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?

Question

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?

Accepted Answer

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

[latex]\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[/latex]

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:

[latex]\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[/latex] (11.23)

[latex]\frac{p_1}{\rho }+\frac{1}{2}V^2_1+gz_1=\frac{p_2}{\rho }+\frac{1}{2}V^2_2+gz_2[/latex]

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 [latex]V^2 _1 = A^2(x^2 _1 +y^2_ 1) = A^2(1^2 +1^2) =2A^2 (\mathrm{ft} ^2)[/latex], and at the point (2, 2, 0) we have [latex]V^2 _2 = A^2 (x^2 _2 +y^2 _2) = A^2 (2^2 +2^2) =8A^2 (\mathrm{ft} ^2)[/latex]. In this case the Bernoulli equation becomes [latex]p_1/ρ +\frac{1}{2} 2A^2 (\mathrm{ft} ^2) = p_2/ρ +\frac{1}{2} 8A^2 (\mathrm{ft} ^2)[/latex], and the pressure difference is [latex]p_1 − p_2 =3(\mathrm{ft} ^2)ρA^2[/latex]. Inserting the data we ﬁnd

p₁ − p₂ = 3(ft²)ρA² = (3 ft²)(0.002378 slug/ft³)(1.5 s⁻¹)²(lb_f-s)/(slug-ft) [latex]\left(\frac{1\ \mathrm{ft}^2}{144\ in.^2} \right)[/latex]
= 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

[latex]\frac{dy}{dx}=\frac{v(x, y, t) }{u(x, y, t)} [/latex] (10.30)

[latex]\frac{dy}{dx}=\frac{v}{u}=\frac{-Ay}{Ax}=-\frac{y}{x}[/latex]

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 [latex]p/ρ +\frac{1}{2} V^2 = p_0/ρ +\frac{1}{2} V^2_ 0[/latex] , where the subscript zero denotes the origin. Since [latex]V^2 =u^2 +v^2 = A^2(x^2 +y^2)[/latex], and [latex]V^2 _0 =0[/latex], we ﬁnd [latex]p_0 − p =\frac{1}{2} ρA^2(x^2 +y^2)[/latex], which shows that the pressure is constant on circles centered on the origin.

Question 11.12: A 2D steady, constant density, inviscid ﬂow of air is descri...

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