The Museum of HP Calculators


Exponential Growth or Decay for the HP-33E/C

This program is Copyright © 1978 by Hewlett-Packard and is used here by permission. This program was originally published in "HP-33E Student Engineering Applications".

This program is supplied without representation or warranty of any kind. Hewlett-Packard Company and The Museum of HP Calculators therefore assume no responsibility and shall have no liability, consequential or otherwise, of any kind arising from the use of this program material or any part thereof.

Overview

Many growth or decay phenomena encountered in science and engineering obey an exponential law of the general form:

Xt = Xss - (Xss - X0)e-t/K

where:
Xt = Value at any time, t, (i.e., the instantaneous value)
Xss = Steady state value (i.e., at t = infinity)
X0 = Initial Value (i.e., at t = 0)
t = Elapsed time (time after t = 0)
K= Exponential time-constant for specific phenomena

This program provides interchangeable solutions for any one of the four variables Xt, Xss, X0, or t provided three variables and K are known.

Instructions

Step

Instructions

Input Data/Units

Keys

Output Data/Units

1

Enter program

     

2

Store time constant

K

STO 4

 

3

Store variables:

 

 

 

 

 Initial value

X0

STO 0

 

 

 Instantaneous value

Xt

STO 1

 

 

 Steady state value

Xss

STO 2

 

 

 Elapsed time

t

STO 3

 

 

 (Store any 3 of the 4 variables)

 

 

 

4

To Calculate:

 

 

 

 

 X0, Initial value

 

GSB 21

X0

 

 Xt, Instantaneous value

 

GSB 01

Xt

 

 Xss, Steady state value

 

GSB 09

Xss

 

 t, Elapsed time 

 

GSB 29

t

5

For a new case go to step 2.     

 

Example 1

Given a 5uF capacitor in series with a 1 megohm resistor. 1.5 seconds after the circuit is completed 125 volts are measured across R. To what voltage was the capacitor originally charged?

Note: K = the RC time-constant, and the voltage at t = infinity is zero.

Keystrokes         Display
5 EEX CHS 6
ENTER EEX 6
x STO 4
125 STO 1
0 STO 2
1.5 STO 3
GSB 21             168.7324    (volts)

Example 2

A cobalt 60 source (half life = 5.26 years) had an activity of 3.54 curies when purchased 8.5 years ago. What is its present activity?

Note: Activity at t = infinity will be zero, K = half life/ln 2.

Keystrokes         Display
5.26 ENTER
2 f LN ÷
STO 4
3.54 STO 0
0 STO 2
8.5 STO 3
GSB 01             1.1549   (curies)

The Program

LINE    CODE    KEYS
 00             f CLEAR PRGM
 01     24  0   RCL 0
 02     24  2   RCL 2
 03        41   -
 04     12 40   GSB 40
 05        71   ÷
 06     24  2   RCL 2
 07        51   +
 08     13 00   GTO 00
 09     12 40   GSB 40
 10        31   ENTER
 11        31   ENTER
 12     24  1   RCL 1
 13        61   x
 14     24  0   RCL 0
 15        41   -
 16        21   x<>y
 17         1   1
 18        41   -
 19        71   ÷
 20     13 00   GTO 00
 21     12 40   GSB 40
 22     24  1   RCL 1
 23     24  2   RCL 2
 24        41   -
 25        61   x
 26     24  2   RCL 2
 27        51   +
 28     13 00   GTO 00
 29     24  0   RCL 0
 30     24  2   RCL 2
 31        41   -
 32     24  1   RCL 1
 33     24  2   RCL 2
 34        41   -
 35        71   ÷
 36     14  1   f LN
 37     24  4   RCL 4
 38        61   x
 39     13 00   GTO 00
 40     24  3   RCL 3
 41     24  4   RCL 4
 42        71   ÷
 43     15  1   g ex
 44     15 12   g RTN

Register Use

R0  X0
R1  Xt
R2  Xss
R3  t
R4  K

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