Chapter1, Lesson B - Dimensionless Groups and Equations

1B-10	Dimensionless Groups and Equations						5 pts

The following dimensionless equation correlates the values of variables in a system in which solid particles are suspended in a flowing gas.

							Eqn 1

Both of the terms in parentheses are dimensionless groups. kg is a coefficient that expresses the rate at which a particular chemical is transferred from the gas to the surface of the solid particles. The value of kg is needed to design a catalytic reactor. Since this coefficient is difficult to determine directly, values of the other variables are measured and kg is calculated from the given equation.
The following values were measured :

dp	5.00	mm		μ	1.00E-05	N-s/m2
y	0.100	(dimensionless)		ρ	1.00E-03	g/cm3
D	0.100	cm2/s		v	10.0	m/s

What is the estimated value of kg ? What are the units of kg ? Show your work.

Read :	The key here is that the equation and the groups in parentheses are dimensionless.
	This tells us that the constants 2.00 and 0.600 are also dimensionless.
	So, we can plug numbers and units into the right-hand side of the equation to determine the value of the dimensionless group on the left-hand side. Then, we use the values and units of dp, y and D to determine the value and units of kg.


	It is probably wise to begin the problem by converting all of the given values to one consistent system of units. I chose the MKS system.


Given :	dp	5.00	mm	μ	1.00E-05	N-s/m2
	y	0.100	(dimensionless)	ρ	1.00E-03	g/cm3
	D	0.100	cm2/s	v	10.0	m/s

Solution :	Once all the given values are in a consistent set of units, we can evaluate each of the dimensionless groups in Eqn 1.


			5000		1.00

	Let's double check the units on these groups using the MKS units.

							Eqn 2

							Eqn 3

	All the units cancel, so we can conclude that both of these groups are indeed dimensionless.

				44.426	(dimensionless)

	First, let's determine the units for kg.

							Eqn 4

			0.889

Answer :	kg	0.889	m/s
		88.9	cm/s

Example Problem with Complete Solution