Flow in an Air Intake (C) The device shown in Fig. E8.1 is used for measuring the flow of air of density ρ into the circular intake port of an engine. Basically, it uses Bernoulli’s equation between two points: (a) far upstream of the intake, where the velocity is essentially zero and the pressure

Question

Flow in an Air Intake (C)

The device shown in Fig. E8.1 is used for measuring the flow of air of density [latex] ho[/latex] into the circular intake port of an engine. Basically, it uses Bernoulli's equation between two points: (a) far upstream of the intake, where the velocity is essentially zero and the pressure is [latex]p_{1}[/latex], and (b) the pressure tapping at a distance [latex]x[/latex] from the inlet, where the pressure is measured to be [latex]p_{2}[/latex].

The situation is complicated by the formation of a boundary layer as the air impinges against the inlet of the port, so the velocity profile at the location of the tapping consists of: (a) a central core, in which the velocity is uniformly [latex]u_{2}[/latex], and (b) a boundary layer in which the velocity declines from its mainstream value [latex]v_{x \infty}=u_{2}[/latex] to zero at the wall.

Bernoulli's equation can be applied, starting from the outside air (where [latex]u_{1} \doteq 0[/latex] ) and continuing into the core (where viscosity is negligible), leading to:

[latex]u_{2}=\sqrt{\frac{2\left(p_{1}-p_{2} ight)}{ ho}} .[/latex] (E8.1.1)

However, the ultimate goal is to be able to determine the mean velocity [latex]u_{\mathrm{m}}[/latex] (defined as the total flow rate divided by the total area), which necessitates the introduction of a discharge coefficient [latex]C_{\mathrm{D}}[/latex], so that:

[latex]u_{\mathrm{m}}=C_{D} \sqrt{\frac{2\left(p_{1}-p_{2} ight)}{ ho}}.[/latex] (E8.1.2)

The total mass flow rate of air is then:

[latex]m= ho u_{\mathrm{m}} A,[/latex] (E8.1.3)

where [latex]A[/latex] is the cross-sectional area of the pipe, which has an internal diameter [latex]D[/latex]. If [latex]D=2[/latex] in., [latex]x=1 \mathrm{ft}, u=1.59 imes 10^{-4} \mathrm{ft}^{2} / \mathrm{s}[/latex], and [latex]u_{\mathrm{m}}=20 \mathrm{ft} / \mathrm{s}[/latex], estimate [latex]C_{D}[/latex].

Accepted Answer

First, estimate the properties of the boundary layer. The largest value of the Reynolds number occurs opposite the pressure tapping, where [latex]x=1 \mathrm{ft}[/latex]. Also, a rough estimate of the mainstream velocity is [latex]u_{\mathrm{m}}[/latex], so that [latex]v_{x \infty} \doteq 20 \mathrm{ft} / \mathrm{s}[/latex]. Thus, to a first approximation:

[latex]\operatorname{Re}_{x} \doteq \frac{v_{x \infty} x}{ u} \doteq \frac{20 imes 1}{1.59 imes 10^{-4}}=1.26 imes 10^{5}.[/latex] (E8.1.4)

From Section 8.5, the flow in the boundary layer is therefore laminar all the way from the inlet to the pressure tapping, and the result of Eqn. (8.16), which was based on the assumed velocity profile [latex]v_{x} / v_{x \infty}=\sin (\pi y / 2 \delta)[/latex], applies:

[latex]\frac{\delta}{x}=\sqrt{\frac{\pi}{c}} \sqrt{\frac{\mu}{ ho v_{x \infty} x}}=\frac{4.79}{\sqrt{\mathrm{Re}_{\mathrm{x}}}}.[/latex] (8.16)

[latex]\frac{\delta}{x}=\frac{4.79}{\sqrt{\mathrm{Re}_{\mathrm{x}}}}=\frac{4.79}{1.26 imes 10^{5}}=0.0135,[/latex] (E8.1.5)

giving a first approximation of the thickness of the boundary layer:

[latex]\delta \doteq 0.0135 imes 12=0.162 ext { in }.[/latex] (E8.1.6)

Thus, the boundary layer has an outer diameter of [latex]2.000[/latex] in. and an inner diameter of [latex](2.000-2 imes 0.162)=1.676[/latex] in., for a mean diameter of [latex]D_{\mathrm{m}}=(2.000+1.676) / 2=[/latex] [latex]1.838[/latex] in.

Although the intake has circular geometry, the boundary layer can be approximated as that on a flat plate of width [latex]\pi D_{\mathrm{m}}[/latex]. The corresponding mass flow rate in the boundary layer is:

[latex]m \doteq \pi ho D_{\mathrm{m}} \int_{0}^{\delta} v_{x} d y=\pi ho D_{\mathrm{m}}\left(\frac{2 \delta v_{x \infty}}{\pi} ight)=2 ho D_{\mathrm{m}} \delta v_{x \infty}.[/latex] (E8.1.7)

The total mass flow rate, which is based on the mean velocity, equals the sum of the mass flow rates in the core and in the boundary layer:

[latex]\underbrace{\left(\frac{\pi imes 2.000^{2}}{4} ight) ho imes 20.0}_{ ext {Total flow rate }}=\underbrace{\left(\frac{\pi imes 1.676^{2}}{4} ight) ho v_{x \infty}}_{ ext {Flow rate in core }}+\underbrace{2 ho imes 1.838 imes 0.162 v_{x \infty}}_{ ext {Flow rate in boundary layer }}, (E8.1.8) [/latex]

which gives the improved estimate [latex]v_{x \infty} \doteq 22.4 \mathrm{ft} / \mathrm{s}[/latex]. The calculations can be repeated iteratively; convergence essentially occurs after just one more iteration, giving a final value [latex]v_{x \infty} \doteq 22.29 \mathrm{ft} / \mathrm{s}[/latex]. Since the pressure drop based on Bernoulli's equation gives [latex]v_{x \infty} \doteq 22.29[/latex], and the actual mean velocity is a lower value, [latex]u_{\mathrm{m}}=[/latex] [latex]20.0[/latex], the required discharge coefficient is:

[latex]C_{D}=\frac{20.0}{22.29}=0.897 .[/latex] (E8.1.9)

Question 8.1: Flow in an Air Intake (C) The device shown in Fig. E8.1 is u...

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