p0867 - u = M a = a = RT = 2 + 1 RT t Hence, we can write...

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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 .H en c e , 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
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Unformatted text preview: u = M a = a = RT = 2 + 1 RT t Hence, we can write the mass flux as follows. m = u A = t X 2 + 1 ~ 1 / ( 1) 2 + 1 RT t A Now, using the perfect-gas law to replace t by p t / ( RT t ) , a little rearrangement of terms yields m = p t A RT t > : X 2 + 1 ~ 2 / ( 1) X 2 + 1 ~ RT t = p t A T t > : R X 2 + 1 ~ 1+2 / ( 1) Therefore, since 1 + 2 / ( 1) = ( + 1) / ( 1) , the mass-flow rate for flow through a choked nozzle is m = p t A T t > : R X 2 + 1 ~ ( +1) / ( 1)...
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This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

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