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Basics of Fluid Mechanics, 2014a

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460 CHAPTER 11. COMPRESSIBLE FLOW ONE DIMENSIONAL<br />

Å ½ÑÜ<br />

1<br />

4fL<br />

D<br />

∣<br />

∣ max ∞<br />

Ä<br />

<br />

Fig. -11.32. The maximum entrance Mach number, M 1 to the tube as a function <strong>of</strong> 4fL<br />

D<br />

supersonic branch.<br />

Let denote<br />

( ) 4 fL<br />

Δ<br />

D<br />

4 fL<br />

= ¯<br />

D actual<br />

− 4 fL<br />

D<br />

∣ (11.204)<br />

sup<br />

Note that 4 fL<br />

D<br />

∣ is smaller than 4 fL<br />

D<br />

∣ . The requirement that has to be satisfied<br />

is that denote 4 fL<br />

D<br />

sup max∞<br />

∣ as difference between the maximum possible <strong>of</strong> length<br />

retreat<br />

in which the supersonic flow is achieved and the actual length in which the flow is<br />

supersonic see Figure 11.32. The retreating length is expressed as subsonic but<br />

4 fL<br />

D<br />

∣ = 4 fL<br />

retreat<br />

D ∣ − 4 fL<br />

max∞<br />

D ∣ (11.205)<br />

sup<br />

Figure 11.32 shows the entrance Mach number, M 1 reduces after the maximum<br />

length is exceeded.<br />

Example 11.20:<br />

Calculate the shock location for entrance Mach number M 1 =8and for 4 fL<br />

D =0.9<br />

assume that k =1.4 (M exit =1).<br />

Solution

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