[How To] Perform Energy Balance

Hiiii everyone.!!

Hope everyone is fine, this post is about Energy Balance which is more basic.

Usually energy balance is used to calculate the unknown parameters such as time, heat loads, utility line sizes, quantities etc.

The basis for energy balance is mass balance.

Energy balance will vary from equipment to equipment and process to process based on the in and out streams. for example the energy balance for a reactor can be explained in one way during reaction and during distillation in other. But the basis will remain same and the most important thing is that we have to stick to basics while performing these calculations.




There wont be anything difficult in these but just need to look with common sense.

Let me explain you with a simple example of energy balance for a distillation operation. Distillation is being performed in a 5 KL reactor and the reaction mass is having ~3 KL toluene and the utility used in reactor jacket is steam pressure of 1.5 bar.

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Energy Balance for Reactor :

Reaction mass in + Hot water in = Distillate out + Hot water out

M  =  C + D - Material Balance [In this case, i've considered pure solvent in reactor],

C is concentrate, D is distillate.

Coming to energy balance, the energy supplied in form of steam will be used to evaporate the toluene. Hence, the equation will be 

( S x Hs ) = ( D x Hd ) + ( S x Hc ),

S x (Hs - Hc) = ( D x Hd ),

S    - Qty. of steam (Kgs),
Hs - Steam enthalphy (Kcal),
Hd - Condensate enthalphy (Kcal),
D    - Qty. of distillate (Kgs),
Hd - Enthalphy of toluene (Heat of vaporation) 




So now we made the energy balance equation for distillation reactor and the equation will suffice two things. 
One is to calculate the steam required to evaporate known quantity of toluene, 
Second is to calculate the amount of toluene that can be evaporated with known amount of steam.

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Case-1: Steam required for evaporating 400 Kgs toluene

S x (2717.13 - 536.27) / 4.178 = ( 400 x 86.8 );

S = ( 400 x 86.8 x 4.178) / (2717.13 - 536.27) = 66.51 Kgs

For the values of steam enthalphy and condensate enthalphy at 1.50 bar pressure, 
refer this link: Steam Tables 





4.178 is the conversion factor from joules to calories, 86.8 is latent heat for toluene in KCal.

Case-2: Amount of toluene to be evaporated with 20 Kgs of steam.

( 20 x 540 ) = ( D x 86.8 ),

D = 20 x 540 / 86.8 = 124.42 Kgs.

Here we got the 2nd thing i.e., amount of toluene that can be vaporated with 20 Kgs od steam.

Note: I haven't considered sensible heat.


Energy Balance for Condenser:

Now lets take the case of condenser.

Condenser will have 5 nozzles.
1. Vapour in,
2. Condensate out,
3. Vent out,
4. Coolant in,
5. Coolant out.

Lets consider an atmospheric distillation.

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Material Balance :

Vapour in = Condensate out + Vent out

V = C + VO

we will consider above mentioned case only where 3KL toluene is being taken in reactor for distillation.

Vapour will be 868 Kgs of toluene, Utility supplied is RT water with an inlet temperature of 25℃ and say the outlet be at 33℃ with a flowrate of 2 m3/hr.




Now the equation will be like,

( V x ƛ ) + ( C x Cp x dT ) = m x Cp x dT.

So above is the Energy balance equation that we prepared.

Now lets prepare a energy balance equation for reaction step.

Let there be a reaction between A & B chemicals and we have to evaluate the utility supply rate through energy balance. I can distinguish this into two types.
1. with available RC1e study,
2. without RC1e study.

Again in without RC1e case, we can get two more types:
1. Reaction above ambient temperature,
2. Reaction below ambient temperature.

Lets discuss about the case if, 

RC1e study is available:

Heat of reaction as per RC1 report is found to be 483 KJ/Kg of batch size,

Batch size be 500 Kgs. 

So the total energy liberation would be 500 x 483 = 241500 KJ = 57802.77 Kcal.

Now in order to host the reaction safely we need to supply 57802.77 Kcal of energy as a counter part.




That means, Energy liberated during reaction = Energy gained by utility.

i.e., 𝝙Hr = U x A x LMTD.

We got our energy balance equation.

So next case we don't have heat of reaction or RC1e study, we have to calculate tentative energy.

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RC1e Study not available:

The method i select for this task is performing group contribution method / using bond energy calculation.

already in one of the previous post i've discussed about bond energy calculation.
Pl refer the following link: Bond Energy Calculations

Let the reactants be A & B, the quantities of A & B are 30 Kgs & 45 Kgs respectively.

Let the derived energy liberation during the reaction be 58319 KCal.

And the reaction temperature be 0-5℃, we have to slowly dose a reagent in the temperature for completing the reaction.

Consider the reaction as spontaneous and propagates in-line to the charged reagent.





So the energy balance equation would be,

𝝙Hr - M x Cp x dT= U x A x LMTD,

𝝙Hr - heat of reaction from group contribution method,
M - reaction mass weight,
Cp - Specific heat of reaction mass,
dT - operating temperature difference,
U - Overall heat transfer coefficient,
A - Heat transfer area,
LMTD - Log mean temperature difference.

The above equation can be written as,

𝝙Hr x t - m x Cp x dT= U x A x LMTD x t,

Since, M = m / t.

As the heat of reaction, reaction mass, area will vary with time, integrals shall be used.

The equation will look like:

In the above equation you can see that we have just applied integrals on both sides to the varying quantities.





As area is dependent on length of cylinder, we can simplify the integral of area as

Area of cylinder = Area of torispherical dish + Area of cylinder.

Let the volume of reactor is 5 KL and the total heat transfer area of the reactor is 12.68 Sq.m.
out of which the torispherical dish will attribute to 2.64 Sq.m and 
the remaining 10.04 Sq.m will be of cylinder,

tentative dimensions of the reactor cylinder would be 
Length of cylinder - 2.00 m, Dia of cylinder - 1.60 m

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Calculate Overall heat transfer coefficient

Lets say before the addition of reagent B, the volume of the reaction mass is completely filled in torispherical dish and the cylinder filling would start by the start of reagent solution addition.

So the initial length would be zero (0) and lets calculate the end length based on our general calculation. presented below:

Let the reagent quantity be 1000 L, for a cylindrical vessel 1000 L will fill upto a height of

1 = (3.141 / 4) x (1.60 ^ 2) x L,

L = 0.497 m = 0.5 m.

so the final height would be 0.5 m.

So after including all the limits for integrals, the above equation would turn to,

You may get a doubt that how Area transformed to length and from where that d came from.

So for those, usually a cylinder would turn into a rectangle if we cut at one point, below is for your understanding:

Hope you understand from above, what i delivered.

Let the reactor be a SSR, and the overall heat transfer coefficient be 300 KCal/Sq.m.hr.C

and the LMTD = ((5 - (-10)) - (0 - (-2))) / ln((5 - (-10)) / (0 - (-2)) = 14 C.

Now include everything in the above integral equation, it will turn to:

1.2 is the specific heat of reaction mass in KCal/Kg.C

Lets start simplifying the above equation,

That's it......!!!!! This is the first case and from this we can derive the tentative addition time of reagent even if we don't have the heat of reaction from RC1e study.

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Lets consider a case where the reaction(addition of reagent) is at temperature higher than ambient temperature.

That case we need to consider the M x Cp x dT as a positive i.e., the equation will look like,

Hope you understand, what i delivered above.


Still any queries feel free to comment / message me.

Comments are most appreciated.........!!!!



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Finally dear engineers, i would like to add one thing..,
Peak Environmental temperature is raising year by year, and as an engineer we are playing a major role in that imbalance, some of you may not accept this but its true.

Most of the imbalance is due to petrochemical and pharmacuetical industries, where we use the chilled water & chilled brine and liquid nitrogen. The more cooling we consume for process the more heat we are generating. 

However most of the times we cannot stop ourselves in stopping all these but try to reduce something which we can. As most of these can be done during process optimization time.

Apart from these try to reduce the usage of AC's and fuels which will generate more heat than we think. Below is the picture at kuwait:

Kuwait this year, may be ours next year. Preserve it, encourage greenary.

Look at China's 'Forest City' which is first in world:

About The Author

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Hi! I am Ajay Kumar Kalva, Currently serving as the CEO of this site, a tech geek by passion, and a chemical process engineer by profession, i'm interested in writing articles regarding technology, hacking and pharma technology.
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