STEM Problem Detail

Problem Statement

Prompt

give the answer of the example 25.1

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**Solution.** The total water available for hydrogenerator is
\begin{equation*}
120 - 50 = 70 \text{ m}^3\text{/sec}.
\end{equation*}

Since there are three subintervals, there are two independent variables $$q^2$$ and $$q^3$$.
Let
\begin{equation*}
q^2 = 25 \quad \text{and} \quad q^3 = 25 \text{ m}^3\text{/sec.}
\end{equation*}
Hence
\begin{equation*}
q^1 = 70 - (25 + 25) = 20 \text{ m}^3\text{/sec.}
\end{equation*}
It is given that $$W^0 = 120 \text{ m}^3\text{/sec.}$$ and $$W^3 = 50 \text{ m}^3\text{/sec.}$$

From equation (25.7) and since $$W_i$$ water inflow is zero
\begin{align*}
W^1 &= W^0 + W_i^1 - q^1 = 120 + 0 - 20 = 100 \text{ m}^3\text{/sec} \\
W^2 &= W^1 + W_i^2 - q^2 = 100 + 0 - 25 = 75 \text{ m}^3\text{/sec.}
\end{align*}

Now using equation (25.8) hydrogeneration for the subintervals can be calculated as follows
\begin{align*}
P_H^1 &= 9.81 \times 10^{-3} \times 20 \, [1 + 0.5 \times 0.006 \, (120 + 100)] \, (20 - 2) \\
&= 9.81 \times 10^{-3} \times 20 \times 1.66 \times 18 = 5.86 \text{ MW} \\
P_H^2 &= 9.81 \times 10^{-3} \times 20 \, [1 + 0.5 \times 0.006 \, (100 + 75)] \times 23 \\
&= 9.81 \times 10^{-3} \times 20 \times 1.825 \times 23 = 8.235 \text{ MW} \\
P_H^3 &= 9.81 \times 10^{-3} \times 20 \, [1 + 0.5 \times 0.006 \, (75 + 50)] \times 23 \\
&= 9.81 \times 10^{-3} \times 20 \times 1.325 \times 23 = 6.2 \text{ MW}
\end{align*}

Therefore, thermal generation during three intervals is
\begin{align*}
P_T^1 &= 8 - 5.86 = 2.14 \text{ MW} \\
P_T^2 &= 12 - 8.23 = 3.77 \text{ MW} \\
P_T^3 &= 7 - 6.2 = 0.8 \text{ MW}
\end{align*}

Now
\begin{equation*}
\frac{dF_T}{dP_T} = P_T + 20 = \lambda_1^m
\end{equation*}
The value of $$\lambda_1$$ for the three intervals
\begin{align*}
\lambda_1^1 &= 20 + 2.14 = 22.14, \\
\lambda_1^2 &= 20 + 3.77 = 23.77, \\
\lambda_1^3 &= 20.8
\end{align*}

Since transmission losses are neglected, $$\frac{dP_L}{dP_H}$$ is zero and hence from equation (25.12)
\begin{align*}
\lambda_1^1 &= \lambda_3^1 = 22.14, \quad \lambda_1^2 = \lambda_3^2 = 23.77 \\
\lambda_1^3 &= \lambda_3^3 = 20.8
\end{align*}

\newpage

Now from equation (25.14)
\begin{align*}
\lambda_2^1 &= \lambda_3^1 \, h_0 [1 + 0.5e \, (2W^0 + W_i^1 - 2q^1 + \rho)] \\
&= 22.14 \times 0.1962 \, [1 + 0.5 \times 0.006 \, (240 + 0 - 40 + 2)] \\
&= 22.14 \times 0.1962 \times 1.6 = 6.976
\end{align*}

From equation (25.13) for $$m = 1$$ and $$2$$, we have
\begin{equation*}
\lambda_2^1 - \lambda_2^2 - \lambda_3^1 \{0.5 \, h_o e \, (q^1 - \rho)\} - \lambda_3^2 \{0.5 \, h_o e \, (q^2 - \rho)\} = 0
\end{equation*}
or
\begin{align*}
\lambda_2^2 &= \lambda_2^1 - \lambda_3^1 \{0.5 \, h_o e \, (q^1 - \rho)\} - \lambda_3^2 \{0.5 \, h_o e \, (q^2 - \rho)\} \\
&= 6.976 - 22.14 \, \{0.5 \times 0.1962 \times 0.006 \times 18\} - 23.77 \, \{0.5 \times 0.1962 \times 0.006 \times (25 - 2)\} \\
&= 6.976 - 22.14 \times 0.01 - 23.77 \times 0.0135 \\
&= 6.976 - 0.22 - 0.32 = 6.4
\end{align*}

Similarly for $$m = 2$$
\begin{align*}
\lambda_2^3 &= \lambda_2^2 - \lambda_3^2 \{0.5 \, h_o e \, (q^2 - \rho)\} - \lambda_3^3 \{0.5 \, h_o e \, (q^3 - \rho)\} \\
&= 6.4 - 23.77 \, \{0.5 \times 0.1962 \times 0.006 \times 23\} - 20.8 \, \{0.5 \times 0.1962 \times 0.006 \times 23\} \\
&= 6.4 - 0.32 - 0.28 = 5.8
\end{align*}

The gradient vector is obtained using equation (25.15) for $$m = 2$$ and $$m = 3$$
\begin{align*}
\frac{\partial F}{\partial q^2} &= \lambda_2^2 - \lambda_3^2 h_0 \, \{1 + 0.5 \times 0.006 \, (2 \times 100 - 2 \times 25 + 2)\} \\
&= 6.4 - 23.77 \times 0.1962 \, \{1 + 0.456\} = 6.4 - 6.79 = -0.39 \\
\frac{\partial F}{\partial q^3} &= \lambda_2^3 - \lambda_3^3 h_0 \, \{1 + 0.5 \times 0.006 \, (2W^2 + 0 - 2q^3 + \rho)\} \\
&= 5.8 - 20.8 \times 0.1962 \, \{1 + 0.5 \times 0.006 \, (2 \times 75 - 2 \times 25 + 2)\} \\
&= 5.8 - 5.3 = 0.5
\end{align*}

Since the gradient vector is not a null or near null vector, we will have to go through second iteration. However, the values of the independent variables at the start of the second iteration will be obtained using equation (25.16). Usually an optimal value of $$\alpha$$ should be selected which comes only after some experience of working on the system. Let us take $$\alpha = 0.4$$
\begin{align*}
q_{\text{new}}^2 &= q_{\text{old}}^2 - \alpha \frac{\partial F}{\partial q^2} = 25 - 0.4 \, (-0.39) = 25.156 \text{ m}^3\text{/sec} \\
q_{\text{new}}^3 &= q_{\text{old}}^3 - \alpha \frac{\partial F}{\partial q^3} = 25 - 0.4 \times 0.5 = 24.8 \text{ m}^3\text{/sec.}
\end{align*}
Hence $$q^1 = 120 - 50 - (25.156 + 24.8) = 20 \text{ m}^3\text{/sec.}$$

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Problem #66