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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
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/ft^{3} 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/lb_{m})(1 kg/1000 g)(35.3145 ft^{3}/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
where
What are the units of h, knowing
that h/(C_{P}G) is unitless?
Answer
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 [=] C_{P}G [=] [Btu/(lb_{m}F)][lb_{m}/(hr*ft^{2})] [=] Btu/(hr*ft^{2}F)
Example 3: Changing
an equation to another set of units. What would the equation in
Example
2 look like in another system of units?
Answer
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.
Answer
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 [=] e^{x
}s^{y
}m^{z}
Step2: Substitute in all of the units
e [=] ergs s
[=] cm m [=] g
D [=] cm^{2}/s
D [=] cm^{2}/s [=] e^{x }s^{y }m^{z} [=] (g*cm^{2}/s^{2})^{x }(cm)^{y }(g)^{z} [=] g^{x+z }cm^{2x+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.