Homework 10 Solution Manual
Problem 1 (6points)
SI (kg /m*s)
1.79E-5
1.12E-3
1
Air
Water
10W30
English (lb*s/ft^2)
3.74E-7
2.34E-5
0.02
Munson, Bruce, Donald Young, and Theodore Okiishi. Fundamentals of fluid mechanics. 4. Appendix B.
John Wiley & Sons Inc, 2002. Print.
Problem 2a (5 points)
The sheer force is the only force in the tangential direction.
𝑚
1𝑠
𝑑𝑢
Ns
𝑚𝑚
𝐹𝑡 = 𝜏𝐴 = (𝜇 ) 𝐴 = (1.79E − 5 2 ) (
) (1000
) ( 1𝑚2 ) = 0.0179 𝑁
𝑑𝑦
m
1 𝑚𝑚
𝑚
Problem 2b (5 points)
Stagnation pressure creates the normal force on the plate. The stagnation pressure can be calculated
using be Bernoulli equation.
1
1 1.23𝑘𝑔 1𝑚 2
𝐹𝑛 = 𝑃𝐴 = ( 𝜌𝑉 2 ) 𝐴 = ( (
) ( ) ) (1𝑚2 ) = 0.615 𝑁
2
2
𝑚3
𝑠
Problem 3 (5 points)
The ratio of normal to tangential force is presented bellow.
1
1 2
1
𝜌𝑉 (2 𝐿)
𝐹𝑛 2 𝜌𝑉 𝐴 2 𝜌𝑉𝐿
=
=
=
= 𝑅𝑒1
𝐿
𝑉
𝐹𝑡
𝜇
𝜇
2
𝜇𝐿 𝐴
Problem 4 (10 points)
The force balance will yield the equation shown bellow. Note that positive forces were in negative x
direction.
(𝑃2)𝑑𝑦𝑑𝑧 − (𝑃1)𝑑𝑦𝑑𝑧 = (𝜏2 )𝑑𝑥 𝑑𝑧 − (𝜏1 )𝑑𝑥 𝑑𝑧
Next note the following:
𝜏2 = 𝜇
𝜏1 = 𝜇
𝑑𝑢
| 𝑦 = 𝑦2
𝑑𝑦
𝑑𝑢
| 𝑦 = 𝑦1 = 𝑦2 + 𝑑𝑦
𝑑𝑦
𝜏1 = 𝜇 (
𝑑𝑢 𝑑 2 𝑢
+
𝑑𝑦)
𝑑𝑦 𝑑𝑦 2
(P2-P1) =dP
Substituting these equations into the force balance and canceling dz will yield
𝑑𝑃 𝑑𝑦 = −𝜇
𝑑2 𝑢
𝑑𝑦 𝑑𝑥
𝑑𝑦 2
𝑑2 𝑢
1 𝑑𝑃
= −
2
𝑑𝑦
𝜇 𝑑𝑥
𝑑𝑢
−1 𝑑𝑃
=
𝑦 + 𝐶1
𝑑𝑦
𝜇 𝑑𝑥
𝑢=
−1 𝑑𝑃 2
𝑦 + 𝐶1𝑦 + 𝐶2
2𝜇 𝑑𝑥
C1 and C2 could be obtained from the boundary conditions:
1) No slip at lower wall  U(y=0) = 0  C2 = 0
1 1 𝑑𝑃
2) No Slip at top wall  U (y = l) = 0  C1 = 2 𝜇 𝑑𝑥 𝑙
Therefore:
𝑢=
−1 𝑑𝑃 2
1 𝑑𝑃
𝑦 +
𝑙𝑦
2𝜇 𝑑𝑥
2𝜇 𝑑𝑥
Maximum velocity occurs in the center at y = l/2
𝑙
−1 𝑑𝑃 2 −1
1
−0.06𝑃𝑎
1𝑚 2
𝑢 (𝑦 = ) =
𝑙 =
(
)(
) (1𝑐𝑚)2 (
) = 0.00067𝑚/𝑠
2
8𝜇 𝑑𝑥
8 1.12E − 3Pa ∗ s
1𝑚
100𝑐𝑚
Problem 5 (10 points)
Drag on a flat plat could be found from Eq 18.22
𝐶𝑓 =
1.328
√𝑅𝑒
1.328
=
√(1.23 ∗ 1 ∗
1
)
1.79E − 5
= 0.005
1
𝐷𝑟𝑎𝑔 = 𝐶𝑓 𝜌𝑉 2 𝐴 = 0.015𝑁
2
Since the plate has 2 faces
𝑁𝑒𝑡 𝐷𝑟𝑎𝑔 = 2 ∗ 𝑃𝑙𝑎𝑡𝑒 𝐷𝑟𝑎𝑔 = 0.03 𝑁
Boundary layer thickness could be calculated using Eq. 18.23
𝛿=
5𝑥
√𝑅𝑒
Boundary Layer Thickness
0.02
BL Thickness (m)
0.015
0.01
0.005
0
-0.005
0
0.2
0.4
0.6
-0.01
-0.015
-0.02
Length Downstream (m)
0.8
1
Problem 6 (5 points)
Following steps in problem 5 yields:
𝐶𝑓 = 0.0014
𝐷𝑟𝑎𝑔 = 3.5 𝑁
𝑁𝑒𝑡 𝐷𝑟𝑎𝑔 = 7.03𝑁
Boundary Layer Thickness in Water
0.006
BL Thickness (m)
0.004
0.002
0
0
0.2
0.4
0.6
0.8
-0.002
-0.004
-0.006
Length Downstream (m)
Boundary Layer thickness Decreases but the drag increase.
1
1.2