Download MAE221 Fluid Mechanics

MAE221 Fluid Mechanics
Homework #1 (8 problems)
Due: 10:00pm on September18, 2015.
-----------------------------------------------------------------------------------------------1. Explain the flowing terms in 1 or 2 sentences.
a) Continuum
b) Viscosity
c) Surface tension
d) Streamline
e) Newtonian fluid
f) No-slip condition
[solution]
a) Continuum: It’s variation in properties is so smooth that differential calculus can be used to
analyze the substance.
b) Viscosity: Viscosity is a measure of the stickness or resistance to deformation of a fluid. It is
due to the internal frictional force that develops between different layers of fluids as they are forced
relative to each other. Viscosity is caused by the cohesive forces between the molecules in liquids,
and by the molecules collisions in gases. In general, liquids have higher dynamic viscosities than
gases.
c) Surface tension: Surface tension is a contractive tendency of the surface of a liquid that allows it
to resist an external force. It is caused by the “inward” attraction at the interface of two fluids, due to
various intermolecular forces.
d) Streamline: Streamline is a line everywhere tangent to the velocity vector at a given instant
(Eulerian) and represents flow directions of fluid particles at a particular time.
e) Newtonian fluid: Shear stress is linearly proportional to the rate of shearing strain. (e.g. water,
air, and oil). Newtonian fluid has constant viscosity across all shear rates.
f) No-slip condition: Viscous fluid always tends to cling to a solid surface in contact with it. A thin
boundary layer of fluid is formed near a surface (in which the fluid is nearly at rest with respect to the
surface).
MAE221 Fluid Mechanics
Homework #1 (8 problems)
Due: 10:00pm on September18, 2015.
-----------------------------------------------------------------------------------------------2. The pressure difference, ∆𝑝 , across a partial blockage in an artery (called a stenosis) is
approximated by the equation
∆𝑝 = 𝐾!
𝜇𝑉
𝐴!
+ 𝐾!
−1
𝐷
𝐴!
!
𝜌𝑉 !
where V is the blood velocity, 𝜇 the blood viscosity (𝐹𝐿!! 𝑇), 𝜌 the blood density (𝑀𝐿!! ), D the
artery diameter, 𝐴! the area of the unobstructed artery, and 𝐴! the area of the stenosis. Determine
the dimensions of the constants 𝐾! and 𝐾! .
[solution]
∆𝑝 = 𝐾!
!!
[𝐹𝐿 ] ≐ [𝐾! ]
𝐹𝑇
𝐿!
𝜇𝑉
𝐴!
+ 𝐾!
−1
𝐷
𝐴!
𝐿
𝑇
1
𝐿
+ 𝐾!
!
𝜌𝑉 !
𝐿!
−1
𝐿!
𝐹𝐿!! ≐ 𝐾! 𝐹𝐿!! + 𝐾! 𝐹𝐿!!
since each term must have the same dimensions,
𝐾! and 𝐾! are dimensionless.
!
𝐹𝑇 ! 𝐿
𝐿! 𝑇
!
MAE221 Fluid Mechanics
Homework #1 (8 problems)
Due: 10:00pm on September18, 2015.
-----------------------------------------------------------------------------------------------3. A layer of water flows down an inclined fixed surface with the velocity profile shown in Fig. P3.
Determine the magnitude of the shearing stress that the water exerts on the fixed surface for
U = 2 m/s and h = 0.1 m .
Figure. P3
[solution]
From definition of shear stress in case of newtonian fluids.
τ =µ
du
dy
From given velocity,
"2 2y %
du
=U $ − 2 '
dy
#h h &
at the fixed surface
( y = 0)
⎛ du ⎞
2U
⎜ ⎟ =
h
⎝ dy ⎠ y =0
Finally,
τ =µ
2U
⎛ 2 ⋅ 2 m/s ⎞
= (1.003 × 10−3 N ⋅ s/m 2 ) ⋅ ⎜
⎟
h
⎝ 0.1 m ⎠
= 0.04012 N/m 2
MAE221 Fluid Mechanics
Homework #1 (8 problems)
Due: 10:00pm on September18, 2015.
-----------------------------------------------------------------------------------------------4. The viscosity of liquids can be measured through the use of a rotating cylinder viscometer of the
type illustrated in Fig. P4. In this device the outer cylinder is fixed and the inner cylinder is rotated
with an angular velocity, Ω . The torque M required to develop Ω is measured and the viscosity
is calculated from these two measurements.
Derive this relation, M =
2π Ri 3l µΩ
, by developing an equation relating µ , Ω , M , l , R0 , and
R0 − Ri
Ri . Neglect end effects and assume the velocity distribution in the gap is linear. (Hint: see Ex. 1.10 in
the text book.)
Figure. P4
[solution]
For a linear velocity distribution in the gap is
τ =µ
Ri Ω
R0 − Ri
Torque due to shear stress on inner cylinder is equal to
dM = τ ( dA) r
where dA = ( Ri ⋅ dθ ) l thus,
dM = Ri 2 ⋅ l ⋅τ dθ
Torque required to rotate inner cylinder is
2π
M = Ri 2 ⋅ l ⋅τ ∫ dθ = 2π Ri 2 ⋅ l ⋅ τ
0
Finally,
M=
2π Ri 3 ⋅ l ⋅ µ ⋅ Ω
R0 − Ri
MAE221 Fluid Mechanics
Homework #1 (8 problems)
Due: 10:00pm on September18, 2015.
-----------------------------------------------------------------------------------------------5. Derive an expression for the capillary height change ℎ for a fluid of surface tension Υ and
contact angle 𝜃 between two vertical parallel plates a distance 𝑊 apart, as in Fig. P5. What will ℎ
be for water at 20℃ if 𝑊 = 0.5 mm?
Figure. P5
[solution]
Set b as the Width into paper.
With b the width of the plates into the paper, the capillary forces on each wall together balance the
weight of water held above the reservoir free surface:
𝜌g𝑊ℎ𝑏 = 2 𝑌𝑏cos𝜃 ,
𝑜𝑟 ∶ ℎ ≈
2𝑌cos𝜃
𝜌g𝑊
For water at 20℃, 𝑌 ≈ 0.0728N/m, 𝜌g ≈ 9790N/m3, and 𝜃 ≈ 0°. Thus, for 𝑊=0.05mm,
ℎ=
2 0.0728N/m 𝑐𝑜𝑠0°
≈ 0.030m ≈ 30mm
9790N/m3 (0.0005m)
MAE221 Fluid Mechanics
Homework #1 (8 problems)
Due: 10:00pm on September18, 2015.
-----------------------------------------------------------------------------------------------6. A solid cylindrical needle of diameter d, length L, and density 𝜌! may float in liquid of surface
tension Υ. Neglect buoyancy and assume a contact angle of 0°. Derive a formula for the maximum
diameter 𝑑!"# able to float in the liquid.
Figure. P6
[solution]
The needle dents the surface downward and the surface tension forces are upward, as shown in Figure.
If these tensions are nearly vertical, a vertical force balance gives:
𝜋
ΣFz = 0 = 2𝑌𝐿 − 𝜌! g 𝑑 ! 𝐿,
4
𝑜𝑟: 𝑑!"# ≈
8𝑌
𝜋𝜌! g
MAE221 Fluid Mechanics
Homework #1 (8 problems)
Due: 10:00pm on September18, 2015.
-----------------------------------------------------------------------------------------------7. For the three-liquid system shown in Fig. P7, find h1 and h2 . Neglect the air density. Specific
weights are
γ water = 9790 N/m3
and
γ mercury = 133,100 N/m3 .
Figure. P7
[solution]
(9790 N/m3 ) × 0.27 m + (133,100 N/m3 ) × h1 = (133,100 N/m3 ) × 0.1 m
=(8811 N/m3 ) ⋅ h2 + (133,100 N/m3 ) × 0.05 m
Solve for h1 , h2
h1 = 0.08 m , h2 = 0.755 m
MAE221 Fluid Mechanics
Homework #1 (8 problems)
Due: 10:00pm on September18, 2015.
-----------------------------------------------------------------------------------------------8. In Fig. P8 pipe A contains gasoline (SG = 0.7), pipe B contains oil (SG = 0.9), and the manometer
fluid is mercury. Determine the new differential reading if the pressure in pipe A is decreased 25 kPa,
and the pressure in pipe B remains constant. The initial differential reading is 0.30 m as shown.
[solution]
For the initial configuration :
𝑃! + 𝛾!"# (0.3m) − 𝛾!" (0.3m) − 𝛾!"# (0.4m) = 𝑃! (1)
with a decrease in 𝑃! to 𝑃! ′ gage fluid levels change as shown
on figure.
Thus, for final configuration :
𝑃! ′ + 𝛾!"# (0.3 − 𝑎) − 𝛾!" (Δℎ) − 𝛾!"# (0.4 + 𝑎) = 𝑃! (2)
where all lengths are in m. Subtract Eq. (2) from Eq. (1) to obtain,
𝑃! − 𝑃! ′ + 𝛾!"# (𝑎) − 𝛾!" (0.3 − Δℎ) + 𝛾!"# (𝑎) = 0 (3)
since
2𝑎 + Δℎ = 0.3 (see figure)
Then,
𝑎=
0.3 − Δℎ
2
and from Eq. (3)
𝑃! − 𝑃! ′ + 𝛾!"#
0.3 − Δℎ
0.3 − Δℎ
− 𝛾!" (0.3 − Δℎ) + 𝛾!"#
=0
2
2
Thus,
Δℎ =
𝑃! − 𝑃! ′ + 𝛾!"# 0.15 − 𝛾!" 0.3 + 𝛾!"# 0.15
𝛾!"# 𝛾!"#
−𝛾!" +
+
2
2
And with 𝑃! − 𝑃! ′ = 25𝑘𝑃𝑎
Δℎ =
25
kN
+ 0.7
m3
9.81
kN
m3
−133
= 0.100 m
kN
kN
0.3m + 0.9 9.81 3
m3
m
kN
0.9
kN
9.81 3 +
9.81 3
2
m
m
0.15m − 133
kN
0.7
+
2
m3
0.15m