Linear method for pipe network analysis

5.1   True relation (for $n = 2$)

\[ \DeclareMathOperator{\sgn}{sgn} Q(H_i - H_j) = \sgn(H_i - H_j) \sqrt{\frac{|H_i - H_j|}{K}} \]

Let $\Delta H = H_i - H_j$

\[ Q(\Delta H) = \sgn(\Delta H) \sqrt{\frac{|\Delta H|}{K}} = \frac{\Delta H}{\sqrt{K |\Delta H|}},\quad \Delta H \neq 0 \]

5.2   Taylor expansion about the initial estimate

Let the initial estimate \[ \Delta H_0 = \Delta H_{ij}^{(0)} = H_i^{(0)} - H_j^{(0)} \]

\[ Q(\Delta H) \approx Q(\Delta H_0) + \left.\frac{dQ}{d(\Delta H)}\right|_{\Delta H_0} (\Delta H - \Delta H_0) \]

Compute derivative \[ \frac{dQ}{d(\Delta H)} = \frac{1}{2\sqrt{K |\Delta H|}},\quad \Delta H \neq 0 \]

Evaluate at the current estimate $\Delta H_0$ \[ Q(\Delta H) \approx \frac{\Delta H_0}{\sqrt{K |\Delta H_0|}} + \frac{1}{2\sqrt{K |\Delta H_0|}} (\Delta H - \Delta H_0) \]

5.3   Rearranged form

\[ Q(\Delta H) \approx \frac{1}{2\sqrt{K |\Delta H_0|}} \Delta H + \left(\frac{\Delta H_0}{\sqrt{K |\Delta H_0|}} - \frac{\Delta H_0}{2\sqrt{K |\Delta H_0|}}\right) \]

Simplify the constant term \[ Q(\Delta H) \approx \frac{1}{2\sqrt{K |\Delta H_0|}} \Delta H + \frac{\Delta H_0}{2\sqrt{K |\Delta H_0|}} \]

5.4   Linear approximation used in the method

Define slope \[ C_{ij} = \frac{1}{2\sqrt{K |\Delta H_0|}} \]

Then \[ Q_{ij} \approx C_{ij} (H_i - H_j) + D_{ij} \]

Where \[ D_{ij} = \frac{\Delta H_0}{2\sqrt{K |\Delta H_0|}} \]

Substituting into continuity \[ \sum C_{ij}(H_i - H_j) = -\sum D_{ij} \]

Interpretation

• $\Delta H_0$ is defined per pipe (i–j)
• Left-hand side → matrix (A)
• Right-hand side → includes constants $D_{ij}$ and external demands (if present)

5.5   Key insight

• Linear method = first-order Taylor expansion
• We approximate the nonlinear curve locally
• Update point → recompute slope → repeat

6.1   Step 1: Initial guess

• Assume initial heads
  • $H_B^{(0)} = 95 \ \text{m}$
  • $H_C^{(0)} = 90 \ \text{m}$
• Head differences based on the current estimate
  • $\Delta H_{AB}^{(0)} = H_A - H_B^{(0)} = 100 - 95 = 5$
  • $\Delta H_{BC}^{(0)} = H_B^{(0)} - H_C^{(0)} = 95 - 90 = 5$
  • $\Delta H_{AC}^{(0)} = H_A - H_C^{(0)} = 100 - 90 = 10$

6.2   Step 2: Compute coefficients and constants

• Linearization about the current head difference $H_0$
  • $Q(\Delta H) \approx C_{ij}(H_i-H_j) + D_{ij}$
• Coefficient
  • $C_{ij} = \frac{1}{2\sqrt{K |\Delta H_0|}}$
• Constant term
  • $D_{ij} = \frac{\Delta H_0}{2\sqrt{K|\Delta H_0|}}$
• A–B ($K=2$, $\Delta H_0=5$)
  • $C_{AB} = \frac{1}{2\sqrt{2|5|}} \approx 0.158$
  • $D_{AB} = \frac{5}{2\sqrt{2|5|}} \approx 0.791$
• B–C ($K=1$, $\Delta H_0=5$)
  • $C_{BC} = \frac{1}{2\sqrt{1|5|}} \approx 0.224$
  • $D_{BC} = \frac{5}{2\sqrt{1|5|}} \approx 1.118$
• A–C ($K=3$, $\Delta H_0=10$)
  • $C_{AC} = \frac{1}{2\sqrt{3|10|}} \approx 0.091$
  • $D_{AC} = \frac{10}{2\sqrt{3|10|}} \approx 0.913$

6.3   Step 3: Continuity equations

• Linearized pipe flows
  • $Q_{AB} \approx C_{AB}(H_A-H_B) + D_{AB}$
  • $Q_{BC} \approx C_{BC}(H_B-H_C) + D_{BC}$
  • $Q_{AC} \approx C_{AC}(H_A-H_C) + D_{AC}$
• Node B
  • $Q_{AB} - Q_{BC} = 0$
  • $C_{AB}(H_A-H_B) + D_{AB} - C_{BC}(H_B-H_C) - D_{BC} = 0$
  • $0.158(100-H_B) + 0.791 - 0.224(H_B-H_C) - 1.118 = 0$
• Node C
  • $Q_{AC} + Q_{BC} = 10$
  • $C_{AC}(H_A-H_C) + D_{AC} + C_{BC}(H_B-H_C) + D_{BC} = 10$
  • $0.091(100-H_C) + 0.913 + 0.224(H_B-H_C) + 1.118 = 10$

6.4   Step 4: Matrix form

• Rearranged system
  • $-0.382H_B + 0.224H_C = -15.473$
  • $0.224H_B - 0.315H_C = -1.131$
• Matrix form \[ \begin{bmatrix} -0.382 & 0.224 \\ 0.224 & -0.315 \end{bmatrix} \begin{bmatrix} H_B \\ H_C \end{bmatrix} = \begin{bmatrix} -15.473 \\ -1.131 \end{bmatrix} \]
• Solve for updated heads after iteration 1
  • $H_B^{(1)} \approx 73.09 \ \text{m}$
  • $H_C^{(1)} \approx 55.56 \ \text{m}$
• These are updated heads, not yet the final solution
  • Use $H_B^{(1)}$ and $H_C^{(1)}$ as the new estimate for the next iteration

7.1   Step 1: Initial guess

• Assume initial heads
  • $H_B^{(0)} = 110 \ \text{m}$
  • $H_C^{(0)} = 100 \ \text{m}$
• Head differences based on the current estimate
  • $\Delta H_{AB}^{(0)} = H_A - H_B^{(0)} = 120 - 110 = 10$
  • $\Delta H_{BC}^{(0)} = H_B^{(0)} - H_C^{(0)} = 110 - 100 = 10$
  • $\Delta H_{AC}^{(0)} = H_A - H_C^{(0)} = 120 - 100 = 20$

7.2   Step 2: Compute coefficients and constants

• Linearization about the current head difference $H_0$
  • $Q(\Delta H) \approx C_{ij}(H_i-H_j) + D_{ij}$
• Coefficient
  • $C_{ij} = \frac{1}{2\sqrt{K |\Delta H_0|}}$
• Constant term
  • $D_{ij} = \frac{\Delta H_0}{2\sqrt{K|\Delta H_0|}}$
• A–B ($K=4$, $\Delta H_0=10$)
  • $C_{AB} = \frac{1}{2\sqrt{4|10|}} \approx 0.0791$
  • $D_{AB} = \frac{10}{2\sqrt{4|10|}} \approx 0.7906$
• B–C ($K=2$, $\Delta H_0=10$)
  • $C_{BC} = \frac{1}{2\sqrt{2|10|}} \approx 0.1118$
  • $D_{BC} = \frac{10}{2\sqrt{2|10|}} \approx 1.1180$
• A–C ($K=5$, $\Delta H_0=20$)
  • $C_{AC} = \frac{1}{2\sqrt{5|20|}} = 0.0500$
  • $D_{AC} = \frac{20}{2\sqrt{5|20|}} = 1.0000$

7.3   Step 3: Continuity equations

• Linearized pipe flows
  • $Q_{AB} \approx C_{AB}(H_A-H_B) + D_{AB}$
  • $Q_{BC} \approx C_{BC}(H_B-H_C) + D_{BC}$
  • $Q_{AC} \approx C_{AC}(H_A-H_C) + D_{AC}$
• Node B
  • $Q_{AB} - Q_{BC} = 0$
  • $C_{AB}(H_A-H_B) + D_{AB} - C_{BC}(H_B-H_C) - D_{BC} = 0$
  • $0.0791(120-H_B) + 0.7906 - 0.1118(H_B-H_C) - 1.1180 = 0$
• Node C
  • $Q_{AC} + Q_{BC} = 8$
  • $C_{AC}(H_A-H_C) + D_{AC} + C_{BC}(H_B-H_C) + D_{BC} = 8$
  • $0.0500(120-H_C) + 1.0000 + 0.1118(H_B-H_C) + 1.1180 = 8$

7.4   Step 4: Matrix form

• Rearranged system
  • $-0.1909H_B + 0.1118H_C = -9.1646$
  • $0.1118H_B - 0.1618H_C = -0.1180$
• Matrix form \[ \begin{bmatrix} -0.1909 & 0.1118 \\ 0.1118 & -0.1618 \end{bmatrix} \begin{bmatrix} H_B \\ H_C \end{bmatrix} = \begin{bmatrix} -9.1646 \\ -0.1180 \end{bmatrix} \]
• Solve for updated heads after iteration 1
  • $H_B^{(1)} \approx 81.36 \ \text{m}$
  • $H_C^{(1)} \approx 56.95 \ \text{m}$
• These are updated heads, not yet the final solution
  • Use $H_B^{(1)}$ and $H_C^{(1)}$ as the new estimate for the next iteration