TI65 Programs Part II: Reynolds Number/Hydraulic Diameter, Escape
Velocity, Speed of Sound/Resonant Frequencies in an Open Pipe
This is the second part of programs for the TI65 this
Fourth of July.
Click here for Part I: Digital Root, Complex Number Multiplication, Dew Point
Click here for Part III: Impedance and Phase Angle of a Series RLC Circuit, 2 x 2 Linear System Solution, Prime Factorization
TI65 Reynolds
Number/Hydraulic Diameter
This program utilities the two keyboard labels:
[F1]: Calculates
the Reynolds Number
[F2]: Calculates
the Hydraulic Diameter of a Rectangular Duct
Formula for the Reynolds Number:
Re = (v * DH)/w
v = velocity of the fluid (liquid or gas)
DH = hydraulic diameter
w = kinematic viscosity
The hydraulic diameter of the following ducts:
Tubular pipes: DH
= diameter of the tube
Annulus: DH =
large radius – small radius
Square Duct: DH =
length of one side
Rectangular Duct:
DH = (2*a*b)/(a + b); a, b are
the lengths of the sides
Program:
CODE

STEP

KEY

COMMENT

2^{nd}
53.53

00

LBL F1

Starts F1, have DH on the display (meters)

38

01

*


51

02

R/S

Prompt: velocity of fluid (m/s)

28

03

÷


51

04

R/S

Prompt: kinematic viscosity (m/s)

39

05

=


15

06

INV EE

Remove Engineering notation

2^{nd}
52

07

RTN

End F1

2^{nd}
53.54

08

LBL F2

Starts F2: have a on display (m)

12.0

09

STO 0


38

10

*


51

11

R/S

Prompt for b

12.1

12

STO 1


38

13

*


2

14

2


28

15

÷


16

16

(


13.0

17

RCL 0


59

18

+


13.1

19

RCL 1


17

20

)


39

21

=

DH of rectangular duct

2^{nd}
52

22

RTN

End F2

TI65 Reynold’s
Number
Input: hydraulic diameter (m) [F1], velocity of the fluid
(m/s) [R/S], kinematic viscosity (m/s) [R/S]
Output: Reynolds
number (dimensionless)
Hydraulic Diameter of a Rectangular Duct
Input: a (m) [F2],
b (m) [R/S]
Output: DH (m)
Test 1: Tubular
Pipe Duct of hydraulic diameter of 3.5 in.
The fluid is water at 68°F (20°C), flowing at 0.5 m/s. The kinematic viscosity of water of 20°C is
1.004 *10^6 m/s.
Input: 3.5 [3^{rd}]
[incm] [ ÷ ] 100 = [ F1 ], 0.5 [R/S], 1.004 [EE] 6 [+/] [R/S]
Output:
44,272.90837
Test 2:
Rectangular Duct where a = 1.27 m and b = 0.508 m (50 in x 20 in). The fluid is air at 60°F (about 15.6°C),
flowing at 0.5 m/s. The kinematic
viscosity of air at 15.6°C is 1.58 * 10^4 m/s.
Input: 1.27 [F2],
0.508 [R/S]
Result:
0.725714286 m (hydraulic
diameter), keep this number in the display
Input: [F1], 0.5
[R/S], 1.58 [EE] 4 [+/] [R/S]
Result: 2296.564195
(Reynolds Number)
TI65 Escape
Velocity
The formula for the escape velocity from a planet is:
v = √(2*G*m/r)
v = escape velocity (m/s)
G = University Gravitational Constant = 6.67384 * 10^11
m^3/(kg*s^2)
m = mass of the planet (kg)
r = radius of the planet (m)
Note that 2*G = 1.334768 * 10^10 m^3/(kg*s^2)
Program:
CODE

STEP

KEY

COMMENT

2^{nd}
16

00

2^{nd}
ENG

Start with mass,
set Engineering mode

38

01

*


1

02

1

Enter 2*G

57

03

.

Decimal Point

3

04

3


3

05

3


4

06

4


7

07

7


6

08

6


8

09

8


15

10

EE


1

11

1


0

12

0


58

13

+/


28

14

÷


51

15

R/S

Prompt for
radius

39

16

=


33

17

√


51

18

R/S

Display escape
velocity

Input: mass of the
planet (kg) [RST] [R/S], radius of the planet (m) [R/S]
Output: escape
velocity (m/s)
Test 1: Earth (m =
5.97219 * 10^24 kg, r = 6.378 * 10^6 m)
Input: 5.97219 [EE] 24 [RST] [R/S], 6.378 [EE] 6 [R/S]
Result: ≈ 11.179E3
(about 11,179 m/s)
Test 2: Jupiter (m
= 1.89796 * 10^27 kg, r = 71.492 * 10^6 m)
Result: ≈ 59.528E3
(about 59,528 m/s)
TI65 Speed of
Sound/Resonant Frequencies in an Open Pipe
Formulas:
Speed of Sound (m/s):
v = t*0.6 + 331.4
Where t = temperature (°C)
Resonant Frequencies in an Open Pipe: fn = n*v/(2*L)
Where fn = frequency (Hz), v = speed of sound (m/s), L =
length of pipe (m), n = 1, 2, 3…
If n = 1, then fn is the fundamental frequency
Program:
CODE

STEP

KEY

COMMENT

2^{nd}
53,53

00

LBL F1

Start Label F1

38

01

*


57

02

.

Decimal Point

6

03

6


59

04

+


3

05

3


3

06

3


1

07

1


57

08

.

Decimal point

4

09

4


39

10

=


2^{nd}
52

11

RTN

End F1

2^{nd}
53, 54

12

LBL F2

Start label F2

28

13

÷


13.1

14

RCL 1


28

15

÷


2

16

2


39

17

=


12.0

18

STO 0


1

19

1


12.3

20

STO 3

Counter

2^{nd}
53.0

21

LBL 0

Start loop

13.3

22

RCL 3


2^{nd}
51

23

PAUSE


38

24

*


13.0

25

RCL 0


39

26

=


51

27

R/S

Display fn

1

28

1


12.59

29

STO+


3

30

3

STO+ 3

13.3

31

RCL 3


3^{rd}
44

32

INV 3^{rd}
x>m

x≤m?

2

33

2

x≤R2?

2^{nd}
54.0

34

GTO 0

If x≤R2, GTO LBL
0

13.2

35

RCL 2


2^{nd}
52

36

RTN

End F2

Speed of Sound in Dry Air:
Input: Enter
temperature in °C [F1]
Result: Speed of
sound (m/s)
Resonant Frequencies:
Store the length of the pipe (m): L [STO] 1
Store the upper limit:
n [STO] 2
Input: speed of sound (m/s) [F2], n flashes before
frequency (Hz), press [R/S] to see other frequencies
The program finishes when n is displayed a second time.
Test:
Open pipe of 0.45, where the temperature of the air is 39°C
(102.2°F). Find out the first 3 resonant
frequencies.
We’ll need the speed of air, but first, store the
required constants:
0.45 [STO] 1, 3 [STO] 2
Next find the speed of air:
Input: 39 [F1]
Result: 354.8 m/s
Find the 3 resonant frequencies:
Input: (with 354.8 in the display) [F2]
Result: 1,
394.2222222 Hz [R/S] \\ fundamental frequency
2, 788.4444444 Hz [R/S]
\\ 2^{nd }frequency
3, 1182.666667 Hz [R/S]
\\ 3^{rd} frequency
Source: Browne Ph.
D, Michael. “Schaum’s Outlines: Physics for Engineering and Science” 2^{nd} Ed. McGraw Hill: New York, 2010
This blog is property of Edward Shore, 2016.