## 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
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.

### Casio fx-CG50: Sparse Matrix Builder

Casio fx-CG50: Sparse Matrix Builder Introduction The programs can create a sparse matrix, a matrix where most of the entries have zero valu...