## 物理代写|流体力学代写Fluid Mechanics代考|Dimensionless Stage Parameters

Equation (5.124) exhibits a direct relation between the specific stage shaft power $l_m$ and the kinetic energies. The velocities from which these kinetic energies are built can be taken from the corresponding stage velocity diagram. The objective of this chapter is to introduce dimensionless stage parameters that completely determine the stage velocity diagram. These stage parameters exhibit a set of unified relations for compressor and turbine stages respectively.

Starting from a turbine or compressor stage with constant mean diameter and axial components, shown in Fig. 5.23, we define the dimensionless stage parameters that describe the stage velocity diagram of a normal stage.

A normal stage is encountered within the high pressure (HP) part of multi-stage turbines or compressors and is characterized by $\boldsymbol{U}3=\boldsymbol{U}_2, \boldsymbol{V}_3=\boldsymbol{V}_1, V{m 1}=V_{m 3}$, and $\alpha_1=\alpha_3$. The similarity of the velocity diagrams allows using the same blade profile throughout the HP-turbine or compressor, thus, significantly reducing manufacturing costs.

We define the stage flow coefficient $\phi$ as the ratio of the meridional velocity component and the circumferential component. For this particular case, the meridional component is identical with the axial component:
$$\phi=\frac{V_{m 3}}{U_3} .$$
The stage flow coefficient $\phi$ in Eq. (5.127) is a characteristic for the mass flow behavior through the stage. The stage load coefficient $\lambda$ is defined as the ratio of the specific stage mechanical energy $l_m$ and the exit circumferential kinetic energy $U_3^2$. This coefficient directly relates the flow deflection given by the velocity diagram with the specific stage mechanical energy:
$$\lambda=\frac{l_m}{U_3^2} .$$
The stage load coefficient $\lambda$ in Eq. (5.128) describes the work capability of the stage. It is also a measure for the stage loading. The stage enthalpy coefficient $\psi$ represents the ratio of the isentropic stage mechanical energy and the exit circumferential kinetic energy $U_3^2$.
$$\psi=\frac{l_s}{U_3^2}$$

## 物理代写|流体力学代写Fluid Mechanics代考|Simple Radial Equilibrium to Determine r

Expressing the relationship between the degree of reaction and the blade height requires the knowledge of the radial equilibrium condition within the axial gaps between the stator and rotor blades. In a fully three-dimensional turbomachinery flow, describing the radial equilibrium condition is a complicated issue. Attempts to numerically analyze the issue of the radial equilibrium have encountered divergence problems. The streamline curvature method based on an axisymmetric assumption exhibits a reasonable and practical solution [23]. For the simple cases we discuss in this Chapter, we further simplify the radial equilibrium condition to arrive at simple relationships between the degree of reaction and the blade height.

The fluid particles in compressors and turbines experience a rotational and translational motion. For the simple turbine and compressor cases case we discussed in this Chapter the rotating fluid is subjected to centrifugal forces that must be balanced by the pressure gradient in order to maintain the radial equilibrium. Consider an infinitesimal sector of an annulus with unit depth containing the fluid element which is rotating with tangential velocity $V_u$ in an absolute frame of reference. The centrifugal force acting on the element is shown in Fig. 5.24. Since the fluid element is in radial equilibrium, the centrifugal force per unit width is obtained from:
$$d F=d m \frac{V_u^2}{R}$$
with $d m=\rho R d R d \phi$. The centrifugal force is kept in balance by the pressure forces:
$$\frac{d p}{d R}-\rho \frac{V_u^2}{R} .$$
This result can also be obtained by decomposing the Euler equation of motion (4.51) for inviscid flows in its three components in a cylindrical coordinate system. The assumptions needed to arrive at Eq. (5.133) are:
$$\frac{\partial V_r}{\partial R} \simeq 0, \text { Axial symmetric: } \frac{\partial V_r}{\partial \phi}=0, \frac{\partial V_r}{\partial z} \simeq 0 .$$
With these assumptions, Eq. (5.133) yields:
$$\frac{1}{\rho} \frac{\partial p}{\partial R}=\frac{V_u^2}{R}$$

## 物理代写|流体力学代写Fluid Mechanics代考|Dimensionless Stage Parameters

$$\phi=\frac{V_{m 3}}{U_3} .$$

$$\lambda=\frac{l_m}{U_3^2} .$$

$$\psi=\frac{l_s}{U_3^2}$$

## 物理代写|流体力学代写Fluid Mechanics代考|Simple Radial Equilibrium to Determine r

$$d F=d m \frac{V_u^2}{R}$$

$$\frac{d p}{d R}-\rho \frac{V_u^2}{R} .$$

$$\frac{\partial V_r}{\partial R} \simeq 0, \text { Axial symmetric: } \frac{\partial V_r}{\partial \phi}=0, \frac{\partial V_r}{\partial z} \simeq 0 .$$

$$\frac{1}{\rho} \frac{\partial p}{\partial R}=\frac{V_u^2}{R}$$

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