Problem Description:
This exercise examines the subsonic flow over a NACA 5012 airfoil at Reynolds numbers of
Re = 100 and 1000. The unit chord length is 𝑐 = 1 and the free-stream velocity is 𝑈∞ =
1 𝑚𝑠 −1 .
Results and Discussion:
(1a)
By examining the graph plot of time versus 𝐶𝐿 & 𝐶𝐷 for 𝑅𝑒 = 100 & 1000, it is seen that
steady state is achieved around 𝑡 = 2𝑠 for 𝑅𝑒 = 100 and 𝑡 = 4.5𝑠 for 𝑅𝑒 = 1000. Steady
state takes longer to achieve for flows of higher Reynolds numbers.
Figure 1a.1: Graph plot of time versus lift and drag coefficients for Re = 100.
Figure 1a.2: Graph plot of time versus lift and drag coefficients for Re = 1000.
(1b)
The graph plot of pressure for Re = 100 and 1000 are as follows.
Figure 1b.1: Graph plot of pressure at t = 10s for Re = 100.
Figure 1b.2: Graph plot of pressure at t = 10s for Re = 1000.
(1c)
The integral equation can be approximated with Riemann Sum. Since we are dealing with
more than one thousand points for each surface pressure, choosing between the different
kinds of Riemann Sums would not yield much of a difference. I opted for Right Hand
Riemann Sum for simplicity.
The integral equation can be decomposed as such:
𝑥=𝑐
𝑥=𝑐
𝑥=𝑐
𝐿 1
1
= ∫ 𝑝𝑝𝑠 − 𝑝𝑠𝑠 𝑑𝑥 = ( ∫ 𝑝𝑝𝑠 𝑑𝑥 − ∫ 𝑝𝑠𝑠 𝑑𝑥)
𝑆 𝑐
𝑐
𝑥=0
𝑁
𝑥=0
𝑁
𝑥=0
≈ ∑ 𝑝𝑝𝑠,𝑖 (𝑥𝑝𝑠,𝑖 − 𝑥𝑝𝑠,𝑖−1 ) − ∑ 𝑝𝑠𝑠,𝑖 (𝑥𝑠𝑠,𝑖 − 𝑥𝑠𝑠,𝑖−1 )
𝑖=0
𝑖=0
In (1b), we have already separated the pressure points to 𝑝𝑝𝑠 and 𝑝𝑠𝑠 , respectively. All there
is left to do is to perform the sums separately, then minus the values at the end to obtain the
lift force per unit area. The results are as follows.
𝐿
( )
= 9.179 × 10−2 𝑁
𝑆 𝑅𝑒=100
𝐿
( )
= 8.936 × 10−2 𝑁
𝑆 𝑅𝑒=1000
This result is expected from the two tested Reynolds numbers. With every other parameter
kept constant, the higher the Reynolds number, the lower the viscosity of the fluid. This
causes a thicker boundary layer to form around the airfoil and thus, creating more drag force.
This ultimately causes the airfoil to have less lift over higher Reynolds numbers. This can be
seen from the graph plots in (1b) where the 𝑝𝑝𝑠 for Re = 1000 takes a downward-bowing
shape than the one for Re = 100.