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Network: unknown discharges and head losses

Short answer

Applying continuity at nodes and equating head losses along alternative paths yields: Q_B = 50, Q_2 = 20 (C to B), Q_4 = 60 (D to A), Q_5 = 10 (A to C); head losses h_{f4} = 100, h_{f5} = 20.

Figure below shows a network in which Q and hfh_f refer to discharges and pressure drops respectively. Subscripts 1, 2, 3, 4 and 5 designate respective values in pipe length AB, BC, CD, DA and AC. Subscripts A, B, C and D designate discharges entering or leaving the junction points A, B, C and D respectively. By sticking to the values given in the figure find the following discharges QBQ_B, Q2Q_2, Q4Q_4 and Q5Q_5 and the pressure drops hf4h_{f4} and hf5h_{f5} and give these computed values at their respective places on a neat sketch of the network along with flow directions. [Figure: square network ABCD with diagonal AC; QAQ_A = 20 leaves at A; Q1Q_1 = 30, hf1h_{f1} = 60 in AB; QBQ_B = ? leaves at B; Q2Q_2 = ?, hf2h_{f2} = 40 in BC; QCQ_C = 30 leaves at C; Q3Q_3 = 40, hf3h_{f3} = 120 in CD; QDQ_D = 100 enters at D; Q4Q_4 = ?, hf4h_{f4} = ? in DA; Q5Q_5 = ?, hf5h_{f5} = ? in AC]

Method. Apply continuity at every node, then use the fact that the head (pressure) drop between any two nodes is the same along every path joining them. External flows: QA=20Q_A = 20 out, QC=30Q_C = 30 out, QD=100Q_D = 100 in; QBQ_B is unknown (out).

Directions of the known flows

Water enters the network only at D (100 units), so D has the highest head in the network and every pipe connected to D carries water away from it. The flow in CD is therefore from D to C, and the flow in DA is from D to A. The flow Q1=30Q_1 = 30 in AB is taken from A to B, as shown in the figure (A is fed from D).

Discharges

  • Node D (100 in): Q3+Q4=100⇒Q4=100−40=60Q_3 + Q_4 = 100 \Rightarrow Q_4 = 100 - 40 = \mathbf{60} (D to A)
  • Node A: inflow Q4=60Q_4 = 60; outflow =QA+Q1+Q5=20+30+Q5⇒Q5=60−50=10= Q_A + Q_1 + Q_5 = 20 + 30 + Q_5 \Rightarrow Q_5 = 60 - 50 = \mathbf{10} (A to C)
  • Node C: inflow Q3+Q5+Q2=40+10+Q2Q_3 + Q_5 + Q_2 = 40 + 10 + Q_2; outflow QC=30⇒Q2=30−50=−20Q_C = 30 \Rightarrow Q_2 = 30 - 50 = -20, i.e. Q2=20Q_2 = \mathbf{20} flowing from C to B
  • Node B: inflow Q1+Q2=30+20=50Q_1 + Q_2 = 30 + 20 = 50, so the outflow is QB=50Q_B = \mathbf{50}

Check for the whole network: inflow 100 = outflows 20+50+30=10020 + 50 + 30 = 100.

Head losses (pressure drops)

Let HH denote the head at a node.

  • Path A to B to C: HA−HB=hf1=60H_A - H_B = h_{f1} = 60. The flow in BC is from C to B, so HC−HB=hf2=40H_C - H_B = h_{f2} = 40. Hence
HA−HC=60−40=20  ⇒  hf5=20H_A - H_C = 60 - 40 = 20 \;\Rightarrow\; h_{f5} = \mathbf{20}
  • Path D to C: HD−HC=hf3=120H_D - H_C = h_{f3} = 120. Path D to A to C: HD−HC=hf4+hf5H_D - H_C = h_{f4} + h_{f5}, so
hf4=120−20=100h_{f4} = 120 - 20 = \mathbf{100}

(Check with the square law hf∝Q2h_f \propto Q^2: hf1/Q12=0.0667h_{f1}/Q_1^2 = 0.0667, hf2/Q22=0.1h_{f2}/Q_2^2 = 0.1, hf3/Q32=0.075h_{f3}/Q_3^2 = 0.075, hf4/Q42=0.0278h_{f4}/Q_4^2 = 0.0278, hf5/Q52=0.2h_{f5}/Q_5^2 = 0.2; all the resistances are positive, so the values are consistent.)

            Q1=30 (hf1=60)
   Qa=20 <- A ----------> B -> QB=50
            |\             ^
 Q4=60 (100)| \ Q5=10 (20) |  Q2=20 (hf2=40)
            |  \           |
   Qd=100 ->D ----------> C -> QC=30
            Q3=40 (hf3=120)

Arrows: D to A (Q4), A to B (Q1), A to C (Q5, diagonal), C to B (Q2), D to C (Q3).

Answer: QB=50Q_B = 50, Q2=20Q_2 = 20 (C to B), Q4=60Q_4 = 60 (D to A), Q5=10Q_5 = 10 (A to C); hf4=100h_{f4} = 100, hf5=20h_{f5} = 20.

✓ Answer checked by Claude Opus 5.5 (high effort)

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