Turbines and Governors

Steam Governor

The control diagram of this model is depicted in Fig. 8. This model receives as input the frequency deviation $\Delta\omega = \omega_{ref} - \omega$ from the nominal frequency (normally $50,\text{Hz}$ or $60,\text{Hz}$) and produces the valve opening signal $p_{gv}$ for the turbine. $p_{ref}$ is the mechanical power produced at nominal frequency. The governor implements a lead-lag controller $\frac{K(1+sT_2)}{(1+sT_1)}$ where $K=1/R$ and $R$ is the droop coefficient, followed by a PT1 integrator with embedded rate limiters and an anti-windup loop. To avoid unnecessary dead-beat behaviour, complex transfer functions with more than one pole and zero are decomposed via partial fraction expansion into parallel PT1 elements, as shown in Fig. 9.

Analogous to the static exciter model, the integrator uses an anti-windup strategy as shown in Fig. 10.

The forward-Euler discretised equations are: $$ \Delta \omega (k-\Delta t) = \omega_{ref} - \omega (k-\Delta t), $$ $$ p_{1}(k) = p_{1}(k-\Delta t) + \frac{\Delta t}{T_{1}} \left(\Delta \omega (k-\Delta t) \cdot \frac{T_{1} - T_{2}}{T_{1}} - p_{1}(k-\Delta t)\right), $$ $$ p(k-\Delta t) = \frac{1}{R} \left(p_{1}(k-\Delta t) + \Delta \omega(k-\Delta t) \cdot \frac{T_{2}}{T_{1}}\right), $$ $$ \dot{p}(k-\Delta t) = \frac{1}{T_{3}}\left(p(k-\Delta t) + p_{ref} - p_{gv}(k-\Delta t)\right) - K_{bc} \left(p_{gv}^{}(k-\Delta t) - p_{gv}(k-\Delta t)\right), $$ $$ p_{gv}^{}(k) = p_{gv}^{*}(k-\Delta t) + \Delta t \cdot \dot{p}(k-\Delta t), $$

and

$$ p_{gv}(k) = p_{gv}^{}(k) \quad \text{if} \quad P_{m,\min} \leq p_{gv}^{}(k) \leq P_{m,\max}, \ p_{gv}(k) = P_{m,\max} \quad \text{if} \quad p_{gv}^{}(k) > P_{m,\max}, \ p_{gv}(k) = P_{m,\min} \quad \text{if} \quad p_{gv}^{}(k) < P_{m,\min}. $$

If $T_1 = 0$ the $p_1(k)$ equation is skipped and $p(k)$ is instead: $$ p(k-\Delta t) = \frac{1}{R} \left(\Delta \omega(k-\Delta t) + \frac{T_{2}}{\Delta t} \left(\Delta \omega(k-\Delta t) - \Delta \omega(k-2\Delta t)\right)\right). $$

Assuming the simulation starts in steady state (all derivatives zero, $\Delta\omega(0)=0$), the initial values are: $$ p_{1}(t=0) = 0, \quad p(t=0) = 0, \quad p_{ref} = p_{gv}^{*}(t=0) = p_{gv}(t=0). $$

Steam Turbine

The steam turbine receives the valve opening signal $p_{gv}$ from the Steam Governor and outputs mechanical power $p_m$ to the synchronous generator. It is divided into high-pressure (HP), intermediate-pressure (IP), and low-pressure (LP) stages, each modelled as a first-order lag with time constants $T_{CH}$, $T_{RH}$, $T_{CO}$ respectively. Setting a time constant to zero disables that lag element. The total mechanical power is a weighted sum of each stage: $F_{HP} + F_{IP} + F_{LP} = 1$ must hold. The forward-Euler discretised equations are:

$$ p_{hp}(k) = p_{hp}(k-\Delta t) + \frac{\Delta t}{T_{CH}} \left(p_{gv}(k-\Delta t) - p_{hp}(k-\Delta t)\right), $$ $$ p_{ip}(k) = p_{ip}(k-\Delta t) + \frac{\Delta t}{T_{RH}} \left(p_{hp}(k-\Delta t) - p_{ip}(k-\Delta t)\right), $$ $$ p_{lp}(k) = p_{lp}(k-\Delta t) + \frac{\Delta t}{T_{CO}} \left(p_{ip}(k-\Delta t) - p_{lp}(k-\Delta t)\right), $$ $$ p_{m}(k) = F_{HP} \cdot p_{hp}(k) + F_{IP} \cdot p_{ip}(k) + F_{LP} \cdot p_{lp}(k). $$

Assuming the simulation starts in steady state (all derivatives zero), the initial values are: $$ p_{hp}(t=0) = p_{gv}(t=0), \quad p_{ip}(t=0) = p_{hp}(t=0), \quad p_{lp}(t=0) = p_{ip}(t=0), $$ $$ p_{m}(t=0) = F_{HP} \cdot p_{hp}(t=0) + F_{IP} \cdot p_{ip}(t=0) + F_{LP} \cdot p_{lp}(t=0). $$

Hydro Turbine Governor

The Hydro Turbine Governor receives the frequency deviation $\Delta\omega = \omega_{ref} - \omega$ as input and produces the valve/gate opening signal $p_{gv}$ for the turbine. $p_{ref}$ is the mechanical power produced at nominal frequency. The controller transfer function is $K\frac{1+sT_2}{(1+sT_1)(1+sT_3)}$, where $K=\frac{1}{R}$ and $R$ is the droop coefficient. The transfer function is decomposed into two parallel PT1 blocks as shown in Fig. 13.

The forward-Euler discretised equations are: $$ x_{1}(k) = x_{1}(k-\Delta t) + \frac{\Delta t}{T_{1}} \left(\Delta\omega(k-\Delta t) - x_{1}(k-\Delta t)\right), $$ $$ x_{2}(k) = x_{2}(k-\Delta t) + \frac{\Delta t}{T_{3}} \left(\Delta\omega(k-\Delta t) - x_{2}(k-\Delta t)\right), $$ $$ p^{*}{gv}(k) = \frac{1}{R}\left(A \cdot x{1}(k) + B \cdot x_{2}(k)\right) + p_{ref}, $$

where $$ A = \frac{T_{1}-T_{2}}{T_{1}-T_{3}}, \qquad B = \frac{T_{2}-T_{3}}{T_{1}-T_{3}}, $$

and the output limiter is applied as: $$ p_{gv}(k) = \begin{cases} P_{m,\max} & \text{if } p^{}{gv}(k) > P{m,\max}, \ P_{m,\min} & \text{if } p^{}{gv}(k) < P{m,\min}, \ p^{*}_{gv}(k) & \text{otherwise.} \end{cases} $$

Assuming the simulation starts in steady state (all derivatives zero, $\Delta\omega(t=0)=0$), the initial values are: $$ x_{1}(t=0) = 0, \quad x_{2}(t=0) = 0, \quad p_{ref} = p_{gv}(t=0). $$

Hydro Turbine

The Hydro Turbine receives the gate opening signal $p_{gv}$ from the Hydro Turbine Governor and outputs mechanical power $p_m$ to the synchronous generator. The transfer function is specified by the water starting time $T_W$ and can be represented as the sum of two parallel blocks as shown in Fig. 15.

The forward-Euler discretised equations are: $$ x_{1}(k) = x_{1}(k-\Delta t) + \frac{\Delta t}{0.5,T_{W}} \left(p_{gv}(k-\Delta t) - x_{1}(k-\Delta t)\right), $$ $$ p_{m}(k) = 3,x_{1}(k) - 2,p_{gv}(k). $$

Assuming the simulation starts in steady state (all derivatives zero), the initial values are: $$ x_{1}(t=0) = p_{gv}(t=0), \quad p_{m}(t=0) = p_{gv}(t=0). $$