Monday, July 4, 2016

TI-65 Programs Part II: Reynolds Number/Hydraulic Diameter, Escape Velocity, Speed of Sound/Resonant Frequencies in an Open Pipe

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. 



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.



No comments:

Post a Comment

Expanding Linear Regression

Expanding Linear Regression Introduction – Linear Regression All scientific calculators that have two-variable dimensions has cu...