1946
DOI: 10.2514/8.11443
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On the Steady Flow of a Gas Through a Tube with Heat Exchange or Chemical Reaction

Abstract: This paper is concerned with a simple phenomenological discussion of gaseous combustion through a tube and related phenomena involving hydrodynamic, thermodynamic, and chemical considerations. It is also indicated how the discussion can be conveniently developed into a treatment from the point of view of microscopic chemical kinetics. Indeed, some interesting results are already obtained without such an extension. For instance, when there is no chemical reaction, it is found that for the range of Mach Numbers … Show more

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Cited by 21 publications
(4 citation statements)
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“…Also S = S o on the line OL. In the other regions the entropy can be found by integrating (1)(2)(3)(4) and (2 in region It is easily shown from the formulae just obtained that in these regions p>p 0 , M <M 0 and a>a 0 (and hence T > T o ). Thus it is clear that an addition of heat cannot in any case result in a drop in temperature.…”
Section: (14)mentioning
confidence: 93%
See 1 more Smart Citation
“…Also S = S o on the line OL. In the other regions the entropy can be found by integrating (1)(2)(3)(4) and (2 in region It is easily shown from the formulae just obtained that in these regions p>p 0 , M <M 0 and a>a 0 (and hence T > T o ). Thus it is clear that an addition of heat cannot in any case result in a drop in temperature.…”
Section: (14)mentioning
confidence: 93%
“…0 , (1-2) dt dx pdx where p(x, t), u(x, t) and p(x, t) are respectively the density, velocity and pressure of the fluid at the point x at time t. Two further equations must be satisfied. One is the equation of state, which is taken to be p=prexp(S/c e ), (1)(2)(3) where S(x, t) is the specific entropy and y = c v jc v is the ratio of the specific heats. The other relation required is that which relates the entropy S to its initial value; this is the equation of conservation of energy.…”
mentioning
confidence: 99%
“…By the Third Law of Motion the force exerted on the fluid by the walls, Fw', is equal and opposite to that exerted by the fluid on the walls, Fw. Reserving positive values for the latter, Fw = -Fw', and Equation 2 becomes: Fa = (Fs -Fi) + p3A3 -PlA, -po (As -AJ (5) which, when pi = p3 = p" is identical with Equation 1. For steady flight Fs must be positive and equal to the total external drag of the ram-jet derived in like fashion from the pressure distribution on the outer surfaces due to its motion through the air.…”
Section: Components Of a Ram-jetmentioning
confidence: 97%
“…The actual performance to be expected of the burner was, however, a major unknown. A start had been made on the theory of reactions in flow systems (26) and of the effect of adding heat to gas streams within ducts under idealized conditions (1,5,19,22,55), but there existed no body of experimental material on combustion as a continuous high velocity flow process to show the extent to which the thermodynamic limits would be achieved. A similar though less critical problem existed in the development of the turbo-jet burner (SO, 89).…”
Section: Crux Of the Burner Problemmentioning
confidence: 99%