Dynamic Modeling, Simulation and Control of Energy by Ranjan Vepa

By Ranjan Vepa

This ebook addresses the middle matters desirous about the dynamic modeling, simulation and regulate of a variety of power structures corresponding to gasoline generators, wind generators, gasoline cells and batteries.

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Thus, states beyond the point of tangency where M ¼ 1 on the H-curve for the critical heat input are not accessible. On the other hand, if the heat added is less than the critical value, the Rayleigh line intersects the H-curve at two points. Only the first point is accessible through continuous heat addition; although in combustion when local heat releases are possible, the second point is also accessible under certain conditions. One way of operating such that the Rayleigh line intersects the H-curve at two points is to reduce G and consequently reduce the slope of the Rayleigh line.

It therefore lies entirely in the fourth quadrant of the diagram as shown in Fig. 2. On the other hand, in supersonic flow, it lies entirely in the second quadrant. Consequently, the other two quadrants may be considered to be forbidden regions as any states that lie in these quadrants are not physically realizable. The H-curve with no heat addition passes through the origin, while the H-curve with a certain fixed heat addition represents a parallel curve above the H-curve passing through the origin.

In particular, one generally considers the area A, the Mach number M, the density q, the velocity u, and the temperature T. All these quantities are denoted with the superscript ‘‘*’’ at the sonic state and M à ¼ 1. The mass flow rate satisfies m_ ¼ quA ¼ qà uà Aà : ð1:1:83Þ 22 1 Introduction to Energy Generation Principles Consequently, A qà uà q0 qà uà ¼ : ¼ Aà qu q q0 u ð1:1:84Þ From Eq. 45), the ratio of the density to its stagnation value is 1   cÀ1  1c  q0 P0 cÀ1 2 ¼ ¼ 1þ : M 2 q p 1 À ÁcÀ1 When M ¼ M à ¼ 1, qq0à ¼ cþ1 , 2 qà q0 qà ¼ ¼ q q q0  2 cþ1 1    cÀ1 cÀ1 2 : 1þ M 2 ð1:1:85Þ ð1:1:86aÞ Similarly, T à T à T0 ¼ ¼ T T0 T    2 cÀ1 2 1þ M : ðc þ 1Þ 2 ð1:1:86bÞ pffiffiffiffiffiffiffiffiffi Since u ¼ M cRT uà 1 ¼ M u rffiffiffiffiffi   12 Tà 1 2 cÀ1 2 1þ M : ¼ M ð c þ 1Þ 2 T ð1:1:86cÞ Hence, A 1 ¼ Aà M  2 cþ1     cþ1 2ðcÀ1Þ cÀ1 : 1þ M2 2 ð1:1:86dÞ When the ratio of the area A to the throat area when the flow is choked Aà is known, the Mach number corresponding to the area A may be found.

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