Materials Balance | Mass Balance |law of conservation of mass

As we know that Material and Energy

Balance required  in various processes. as we know that materials and energy

balance are the routine plant exercise.

         Material balance is depends on law of conservation of mass.you must know what is law of conservation of mass.
It is nothing new that I have to tell you,

because every businessman or every shopkeeper or every plant owner every

banker every person, have to do auditing or checking of material and energy

from the point of view ‎of the economics.both terms are very important from

economics point of view.He would like to know, what the inputs are and what

are the outputs of material and where are the losses. He would also like

to know, what is the energy in into the system and what the energy out is;

so that he can assess the energy consumption because energy consumption

acts directly to the cost of the product. He would like to introduce measures

like saving of energy or utilization of energy, if happens to know how the

energy is being utilized. Let us see, first of all, what are the basics of materials

and energy balance.Basic is most important factor in material balance .

I mean it is nothing very great, but it is very simple. without basic we can not

do material and energy balance . we can explain this with a very simple example ,

We know that you also do a sort of balance,whenever you go to the market

and do some expenditure. You are given 50 rupees.When you come back,

you will like to know how you have spent 50 rupees. So, you keep an

account of int and out. So, if it is possible, you can introduce the measure of.

Now, similar to it; here also, you have to think in terms of inputs and output

of material, energy. Now, in fact, the basic is – it is based on law of conversation

of mass. Now, mass of an isolated system remains constant, irrespective

of the changes occurring within the system. Now, let us consider an open system.

                         For example, consider an open system at this point, which is

point 1 and this point 2. Here, we have in and at point 2 we have out. Now,

let us consider the volume as V’, mass is transferred in and out of the system.

Mass will accumulate, when the input and output rates are unequal. So, in

this case, we can write down rate of mass in is equal to rate of mass out

 plus rate of rate of accumulation of mass. Now, writing down in terms

of mathematical expression – m is the mass, dm/ dt = m in – m out. Dot

represents the rate of change of mass with reference to time. So, this is

in fact valid for unsteady state operation. Now, this material balance

equation is written for each and every component of this system. let us

 consider in a system, there are 10 different elements entering or five

 different elements entering and 5 different elements or 10 different

elements leaving the system. Then we have to write down the mass

balance for each and every component that is entering in.

let us consider one example for it  example, if we have component 1 ,

dm1 /dt = m 1 in – m 1 out.

 Similarly, for component 2,

dm2 /dt = m 2 in – m 2 out.

for third component ,

dm3 /dt = m3 in – m3  .

now  for i th component that becomes,

dm / dt il = m’ in  -m’ out

           Now, for a system with multiple inputs and output streams, the material balance equation for i th component becomes dm/dt=sigma m i in all components minus sigma m dot out of all components.

 First law is law of definite proportions. Now, a given chemical compound always contains the same constitutional elements in the same weight proportions. This is only true for stoichiometric compounds and for non-stoichiometric compounds, this may not hold good.

               Now we can discus with an a example,   FeO has Fe one and O one atom and it is also available as Fe 0.95OSo, in that case this law of definite proportion will not be valid. So, u will just write down a given chemic compound always contains the same constitutional elements in the same weight proportions. For example, in Fe2O3, there will be 2 atoms of iron combining with 3catoms of oxygen. For example, in Cu2S, there are 2 atoms of copper combining with 1 atom of sulfur. However, copper and sulfur are the same constitutional elements, but CuS is also formed. So, in that case, it not be valid and it means, it is true for stoichio compounds.

                     For thos compound which are  non- stoichiometic compound, you have to find out in what proportion the elements are combined. another  law is nothing but law of multiple proportions. it clearly say that, if two elements can form more than 1 compound, then the respective weights of 1 element that combines with a given weight of the other are in the ratio of a small whole number. I think, I should write it. If two elements can form more than one, then the respective weights respective of one element combines with other in the ratio of small whole numbers. So, you will very often come across the various compounds, stoichiometic as well non-stoichiometic in nature. In case of stoichiometic compounds, the combination of elements is clear, but in case of non- stoichiometic compound, you must know in what proportion the elements are combined. So, we can perform the material balance.

               Now, I mean this is basic and I think all of you are aware that in = out, when there is a steady state and this is a very simple thing. Now, about the energy balance; one important thing for energy balance is energy balance cannot be done without material balance; whenever,you are required to solve a problem of energy balance without performing material balance energy, balance cannot be done. So that is one important thing. Now, here are some tips. For example, one can perform elemental balance either in kg or kg mole. Now, it is a matter of practice and matter of convince about how you perform the material balance.
You can perform in kg or gram or in gram mole or in kg mole also. If you ask me, I will prefer or I will do the material balance by considering kg mole because I find it very convenient by determining the input and output of masses in terms of kmole because later on the thermodynamic values, which I will be getting in the literature are given in per kg mole and they are also given in per kg. So, what I wanted to say is that you have to develop your own habit, whether kg is suitable for you or kg mole is suitable for you. The thermodynamic values are available in kg, kg mole, gram and as well as in gram mole and there is no problem at all. It is just a matter of convenience; the material balance in kg mole becomes little bit easier. Say, if you write down the stoichiometric equations, then immediately, it is apparent how many moles are entering in and how many moles are entering out. So, I find it little easier, but you have to make your own style of solving these problems. Once you have done the material balance of all inputs, it is very important to know where the input of materials is going in. Say, you have y kg or y kg mole of certain material and you have to see how much of that is going into the products.

The products could be metal, it could be slag, waste product or in the gaseous form. So, to collect all such information, make a box to determine material balance and then proceed to solve the heat balance and that is very important. Another important thing for the energy balance is to make the basis of calculation.

            we can go through a example, 1 kg mole and then you follow that kg mole until the end of your heat balance or you can do 100 kg mole or 10 kg mole, whatever is convenient to you, but do not change the basis till the end of the problem number. Number 2: the several energy balance is done by considering a reference temperature of 25 DegC . Now, this is an advantage because the values or the thermodynamic values like specific heat content, heat of reaction, heat of formation are all available at 25 DegC .  Now, once you have done this and then perform heat balance. Now, needless to mention heat input is equal to heat output because we are talking of a steady state. We are not doing  heat balance under unsteady state,

further we can discuss in next blog , to see in next blog , visit our next post , or give comment so we will give you a link.

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