1. Consider the flow of air over a small plate that is 5cm long in the flow direction and 1cm wide.
The free stream conditions correspond to standard sea level and the flow velocity is 120 m
s .
Assuming laminar flow, calculate the boundary layer thickness at the trailing edge and the drag
2
kg
force on the plate. (υair = 1, 5 · 10−5 ms , % = 1, 2 · m
3)
Solution:
Boundary layer thickness for laminar flows using Blasius eq.
r
4, 91x
υ·x
δ ≈ 4, 91
=√
u0
Rex
u∞ x
Rex =
υs
2
1, 5 · 10−5 ms · 0, 05m
≈ 3, 88 · 10−4 m ≈ 0, 04cm
120 m
s
δ ≈ 4, 91 ·
Drag force (blasius, laminar flow) with ReL =
0,664
√
Rex
local drag coefficient
Cf =
global (integral)
drag coefficient
Cd = Cw =
Fd =
=
τw
%u2
0
2
1,328
√
ReL
u0 ·L
υ
depend on position on the plate
(depend on length L of plate)
1
· % · u20 · A · cd
2
cd = r
1, 328
120 m
s ·0,05m
1,5·10−5
= 2, 099 · 10−3
m2
s
1
kg m 2
· 1, 2 3 · 120
· (1m · 0, 05m) · 2, 099 · 10−3
2
m
s
= 0, 907N
=
2 − sides of plate:
Fd = 2 · 0, 907N= 1, 814N
2. For the case of exercise 1, compare the local shear stress at 1cm and 5cm from the leading edge.
Solution:
wall shear stress for laminar flow
τw = 0, 332(Rex )−0,5 · %u20
τw (x = 0, 01m) = 0, 332 · 80000−0,5 · 1, 2
kg m 2
·
120
m3
s
≈ 20, 28Pa
τw (x = 0, 05m) = 0, 332 · 400000−0,5 · 1, 2
kg m 2
· 120
3
m
s
≈ 9, 07Pa
3. Assuming same parameters as in exercise 1, calculate the boundary layer thickness at the trailing
edge and the drag force on the plate but assuming now that the flow is fully turbulent from the
leading edge.
Solution:
Boundary layer thickness for turbulent flow


1/7
υ
6/7
 δ = 0, 217x

u0




!
1/7
2


−5 m
1,
5
·
10


6/7
s
δ = 0, 217 · (0, 05m) ·
=
120 m
s
Schlichting; 1/7 power law profile
δ=
0, 382x
1/5
R ex
=
0, 382 · 0, 05m
= 1, 44 · 10−3 m = 14, 4cm
4000000,2
Drag force: fully turbulent flow (estimation)
1
(a) cd =
(b) cd =
(c) cd =
0,074
5 · 105 ≤ ReL < 107
1/5
ReL
0,455
ReL ≥ 106
turbulent
(logReL )2,58
0,455
1700
transitional with Rexcr
(logReL )2,58 − ReL
= 5 × 105
Fd = 1/2 · % · u20 · A · cd
kg
m 2
(a) Fd = 1/2 · 1, 2 m
· (0, 05m2 ) · 5, 608 · 10−3 = 2, 42N
3 · 120 s
2 − sides of plate:
Fd = 2 · 2, 42N = 4, 84N
kg
m 2
· (0, 05m2 ) · 5, 33 · 10−3 = 2, 30N
(b) Fd = 1/2 · 1, 2 m
3 · 120 s
2 − sides of plate:
Fd = 2 · 2, 3N = 4, 6N
4. The boundary layer thickness δ and wall shear stress τw along a turbulent flat plate is described by
u0 x
0, 382x
,
Rex =
δ=
1/5
υ
Rex
r
τw
0, 074
τw
cd = %u2 ,
cd = 1/5 ,
uτ =
0
%
ReL
2
1
uτ
u+ = lmy + + c
y+ = y
κ
ν
m2
kg
)
(u0 = 10m/s, %air = 1, 2 3 , υair = 1, 5 · 10−5
m
s
(a) calculate the thickness of the viscous sublayer (y + ≤ 5) of the transition region (y + ≤ 30) and
of the logarithmic layer at the distance x = 1m from the leading edge of the plate.
Solution:
10m/s · 1m
= 666666, 6
1, 5 · 10−5 m2 /s
0, 382 · 1m
δ=
= 0, 0261m
(666666, 6)1/5
0, 074
kg
2
= 0, 303Pa
τw = %u20 · 1/2 · cd = 1, 2 3 · (10m/s) · 0, 5 ·
m
(666666, 6)0,2
s
0, 303Pa
m
uτ =
= 0, 503
kg
s
1, 2 m
3
υ
y = y+ ·
uτ
Rex =
2
1, 5 · 10−5 ms
(y = 5) → y = 5 ·
= 1, 49 · 10−4 m
0, 503 m
s
+
2
1, 5 · 10−5 ms
(y = 30) → y = 30 ·
= 8, 94 · 10−4 m
0, 503 m
s
+
(b)
u0 = 100m/s
100 m
s · 1m
Rex =
2 = 6666666, 6̄
1, 5 · 10−5 ms
0, 382 · 1m
δ=
= 0, 0165m
(6666666, 6̄)1/5
τw = 19, 17Pa
s
19, 17Pa
m
uτ =
= 3, 997
kg
s
1, 2 m3
2
1, 5 · 15−5 ms
(y = 5) → y = 5 ·
= 1, 876 · 10−5 m
3, 997 m
s
+
2
1, 5 · 10−5 ms
(y = 30) → y = 30 ·
= 1, 125 · 10−4 m
3, 997 m
s
+
(y + = 1) → y = 3, 75 · 10−6 m
2