CIVL 7020 – Fluid Mechanics
Quiz 5
1. Consider the visualization of flow over a sphere below. Are we seeing streamlines, streaklines,
pathlines, or timelines? Briefly explain. (2 marks)
Pathlines
The picture is a time exposure of air bubbles in water, each
white streak shows the path of an individual air bubble.
Therefore, these are pathlines. Since the outflow (top and
bottom portions of the photo) appears to be steady, these
pathlines are the same as streaklines and streamlines
Visualization is produced by a time exposure of air bubbles in water.
2. Is the control volume shown a Lagrangian or an Eulerian description? Explain. (2 marks)
Eulerian
Field variables (e.g., pressure, velocity) are described at any
location and instant in time
3. Is momentum a vector? If so, in what direction does it point? (1 mark)
since momentum is a product of a vector (velocity) and a scalar (mass), momentum must be a vector
points in the same direction as the velocity vector
4. Under what condition can the momentum-flux correction factor be ignored? (1 mark)
a) Steady state flow
b) Conservation of mass
c) Incompressible fluid
d) Uniform flow
12
5. Consider the following steady, two-dimensional velocity field:
⃗ = (𝑢, 𝑣) = (0.46 + 2.5𝑥)𝑖 + (−2.7 − 2.1𝑦)𝑗
𝑉
Is there a stagnation point? If so, where is it? (2 marks)
𝑢 = (0.46 + 2.5𝑥)
v= (−2.7 − 2.1𝑦)
0= (−2.7 − 2.1𝑦)
0 = (0.46 + 2.5𝑥)
𝑥 = -0.18
𝑦 = -1.29
6. A reducing elbow in a horizontal pipe is used to deflect water flow by an angle 𝜃� = 45° from the
flow direction while accelerating it. The elbow discharges water into the atmosphere. The crosssectional area of the elbow is 150 cm2 at the inlet and 25 cm2 at the exit. The elevation difference
between the centres of the exit and the inlet is 30 cm. The mass of the elbow and the water in it is
40 kg. Determine the anchoring force needed to hold the elbow in place. Take the momentum-flux
correction factor to be 1.05 at both the inlet and outlet. (4 marks)
A1 = 0.015 m2
V1 = 2.0 m/s
A2 = 0.0025 m
𝑚̇ = 𝜌𝑉𝑎𝑣𝑔 𝐴𝑐
2
V2 = 12 m/s
𝑃1
𝑉12 𝑃2
𝑉22
+ 𝑧1 +
= + 𝑧2 +
𝛾
2𝑔
𝛾
2𝑔
𝑉22
𝑃1 − 𝑃2 = 𝜌𝑔 ( 2 + 𝑧2 )
𝑉1
Where P2 = Patm and P1 = Pgauge
rearrange equation
P1 = 72943 Pa = 72.9 kPa
∑ 𝐹 = ∑ 𝛽𝑚̇ 𝑉 − ∑ 𝛽𝑚̇ 𝑉
𝑜𝑢𝑡
𝑖𝑛
𝐹𝑅𝑥 = −890�𝑁 = � −0.89�𝑘𝑁
𝐹𝑅𝑥 + 𝑃1 𝐴1 = 𝛽𝑚̇𝑉2 𝑐𝑜𝑠𝜃 − 𝛽𝑚̇𝑉1
𝐹𝑅𝑧 − 𝑊 = 𝛽𝑚̇𝑉2 𝑠𝑖𝑛𝜃
𝐹𝑅𝑧 = 660�𝑁 = 0.66�𝑘𝑁