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Basics of Fluid Mechanics, 2014a

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11.10. INTRODUCTION 481<br />

Example 11.24:<br />

Heat is added to the air until the flow is choked in amount <strong>of</strong> 600 [kJ/kg]. The exit<br />

temperature is 1000 [K]. Calculate the entrance temperature and the entrance Mach<br />

number.<br />

Solution<br />

The solution involves finding the stagnation temperature at the exit and subtracting the<br />

heat (heat equation) to obtain the entrance stagnation temperature. From the Table<br />

(11.7) or from the Potto-GDC the following ratios can be obtained.<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 />

1.0000 0.83333 0.63394 1.0000 0.52828 0.52828 0.52828<br />

The stagnation temperature<br />

T 02<br />

T 02 = T 2 = 1000<br />

T 2 0.83333 = 1200.0[K]<br />

The entrance temperature is<br />

T 01<br />

T 02<br />

=1− Q/ṁ<br />

T 02 C P<br />

= 1200 −<br />

600<br />

1200 × 1.004 ∼ = 0.5016<br />

It must be noted that T 02 = T ∗ 0 . Therefore with T 0 1<br />

T ∗ 0<br />

(11.7) or by Potto-GDC the following is obtained<br />

=0.5016 either by using Table<br />

M<br />

T<br />

T ∗ T 0<br />

T 0<br />

∗<br />

P<br />

P ∗ P 0<br />

P 0<br />

∗<br />

ρ ∗ ρ<br />

0.34398 0.50160 0.42789 2.0589 1.1805 0.24362<br />

Thus, entrance Mach number is 0.38454 and the entrance temperature can be<br />

calculated as following<br />

T 1 = T ∗ T 1<br />

= 1000 × 0.58463 = 584.6[K]<br />

T<br />

∗<br />

End Solution<br />

The difference between the supersonic branch to subsonic branch<br />

Example 11.25:<br />

Air with Mach 3 enters a frictionless duct with heating. What is the maximum heat<br />

that can be added so that there is no subsonic flow? If a shock occurs immediately at<br />

the entrance, what is the maximum heat that can be added?

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