57:020 Fluids EFD Lab3
Fall 2009
Appendix C. Calculation of Airfoil Drag Force Using the Momentum Integral Method
Consider an experiment in which there is drag force on an airfoil that is immersed in a steady
incompressible flow. The drag force can be determined from the measurement of velocity distributions
far upstream and downstream of the airfoil body (figure below).
1. Velocity far upstream is the uniform flow 𝑈∞ .
2. Velocity in the wake of the body is measured with hotwire/Pitot probe to be 𝑢(𝑦), which is less
than 𝑈∞ due to the drag of the airfoil.
3. Objective: Find the drag force 𝐷 per unit length (span wise) of the airfoil.
Method 1: Choose a control volume following the flow streamline.
Fig. 1 Control volume following the flow streamline.
Solution:
Find relation between 𝐻 and 𝑏 using the mass conservation
̂
Since we choose the streamline as the control volume (CV), there is no mass flow across it. For the CV, 𝒏
is the unit normal vector and it is assumed that the CV has a unit depth.
̂ ) 𝑑𝐴 = 0
𝜌 ∫ (𝑽 ⋅ 𝒏
(1)
𝐶𝑉
𝑦𝑈
𝜌 ∫(−𝑈∞ )𝑑𝑦 + 𝜌 ∫𝑢(𝑦)𝑑𝑦 = 𝜌𝑈∞ 𝐻 + 𝜌 ∫ 𝑢(𝑦)𝑑𝑦 = 0
1
3
(2)
𝑦𝐿
Thus,
𝐻=
1 𝑦𝑈
∫ 𝑢(𝑦)𝑑𝑦
𝑈∞ 𝑦𝐿
(3)
1
57:020 Fluids EFD Lab3
Fall 2009
Momentum conservation
The pressure is uniform and so there is no net pressure force. The flow is assumed incompressible and
steady, so the momentum conservation equation applies only across the sections 1 and 3 without any
unsteady terms.
̂ )𝑑𝐴 = ∑ 𝐹𝑥
𝜌 ∫ 𝑢(𝑽 ⋅ 𝒏
(4)
𝐶𝑉
𝜌 ∫𝑈∞ (−𝑈∞ )𝑑𝑦 + 𝜌 ∫𝑢(𝑦)(𝑢(𝑦))𝑑𝑦 = −𝐷
1
(5)
3
∴𝐷=
2
𝜌𝐻𝑈∞
𝑦𝑈
− 𝜌 ∫ 𝑢(𝑦)2 𝑑𝑦
(6)
𝑦𝐿
Substitute 𝐻 from the mass conservation equation (3) into (6),
𝑦𝑈
𝐷 = 𝜌 ∫ 𝑢(𝑦)(𝑈∞ − 𝑢(𝑦))𝑑𝑦
(7)
𝑦𝐿
Thus,
2 𝑦𝑈 𝑢(𝑦)
𝑢(𝑦)
𝐶𝐷 = 1 2 = ∫
(1 −
) 𝑑𝑦
𝜌𝑈∞ 𝑐
𝑐 𝑦𝐿 𝑈∞
𝑈∞
2
𝐷
(8)
Or, in a non-dimensional form
∗
𝑦𝑈
𝐶𝐷 = 2 ∫ 𝑢∗ (1 − 𝑢∗ )𝑑𝑦 ∗
𝑦𝐿∗
(9)
where,
𝑦∗ =
𝑦
𝑐
𝑢∗ =
𝑢(𝑦)
𝑈∞
; 𝑦𝐿∗ =
𝑦𝐿
;
𝑐
𝑦𝐻∗ =
𝑦𝐻
𝑐
and 𝑐 is the chord length of the airfoil.
2
57:020 Fluids EFD Lab3
Fall 2009
Method 2: Rectangular control volume
Fig. 2 Rectangular control volume
Solution:
Mass conservation
Now, there is outflow of mass (and 𝑥-momentum) across the sections 2 and 4. Then, the mass
conservation equation (1) is written as,
𝜌 ∫(−𝑈∞ )𝑑𝑦 + 𝜌 ∫𝑣(𝑥)𝑑𝑥 + 𝜌 ∫𝑢(𝑦)𝑑𝑦 + 𝜌 ∫(−𝑣(𝑥))𝑑𝑥 = 0
1
2
3
(10)
4
𝑦𝑈
∴ 𝜌 (∫𝑣(𝑥)𝑑𝑥 − ∫𝑣(𝑥)𝑑𝑥 ) = 𝜌𝑏𝑈∞ − 𝜌 ∫ 𝑢(𝑦)𝑑𝑦
2
4
(11)
𝑦𝐿
Momentum conservation
The 𝑥-momentum equation (4) can be expressed as
𝜌 ∫𝑈∞ (−𝑈∞ )𝑑𝑦 + 𝜌 ∫𝑢(𝑥)𝑣(𝑥)𝑑𝑥 + 𝜌 ∫𝑢(𝑦)𝑢(𝑦)𝑑𝑦 + 𝜌 ∫𝑢(𝑥)(−𝑣(𝑥))𝑑𝑥 = −𝐷
1
2
3
(12)
4
If it is assumed that the x-directional velocity at the sections 2 and 4 are nearly same as the free stream
velocity, i.e. u(x)  U, then, the second and fourth integrals at the left hand side of the equation (12) can
be rewritten as
𝜌 ∫𝑈∞ 𝑣(𝑥)𝑑𝑥 + 𝜌 ∫𝑈∞ (−𝑣(𝑥))𝑑𝑥 = 𝑈∞ 𝜌 (∫𝑣(𝑥)𝑑𝑥 + ∫(−𝑣(𝑥))𝑑𝑥 )
2
4
2
(13)
4
By using the relation (11) from the mass conservation, the right hand side of the equation (13) can be
rewritten as
3
57:020 Fluids EFD Lab3
Fall 2009
𝑦𝑈
𝑈∞ ⋅ 𝜌 (∫𝑣(𝑥)𝑑𝑥 + ∫(−𝑣(𝑥))𝑑𝑥 ) = 𝑈∞ (𝜌𝑏𝑈∞ − 𝜌 ∫ 𝑢(𝑦)𝑑𝑦)
2
4
(14)
𝑦𝐿
Then, the x-momentum equation (12) becomes as
𝑦𝑈
𝑦𝑈
2
−𝜌𝑈∞
𝑏 + 𝑈∞ (𝜌𝑏𝑈∞ − 𝜌 ∫ 𝑢(𝑦)𝑑𝑦) + 𝜌 ∫ 𝑢(𝑦)2 𝑑𝑦 = −𝐷
𝑦𝐿
(15)
𝑦𝐿
𝑦𝑈
∴ 𝐷 = 𝜌 ∫ 𝑢(𝑦)(𝑈∞ − 𝑢(𝑦))𝑑𝑦
(16)
𝑦𝐿
Thus,
𝐶𝐷 = 1
𝐷
2𝑐
𝜌𝑈∞
2
=
𝑦𝑈
2
∫
𝑢(𝑦)(𝑈∞ − 𝑢(𝑦))𝑑𝑦
2𝑐
𝑈∞
𝑦𝐿
(17)
Or, in a non-dimensional form
∗
𝑦𝑈
𝐶𝐷 = 2 ∫ 𝑢∗ (1 − 𝑢∗ )𝑑𝑦 ∗
𝑦𝐿∗
(18)
where,
𝑦
𝑦 ∗ = 𝑐 ; 𝑦𝑈∗ =
𝑢∗ =
𝑦𝑈
;
𝑐
𝑦𝐿∗ =
𝑦𝐿
𝑐
𝑢(𝑦)
𝑈∞
4
57:020 Fluids EFD Lab3
Fall 2009
Example
𝑈∞ = 14.4 m/s, 𝑐 = 0.3048 m, AOA = 16
0.8
0.6
0.4
y*
0.2
0.0
-0.2
-0.4
-0.6
-0.8
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
u
*
i
yi (m)
ui (m/s)
yi*
u i*
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
0.200
0.100
0.050
0.025
0.015
0.010
0.005
0.003
0.000
-0.003
-0.005
-0.008
-0.010
-0.015
-0.025
-0.050
-0.100
-0.151
14.4384
13.9520
13.1723
12.8620
12.8298
12.2982
10.5453
9.4002
7.9273
6.6970
8.3346
10.9333
13.0791
13.2519
13.1977
13.3596
13.5565
13.6128
0.65617
0.32808
0.16404
0.08202
0.04921
0.03281
0.01640
0.00984
0.00000
-0.00984
-0.01640
-0.02625
-0.03281
-0.04921
-0.08202
-0.16404
-0.32808
-0.49541
1.00000
0.96631
0.91231
0.89082
0.88859
0.85178
0.73037
0.65106
0.54904
0.46383
0.57725
0.75724
0.90586
0.91783
0.91407
0.92529
0.93892
0.94282
Pitot measured velocity profile (left) and the measurement data (right), where y* = 0 is at the trailing edge
(TE) of the wing model, and the measurement is at about one inch behind the TE.
Drag coefficient 𝐶𝐷 using the momentum integral method
𝐶𝐷 = 2∫ 𝑢∗ (1 − 𝑢∗ )𝑑𝑦 ∗
The integration may be evaluated numerically such that
17
𝐶𝐷 = 2 × ∑ (
𝑖=1
𝑓𝑖 + 𝑓𝑖+1
) ⋅ Δ𝑦𝑖∗ = 0.1398
2
where,
𝑓𝑖 = 𝑢𝑖∗ ⋅ (1 − 𝑢𝑖∗ )
∗
Δ𝑦𝑖∗ = 𝑦𝑖+1
− 𝑦𝑖∗
5
57:020 Fluids EFD Lab3
Fall 2009
0.5
Loadcell measurements (Fall09)
Momentum integral method (Fall 09)
0.4
CD
0.3
0.2
0.1
0.0
0
5
10
15
20
25
 ()
30
Comparisons of the drag coefficient 𝐶𝐷
6