soon mounts up to a significant portion of the mass of B. Edited March 28, 2014 by studiot Link to comment Share on other sites More sharing options...
dr2much Posted March 28, 2014 Author Share Posted March 28, 2014 Studiot, thank you for your reply. However, doesn't the calorific value, being the heat released when a substance is combusted, already take the contribution of the reaction with oxygen (including the energy released when breaking the chemical bond of the oxygen atoms in an oxygen molecule) into account? Therefore, my question remains whether based on the calorific values and in this case electric energy is it possible that the energy output of this machine seems higher than the energy input? How to account for the positive difference? Link to comment Share on other sites More sharing options...
studiot Posted March 28, 2014 Share Posted March 28, 2014 (edited) Read this part of my post and what follows again. That is you need to take the mass of oxygen into account in your reckoning. The product B is composed of some mass from A plus the mass of the oxygen needed for combustion. For example burning 1kg of A produces between 5 and 15 kg of B depending upon the fuel. Edited March 28, 2014 by studiot Link to comment Share on other sites More sharing options...
imatfaal Posted March 28, 2014 Share Posted March 28, 2014 (edited) ...1 kg of product A yields 1 kg of product B and C kWh of electricity.... This is the line that is causing consternation above - not the energy balance Let's assume a machine with substance A as the fuel. The output of the machine is electricity plus a product B. 1 kg of product A yields 1 kg of product B and C kWh of electricity. If the calorific value of product A is for example 20 MJ/kg and of product B it is 15 MJ/kg The electric output C of the machine from 1 kg of fuel A is 2,777 KWh. 1 kg A ---> 1 kg B + 2,777 kWh electricity A simple attempt to do an energy balance based on the calorific values gives the following: 20 MJ ----> 15 MJ + 2,7777 kWh 20 MJ ----> 15 MJ + 10 MJ ... Your conversions are also wrong. A watt is one joule per second - work though the maths and you will realise your conversions are askew Always convert all units to the same thing straight away - unless you have specific trick in mind it will avoid confusion and you can spot errors early on. That is unless you are using a comma to signify the decimal point - which is possible and would explain why my sums are nowhere near yours. I will double check. Ok so you are using comma as decimal point. Your sums and conversions are correct. The mass statement is still confusing and almost certainly wrong. The energy balance is wrong - even with impossible 100pct efficiency you do get to input a kilogram of 20MJ/kilo fuel and get in the end 25 MJ of energy out. Edited March 28, 2014 by imatfaal realised the confusion Link to comment Share on other sites More sharing options...
dr2much Posted March 28, 2014 Author Share Posted March 28, 2014 @ studiot and imatfaal Yes, the line: 1 kg of product A yields 1 kg of product B and C kWh of electricity. is important as to the mass argument. My confusion remains the energy. Maybe I should rephrase the questions as follows: Can a system have more energy output than input when using mainly calorific values as the measure of energy? What should I factor in, to obtain a balance in the energy input and output? Link to comment Share on other sites More sharing options...
studiot Posted March 28, 2014 Share Posted March 28, 2014 2.7777 x 3600 kwhrs = 10 thousand kJ = 10 MJ I am taking the comma as a continental decimal point. But, dr2much The products of combustion always weigh more than the weight of the fuel. Link to comment Share on other sites More sharing options...
imatfaal Posted March 28, 2014 Share Posted March 28, 2014 @ studiot and imatfaal Yes, the line: 1 kg of product A yields 1 kg of product B and C kWh of electricity. is important as to the mass argument. My confusion remains the energy. Maybe I should rephrase the questions as follows: Can a system have more energy output than input when using mainly calorific values as the measure of energy? What should I factor in, to obtain a balance in the energy input and output? "Can a system have more energy output than input when using mainly calorific values as the measure of energy?" Of course it cannot - if it could our energy problems would be solved. The mass equation is wrong - as you have been told a few times now - that is giving you an incorrect energy balance. I would suggest that it is more likely that you get significantly less thatn 1kg of product that is usable at 15MJ/kg plus waste products that have no signficant value to you (together these must weigh more than 1kg) Link to comment Share on other sites More sharing options...
dr2much Posted March 28, 2014 Author Share Posted March 28, 2014 Your sums and conversions are correct. The mass statement is still confusing and almost certainly wrong. The energy balance is wrong - even with impossible 100pct efficiency you do get to input a kilogram of 20MJ/kilo fuel and get in the end 25 MJ of energy out. Can 2.7777 x 3600 kwhrs = 10 thousand kJ = 10 MJ I am taking the comma as a continental decimal point. But, dr2much The products of combustion always weigh more than the weight of the fuel. There is combustion exhaust gas from the machine that, after heat exchange, is released to atmosphere. Although the residual energy in that gas is minimal it will add to the energy output. In view of the expected low energy content I have not included it in my energy equation. Link to comment Share on other sites More sharing options...
studiot Posted March 28, 2014 Share Posted March 28, 2014 (edited) I'm sorry I don't understand you last remark. As an example 1kg of hydrogen burns with 8kg of oxygen to yield 9kg of water plus C joules of energy. The definition of calorific value (of hydrogen) is C joules per kg. There is no escaping this. Edited March 28, 2014 by studiot Link to comment Share on other sites More sharing options...
dr2much Posted March 28, 2014 Author Share Posted March 28, 2014 I'm sorry I don't understand you last remark. As an example 1kg of hydrogen burns with 8kg of oxygen to yield 9kg of water plus C joules of energy. The definition of calorific value (of hydrogen) is C joules per kg. There is no escaping this. I understand that there is no way to change the calorific value of a given substance. However, is calorific value enough to define the amount of energy that can be obtained from that substance. My guess is: yes if you look only at heat generated by burning the substance. What about (non-heat) energy that can be extracted from the change of the substance from a solid state into the gaseous reaction products of that substance and oxygen? If I'm not mistaken that is enthalpy change. Would that be able to balance my energy equation? Link to comment Share on other sites More sharing options...
studiot Posted March 28, 2014 Share Posted March 28, 2014 (edited) What did you not understand about what I posted. I did not say anything about changing the calorific value. imatfaal has already told you (as I have) that your problem is in the belief that burning 1kg of A produces only 1kg of B. That is not true for any substance in the known universe. Perhaps you do not understand that the 1kg of A refers only to the fuel. But the number of kg of B refers to everything that results from the combustion. So in my example A = 1 kg hydrogen B = 9kg of water The burning chemical reaction adds 8 kg of oxygen to the 1kg of hydrogen. But the calorific value is still calculated on the base of the mass of fuel (hydorgen only. I will be out on site for the rest of the afternoon, so perhaps one of the chemists here might make this more plain for you. Edited March 28, 2014 by studiot Link to comment Share on other sites More sharing options...
dr2much Posted March 28, 2014 Author Share Posted March 28, 2014 (edited) studiot thank you. I do understand that burning 1 kg of a substance results in much more than 1 kg of reaction products due to the reaction with oxygen. As you may have seen my post of 11:58 AM above mentions exhaust gases that I left out of my energy question, but which of course are part of the mass of reaction products versus reactant A. In terms of reaction products that have a (significant) calorific value product B is the only worth mentioning. It is correct that 1 kg product A put into this machine as a fuel gives 1 kg of product B (plus the exhaust gases as mentioned above). So, confirming that I do understand the role of mass and that I have factored that in properly, I come back to my equation: 20 MJ ---> 15 MJ + 10 MJ using the calorific values of product A and product B as the only measures of energy in and out (apart from the 10 MJ electricity) Can the difference be accounted for by enthalpy changes? The concept of enthalpy is one that many laymen like me struggle with, so you're replies/views will be appreciated. Edited March 28, 2014 by dr2much Link to comment Share on other sites More sharing options...
studiot Posted March 28, 2014 Share Posted March 28, 2014 I can only suggest you re read your own thread title, where you asked "How to evaluate.....?" I am trying to tell you how to do just that in accordance with currently accepted definitions; neither you nor I have the power to change these at whim. Your equation is not an equation at since since the left hand side does not, by your own words, equal the right hand side. I live 150 miles from imatfaal. What do you think he would say if I said to him? "just pop over on your bike, it's only 10 miles - the other 140 can be discounted because they are boring and of no consequence." 1 Link to comment Share on other sites More sharing options...
imatfaal Posted March 28, 2014 Share Posted March 28, 2014 I... "just pop over on your bike, it's only 10 miles - the other 140 can be discounted because they are boring and of no consequence." I might try that as motivation - only the last ten miles matter the rest are of no consequence. Link to comment Share on other sites More sharing options...
dr2much Posted March 28, 2014 Author Share Posted March 28, 2014 studiot and imatfaal thank you both for your replies and your patience. Link to comment Share on other sites More sharing options...
studiot Posted March 28, 2014 Share Posted March 28, 2014 (edited) You need to realise that the calorific value is the total amount of heat obtainable by complete burning of the fuel until no burnable material is left. Thus there can be no product B that has any further calorific value. For instance the complete combustion of carbon results in carbon dioxide, which is incombustible. No further heat can be obtained from the dioxide. Partial combustion results in carbon monoxide, which can be burned further to carbon diode with the release of more heat. The total heat avilable is the same whether you do the combustion completely in one stage to dioxide or in two, first to monoxide and then to dioxide. This heat is the calorific value of carbon. Enthalpy is a different concept entirely, although it is often measured in a calorimeter, and takes into account other forms of energy change such as mechanical work. So what would you like to do now? Edited March 29, 2014 by studiot Link to comment Share on other sites More sharing options...
dr2much Posted March 29, 2014 Author Share Posted March 29, 2014 I'll have to go back and see what else is going into the machine in addition to the fuel, product A. I'm certain that product B is coming out of the machine. Once I know what else is going in I will post that here in this thread. Link to comment Share on other sites More sharing options...
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