## 物理代写|流体力学代写Fluid Mechanics代考|Energy Transfer in Relative Systems

Since the rotor operates in a relative frame of reference (relative system), the energy conversion mechanism is quite different from that of a stator (absolute system). A fluid particle that moves with a relative velocity $W$ within the relative system that rotates with the angular velocity $\boldsymbol{\omega}$ has an absolute velocity:
$$V=W+\omega \times R=W+U, \omega \times R=U$$

with $\boldsymbol{R}$ in Eq. (5.117) as the radius vector of the particle in the relative system. Introducing the absolute velocity vector $V$ in the equation of motion and multiplying the results with a relative differential displacement $d R$, we obtain the energy equation for an adiabatic steady flow within a rotating relative system:
$$d\left(h+\frac{1}{2} W^2-\frac{\omega^2 R^2}{2}+g z\right)=0$$
or the relative total enthalpy:
$$H_r=h+\frac{1}{2} W^2-\frac{\omega^2 R^2}{2}+g z=\text { const. }$$
Negglecting thẽ gravitational term, $g z \approx 0$, Eq. (5.122) cãn be writteñ âs:
$$h_1+\frac{1}{2} W_1^2-\frac{1}{2} U_1^2=h_2+\frac{1}{2} W_2^2-\frac{1}{2} U_2^2$$
Equation (5.120) is the energy equation transformed into a relative system. As can be seen, the transformation of kinetic energy undergoes a change while the transformation of static enthalpy is frame indifferent. With these equations in connection with the energy balance, we can analyze the energy transfer within an arbitrary turbine or compressor stage.

## 物理代写|流体力学代写Fluid Mechanics代考|Unified Treatment of Turbine and Compressor Stages

In this chapter, compressor and turbine stages are treated from a unified physical point of view. Figures $5.19$ and $5.20$ show the decomposition of a turbine and a compressor stage into their stator and rotor rows. The primes ” $”$ ” and ” $” / \prime \prime$ refer to stator and rotor rows, respectively. As seen, the difference between the isentropic and the polytropic enthalpy difference is expressed in terms of dissipation $\Delta h_d^{\prime}=$ $\Delta h_s^{\prime}-\Delta h^{\prime}$ for turbines and $\Delta h_d^{\prime}=\Delta h^{\prime}-\Delta h_s^{\prime}$ for compressors. For the stator, the energy balance requires that $H_2=H_1$. This leads to:
$$h_1-h_1=\Delta h^{\prime}=\frac{1}{2}\left(V_2^2-V_1^2\right)$$
Moving to the relative frame of reference, the relative total enthalpy $H_{r 2}=H_{r 3}$ remains constant. Thus, the energy equation for the rotor is according to Fig. 5.20:
$$h_2-h_3=\Delta h^{\prime \prime}=\frac{1}{2}\left(W_3^2-W_2^2+U_2^2-U_3^2\right)$$
The stage specific shaft power balance requires:
$$1_m=H_1-H_3=\left(h_1-h_2\right)-\left(h_3-h_2\right)+\frac{1}{2}\left(V_1^2-V_3^2\right) .$$
Insërting Eq̣s. (5.121) and (5.122) intō Eq. (5.123) yiêlds :
$$1_m=\frac{1}{2}\left[\left(V_2^2-V_3^2\right)+\left(W_3^2-W_2^2\right)+\left(U_2^2-U_3^2\right)\right] .$$

Equation (5.124), known as the Euler Turbine Equation, indicates that the stage work can be expressed simply in terms of absolute, relative, and rotational kinetic energies. This equation is equally applicable to turbine stages that generate shaft power and to compressor stages that consume one. In the case of a turbine stage, the sign of the specific mechanical energy $l_m$ is negative, which indicates that energy is removed from the system (power generation). In compressor cases, it is positive because energy is added to the system (power consumption). Figures $5.21$ and $5.22$ show the stage configuration, the velocity diagram and the expansion, compression process within a single stage turbine and compressor.

## 物理代写|流体力学代写Fluid Mechanics代考|Energy Transfer in Relative Systems

$$V=W+\omega \times R=W+U, \omega \times R=U$$

$$d\left(h+\frac{1}{2} W^2-\frac{\omega^2 R^2}{2}+g z\right)=0$$

$$H_r=h+\frac{1}{2} W^2-\frac{\omega^2 R^2}{2}+g z=\text { const. }$$

$$h_1+\frac{1}{2} W_1^2-\frac{1}{2} U_1^2=h_2+\frac{1}{2} W_2^2-\frac{1}{2} U_2^2$$

## 物理代写|流体力学代写Fluid Mechanics代考|Unified Treatment of Turbine and Compressor Stages

$$h_1-h_1=\Delta h^{\prime}=\frac{1}{2}\left(V_2^2-V_1^2\right)$$

$$h_2-h_3=\Delta h^{\prime \prime}=\frac{1}{2}\left(W_3^2-W_2^2+U_2^2-U_3^2\right)$$

$$1_m=H_1-H_3=\left(h_1-h_2\right)-\left(h_3-h_2\right)+\frac{1}{2}\left(V_1^2-V_3^2\right) .$$

$$1_m=\frac{1}{2}\left[\left(V_2^2-V_3^2\right)+\left(W_3^2-W_2^2\right)+\left(U_2^2-U_3^2\right)\right] .$$

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