Finding a dimensionless time in a very basic problem

  • Thread starter tomgill
  • Start date
  • Tags
    Time
In summary: RHS has units of c/time too, so (c_i - c) is dimensionless. k has units of 1/time. The end.In summary, the conversation discusses a chemical reaction problem with a given reaction rate and asks for the dimension of the reaction constant, as well as the use of dimensionless variables and reformulation of the problem. The solution involves equalizing the dimensions and finding a dimensionless concentration and time. However, there is a discrepancy in the solution manual's suggested dimensionless time, leading to confusion and uncertainty about the solution.
  • #1
tomgill
2
0

Homework Statement



In the chemical reaction problem, assume that the reaction is C + C --> Product, and the chemical reaction rate is [itex]r=kc^{2}[/itex], 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.

Homework Equations


We are assuming temperature is not relevant.

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

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

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.


The Attempt at a Solution



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

However, I am being asked to dimensionalize the problem. Finding a dimensionless concentration is easy, [itex]C=\frac{c}{c_{i}}[/itex]. I am trying to find a dimensionless time. I have gotten as far as [itex]\tau=\LARGE{\frac{t}{\frac{1}{c_{0}k}}}[/itex], since [itex]c_{0} * k[/itex] has units of [itex]c_{0}=\frac{M}{V}[/itex], and [itex]k=\frac{V}{MT}[/itex] This much is easy. However I look in the solution manual. It is telling me that the dimensionless time [itex]\tau=\LARGE{\frac{t}{\frac{V}{Q}}}[/itex]. When I work out the units, I get [itex]t[/itex] on the top, but the bottom is [itex]\LARGE{\frac{V}{Q}=\frac{V}{\frac{M}{T}}=T\frac{M}{V}}[/itex]. But [itex]\frac{M}{V}[/itex] is not dimensionless, so [itex]\tau=\frac{M}{V}[/itex]. 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!
 
Physics news on Phys.org
  • #2
tomgill said:
[itex]\frac{dc}{dt} = \frac{q}{V}(c_{i}-c)-k^{2}c , t>0 [/itex]

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.

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.
 

1. What is a dimensionless time and why is it important in basic problems?

A dimensionless time is a time scale that has no units and is used to simplify mathematical equations and make them easier to analyze. In basic problems, it is important because it allows for a more universal understanding of the problem and can be applied to different scenarios.

2. How do I calculate a dimensionless time in a basic problem?

To calculate a dimensionless time, you need to divide the actual time by a characteristic time scale. This characteristic time scale can be determined by the problem's physical properties or by trial and error. The resulting dimensionless time will have no units and can be used for further analysis.

3. Can a dimensionless time be negative in a basic problem?

No, a dimensionless time cannot be negative. It is a ratio of two positive values, and since it has no units, it cannot have a negative value. If your calculation results in a negative dimensionless time, it is likely that there was an error in your calculations.

4. What are some common uses of dimensionless time in basic problems?

Dimensionless time is commonly used in fluid mechanics, heat transfer, and other areas of physics and engineering. It is especially useful in problems involving transient phenomena, where the time scale can vary significantly.

5. Can a dimensionless time be greater than 1 in a basic problem?

Yes, a dimensionless time can be greater than 1. This indicates that the actual time scale is larger than the characteristic time scale, and the process is taking longer to occur. It is important to carefully interpret the results and determine if they are reasonable for the given problem.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
575
  • Calculus and Beyond Homework Help
Replies
3
Views
492
  • Calculus and Beyond Homework Help
Replies
8
Views
234
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
299
  • Calculus and Beyond Homework Help
Replies
7
Views
282
  • Calculus and Beyond Homework Help
Replies
0
Views
163
  • Calculus and Beyond Homework Help
Replies
3
Views
914
  • Calculus and Beyond Homework Help
Replies
16
Views
564
Replies
1
Views
1K
Back
Top