ME221 Fluid Mechanics (Fall 2020)
Homework 6 (Total 100 points)
Monday, November 23, 2020 (Out)
DUE: Sunday (11:59 PM), December 6, 2020
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*If there are no detail information on the physical properties of liquid or gas, please define them yourself.
Furthermore, please show all your work to receive any partial credit when solving your problem. If you
only write down the answers without proper step by step solution, no points will be given.
Problem 1 (15 points) Water at 20°C is to be siphoned through a tube 1 m long and 2 mm in
diameter, as shown in the Figure. Neglect the tube curvature and minor losses. Use the following
assumptions to begin solving the problem.
- Steady, laminar pipe flow
- Flow velocity at point 1 is negligible
- Fully-developed flow at point 2
(a) (8 points) Is there any height H for which the flow might not be laminar?
(b) (3 points) What is the flow rate if H = 50 cm?
(c) (4 points) Verify if flow is fully developed at point 2
Solution:
(a) The properties of water at 20℃ are ρ = 998 kg/m3 and μ = 0.001 kg/m∙s.
Steady flow energy equation:
p1 V12 p2 V22
+ α1 + z1 = +α2 + z2 + h f (1)
ρg 2g ρg 2g
( +2 )
Since p1 = p2 = patm and V1 = 0, Eq. (1) becomes
V22 V2 (2)
h f = ( z1 − z2 ) − α 2
=H− 2
2g g ( +2 )
if we set α2 = 2.0 for laminar fully-developed flow.
It is also known that
32 µLV2 (3)
hf =
ρgd 2
2
Equating (2) and (3) yields a quadratic expression for V2 as
V22 32µL
+ V2 − H = 0 (4)
g ρgd 2
( +2 )
The only positive solution for V2 will thus be
2
16µL 16µL (5)
V2 = − + + gH
2
ρd 2 ρd
Some possible values of H and their corresponding values of V2 are tabulated below:
H (m) V2 (m/s) Q (m3/s) ( = πd2V2/4) Re ( = ρV2d/μ)
0.1 0.120 3.78E-7 240
0.2 0.237 7.46E-7 474
0.3 0.351 1.10E-6 701
0.4 0.462 1.45E-6 923
0.5 0.571 1.79E-6 1139
0.6 0.676 2.13E-6 1350
0.7 0.780 2.45E-6 1557
0.8 0.881 2.77E-6 1759
0.9 0.980 3.08E-6 1957
1.0 1.078 3.39E-6 2151
Since L=1 m, the maximum value of H will also be 1.0 m, as a rough approximation.
Even at this value of H, Re does not exceed 2300, the pipe flow transition value.
∴ The Flow is always laminar. ( +2 )
(b) As can be seen from the Table, the flow rate at H=0.5m is Q=1.79E-6 m3/s. ( +3 )
However, the value of Re=2151 at H=1.0 m might induce some intermittent bursts of turbulence.
(C) Check whether the flow becomes fully-developed at point 2:
Entrance region Le = 0.06Re×d = 0.06×2151×0.002m = 0.258 m < L = 1.0 m
∴ The flow becomes fully-developed at point 2 (+4)
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Problem 2 (10 points)
Kerosine at 20oC (ρ=804kg/m3, μ=1.92x10-3kg/m•s) is pumped at 0.15m3/s through a 20km of
16-cm-diameter cast-iron (ε=0.26mm) horizontal pipe. Compute the input power required if the
pumps are 85 percent efficient.
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Problem 3 (15 points)
The tank-pipe system of the figure below is to deliver water at 20oC (ρ=998kg/m3,
μ=0.001kg/m⋅s) to the reservoir. The pipe is made of cast iron (ε = 0.26mm). Take the kinetic
energy correct factor α = 1. Estimate the flow rate.
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Problem 4 (20 points)
4-A 4-B
Part A. (10 points)
For the configuration shown in figure 4-A, the fluid is ethyl alcohol at 20°C, and the tanks are
very wide. Find the flow rate which occurs in m3/h.
Is the flow laminar? Assume steady flow inside the pipe.
ρethyl alcohol =789 kg/m3 and μethyl alcohol = 0.0012 kg/m∙s
Part B. (10 points)
Let us attack Problem Part A in symbolic fashion, using figure 4-B. All parameters are constant
except the upper tank depth Z(t).
Find an expression for the flow rate Q(t) as a function of Z(t). Set up a differential equation, and
solve for the time t0 to drain the upper tank completely.
Assume quasi-steady laminar flow.
PART A
Assumptions:
- Steady pipe flow
- Flow velocities at points 1 and 2 are negligible
- Ethanol surface areas are very large at both tanks
Steady flow energy equation:
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p1 V2 p V2
+ α 1 1 + z1 = 2 + α 2 2 + z 2 + h f (1)
ρg 2g ρg 2g
( +2 )
Since p1 = p2 = patm and V1 = V2=0, it becomes
h f = ( z1 − z 2 ) = 0.90 m (2)
( +1)
It is also known that
32 µLV p 128µLQ (3)
hf = =
ρgd 2
πρgd 4 ( +1 )
where Vp is the flow velocity in the pipe.
Equating (2) and (3) enables us to solve for Q as:
πρgd 4 π (789kg/m 3 )(9.8m/s 2 )(0.002m) 4
Q= hf = (0.9m) = 1.90E − 6 m 3 /s
128µL 128(0.0012kg/m ⋅ s)(1.20m)
( +2 )
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or, in m /h,
Q = (1.90E − 6 m 3 /s)(3600 s/h) = 6.83E - 3 m 3 /h (Answer)
( +2)
The Reynolds number will thus be
4Qρ 4(1.90 E − 6 m 3 /s)(789 kg/m 3 )
Re = = = 795
πµd π (0.0012 kg/m ⋅ s)(0.002 m)
( +2 )
∴ The flow is laminar. (Answer)
PART B.
Assumptions:
- Quasi-steady laminar flow in pipe
- Upper tank is in cylindrical form, with diameter D
- Lower tank is very large so that H and h remains constant
We recall Eq. (2) from Part A, but now in symbolic form 1:
h f = ( z1 − z 2 ) = h + Z (t ) (4)
( +2 )
1
Or, according to the physical nature of the terminology of head loss, one can intuitively know that the head loss
required to drive the pipe flow would be the same as the height difference of the surfaces, which has the same result
as in Eq. (4) above.
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it is also known that
32 µLV p (5)
hf =
ρgd 2
( +1 )
combining (4) and (5) yields an expression for Vp:
ρgd 2
Vp = (h + Z (t )) (6)
32µL ( +1)
And, from the mass balance
πd 2 πD 2 dZ (t ) (7)
Vp = Q = −
4 4 dt (Answer)
( +1 )
Substituting Eq. (6) to Eq. (7) yields
dZ ρgd 4
= dt (8)
h + Z 32 µLD 2 ( +1 )
Integrating both sides of Eq. (8) yields
ρgd 4
Z (t ) = (h + Z 0 ) exp − t −h
2
32 µLD ( +2 )
where Z0 is the initial value of Z(t).
Letting Z(t) = 0 and solving for t yields
32 µLD 2 Z 0 (9)
t0 = ln1 +
ρgd 4 h (Answer)
( +2 )
the time to drain the upper tank completely.
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Problem 5 (10 points)
Hydrogen at 20°C and approximately 1atm (ρ=0.0839kg/m3, μ=9.05x10-6 kg/m∙s) is to
be pumped through a smooth rectangular duct 85m long of aspect ratio 5:1. If the flow rate is
0.7m3/s and the pressure drop is 80Pa. What should the width and height of the duct cross
section be?
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Problem 6 (10 points) The 6-cm-diameter pipe contains glycerin at 20oC (ρ=1260kg/m3,
μ=1.49kg/m∙s) flowing at a rate of 6m3/h.
(a) (2pt) Verify that the flow is laminar.
(b) (5pt) For the pressure measurements shown, is the flow up or down? Why
(c) (3pt) If the bend losses are neglected, how long is the section of pipe between A and B?
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Problem 7 (8 points) Compute the position (r/R) in a turbulent pipe flow where local velocity u
equals average velocity V, independent of the Reynolds number.
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Problem 8 (12 points) Water at 20oC (ρ=998kg/m3, μ=10-3kg/m∙s) flows in a 9-cm-diameter
pipe under fully developed conditions. The centerline velocity is 10m/s.
(a) (7pt) Compute τw
(b) (3pt) Compute Q
(c) (2pt) Compute Δp for a 100-m pipe length.
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