Supplementary Material, Appendix 0 – A0.49

APPENDIX 0 – Pre-requisite: Steady State Hydraulics

Main – Previous – Next

Surge Analysis and the Wave Plan Method

Supplementary Material: Example Problems and Solutions

Appendix 0 – Problem 49

A0.49 Consider the classical three-reservoir problem as shown in the following.

Length, diameter, and Hazen-William roughness coefficient of the pipeline are:

•  From A to J are 200m, 500 mm, and 100
• From J to B are 2000m, 500 mm, and 100
• From J to C are 400m, 200 mm, and 140

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

Solution:

Assuming that the flow takes place from reservoir A to junction node J, and from there to reservoirs B and C, there are three unknown flowrates: Q_AJ, Q_JB, and Q_JC. We need to setup 3 non-redundant equations to solve for these 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 3 equations are Q_AJ, Q_JB, and Q_JC.

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

---

Summary of revisions to this page:

Date/Revision