Engineering 25
Problem
2.23
Solution
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering-25: Computational Methods
1
Bruce Mayer, PE
ENGR-25_Prob_2-23_HalfLife.pptx
Analysis
 Given:
C t   C 0 e
 kt
 To Analyze this Problem we
may have to a bit of Google
Research to learn the
meaning of “Half Life”
 After “Googling” Find that at
the Half Life time, th:
C t h   C 0  2  0.5  C 0 
 Now Analyze by HAND to set
up MATLAB plot
Engineering-25: Computational Methods
2
Bruce Mayer, PE
ENGR-25_Prob_2-23_HalfLife.pptx
Engineering-25: Computational Methods
3
Bruce Mayer, PE
ENGR-25_Prob_2-23_HalfLife.pptx
MATLAB Code
 Using
t h  ln 2  k
 The MATLAB code:
% Bruce Mayer, PE
% ENGR25 * 05Sep12
% file = ENGR25_P2_23_HalfLife_1209.m
%
k = linspace(0.047,0.107,500);
th = log(2)./k;
plot(k, th, 'LineWidth',2), grid,...
xlabel('Elimination Rate Constant, k (1/hr)'),...
ylabel('HalfLife, t_h (hr)')
Engineering-25: Computational Methods
4
Bruce Mayer, PE
ENGR-25_Prob_2-23_HalfLife.pptx
MATLAB Plot
15
14
13
h
HalfLift, t (hr)
12
11
10
9
8
7
6
0.04
0.05
0.06
0.07
0.08
0.09
Elimination Rate Constant, k (1/hr)
Engineering-25: Computational Methods
5
Bruce Mayer, PE
ENGR-25_Prob_2-23_HalfLife.pptx
0.1
0.11