Mathcad - Problem 2.50.xmcd

Problem 2.50 [Difficulty: 3]
Given:
Block on oil layer pulled by hanging weight
Find:
Solution:
Expression for viscous force at speed V; differential equation for motion; block speed as function of time; oil viscosity
Mg
x
Governing equations: τ yx = μ⋅
du
ΣF
dy
x = Ma ⋅ x
F v
F t
y
F t
Assumptions: Laminar flow; linear velocity profile in oil layer
N
mg
The given data is M = 5⋅kg
W = m⋅g
= 9.81⋅N
A = 25⋅cm 2 h = 0.05⋅mm
Equation of motion (block) ΣF x = Ma ⋅ x
so F t − F v
Equation of motion (block) ΣF y = ma ⋅ y
so mg ⋅
= M⋅
dV
( 1)
dt
− F t = m⋅
dV
( 2)
dt
Adding Eqs. (1) and (2)
mg ⋅
− F v = ( M + m) ⋅
dV
dt
The friction force is F v = τ yx ⋅A
= μ⋅
du ⋅A
= μ⋅
V ⋅A
dy h
Hence
mg ⋅
μ⋅A
− ⋅V = ( M + m) ⋅
dV
h
dt
To solve separate variables
dt =
M + m
⋅dV
μ⋅A
mg ⋅ − ⋅V
h
⎛
⎝
⎛
⎝
( M + m) ⋅h
t = − ⋅⎜ln⎜m⋅g
−
μ⋅A
μ⋅A
h
⋅V
⎞
⎠
−
ln( m⋅g)
⎞
⎠
⎛
⎝
( M + m) ⋅h
= − ⋅ln⎜1
−
μ⋅A
μ⋅A
mg ⋅ ⋅h
⋅V
⎞
⎠
Hence taking antilogarithms 1 −
μ⋅A
mg ⋅ ⋅h
⋅V = e
−
μ⋅A
⋅t
( M+
m) ⋅h

⎡
μ⋅A
Finally V mg ⋅ ⋅h
−
( M
1 e
+ m) ⋅
⋅t
⎢
h
= ⋅⎣
−
The maximum velocity is V =
mg ⋅ ⋅h
μ⋅A
μ⋅A
In Excel:
⎤
⎥⎦
The data is M = 5.00 kg To find the viscosity for which the speed is 1 m/s after 1 s
m = 1.00 kg use Goal Seek with the velocity targeted to be 1 m/s by varying
g = 9.81 m/s 2 the viscosity in the set of cell below:
0 = 1.30 N.s/m 2
A = 25 cm 2 t (s) V (m/s)
h = 0.5 mm 1.00 1.000
Speed V of Block vs Time t
t (s) V (m/s)
0.00 0.000
0.10 0.155
0.20 0.294
0.30 0.419
0.40 0.531
0.50 0.632
0.60 0.722
0.70 0.803
0.80 0.876
0.90 0.941
1.00 1.00
1.10 1.05
1.20 1.10
1.30 1.14
1.40 1.18
1.50 1.21
1.60 1.25
1.70 1.27
1.80 1.30
1.90 1.32
2.00 1.34
2.10 1.36
2.20 1.37
2.30 1.39
2.40 1.40
2.50 1.41
2.60 1.42
2.70 1.43
2.80 1.44
2.90 1.45
3.00 1.46
1.6
1.4
1.2
1.0
V (m/s) 0.8
0.6
0.4
0.2
0.0
0.0 0.5 1.0 1.5 2.0 2.5 3.0
t (s)

Problem 2.63 [Difficulty: 4]

Problem 2.73 [Difficulty: 4]
Given:
Find:
Conical bearing geometry
Expression for shear stress; Viscous torque on shaft
Solution:
ds
dz
Basic equation τ = μ⋅
du dT = r⋅τ⋅dA
Infinitesimal shear torque
dy
Assumptions: Newtonian fluid, linear velocity profile (in narrow clearance gap), no slip condition
tan( θ)
r
= so r = z⋅tan( θ)
z
Then τ = μ⋅
du = μ⋅
∆u = μ⋅
( ω⋅r
− 0)
=
dy ∆y ( a − 0)
μω ⋅ ⋅z⋅tan( θ)
a
a
z
r
Section AA
AA
U= ωr
As we move up the device, shear stress increases linearly (because rate of shear strain does)
But from the sketch dz = ds⋅cos( θ)
dA = 2⋅π⋅r⋅ds
= 2⋅π⋅r⋅
dz
cos( θ)
The viscous torque on the element of area is dT = r⋅τ⋅dA
r μω ⋅ ⋅z⋅tan(
θ)
dz
= ⋅ ⋅2⋅π⋅r⋅
dT =
a
cos( θ)
Integrating and using limits z = H and z = 0
πμ ⋅ ⋅ω⋅tan( θ) 3 ⋅H 4
T =
2a ⋅ ⋅cos( θ)
2⋅π⋅μ⋅ω⋅z 3 ⋅tan( θ) 3
⋅dz
a⋅cos( θ)
Solving for µ
μ =
2a ⋅ ⋅cos( θ)
⋅T
πω ⋅ ⋅tan( θ) 3 ⋅H 4
Using given data H = 25⋅mm
θ = 30⋅deg
a = 0.2⋅mm
ω = 75⋅
rev T = 0.325⋅N⋅m
s
μ
=
2a ⋅ ⋅cos( θ)
T
μ = 1.012⋅
Ns πω ⋅ ⋅tan( θ) ⋅H 4
m 2
From Fig. A.2, at 20 o C, CASTOR OIL has this viscosity!