EGN 3353C – Fluid Mechanics – Fall 2023
HW #1
Problem 1:
Given:
•
A formula for estimating the mean free path,
a)
What are the units of the constant “ 1.26 ”?
ℓ , of a perfect gas is ℓ = 1.26
µ
ρ RT
Find:
b) Using this formula, estimate the mean free path of air at 200 C and 7kPa .
c)
Is the continuum assumption valid when analyzing a full-scale commercial aircraft?
Solution
⎡ L2 ⎤
⎡M⎤
⎡M⎤
ℓ ≡ ⎡⎣ L ⎤⎦ ; µ ≡ ⎢
⎥ ; ρ ≡ ⎢ L3 ⎥ ; R ≡ ⎢ T 2Θ ⎥ ; T ≡ ⎡⎣Θ ⎤⎦
⎣ LT ⎦
⎣ ⎦
⎣
⎦
a)
⎡M⎤
⎡M⎤
⎢ LT ⎥
⎢ LT ⎥
⎣
⎦
⎦ = 1.26 ⎡ L ⎤
⎡⎣ L ⎤⎦ = 1.26
= 1.26 ⎣
⎣ ⎦
2
⎡ M ⎤⎡ L⎤
⎡M⎤ ⎡ L ⎤
⎢ L3 ⎥ ⎢ T ⎥
⎢ L3 ⎥ ⎢ T 2Θ ⎥ ⎡⎣Θ ⎤⎦
⎣ ⎦⎣ ⎦
⎣ ⎦ ⎣
⎦
∴ 1.26 is unitless
ℓ = 1.26
b)
µ
ρ RT
→ p = ρ RT → ℓ = 1.26
µ RT
p
⎛
Ns ⎞ ⎛
m2 ⎞
1.80E
−
5
287
( 293K )
⎜⎝
m2 ⎟⎠ ⎜⎝
s 2 K ⎟⎠
ℓ = 1.26
→ ℓ = 9.4E − 7m
N
7E3 2
m
Use Appendix fig A.3 to obtain viscosity.
c)
Yes, the continuum assumption is valid as the length scale for an aircraft is 10s to 100s of meters, which is much larger
than the mean free path.
1
EGN 3353C – Fluid Mechanics – Fall 2023
Problem 2:
Given:
•
One definition of a “rarefied gas” is if it contains less than
•
A formula for estimating the mean free path,
1012
molecules
mm3
ℓ , of a perfect gas is ℓ = 1.26
µ
ρ RT
0
•
Assume an air temperature of 20 C
a)
What is the density under these conditions? Recall how to obtain mass of a single molecule.
Find:
b) What is the pressure under these conditions, assuming a perfect gas?
c)
What is the mean free path under these conditions?
Solution
molecular weight
mass of one molecule is, m =
=
Avogadro's number
a)
1
mol
= 4.81E − 23 g
molecules
6.023E23
g mol
28.97
⎛
molecules ⎞ ⎛ 4.81E − 23 g ⎞ ⎛ 1kg ⎞ ⎛ 1000mm ⎞
ρ = ( molecular density ) ( molecular mass ) = ⎜ 1012
⎝
mm3 ⎟⎠ ⎜⎝ molecule ⎟⎠ ⎜⎝ 1000g ⎟⎠ ⎜⎝ 1m ⎟⎠
ρ = 4.81E − 5
3
kg
m3
b)
⎛
kg ⎞ ⎛
m2 ⎞
p = ρ RT = ⎜ 4.81E − 5 3 ⎟ ⎜ 287 2 ⎟ ( 293K ) → p = 4Pa
⎝
m ⎠⎝
s K⎠
c)
⎛
Ns ⎞ ⎛
m2 ⎞
1.80E
−
5
287
( 293K )
⎜⎝
m2 ⎟⎠ ⎜⎝
s 2 K ⎟⎠
µ RT
ℓ = 1.26
= 1.26
→ ℓ = 1.64mm
N
p
4 2
m
2
EGN 3353C – Fluid Mechanics – Fall 2023
Problem 3:
Given:
⎛
kg ⎞
⎜⎝ ρ Al = 2700 m3 ⎟⎠ and steel
⎛
kg ⎞
⎜⎝ ρ St = 7800 m3 ⎟⎠ balls
•
Consider aluminum
•
Assume a water temperature of 200 C
a)
What is the maximum diameter of each of the balls that can float on a clean water surface?
Find:
Please neglect buoyancy in this calculation.
Solution: Consider a force balance between the weight of the ball and surface tension force
∑ F = 0 → W = Fsurface → r ∀g = sp D → r
tension
a)
p D3
6s
g =s D→ D =
6
rg
DAl =
⎛
N⎞
6 ⎜ 0.073 ⎟
m⎠
⎝
⎡ N ⎤ ⎡ m 1 ⎤ ⎡ kg ⎤
6s
=
, ⎢ ⎥ = ⎢ kg 2 ⎥ = ⎢ 2 ⎥ → DAl = 4.1mm
⎛
r Al g
kg ⎞ ⎛
m⎞ ⎣ m ⎦ ⎣ s m⎦ ⎣ s ⎦
⎜⎝ 2700 m3 ⎟⎠ ⎜⎝ 9.81 s 2 ⎟⎠
DSt =
⎛
N⎞
6 ⎜ 0.073 ⎟
m⎠
⎝
⎡ N ⎤ ⎡ m 1 ⎤ ⎡ kg ⎤
6s
=
, ⎢ ⎥ = ⎢ kg 2 ⎥ = ⎢ 2 ⎥ → DSt = 2.4mm
⎛
r Al g
kg ⎞ ⎛
m⎞ ⎣ m ⎦ ⎣ s m⎦ ⎣ s ⎦
⎜⎝ 7800 m3 ⎟⎠ ⎜⎝ 9.81 s 2 ⎟⎠
3
EGN 3353C – Fluid Mechanics – Fall 2023
Problem 4:
Given:
•
!
!
!
V = Ax 2 i + Bxy j , is a velocity vector, where A, B are constants
Find:
a) Is this a system or a field?
b) Is this a Lagrangian or Eulerian description?
c)
d)
e)
f)
g)
!
V unsteady or steady? !
What is the dimension of V ?
What is the velocity component in the y-direction?
!
What is a ?
What is the equation of the streamline through ( C , D )?
Is
Answers:
a)
field, spatial dependence
b) Eulerian, spatial dependence
c)
Steady, no time dependence
d) 2D
e)
f)
Bxy
"
"
"
"
"
"
∂V
∂V
! ∂V
a=
+u
+v
= 0 + Ax 2 ( 2 Ax i + By j ) + Bxy ( Bx j )
∂t
∂x
∂y
"
"
!
a = 2 A2 x 3 i + Ax 2 By + B 2 x 2 y j
(
)
g)
y
x
dy v Bxy By
A dy
dx
= =
=
→ ∫
=∫
2
dx u Ax
Ax
By y x x
0
0
A
A
ln ( y y0 ) = ln ( x x0 ) → ( y y0 ) B = ( x x0 )
B
A
( y D)B = ( x C )
4
EGN 3353C – Fluid Mechanics – Fall 2023
Problem 5:
Given:
•
•
•
φ ( x, y ) is a scalar
!
!
!
ds = dxi + dy j is a vector lying on the constant scalar surface
Hint: recall definition of total differential and the geometric interpretation of a dot
product
Find:
a)
Mathematically prove that the gradient of a scalar is normal to a constant scalar
surface, ∇φ ⊥
scalar surface.
!
!
!
φconst , where ds = dxi + dy j is a vector lying on the constant
Answer:
!
!
!
ds = dxi + dy j ⇒ "a vector lying on φconst surface"
φ = φ ( x, y )
!
∂φ
∂φ
dx +
dy = ∇φ ⋅ ds = 0, for φ ≡ const
∂x
∂y
!
∇φ ds cos (θ ) = 0, θ = 90°
!
∴∇φ ⊥ ds
dφ =
5
EGN 3353C – Fluid Mechanics – Fall 2023
Problem 6:
Given:
•
•
Consider a cylinder of length L and radius
The uncertainty in the length is uL = 2%
R.
•
The uncertainty in the radius is uR = 3%
•
Determine the uncertainty in the volume uV .
Find:
Answer:
∂V
∂V
d R; dVL =
d L → dVR = 2p RLd R; dVL = p R 2d L →
∂R
∂L
2
dVR
dVL
2p RL
pR
=
d R;
=
d L → uV = 2uR ; uV = uL
2
2
2
R
L
pR L pR L
p R L p R2 L
V = p R 2 L → dVR =
∴uV =
(2u ) + (u )
2
VR
VL
2
=
( 2 (.03)) + (.02)
2
2
→ uV = 4.7%
6
EGN 3353C – Fluid Mechanics – Fall 2023
Problem 7:
Problem 2.1 from Textbook
7
EGN 3353C – Fluid Mechanics – Fall 2023
Problem 8:
Problem 2.8 from Textbook
8