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Additional resources for Gasdynamics : theory and applications

Example text

Since higher-order terms are positive, we observe that pQ > Q. 3 AREA AND MASS FLOW RATE Our immediate task is to relate the area A(x) to the Mach number and to determine the mass flow rate m. T&TM (4-11) For Eqs. 10), the reference condition is a stagnation condition, where M = 0. Here, we obtain a more convenient reference condition by setting Af = 1. , when M = 1. Since the mass flow rate is a constant, it is not altered by this choice. Hence, at M = 1, we have, from Eq. 11), We eliminate m from Eqs.

Based on the earlier discussion, substitute 1 p e = ——r y-1 — p A = const, into Eqs. 10a) to obtain f\ , _ V I at f\ ox _ /? 5^ /? 5x I r\ p ox _ _ __ _ p dx p 3/ __ for the equations of motion. 13) are now introduced, to yield 3V dV dt "*" v dx I dp' V dp' 1 dp' px + p' dx y =o dp' These relations can be simplified, since the product of two primed quantities, for instance, is negligibly small. Thus, we have dt 1 PX+P' _ dx . dv dp' _ 1 dp' P* dt " ONE-DIMENSIONAL CONSERVATION EQUATIONS 33 ,duct diaphragm / Fig.

And p change from their initial values. We assume that the magnitude of the displacement of the diaphragm from its neutral position is small. As a consequence, the gas experiences a small perturbation of its state, and we can write V=V'(x,t), >'(*, 0, where a prime denotes a perturbation quantity. Based on the earlier discussion, substitute 1 p e = ——r y-1 — p A = const, into Eqs. 10a) to obtain f\ , _ V I at f\ ox _ /? 5^ /? 5x I r\ p ox _ _ __ _ p dx p 3/ __ for the equations of motion. 13) are now introduced, to yield 3V dV dt "*" v dx I dp' V dp' 1 dp' px + p' dx y =o dp' These relations can be simplified, since the product of two primed quantities, for instance, is negligibly small.