Dimensional Analysis 2

DJS 20130530

Following on directly from the previous page......

Of course, my list of dimensions is not complete: we could or should add Q for electric charge, Θ for absolute temperature, N for amount of substance and J for luminous intensity, according to the SI standard. There is some dispute, understandably, since some of this later set can be expressed in the others.

Notice, please, that while you may have units in mind for these quantities, no particular units are required by the method.

Students who have read about Groups (next in the list as I write) may recognise that each of the MLT dimensions form a group and that we have a vector space where (i,j,k) corresponds with MiLjTk. Generally we assume that i,j,k are integers. There are consequences to this observation, largely built around recognising the space of possible combinations of the elements of a problem, among which include writing all the equivalent (‘commensurate’) equations.

Note also that in Mechanics (the MLT field) we could argue that the choice of those three as base dimensions is arbitrary and we could make force a base dimension, T=(ML/F)1/2.


Exercise:

1. Show that Length, Velocity and Time do not span the space, but that momentum length and time do. Discover whether force and momentum and one other (conceptual unit) will span the mechanics space.

You might want to look at Affine Spaces.                                             DJS 20130530


2.  Consider the speed of water waves and assume they are affected by gravity, depth of water and wavelength. Clearly, since mass is excluded, we can only solve for two unknowns. There is a distinction between waves in shallow water and waves in deep water. We might assume that depth affects shallow waves but not deep water waves (or that any effect in minimised). Experimentation shows that once the depth exceeds half the wavelength, we should treat the waves as being ‘deep’. Show that v²=gh or gλ, depending upon the case.


3.  Consider circular-section pipes and the problems of pushing water through them. The possible factors to include are:  the pressure drop (Δp), pipe length (l), pipe diameter (D), fluid velocity (V), fluid density (ρ), fluid viscosity (μ) and pipe surface roughness (ε). Using the MLT dimension set, these have dimensions as follows:
pressure drop (Δp),                           ML⁻¹ T⁻²
pipe length (l), pipe diameter (D), pipe surface roughness (ε).                  all    L
fluid velocity (V),                               LT
⁻¹
fluid density (ρ),                                ML⁻³
fluid viscosity (μ)                               M L⁻¹T⁻¹

We could do experiments by choosing a particular pipe size, a particular density and specifying a target velocity. Call these the repeating parameters. So the other four can be mixed to form dimensionless factors, each mixing with the three chosen: I’m going to call these capital Pi for reasons that are to do with academic convention…

Π1 = Δp.Da1.Vb1.ρc1
Π2 = l.Da2.Vb2.ρc2
Π3 = μ.Da3.Vb3.ρc3
Π4 = ε.Da4.Vb4.ρc4

Π1 = Δp.Da1 Vb1 ρc1 = [ML⁻¹T⁻²] [L]a1 [LT⁻¹]b1 [ML⁻³]c1 = M1+ c1 L-1 + a1 + b1 - 3c1 T-2 - b1

This is supposed to be dimensionless, of index equal to zero, so from the M component, c1 = -1, from the T component, b1=-2 and thence a1=0, so Π1 =Δp/ρV².  
Show (easy) that Π
2 = l/D; that Π4 = ε/D and, slightly less easy, that Π3 = μ/ρDV.   We often choose to write this inverted, as ρDV/μ, which is the Reynolds number (Re). The Reynolds number is an example of a dimensionless constant made from a mix of the constituent elements (modelling factors). Then we start being clever, by making Π1 as a function of the others. For example, the pressure drop for flow in a circular pipe is given by 
         Δp/ρV² = f (l/D, ε/D, Re).

4.   Suppose we are worried about the force of wind upon a building.  Take as variables the height h, width w, and the fluid density, viscosity and velocity as in the last example. Choose height velocity and viscosity to be the repeating variables and there will be two mixes to find:
Π1 = w ha1 vb1 μ c1
Π2 = ρ ha2 vb2 μc2

Solve these to show that Π1 = w/h and Π2 = ρhv/μ.    Since both of these are dimensionless, one can be expressed as a function of the other, so perhaps w = h f(ρhv/μ). The relationship expressed in Π1 shows also that a scale model is appropriate.


Some feedback would be appreciated.


DJS 20130531


Additional Notes

1.   I noticed that there is a bit of fuss over economic modelling as lacking dimensional consistency. I’d like to think I saw that immediately.

2.      There’s an article to write on Natural Units, starting from the idea of Action, S, whose dimension is energy.time (unit is joule.second). An original definition was the integral of a particle’s speed (w.r.t. time) along its path. I found the integral idea easy, but the evaluation of the integrand when beyond one dimension remains incomprehensible to me.

Using action as S, and with energy E and velocity v, we can write M=E/v², L=Sv/E and T=S/E. To do this we need to adjust our units so that speed and action are dimensionless; we do this by measuring speed in terms of light speed c and action in terms of the Planck constant h. All this means that we could express all our units in terms of energy.
 So dim(S) = dim(v) = 0;      dim(E) = dim(M) = 1;      dim(T) = dim(L) = -1
Simplest way I see to show this: ∫v dt = ∫dv = v; E=Mv² so dim(M) = dim(E/v²) = dim (E).
v=L/T so dim(T) = dim(L/v) = dim(L).  
 dim(S) = dim(v) = 0 by definition. Clearly dim(E,M) = - dim(T,L).  Choose to measure in terms of energy, i.e that dim(E) is unity, so dim (T,L) is negative.
I don’t know if we have a decision for the appropriate energy unit, but I suppose it depends upon context. I suspect that such units are found to be useful in what the rest of us would call extreme situations, where I’m thinking cosmology and quantum physics.

I note that dim(Eⁿ) = dim(MpLqTr) = dim(Ep-q-r)  is an obvious result.



Answers: 

1  (a) {L,V,T} couldn’t span mechanics, since V=L/T so M is eliminated.
b)  Momentum (Mv) = ML/T, so we could eliminate any one of ML&T     So it spans.
(c)  F=MLT⁻² and  Mv= MLT⁻¹ so we could replace all T as Mv/F, but not either of M&L as F&Mv are dimensionally adjacent (just a factor of T different)—there’d be more success with momentum and viscosity (but it’d be a pretty silly thing to do).

The other questions have the answers included, I think.

 

 However, © David Scoins 2017