I would like to ask what is the difference (if any) between normalization and non-dimensionalization. I will try to explain what I have done until now with an equation that I would like to work with and you can make your comments on this specific example. The mass balance is given by: $$\frac{\partial C_{i}}{\partial t}=-u_s\frac{\partial C_{i}}{\partial z}+\rho_b \sum_{k=1}^{N_{reac}} \nu_{j,k} R_{i,k}$$ If we choose as characteristic values the: $$L_o, F_o, T_o, P_o, m_o$$ $$u_o = \frac{F_o Rg T_o}{P_o L_o^2}$$ $$C_o = \frac{P_o}{Rg T_o}$$ length, molar flowrate, temperature, pressure, mass accordingly, and by substituting then we get: $$\frac{u_o C_o}{L_o}\frac{\partial C^*_{i}}{\partial t^*}=-\frac{u_o C_o}{L_o}u_s^*\frac{\partial C^*_{i}}{\partial z^*}+\frac{m_o}{L_o^3}\rho_b^* \frac{F_o}{m_o}\sum_{k=1}^{N_{reac}} \nu_{j,k} R^*_{i,k}$$ where $^*$ denotes the dimensionless (?normalized?) variables. After some more calculations we get: $$\frac{\partial C^*_{i}}{\partial t^*}=-u_s^*\frac{\partial C^*_{i}}{\partial z^*}+\frac{F_o}{L_o^2 u_o C_o}\rho_b^* \sum_{k=1}^{N_{reac}} \nu_{j,k} R^*_{i,k}$$ BUT $$\frac{F_o}{L_o^2 u_o C_o}=\frac{F_o}{L_o^2 \frac{F_o Rg T_o}{P_o L_o^2} \frac{P_o}{Rg T_o}}=1$$ So is this equation normalized? or non-dimensinalized? Shouldnt there be some Peclet / Reynolds etc number to work with? If for example I try to change the inlet conditions then this number ($\frac{F_o}{L_o^2 u_o C_o}$) will still be equal to 1. Can anyone point out a book or a site where I could read and understand more about this subject?

Thanks in advance!

  • $\begingroup$ I'm not sure what all your units mean. first equation, right side fuirst term looks like difusion? and so on. $\endgroup$
    – mart
    Apr 8, 2016 at 11:19
  • $\begingroup$ The term on the left hand side is accumulation, the first (left) term on the right hand side is for advection and the last (right) term in the right hand side is for reaction. There is no diffusion term in this equation. Also $C=mol/m^3$, $t=s$, $z=m$, $u=m/s$, $\rho_b=kg_{cat}/m^3$, $R_{i,j}=mol/(kg_{cat} s)$, $F=mol/s$ $\endgroup$
    – ASK22
    Apr 8, 2016 at 12:37

2 Answers 2


Sorry with not being familiar with all of your variables. If it is a mass balance as you say, then you can't get a Reynolds or Péclet number. A Reynolds number appears in the dimensional analysis of the impulse balance (the Reynolds number contains viscosity which originates from the stress tensor for Newtonian fluids). I'm also not very familiar with Péclet, but as it contains thermal diffusivity I suppose it appears in the dimensional analysis of the energy equation.

From my understanding, non-dimensionalizing is a special case of normalizing. In dimensional analysis you normalize a variable using a characteristic value of this variable (to receive equations with unitless variables). The idea behind dimensional analysis is that relationships in physics should not depend on your measurement units. This is used by measuring quantities in multiples of the respective characteristic value picked.

And sorry, I could only point you to German books on dimensional analysis. Good books on fluid mechanics usually also have an introduction to dimensional analysis. (I prefer Schade, Strömungslehre, 2013, e-ISBN 978-3-11-029223-7.) an American author book well know as Cengle, Fluid Dynamics.

  • $\begingroup$ I think that, referring to Reynolds number, you are right but there is a Peclet definition for mass where Pe=uL/D (but still i do not have diffusivity, so also correct about Pe). However, shouldnt there be a dimensionless number where I could change this number only and plot e.g. Ci vs {dimensionless number}?? $\endgroup$
    – ASK22
    Apr 7, 2016 at 13:04
  • $\begingroup$ @ASK22: Re-formulating your equation using the material derivative yields $DC_i/Dt=\rho\sum_k \nu_{j,k}R_{i,k}$. The statement of this equation is quite simple: change in concentration linearly depends on density and reaction rate. You have 4 variables ($C_i$, $t$, $\rho$, $R_{i,k}$) and 4 basic units (mol, m, s, kg). There is no degree of freedom left for a function here. (I suppose $\nu$ are the stochiometric coefficients, so they are given numbers only.) $\endgroup$
    – Robin
    Apr 11, 2016 at 13:25
  • $\begingroup$ I think normalizing is a special case of non-dimensionalizing. See answer below. $\endgroup$ May 25, 2016 at 15:05
  • $\begingroup$ @sturgman I agree that "normalization" is missing a definition here. As non-native speaker I translated to German "normieren", which means scaling with a normed value. Wrt you comment, unitless quantities can be scaled (and normalized), but cannot be non-dimensionalized (since they have no physical dimension or base unit). So to me normalizing (= "scaling plus") seems the more general operation, as it is applicable to a wider range of problems. $\endgroup$
    – Robin
    May 27, 2016 at 8:28

When you want to work with non-dimensional variables you should also look at the boundary and initial conditions of the problem. That way you can check whether non-dimensional groups appear in the boundary conditions as well.

That say, to make an equation non-dimensional means to replace dimensional variables like $C$ with non-dimensional ones like $\Theta=\frac{C}{C_0}$. Now, when making the equation non-dimensional you may choose your quantity $C_0$ to be anything at all as long as it has the same units as $C$. There are many ways to make an equation non-dimensional.

When you normalize or scale an equation (I am assuming scaling in my vocabulary is what you mean), you pick $C_0$ so that the new non-dimensional variable has a domain from 0 to 1. This is often accomplished in mass transport problems by also subtracting by a reference value: $\Theta = \frac{C-C_R}{C_0-C_R}$. You decide on the variables $C_0$ and $C_R$ so that your new variable $\Theta$ stays in a range $0 \leq \Theta \leq 1$.

Scaling (or normalizing) has the advantage that it is more likely to reduce the number of parameters in your problem. In addition, it has the advantage of making your derivatives also scaled (hopefully, and with some care). That means that your derivatives will also have order of magnitude one. If this is accomplished (sometimes not trivial), any non-dimensional groups can be probed to see if they are much larger or much smaller than one to verify their importance to the problem.

A good source for information on non-dimensional and scale techniques to solve transport problems is Analysis of Transport Phenomena by Deen.


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