Download Mid term practice problems

NAME _____________________________
ID Number__________________________
Problem 1 (30 points)
1.
(12 points) True/False Questions. Please circle whether the statement is true or false.
If the statement is false, please provide a correction in the space provided. No
partial credit will be given if the reasoning is wrong if the answer is False (i.e. Score
will be either 0 or 2)
(a) (2 points) (T/F) Density of a certain liquid is 1360 kg/m3, then the specific gravity
of the liquid would be 1.36 kg/m3.
_______________________________________________________
____________________________________________________________
(b) (2 points) (T/F) The no-slip boundary condition is an experimental observation
that the velocity of a fluid in contact with a solid surface is equal to the velocity
of the surface, and it is only valid for liquids.
False. Viscous fluid (not only liquids also gases) always tends to cling to a solid
surface in contact with it.
(c) (2 points) (T/F) When an interface is a plane so that the radius of curvature is
infinite, the pressure difference across the interface is zero, i.e., the pressure is
equal on both sides of the plane interface despite the surface tension.
___________________________________________________________________
_______________________________________________________________
(d) (2 points) (T/F) Fluid in which shear stress is not directly proportional to shear
strain rate is called non-Newtonian. If the apparent viscosity decreases with
increasing shear strain rate, such fluid is called dilatant.
Two possible answers
1)Such fluid is called pseudoplastic or shear thinning_________________
2)Apparent viscosity increases with increasing shear strain rate
(e) (2 points) (T/F) The buoyant force is essentially caused by the difference between
the pressure at the top of the object and the pressure at the bottom of the object where
the pressure at the bottom is always greater than that at the top, generating an upward
net force on a submerged object.
____________________________________________________________
____________________________________________________________
(f) (2 points) (T/F) Cavitation occurs when liquid pressure rises above the vapor
pressure due to fluid motion.
Drops below__________________________________________________
____________________________________________________________
NAME _____________________________
ID Number__________________________
2.
(7.5 points) Multiple Choice Questions: Choose only ONE best answer for each
problem.
(1.5 points) The following figure illustrates the differences between solid and fluid
and the arrows indicate shear and normal stresses acting on the surfaces. Which of
the stresses A, B, C, D, has a quantity of zero (0)?
(a)
(b)
(c)
(d)
(e)
A only
B only
C&D
A& D
A, B, & C
(A) (1.5 points) Which of the following is NOT true about fluid properties?
(a) The viscosity of the liquid decreases with increasing temperature because
of weakening of hydrogen bonding at higher temperature.
(b) Contact angle of fluid on a wetting surface is larger than 90°.
(c) Surface tension is caused by the inward attraction at the interface of two
fluids, due to various intermolecular forces.
(d) Fluid viscosity can either increase or decrease with increase in
temperature.
(e) Pressure is one of the most dynamic variables in fluid and its difference
drives a fluid flow.
(B) (1.5 points) A wire is attached to a block of metal that is submerged in a tank of
water as shown below. Which of the following graphs most correctly describes
the relation between the force in the wire and time as the block is pulled slowly
NAME _____________________________
ID Number__________________________
out of the water? (d)
(C) (1.5 points) Consider constant altitude, steady flow along a streamline with a
flow that satisfies the assumptions necessary for Bernoulli’s question shown
below.
1
p + ρV 2 + ρ gz =
const
2
Which of the following must have a constant value along the streamline?
(a)
(b)
(c)
(d)
(e)
Internal energy
Stagnation pressure only
Stagnation and total pressure
Dynamic pressure only
Static and Dynamic pressure
(D) (1.5 points) The laminar velocity profile for a Newtonian fluid is shown above.
Which of the following best describes the variation of shear stress with distance
from the surface? (c)
NAME _____________________________
ID Number__________________________
3.
(10.5 points) Short answer problems
(A) (8 points) Viscosity
(a) (3 points) An infinite plate is moved over a second plate on a layer of liquid
as shown. For small gap width, d, we assume a linear velocity profile in the
liquid. Derive the relations of 𝜏𝜏 as a function of u, y, and µ (dynamic
viscosity) where 𝜏𝜏 is the shear stress applied on the liquid by the plate. Make
appropriate assumptions and a small angle approximation for the shearing
strain.
NAME _____________________________
ID Number__________________________
(b) (1 points) What is the SI unit of the dynamic viscosity µ ?
(c) (4 points) Suppose a layer of water flows down an inclined surface with the
velocity profile shown below. What is the magnitude of the shear stress that
the water exerts on the surface if U = 2 m/s and h = 0.1 m (the dynamic
viscosity of water is 1.0×10-3 kg/m⋅sec)?
𝛍𝛍
𝐝𝐝𝐝𝐝
�
𝐝𝐝𝐝𝐝 𝒚𝒚=𝟎𝟎
𝟐𝟐
𝟐𝟐
= 𝛍𝛍𝛍𝛍( − 𝟐𝟐𝟐𝟐/𝒉𝒉𝟐𝟐 )= 𝛍𝛍𝛍𝛍 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎 ∙ 𝟐𝟐 ∙
𝐡𝐡
𝐡𝐡
𝟐𝟐
𝟎𝟎.𝟏𝟏
= 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎/𝒎𝒎𝟐𝟐
(B) (2.5 points) Surface tension problem
Surface tension force can be strong enough to allow a double-edge steel razor
blade to float on water as shown below. Assume that the surface tension forces act
at an angle θ relative to the water surface as shown.
NAME _____________________________
ID Number__________________________
The mass of the double-edge blade is 0.64 × 10−3 kg and the total length of
its sides is 206 mm, Determine the value of θ required to make the blade float.
Neglect the effect of buoyant force. Surface tension ϒ water / air =
0.073 N/m and
gravitational acceleration is 9.8 m/s2.
𝑚𝑚𝑚𝑚 = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
0.64 × 10−3 × 9.8 = 206 × 10−3 × 0.073 × sin 𝜃𝜃
𝜃𝜃 = 24.65°
혹은0.43(𝑟𝑟𝑟𝑟𝑟𝑟)
NAME _____________________________
ID Number__________________________
Problem 2 (20 points)
(a) (4 points)
Since the atmospheric pressure is applied on both sides, we don’t have to consider the
atmospheric pressure for the net hydrostatic force.
𝐹𝐹 = 𝑃𝑃𝐶𝐶𝐶𝐶 𝐴𝐴 = 𝜌𝜌𝑔𝑔ℎ𝐶𝐶𝐶𝐶 𝐴𝐴 = 9.8 × 1000 × 1.5 sin(60°) × 3 × 2 = 76383
= 𝟕𝟕. 𝟔𝟔 × 𝟏𝟏𝟏𝟏𝟒𝟒 [𝐍𝐍]
The F acts in the direction normal to the gate (and onto the gate).
Partial point: ℎ𝑐𝑐𝑐𝑐 (+1 point)
Deduction: without direction (-1 point)
(b) (3 points)
CG is 1.5 m from the hinge. The original of the coordinate system is on CG.
𝑦𝑦𝐶𝐶𝐶𝐶 =
−𝐼𝐼𝑥𝑥𝑥𝑥 𝜌𝜌𝜌𝜌 sin(60°)
𝑝𝑝𝐶𝐶𝐶𝐶 𝐴𝐴
=
−𝐼𝐼𝑥𝑥𝑥𝑥 𝜌𝜌𝜌𝜌 sin(60°)
𝜌𝜌𝜌𝜌ℎ𝐶𝐶𝐶𝐶 𝐴𝐴
=−
𝐼𝐼𝑥𝑥𝑥𝑥 𝜌𝜌𝜌𝜌 sin(60°)
𝜌𝜌𝜌𝜌1.5 sin(60°)𝐴𝐴
1
×2×33
= − 12
= −0.5 [𝑚𝑚]
1.5×2×3
Therefore, the distance between the hinge and CP is 1 m.
Partial point: 𝑦𝑦𝐶𝐶𝐶𝐶 = −0.5 (+1 point)
Deduction: use miscalculated F(form(a)) (-2 point)
(c) (8 points)
Momentum equilibrium w.r.t the hinge
∑ 𝑀𝑀 = 0
Consider three momentum components caused by the hydrostatic force F, the cable
tension T, and the gate weight W.
∑ 𝑀𝑀 = 𝑇𝑇 × 4 × sin(60°) − 𝑊𝑊 × 2 × cos(60°) − 𝐹𝐹 × 1 = 0
𝐓𝐓 = 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 = 𝟐𝟐. 𝟑𝟑 × 𝟏𝟏𝟏𝟏𝟒𝟒 [𝑵𝑵]
Deduction:
1. Without gate weight (-2 points)
∑ 𝑀𝑀 = 𝑇𝑇 × 4 × sin(60°) − 𝐹𝐹 × 1 = 0 , 𝐹𝐹 = 22050 = 2.2 × 104 [𝑁𝑁]
2. Use miscalculated F(from(a)) or CP(from(b)) (-2 points)
NAME _____________________________
ID Number__________________________
(d) (5 points)
Upward buoyant force by the buoy:
4
𝐹𝐹𝑏𝑏 = 𝜌𝜌𝜌𝜌𝑉𝑉𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 1000 × 9.8 × 0.5 × 3 𝜋𝜋(0.5)3 = 2566 = 2.6 × 103 [𝑁𝑁]
Net upward force by the buoy (𝑊𝑊𝑏𝑏 : weight of the buoy):
𝐹𝐹𝑢𝑢 = 𝐹𝐹𝑏𝑏 − 𝑊𝑊𝑏𝑏 = 2.6 × 103 − 300 = 2.3 × 103 [𝑁𝑁]
Momentum equilibrium:
∑ 𝑀𝑀 = 𝑇𝑇 × 4 × sin(60°) − 𝑊𝑊 × 2 × cos(60°) − 𝐹𝐹 × 1 = 0
(without the buoy)
∑ 𝑀𝑀 = (𝑇𝑇 − 200) × 4 × sin(60°) − 𝑊𝑊 × 2 × cos(60°) − 𝐹𝐹 × 1 + 𝐹𝐹𝑢𝑢 × 𝑑𝑑 ×
cos(60°) = 0
(d: the distance between the hinge and the string)
Form the above two equations,
200 × 4 × sin(60°) − Fb × 𝑑𝑑 × cos(60°) = 0
𝑑𝑑 =
200×4×𝑠𝑠𝑠𝑠𝑠𝑠(60)
𝐹𝐹𝑢𝑢 ×𝑐𝑐𝑐𝑐𝑐𝑐(60°)
= 𝟎𝟎. 𝟔𝟔𝟔𝟔 [𝒎𝒎]
Partial point: buoyance force 𝐹𝐹𝑏𝑏 = 2.6 × 103 [𝑁𝑁] (+1 point)
Deduction:
1. Without buoy weight (-2 points), if you use full equation.
(∑ 𝑀𝑀 = −200 × 4 × sin(60°) + 𝐹𝐹𝑢𝑢 × 𝑑𝑑 × cos(60°) = 0, there is no deduction.)
2. Use miscalculated F(from(a)) or CP(from(b)) (-2 points)
NAME _____________________________
ID Number__________________________
Problem 3 (25 points)
(a) (10 points)
Fy
Fx
Fixed, stationary CV
1) mass conservation
m=
m 2 + m 3
1
+1
+1
ρ=
ρ A2V2 + ρ A3V3
AV
1 1
D2
D12
D22
=
V1
V2 + 3 V3
4
4
4
2
2
D1 V1 − D2 V2 0.122 × 5 − 0.062 × 5
=
= 33.75m / s
V3 =
D32
0.042
+1
2) momentum conservation
Fx m 2V2 − m 1V1
=
2
Fx ρ A2V22 − ρ AV
=
1 1
+1
+ 1.5
 0.062 − 0.122 
D22
D2
2
2
1000 × 
π V22 − ρ 1 π V1=
 ×π × 5
4
4
4


Fx = −212.0575 N
+1
F=
ρ
x
NAME _____________________________
ID Number__________________________
Fy − ( ρVwater g + mg ) =
−m 3V3
+1
Fy =
− ρ A V + ( ρVwater g + mg )
2
3 3
Fy =− ρ
+ 1.5
 0.042 
D32
2
π V32 + ( ρVwater g + mg ) =−1000 × 
 × π × 33.75 + 1000 ×1× 9.8 + 10 × 9.8
4
 4 
Fy = 8466.6 N
+1
(b) (15 points)
moving, stationary CV (IRF on the ground)
a) mass conservation
+1
m=
m 2 + m 3
1
ρ=
ρ A2V2 + ρ A3V3
AV
1 1
2
1
2
2
+2
+2
2
3
D
D
D
=
V1
V2 +
V3
4
4
4
D12V1 − D22V2 0.122 × 5 − 0.062 × 5
=
= 33.75m / s
V3 =
D32
0.042
+1
2) momentum conservation
F=
m 2 (V2 + Vc ) − m 1 (V1 + Vc ) + m 3Vc
x
+3
=
Fx ( ρ A2V2 )(V2 + Vc ) − ( ρ AV
1 1 )(V1 + Vc ) + ( ρ A3V3 )Vc
2
2
2
1
2
3
+4
D
D
D
π V2 (V2 + Vc ) − ρ
π V1 (V1 + Vc ) + ρ
π V3Vc
4
4
4
 0.062 × 5 × 6 − 0.122 × 5 × 6 + 0.042 × 33.75 ×1 
=
1000 × 
 ×π
4


=
Fx ρ
Fx = −212.0575 N
+2
moving, stationary CV (IRF on the pipe)
=
Fx m 2V2 − m 1V1
=
Fx ( ρ A2V2 )(V2 ) − ( ρ AV
1 1 )(V1 )
D22
D2
π V2 2 − ρ 1 π V12
4
4
2
 0.06 − 0.122 
2
= 1000 × 
 ×π × 5
4


=
Fx ρ
Fx = −212.0575 N
NAME _____________________________
ID Number__________________________
1) Without specifications of the CV and IRF −1, respectively.
2) Wrong velocity direction (sign) in the momentum equation −0.5
3) Consider the pressure force in the momentum equation −0.5
4) Omit the gravity terms, mass of water and pipe + plate, −0.5, respectively
NAME _____________________________
ID Number__________________________
Problem 4 (25 points)
Solution)
(a)
=
t0
2m / (0.5 × 9.8m / s 2 )
=
=
v2 15
m / t0 23.48m / s
0.5 ×1000kg / m3 × v2 2 + 1000kg / m3 × 9.8m / s 2 × 2m + 101.3kPa
= 1000kg / m3 × 9.8m / s 2 ×15m + P1
P1 = 249.53kPa
(b)
Q=
A2 × v2
2
A2 (2.5cm) 2 × π
=
Q2 = 0.0461m3 / s
1
Q2
99
0.5 ×1000kg / m3 × v2 2 + 1000kg / m3 × 9.8m / s 2 × 2m + 101.3kPa
Q3 =
= 1000kg / m3 × 9.8m / s 2 ×10m + 101.3kPa + 0.5 ×1000kg / m3 × v32
v3 = 19.86m / s
/ v3 2.34 ×10−5 m 2
A3 Q3 =
=
(c)
=
t0
2m / (0.5 × 9.8m / s 2 )
=
=
v2 20
m / t0 31.30m / s
=
kPa 1000kg / m3 × 9.8m / s × h1 + P1
0.5 × 1000kg / m3 × v2 2 + 1000kg / m3 × 9.8m / s 2 × 2m + 101.3
P1 = 249.53kPa
h1 = 36.87 m
: -4 points subtraction when the mistakes happens in the Bernoulli equation part. When
the mistaken in the physical quantities (likes t0, Q……) make a different answer. The
score of correct process is counted. (ex. Prob 4 in (c), Wrong answer h1 is got due to the
selection of wrong v2. Only -1.5 points subtraction is counted. If the other processes are
correct.)