Question 11.12: A 2D steady, constant density, inviscid flow of air is descri...
A 2D steady, constant density, inviscid flow of air is described by the velocity field u = Ax, v = −Ay, where A = 1.5 s−1 and the coordinates are measured in feet. Find the pressure difference between a point at (1, 1, 0) and a point at (2, 2, 0). Are these two points located on the same streamline?
We are asked to find the pressure difference between two specified points in a flow and to determine whether the points are located on the same streamline. Figure 11.15 serves as a sketch for this flow problem. No additional assumptions are required to solve this problem. Since this is an inviscid flow, we know that the Bernoulli equation is applicable along a streamline. If the flow is also irrotational, then the Bernoulli equation is applicable between any two points in the flow field. We will first check for irrotational flow in this 2D velocity field by using Eq. 10.51c to compute the vorticity. Inserting the known velocity components we find
\omega _z=\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y} \right)=\left[\frac{\partial (-Ay)}{\partial x}-\frac{\partial (Ax)}{\partial y}\right]=0
Thus, the flow is irrotational and we can apply the Bernoulli equation, Eq. 11.23, between points (1, 1, 0) and (2, 2, 0). For this steady constant density flow, Eq. 11.23 reduces to the traditional form of the Bernoulli equation as given by Eq. 8.6:
\int_{1}^{2}\frac{\partial u}{\partial t} • dr+\int_{1}^{2}{\frac{dp}{\rho }}+\frac{1}{2}\left(V^2_2-V^2_1\right)+g(z_2-z_1)=0 (11.23)
\frac{p_1}{\rho }+\frac{1}{2}V^2_1+gz_1=\frac{p_2}{\rho }+\frac{1}{2}V^2_2+gz_2
Since the points are at the same elevation, z1 = z2. To calculate the kinetic energy terms we write
V2 = u2 + v2 = (Ax)2 + (−Ay)2 = A2(x2 + y2)
Thus at (1, 1, 0) we can determine that V^2 _1 = A^2(x^2 _1 +y^2_ 1) = A^2(1^2 +1^2) =2A^2 (\mathrm{ft} ^2), and at the point (2, 2, 0) we have V^2 _2 = A^2 (x^2 _2 +y^2 _2) = A^2 (2^2 +2^2) =8A^2 (\mathrm{ft} ^2). In this case the Bernoulli equation becomes p_1/ρ +\frac{1}{2} 2A^2 (\mathrm{ft} ^2) = p_2/ρ +\frac{1}{2} 8A^2 (\mathrm{ft} ^2), and the pressure difference is p_1 − p_2 =3(\mathrm{ft} ^2)ρA^2. Inserting the data we find
p1 − p2 = 3(ft2)ρA2 = (3 ft2)(0.002378 slug/ft3)(1.5 s−1)²(lbf-s)/(slug-ft) \left(\frac{1\ \mathrm{ft}^2}{144\ in.^2} \right)
= 1.1 × 10−4 psia
To check whether these two points are on the same streamline, we can write the equation for a streamline in this steady flow, using Eq. 10.30 as
\frac{dy}{dx}=\frac{v(x, y, t) }{u(x, y, t)} (10.30)
\frac{dy}{dx}=\frac{v}{u}=\frac{-Ay}{Ax}=-\frac{y}{x}
Solving for the streamline we have dy/y = −dx/x, or x dy + y dx = 0. Thus the streamlines are given by xy = C (see Example 10.9). The streamline through the point (1, 1, 0) is xy = 1, and it is clear that this streamline does not pass through the point (2, 2, 0).
The family of streamlines and the pressure contours for this flow are shown in Figure 11.15. The latter are obtained by realizing that at the origin the velocity is zero; thus the pressure there is the stagnation pressure. Applying the Bernoulli equation between the origin (0, 0, 0) and any point (x, y, 0) we can write p/ρ +\frac{1}{2} V^2 = p_0/ρ +\frac{1}{2} V^2_ 0 , where the subscript zero denotes the origin. Since V^2 =u^2 +v^2 = A^2(x^2 +y^2), and V^2 _0 =0, we find p_0 − p =\frac{1}{2} ρA^2(x^2 +y^2), which shows that the pressure is constant on circles centered on the origin.
