I have to do some calculations to get the drag from an experiment with a wake rake. In the equation I have to enter the total pressure coefficient $C_{pt}$, but in my calculations it seems to always be equal to +1. Is this correct?
[Physics] Is the total pressure coefficient always 1 in incompressible flow
aerodynamics
Related Solutions
According to Sighard Hoerner's Fluid Dynamic Drag, this would be the half-sphere with the open side exposed to the wind. Its drag coefficient is 1.42. A rod with a hemispherical cross section will even have a drag coefficient of 2.3 (right column in the graph below).
If you restrict the competition to solid objects, still the half sphere wins with a drag coefficient of 1.17. In all cases, the reference area is the cross section orthogonal to the flow direction.
Figure 33 from Sighard Hoerner's Fluid Dynamic Drag, Chapter 3.
Note that the difference in drag of half spheres due to their orientation is used in anemometers for measuring the wind speed. When the open face is turned away from the wind, its drag coefficient drops to 0.42.
The reason for the difference, and the high drag when the open side is exposed to the wind, is the massive separation around and behind the sphere. Air flowing out from the inside and over the rim of the sphere will need some space to "turn around", effectively increasing the blocked cross section that the outside flow experiences. When the round side is exposed to the wind, the separation is restricted to the cross section of the sphere itself.
Here is the way in which you should interpret the different symbols:
- $\widehat{\mathbf{i}}$ is the unit vector in the direction of relative motion (of the object relative to the mass of air).
- $\mathbf{n}$ is the unit outer normal to the spherical surface.
- $\widehat{\mathbf{t}}$ is the director vector of the tangential part of the traction vector, whose magnitude is $T_w$. That is, the tangential part of $\mathbf{\sigma} \cdot \mathbf{n}$, with $\mathbf{\sigma}$ the Cauchy stress tensor. Note that the tangential component of an arbitrary vector $\mathbf{v}$ can be calculated as $\mathbf{v}-\mathbf{n} \cdot \mathbf{v}$.
Notice that the first term is the projection along the direction of motion $\widehat{\mathbf{i}}$ of integral of the normal component of the traction vector (the pressure) over the sphere (the rest of components are zero (on average) due to symmetry). This is the pressure drag.
Similarly, the second term represents the integral of the remaining components, also projected along the direction of motion. This is the friction (or skin) drag, which is due to viscous friction.
Best Answer
To get the drag from a wake survey, use the following formula from Fundamentals of Aerodynamics by John D. Anderson:
$D' = \int_{-h}^h\rho u\left(U_\infty - u\right)dy$
I'm assuming you know $U_\infty$ from a freestream Pitot tube. You can find $u$ from the wake rake using Bernoulli's equation. The equation above can be derived using a control volume of the entire test section.