TI-65 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 TI-65 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
TI-65 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
|
2nd
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
|
2nd
52
|
07
|
RTN
|
End F1
|
2nd
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
|
2nd
52
|
22
|
RTN
|
End F2
|
TI-65 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 [3rd]
[in-cm] [ ÷ ] 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)
TI-65 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
|
2nd
16
|
00
|
2nd
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)
TI-65 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
|
2nd
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
|
=
|
|
2nd
52
|
11
|
RTN
|
End F1
|
2nd
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
|
2nd
53.0
|
21
|
LBL 0
|
Start loop
|
13.3
|
22
|
RCL 3
|
|
2nd
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
|
|
-3rd
44
|
32
|
INV 3rd
x>m
|
x≤m?
|
2
|
33
|
2
|
x≤R2?
|
2nd
54.0
|
34
|
GTO 0
|
If x≤R2, GTO LBL
0
|
13.2
|
35
|
RCL 2
|
|
2nd
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]
\\ 2nd frequency
3, 1182.666667 Hz [R/S]
\\ 3rd frequency
Source: Browne Ph.
D, Michael. “Schaum’s Outlines: Physics for Engineering and Science” 2nd Ed. McGraw Hill: New York, 2010
This blog is property of Edward Shore, 2016.