Saturday, April 23, 2022

Applications: Sharp EL-5150

Applications:  Sharp EL-5150 


Note:  spaces included for readability

Fan Laws

AER Equation:
1; f(BCDI) = B * ( C ÷ D ) Y^x ( 1 ÷ I  )

Variables:
Calculate CPM_new
B = CPM_old

1st Fan Law:
I = 1
C = RPM_new
D = RPM_old

2nd Fan Law:
I = 2
C = SP_new
D = SP_old

3rd Fan Law:
I = 3
C = BHP_new
D = BHP_old

Example 1:

Fan Law 2:
CPM_old = B = 4000 CPM
SP_new = C = 48 
SP_old = D = 36
I = 2

Result:  4618.802153

Example 2:

Fan Law 3:
CPM_old = B = 3500 CPM
BHP_new = C = 59
BHP_old = D = 52
I = 3

Result:  3650.488072


Ideal Shockley Diode Equation


I = I0 * e^((VD/(n* VT) - 1)
where VT = K * T/q

I = diode current (amps)
I0 = saturation current (amps)
VT = thermal voltage (V) - see notes below
VD = voltage across the diode (V)
n = ideality factor, in ideal situations, n = 1

Notes:  

*  The equation below assumes the ideal diode, n = 1
*  The equation uses a ratio of scientific constants:  k/q 
*  k = Boltzmann's Constant = 1.380649 * 10^-23 J/K
*  q = Charge of an Electron = 1.602176634 * 10^-19 C  (on some calculators, like the Casio fx-991EX, this constant is labeled e)
*  k/q = 8.617332385 * 10^-5 J/(K*C) = 8.617332385 * 10^-5 V/K  (volts/degrees Kelvin)

AER Equation:
1; f(IDE) = 8.617332385E-5 × E STO A, I ×(e(D ÷ A) - 1)

Calculate VT (stored in A), the I 
I = I0
D = VD
E = temperature in Kelvin

Example 1:

I = 4E-6 A 
D = 0.08 V
E = 280 K

Results:
A = 0.024128531, (I) 0.00001061
 
Example 2:

I = 4E-6 A 
D = 0.06 V
E = 300 K

Results:
A = 0.025852000, (I) 0.00003674

Dot and Cross Product of Two 3D Vectors

For the two vectors [A, B ,C] and [D, E, F]:

AER Equations:
1; f(ABCDEF) = A × D + B × E + C × F ◣
2; B × F - C × E, C × D - A × F, A × E - B × D

Example 1:
[ A, B, C ] and [ D, E, F]

[ 4.5, -2.5, -8 ] and [ 1.6, 3.9, 6 ]

Dot Product:  -50.55
Cross Product:  [ 16.2, -39.8, 21.55 ]

Example 2:
[ A, B, C ] and [ D, E, F]

[ 4, 3, 2 ] and [ 2, 7, 0 ]

Dot Product:  29
Cross Product:  [ -14, 4, 22 ]

Law of Cosines

Sides with lengths A, B, C with D as the angle opposite of A.  Equation 1 finds the length of side A, while Equation 2 finds the angle D.

AER Equations:
1; f(BCD) = √(B^2 + C^2 - 2 × B × C × COS D) STO A ◣
2; f(ABC) = cos^-1 ((B^2 + C^2 - A^2) ÷ (2 × B × C)) STO D

Example 1 - find A:
Degree Mode Set
Input:  B = 4.5, C = 3.7, D = 30°
Run 1:

Result:  2.258617731


Example 2 - find D:
Degree Mode Set
Input:  A = 40, B = 56, C = 38
Run 2:

Result:  45.5579132°


Note:  Due to the incredible amount of spam comments that get sent on this blog, which I moderate so the readers don't see them, I have decided to turn comments off.  I will review whether to turn comments back on at a later time.  My apologizes to those who leave legitimate comments.  

Eddie 


All original content copyright, © 2011-2022.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.