Dimensional Consistency 

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All Equations Must Be Dimensionally Consistent

Units are also important to chemical engineers as computational tools.  Because all equations must be dimensionally consistent, engineers can exploit that constraint to put things in different units and to find the units of variables.  By dimensionally consistent, we mean that an equality, signified by the equals sign, requires not only that the value be identical but that the units be the same on both sides of the equation.  Perhaps we can best illustrate the power of using dimensional homogeneity as an engineering computational tool by giving four examples of its use.

Example 1 - Using dimensional consistency
Example 2 - Engineers use dimensionless groups
Example 3 - Changing an equation to another set of units
Example 4 - Dimensional analysis

Example 1:  Using dimensional consistency
Suppose that it is known that the composition C varies with time t in the following manner:

        C = 0.03 exp(-2.00t)

where C [=] (has units of) kg/L and t [=] s.  What are the units associated with 0.03 and 2.00?

Answer:  Because the argument of an exponential function must be dimensionless, 2.00t must be unitless.  Therefore 2.00 must have units of s-1.

Now suppose that during a particular experiment the data are obtained in units of C [=] lbm/ft3 and t [=] hr.  Change the above equation so that the data can be directly plugged into it in these units.

Answer:  What we will do is write a new expression for the variables C and t in the previous units in terms of new variables C and t in the new units. Thus,

        t[s] = t [hr]·(3600 s/hr) = 3600·t

        C[kg/L] = C[lbm/ft3]·(453.59 g/lbm)(1 kg/1000 g)(35.3145 ft3/1000 L) = 0.0160·C
  So in terms of the new variables (the red ones) we have 0.0160C = 0.03 exp[-(2.0)(3600)t], or finally,

C = 1.9 exp(-7200t)

Example 2: Engineers use dimensionless groups


What are the units of h, knowing that h/(CPG) is unitless?
Note that you do not need to work through all of the equations to get the units of h.  Since each individual group must be unitless or dimensionless, all that we nee to do is work from a single group containing h.  Thus,

 h [=] CPG [=] [Btu/(lbmF)][lbm/(hr*ft2)] [=] Btu/(hr*ft2F)

Example 3: Changing an equation to another set of units. What would the equation in Example 2 look like in another system of units?
The equation and number are identical because within each dimensionless group, all conversion factors must cancel out. Frequently, engineers deal in dimensionless groups because the equations derived then apply to any system of units. It is therefore convenient to be able to express variables as dimensionless variables. To do so, we use a technique called dimensional analysis.

Example 4: Dimensional Analysis.  Find the dependence of the diffusion coefficient on the molecular size s, the intermolecular attraction e, and the molecular mass m of the particles.
Dimensional analysis is a very powerful tool to develop relationships between properties if the independent variables are known.  For example, in this case, we know that the only independent variables are the molecular parameters m, e, and s.  We know that whatever correlation we use must be dimensionally consistent, so what we do is a three-step process.  (1) First we write the dependent variable as a product of the independent variables raised to an unknown power.  (2) Next, we substitute in the units for each of the independent variables and collect terms of like units.  (3)  Then we match the power of each separate unit on both sides of the equation in order to maintain dimensional consistency.  (4) Finally, we solve for the unknown powers.  This is illustrated in the example.

Step 1:  Write dependent variable as product of independent variables raised to unique powers
 D = D(e, s, m)        or       D [=]  ex sy mz

Step2: Substitute in all of the units
e [=] ergs     s [=] cm         m [=] g        D [=] cm2/s

 D [=] cm2/s [=] ex sy mz [=] (g*cm2/s2)x (cm)y (g)z [=] gx+z cm2x+y s-2x

Step 3: Match powers on each unit on both sides of the equation
 g:   x + z = 0

cm:   2x + y = 2

s:   -2x = -1

Step 4: Solve for the unknown powers
Solving for x, y, and z gives:   x = 1/2,     y = 1,  and     z = -1/2.

Thus, D [=] D[s(e/m)0.5].

Without knowing any theory about the diffusion coefficient, we can see that D depends on the molecular diameter s linearly. Thus, for molecules with twice the diameter, the diffusion coefficient will be twice as large. Similarly, all other things being equal, a molecule that is four times as massive will have half the value of D.  These are not the only properties that will affect the diffusion coefficient, but if the others are all constant, this is the way the molecular properties s, e and m would affect D.

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