Supplementary Material 0.19

Chapter 0 – Steady-State Hydraulics

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Supplementary Material: Example Problems and Solutions

Chapter 0 – Problem 0.19

0.19 Consider the classic three-reservoir problem shown in the following figure:

The length, diameter, and Hazen-William roughness coefficients of the pipeline are:

1.  From A to J: 200 m, 500 mm, and 100, respectively
2. From J to B: 2000 m, 500 mm, and 100, respectively.
3. From J to C: 400 m, 200 mm, and 140, respectively.

The hydraulic grades at reservoirs A, B, and C are 200 m, 180 m, and 185 m, respectively. Compute the steady-state flowrates in all three pipelines, neglecting minor losses.

Solution:

Assuming the flow is from reservoir A to junction node J, and from junction J to reservoirs B and C, there are three unknown flowrates: Q_AJ, Q_JB, and Q_JC.This requires three non-redundant equations to solve for the three unknown flowrates:

Eq. 1.   Energy equation between reservoirs A and B:

E_A – ΔH_AJ – ΔH_JB = E_B

Eq.2.   Energy equation between reservoirs A and C:

E_B – ΔH_AJ – ΔH_JC = E_C

Eq. 3.   Continuity equation at junction J:

Q_AJ = Q_JB + Q_JC

Substituting the appropriate Hazen-William expressions for ΔH terms:

E_A – (10.67 L_AJ Q_AJ^1.852)/(C_AJ^1.852 D_AJ^4.87) – (10.67 L_JB Q_JB^1.852)/(C_JB^1.852 D_JB^4.87) = E_B

E_A – (10.67 L_AJ Q_AJ^1.852)/(C_AJ^1.852 D_AJ^4.87) – (10.67 L_JC Q_JC^1.852)/(C_JC^1.852 D_JC^4.87) = E_C

Q_AJ = Q_JB + Q_JC

The only unknowns in the above three equations are Q_AJ, Q_JB, and Q_JC.

Solve these equations iteratively to obtain the following results:  Q_AJ = 0.433 m³/s, Q_JB = 0.346 m³/s, and Q_JC =0.087 m³/s.

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