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Basics of Fluid Mechanics, 2014a

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528 CHAPTER 12. COMPRESSIBLE FLOW 2–DIMENSIONAL<br />

Example 12.16:<br />

A wall is included with 20.0 ◦ an inclination.<br />

A flow <strong>of</strong> air with a temperature<br />

<strong>of</strong> 20 ◦ C and a speed <strong>of</strong> U = 450m/sec<br />

flows (see Figure 12.29). Calculate the<br />

pressure reduction ratio, and the Mach<br />

number after the bending point. If the<br />

air flows in an imaginary two–dimensional<br />

tunnel with width <strong>of</strong> 0.1[m] what will the<br />

width <strong>of</strong> this imaginary tunnel after the<br />

bend? Calculate the “fan” angle. Assume<br />

the specific heat ratio is k =1.4.<br />

x 1 =0.1 [m]<br />

U=450 [m/sec]<br />

μ 1<br />

T=20 ◦ C<br />

μ 2<br />

x 1 =?M=?<br />

Δν =20 ◦<br />

Fig. -12.29. Schematic for Example (12.5).<br />

Solution<br />

First, the initial Mach number has to be calculated (the initial speed <strong>of</strong> sound).<br />

a = √ kRT = √ 1.4 ∗ 287 ∗ 293 = 343.1m/sec<br />

The Mach number is then<br />

M = 450<br />

343.1 =1.31<br />

this Mach number is associated with<br />

M ν<br />

P<br />

P 0<br />

T<br />

T 0<br />

ρ<br />

ρ 0<br />

μ<br />

1.3100 6.4449 0.35603 0.74448 0.47822 52.6434<br />

The “new” angle should be<br />

ν 2 =6.4449 + 20 = 26.4449 ◦<br />

and results in<br />

M ν<br />

P<br />

P 0<br />

T<br />

T 0<br />

ρ<br />

ρ 0<br />

μ<br />

2.0024 26.4449 0.12734 0.55497 0.22944 63.4620<br />

Note that P 01 = P 02<br />

P 2<br />

= P 01 P 2<br />

= 0.12734<br />

P 1 P 1 P 02 0.35603 =0.35766<br />

The “new” width can be calculated from the mass conservation equation.<br />

ρ 1 x 1 M 1 c 1 = ρ 2 x 2 M 2 c 2 =⇒ x 2 = x 1<br />

ρ 1<br />

ρ 2<br />

M 1<br />

M 2<br />

√<br />

T1<br />

T 2

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