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p0867

# p0867 - u ∗ = M ∗ a ∗ = a ∗ = ² γ RT ∗ = ³ 2...

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1206 CHAPTER 8. ONE-DIMENSIONAL COMPRESSIBLE FLOW 8.67 Chapter 8, Problem 67 Problem: In terms of standard compressible-flow notation, verify that the mass-flow rate, ˙ m , for flow of a perfect, calorically-perfect gas through a choked nozzle is ˙ m = p t A T t > ± ± : γ R X 2 γ + 1 ~ ( γ +1) / ( γ 1) Solution: Throughout the nozzle, the mass flux is constant so that ˙ m = ρ uA = ρ u A Because the flow is isentropic, the total density, ρ t , and total temperature, T t , are constant and equal to ρ t = ρ } 1 + γ 1 2 M 2 ] 1 / ( γ 1) , T t = T } 1 + γ 1 2 M 2 ] When the flow is choked, the flow is sonic at the throat so that M = 1 . Hence, ρ t and T t are related to the static density, ρ , and static temperature, T , at the throat by ρ t = ρ w γ + 1 2 W 1 / ( γ 1) , T t = γ + 1 2 T Also, by definition of the Mach number, we know that
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Unformatted text preview: u ∗ = M ∗ a ∗ = a ∗ = ² γ RT ∗ = ³ 2 γ γ + 1 RT t Hence, we can write the mass flux as follows. ˙ m = ρ ∗ u ∗ A ∗ = ρ t X 2 γ + 1 ~ 1 / ( γ − 1) ³ 2 γ γ + 1 RT t A ∗ Now, using the perfect-gas law to replace ρ t by p t / ( RT t ) , a little rearrangement of terms yields ˙ m = p t A ∗ RT t > ± ± : X 2 γ + 1 ~ 2 / ( γ − 1) X 2 γ + 1 ~ γ RT t = p t A ∗ √ T t > ± ± : γ R X 2 γ + 1 ~ 1+2 / ( γ − 1) Therefore, since 1 + 2 / ( γ − 1) = ( γ + 1) / ( γ − 1) , the mass-flow rate for flow through a choked nozzle is ˙ m = p t A ∗ √ T t > ± ± : γ R X 2 γ + 1 ~ ( γ +1) / ( γ − 1)...
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