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solution functions for the momentum-energy problem (3.4)– (3.8). The resulting implicit integral representation (3.9) for the exact solution must be inverted to provide x = x(t), and this inversion is nontrivial from a practical point of view. Hence an approach based on (3.9) is perhaps not best suited for use in obtaining qualitative and quantitative information on the liq- uid hammer. We follow an alternative approach based on a combination of phase plane and perturbation analyses. The zero torque line y = λ1x in the phase plane is a line of equilibrium points for (3.4), which means that any point (x1, y1) on this line in the phase plane is a fixed point solu- tion. That is, the constant functions Hence the rotational speed x(t) of the turbine is a mono- tonic increasing function for t ≥ t0, and we now show that the same is true for the fluid flow rate y(t). Differentiating (with respect to time) the energy equation (3.7), and then us- ing the angular momentum equation (3.4) to replace d x /d t in that result, yields the following equation on the balance of linear momentum, In this case not only is the generator torque equal to zero (cor- responding to disconnected power) but also there is no load because the turbine shaft torque (y − λ1x)(y − λ2x) is also zero. y ✻∗∗ (x ,y ) I y = λ1x y =Iλ 2 x 2 j + 2β2 y + β(γ − α)x ≥ 2 j > 0 everywhere in the wedge region (1.9). These considerations ε dy = dt (y − λ1x)(y − λ2x) [β(α − γ )y + 2αγ x] 2 j + 2β2 y + β(γ − α)x (3.14) x(t) ≡ x1 and y(t) ≡ y1 provide solutions for (3.4) if y1 = λ1x1. for all t (3.11) (3.12) for t ≥ t0. Since the solution (x(t), y(t)) remains in the wedge region (3.6), it follows with (1.4) that there holds (y−λ1x)(y−λ2x)>0 for t ≥t0. Similarly the other factors on the right side of (3.14) are also positive. For example (1.9), (2.2), (2.3) and (2.13) imply β(α − γ)y + 2αγx ≥ β(α − γ)λ1x + 2αγx ≥ [α(α − γ)+2αγ]x =α(α+γ)x >0fort >t0,soallfactorsinthe numerator are positive on the right side of (3.14). A similar calculation shows 2βy + (γ − α)x ≥ 2βλ1x + (γ − α)x ≥ (α + γ )x ≥ 0, so (cf. (2.12)) (3.15) (x0, y0) ❘• I • ✠•✲x with (3.14) imply the stated monotonicity result for y = y(t), dy L2 : slope = −γ /β • • (x, y) dt >0 for t>t0. (3.16) L1: slope = α/β Figure 2 The initial state (x0, y0) lies in the wedge region (3.6) characterized by the dominant shaft coefficient λ1. It follows with (1.4) and (3.4) that the turbine speed satisfies initially dx >0, (3.13) dt so x(t) increases initially with increasing t ≥ t0. Hence the motion is “to the right” along the energy hyperbola (3.7), to- ward the line y = λ1x in the first quadrant, as illustrated in Figure 2. The conditions (2.2), (2.11) and (2.13) guaran- tee that the resulting solution state (x(t), y(t)) remains in the wedge region (3.6), and (3.13) continues to hold, for all t ≥ t0. y2<✒0 A routine calculation shows that the intersection point (x∗, y∗) of the zero torque line y−λ1x = 0 and the hyperbola (3.7)–(3.8) in the first quadrant has coordinates x∗ =x∗(x0,x0)= (3.17) j2λ2 +(βλ −α)(βλ +γ)·C(x ,y )− jλ 111001 (βλ1 −α)(βλ1 +γ) y∗ = y∗(x0, y0) = λ1x∗(x0, y0) with positive constant C(x0, y0) given by (3.8). Note also that (2.13) guarantees the positivity of the quantity (βλ1 −α)(βλ1 +γ)=β2λ21 +β(γ −α)λ1 −αγ >0 (3.18) since the two roots of the quadratic function f (λ) := β2λ2 + β(γ−α)λ−αγ areλ=−γ/βandλ=+α/β,and f(λ)>0 5PDF Image | Transient Characteristics of Radial Outflow Turbine Generators
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