# Homework Help: Finding a dimensionless time in a very basic problem

1. Sep 11, 2012

### tomgill

1. The problem statement, all variables and given/known data

In the chemical reaction problem, assume that the reaction is C + C --> Product, and the chemical reaction rate is $r=kc^{2}$, where k is the reaction constant. What is the dimension of k? Define dimensionless variables and reformulate the problem in dimensionless form. Solve the dimensionless problem to determine the concentration.

2. Relevant equations
We are assuming temperature is not relevant.

The mathematical model given by the initial value of the problem is as follows.

$\frac{dc}{dt} = \frac{q}{V}(c_{i}-c)-k^{2}c , t>0$

In this problem, c is the concentration, t is time, q is the inflow rate (mass/time), V is the volume, and k is the reaction constant.

3. The attempt at a solution

First, I equalized the dimensions on both sides to get that $k=\frac{V}{MT}$ . This part seems simple as the dimensions work now.

However, I am being asked to dimensionalize the problem. Finding a dimensionless concentration is easy, $C=\frac{c}{c_{i}}$. I am trying to find a dimensionless time. I have gotten as far as $\tau=\LARGE{\frac{t}{\frac{1}{c_{0}k}}}$, since $c_{0} * k$ has units of $c_{0}=\frac{M}{V}$, and $k=\frac{V}{MT}$ This much is easy. However I look in the solution manual. It is telling me that the dimensionless time $\tau=\LARGE{\frac{t}{\frac{V}{Q}}}$. When I work out the units, I get $t$ on the top, but the bottom is $\LARGE{\frac{V}{Q}=\frac{V}{\frac{M}{T}}=T\frac{M}{V}}$. But $\frac{M}{V}$ is not dimensionless, so $\tau=\frac{M}{V}$. Can anyone figure this out? Seems beyond obvious and ridiculously simple of a problem, is my solution manual blatantly wrong or am I just completely clueless?
Thank you!!

2. Sep 12, 2012

### clamtrox

Unless you left out some constants, the units don't match up. LHS has units of c/time so to match it, q has to be "volume inflow rate" or something like that, with units volume/time.