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Basics of Fluid Mechanics, 2014a

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

M<br />

4fL<br />

D<br />

P<br />

P ∗ P 0<br />

P 0<br />

∗<br />

ρ<br />

ρ ∗<br />

U<br />

U ∗<br />

T<br />

T ∗<br />

0.90000 0.01451 1.1291 1.0089 1.0934 0.9146 1.0327<br />

So, the total temperature at the exit is<br />

T ∗ | 2<br />

= T ∗<br />

T ∣ T 2 = 300<br />

2<br />

1.0327 = 290.5[K]<br />

To “move” to the other side <strong>of</strong> the tube the 4fL<br />

D<br />

∣<br />

∣<br />

4fL<br />

D<br />

∣ = 4fL<br />

1<br />

D<br />

+ 4fL<br />

D<br />

∣ =<br />

2<br />

4 × 0.004 × 10<br />

0.05<br />

is added as<br />

+0.01451 ≃ 3.21<br />

The rest <strong>of</strong> the parameters can be obtained with the new 4fL<br />

D<br />

either from Table (11.6)<br />

by interpolations or by utilizing the attached program.<br />

M<br />

4fL<br />

D<br />

P<br />

P ∗ P 0<br />

P 0<br />

∗<br />

ρ<br />

ρ ∗<br />

U<br />

U ∗<br />

T<br />

T ∗<br />

0.35886 3.2100 3.0140 1.7405 2.5764 0.38814 1.1699<br />

Note that the subsonic branch is chosen. The stagnation ratios has to be added<br />

for M =0.35886<br />

M<br />

T<br />

T 0<br />

ρ<br />

ρ 0<br />

A<br />

A ⋆<br />

P<br />

P 0<br />

A×P<br />

A ∗ ×P 0<br />

F<br />

F ∗<br />

0.35886 0.97489 0.93840 1.7405 0.91484 1.5922 0.78305<br />

The total pressure P 01 can be found from the combination <strong>of</strong> the ratios as follows:<br />

P 1<br />

{ }} {<br />

P ∗<br />

{ }} { ∣ P ∗<br />

P 01 = P 2 P ∣∣∣1 P 0<br />

P ∣<br />

2<br />

P ∗ P ∣<br />

1<br />

1<br />

=1 ×<br />

1.12913 × 3.014 × 1<br />

0.915 =2.91[Bar]<br />

T 1<br />

{ }} {<br />

T ∗<br />

{ }} { ∣ T ∗<br />

T 01 = T 2<br />

∣<br />

T ∣∣∣1 T 0<br />

∣<br />

T T ∗ T<br />

∣<br />

2<br />

∣<br />

1<br />

=300 × 1<br />

1.0327 × 1.17 × 1<br />

0.975 ≃ 348K =75◦ C

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