HP 32S and HP 32SII Week: Jurin's Law - Capillary Rise

HP 32S and HP 32SII Week:  Jurin's Law - Capillary Rise 

Capillary Motion 

Jurin's law describes the motion of liquid in small tubes, as the height is inversely proportional to the tube's diameter (and radius).  Factors include the contact angle and the density of the liquid.

The height of the liquid is determined by:

h = (2 * σ * cos θ) / (ρ * g * r)

σ = surface tension of the liquid (N/m)  (T)

θ = the angle of liquid in degrees, from adhesive (0° to 90°) to cohesive (90° to 180°) (B)

ρ = density of liquid (kg/m^3) (D)

r = radius of the tube (m) (R)

HP 32S and HP 32SII: Jurin's Law

Size:  33.5 bytes

J01 LBL J

J02 DEG

J03 INPUT T

J04 INPUT B

J05 INPUT D

J06 INPUT R

J07 2

J08 RCL× T

J09 RCL B

J10 COS

J11 ×

J12 RCL D

J13 RCL× R

J14 9.80665

J15 ×

J16 ÷

J17 STOP

Example:

Find the capillary rise of water in a tube with radius of 0.1 m.   

Data:  σ = 0.0728 N/m, θ = 0°, and ρ = 1000 kg/m^3

Inputs:

T = 0.0728

B = 0

D = 1000

R = 0.1

Result:  1.48470681E-4 m  (height)

Source:

"Jurin's law" Wikipedia.  Last updated February 11, 2022.  https://en.wikipedia.org/wiki/Jurin%27s_law  Last Accessed April 1, 2022.  

Lindeburg, Michael R. PE   Civil Engineering Reference Manual for the PE Exam 14th Edition  Professional Publications, Inc:  Belmont, CA.  pp. 14-11 to 14-13

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. 

HP 32S and HP 32SII Week: Total Drag

HP 32S and HP 32SII Week:  Total Drag

Introduction

The following equation calculates the total drag force applied to parallel to an area in the opposite direction of the object's motion.  

Fd = 1/2 * ρ * v^2 * Cd * A,   ρ = P / (R * T)

ρ = air pressure 

P = absolute pressure (101,325 Pa, 14.696 psi, 2116.224 lb/ft^2)

R = specific gas constant (287.03 J/(kg k), 53.3533 (ft lbf/lb °R))

T = temperature (K = °C + 273.15, °R = °F + 459.67)

A = cross area that the drag force is applied (m^2, ft^2)

V = velocity of the object (m/s, ft/s)

Cd = drag coefficient (unitless)

The program gives outputs:

ρ = air pressure (kg/m^3, lb/ft^3)

Fd = total drag (kg*m/s^2, lb*ft/s^2)

HP 32S/32SII Program:  Total Drag, SI Units

Size: 55.5 bytes

F01 LBL F

F02 INPUT C

F03 INPUT A

F04 INPUT V

F05 INPUT T

F06 273.15

F07 +

F08 287.03

F09 ×

F10 1/x

F11 101,325

F12 ×

F13 STOP

F14 RCL× C

F15 RCL× A

F16 RCL V

F17 x^2

F18 ×

F19 2

F20 ÷

F21 STOP

Example:

C = drag coefficient = 0.31

A = area =7.0686 m^2

T = temperature = 18.8 °C

V = velocity = 3 m/s

Results:

ρ = 1.20915184207 kg/m^3

Fd = 11.9230799416 kg*m/s^2

HP 32S/32SII Program:  Total Drag, US Units

Size: 55.5 bytes

F01 LBL F

F02 INPUT C

F03 INPUT A

F04 INPUT V

F05 INPUT T

F06 459.67

F07 +

F08 53.3533

F09 ×

F10 1/x

F11 2116.224

F12 ×

F13 STOP

F14 RCL× C

F15 RCL× A

F16 RCL V

F17 x^2

F18 ×

F19 2

F20 ÷

F21 STOP

Example:

C = drag coefficient = 0.31

A = area =76.0868 ft^2

T = temperature = 65.84 °F

V = velocity = 9.84252 ft/s

Results:

ρ = 7.54778228E-2 lb/ft^3

Fd = 86.232900381 lb*ft/s^2

Source:

Lindeburg, Michael R. PE   Civil Engineering Reference Manual for the PE Exam 14th Edition  Professional Publications, Inc:  Belmont, CA.  pp. 17-41 and 17-42

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.