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CE 3305 Engineering Fluid Mechanics
Spring 2023 Exercise Set 2

LAST NAME, FIRST NAME

R00000000

Purpose :

Apply definition of viscosity in Newtonian fluid. Apply principles of capillarity in a glass-air-water system.

Assessment Criteria :

Completion, plausible solutions, use Jupyter Notebook as a calculator

Problem 1 (Problem 1-30 pg. 34).

The plate is moving at 0.6 mm/s when the force applied to the plate is 4mN. The surface area of the plate in contact with the liquid is 0.5 m$^2$.

Determine:

  1. The approximate viscosity of the liquid, assuming the velocity profile is linear.

sketch here

list known quantities

list unknown quantities

governing principles

solution (step-by-step)

In [19]:
# code cell(s) here
# solution (step-by-step)
force = 4.0e-3 #newtons
#length = 1000 #mm
#width = 500 #mm
#area = length*width/(1000*1000) # in m^2
area = 0.5 #m^2
tau = force/area
du = 0.0006 # m/s
dy = 4/1000 #m
dudy = du/dy
viscosity = tau/dudy
print("Force (shear) = ",round(force,3),"   N ")
print("         Area = ",round(area,3),"     m^2")
print("     Velocity = ",round(du,6),"  m/sec")
print(" Shear stress = ",round(force/area,3),"   N/m^2 ")
print("    Thickness = ",round(dy,6),"   m" )
print("    Viscosity = ",round(viscosity,4),"  N*sec/m^2")

Force (shear) =  0.004    N 
         Area =  0.5      m^2
     Velocity =  0.0006   m/sec
 Shear stress =  0.008    N/m^2 
    Thickness =  0.004    m
    Viscosity =  0.0533   N*sec/m^2

discussion

Application of definition of shear, and shear-viscosity formula for Newtonian fluid. Then a script to generalize.

Problem 2 (Problem 1-49 pg. 37).

The tube rests on a 1.5-mm-thick film of lubricant having a viscosity of $\mu = 0.0586~\frac{N \cdot s}{m^2}$. The tube is rotating at a constant angular velocity of $\omega = 4.5~rad/s$

Determine:

  1. The torque $T$ that must be applied to maintain the motion. (Assume the velocity profile in the lubricant is linear)

sketch here

list known quantities

list unknown quantities

governing principles

solution (step-by-step)




In [12]:
# code cell(s) here
#read in data
viscosity = 0.05860000 #N.sec/m^2
angular_speed = 4.5 #rad/sec
lube_thickness = 1.5e-3 #meters
radius_inner = 40e-3 #meters
radius_outer = 80e-3 #meters
import math
torque = viscosity*angular_speed*2*math.pi*(0.25*radius_outer**4-0.25*radius_inner**4 )/lube_thickness
print("          Viscosity = ",round(viscosity,6),"    N-sec/m^2 ")
print("   Angular Velocity = ",round(torque,6),"  rad/sec ")
print("Lubricant Thickness = ",round(torque,6),"  meters ")
print(" Tube Inner Radius = ",round(torque,6),"   meters ")
print(" Tube Outer Radius = ",round(torque,6),"   meters ")
print("    Applied Torque = ",round(torque,6),"   N-m ")

          Viscosity =  0.0586     N-sec/m^2 
   Angular Velocity =  0.010604   rad/sec 
Lubricant Thickness =  0.010604   meters 
 Tube Inner Radius =  0.010604    meters 
 Tube Outer Radius =  0.010604    meters 
    Applied Torque =  0.010604    N-m

discussion

Application of definitions, recognize need to integrate the annulus to find equivalent applied torque (Force X Lever Arm). Then apply analysis as per calculus.

Then script for the arithmetic.

Problem 3. (Problem 1-59 pg. 39)

Water is at temperature of 30$^o$ C. A pair of glass plates is lowered into the water as shown.

Determine:

  1. The height, $h$, of the water as a function of the gap, $w$, between the two glass plates, for a surface tension of $\sigma = 0.0718~N/m$
  2. Plot the height, $h$, versus the width the gap, $w$, in 0.4 mm increments starting at $w=0.4~mm$
Width $w$ (mm) Height $h$ (mm)
0.4 ??
0.8 ??
1.2 ??
1.6 ??
2.0 ??
2.4 ??

sketch here

list known quantities

  • surface tension $\sigma = 0.0718~N/m$
  • width (t in my drawing) variable, but values given.

list unknown quantities

Also produce a plot to complete the problem

governing principles

solution (step-by-step)

Then build a script to compute $h$ for any width, and produce a plot as prescribed. Can also complete the table.

In [23]:
# code cell(s) here
# solution (step-by-step)
def rise(sigma,gamma,thickness): #prototype function
    rise = (2*sigma)/(gamma*thickness)
    return(rise)

# input
sigma = 0.0718 #N/m
start_thickness = 0.4 #mm
height = [] # empty list
thickness = []
for i in range(6):
    thickness.append(float(i+1)*start_thickness)
    thickness[i] = thickness[i]/1000 #convert mm into meters
    rho = 1000 #kg/m^3
    gravity = 9.81 #m/s^2

# intermediate values
    gamma = rho*gravity #N/m^3

#
    height.append(1000*rise(sigma,gamma,thickness[i]))

#output
    print("Plate width is ",round(thickness[i]*1000,3)," mm.  Capillary Rise is ",round(height[i],3)," mm")

import matplotlib.pyplot as plt
def makeAplot(listx1,listy1,strlablx,strlably,strtitle):
    mydata = plt.figure(figsize = (8,8)) # build a square drawing canvass from figure class
    plt.plot(listx1,listy1, c='blue', marker='o',linewidth=1) # basic data plot
    plt.xlabel(strlablx)
    plt.ylabel(strlably)
    plt.legend(['Computed Values'])# modify for argument insertion
    plt.title(strtitle)
    plt.show()
    return

makeAplot(thickness,height,'plate spacing(mm)','capillary rise (mm)','Plot of capillary rise versus plate spacing')

Plate width is  0.4  mm.  Capillary Rise is  36.595  mm
Plate width is  0.8  mm.  Capillary Rise is  18.298  mm
Plate width is  1.2  mm.  Capillary Rise is  12.198  mm
Plate width is  1.6  mm.  Capillary Rise is  9.149  mm
Plate width is  2.0  mm.  Capillary Rise is  7.319  mm
Plate width is  2.4  mm.  Capillary Rise is  6.099  mm

discussion

Application of capillary definition and surface tension. Essentially application of page 30 in textbook.

In [ ]: