--- title: "Pump-driven circulating pipe network" subject: "Hydraulics" chapter: 3 key: "semester_4/hydraulics:circulating-network-pump-max-discharge" source: https://www.bctengineer.com/q/semester_4/hydraulics/circulating-network-pump-max-discharge site: BCT Engineering --- # Pump-driven circulating pipe network Hydraulics · Chapter 3: Three reservoirs problem and Pipe networks - Asked in: 2080 Chaitra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/circulating-network-pump-max-discharge - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-3 ## Question A circulating pipe flow network as shown in Fig 1 is constructed for a cooling system in Tribhuwan University Main building. If a pump capacity of 500 watts is used at junction I, determine the maximum discharge that the pump can circulate in the system. [Figure: loop network A-B-C-D, then D to E (link DE), then E-F-G-H, then H to I (link HI), pump P at I, returning to A (link AI); discharge Q leaves/returns at the left] | Link | f | L (m) | D (m) | |---|---|---|---| | AB | 0.06 | 10 | 0.1 | | AC | 0.06 | 10 | 0.1 | | BD | 0.06 | 10 | 0.1 | | CD | 0.06 | 10 | 0.1 | | DE | 0.05 | 15 | 0.2 | | EF | 0.06 | 10 | 0.1 | | EG | 0.06 | 10 | 0.1 | | FH | 0.06 | 10 | 0.1 | | GH | 0.06 | 10 | 0.1 | | HI | 0.04 | 50 | 0.2 | | AI | 0.04 | 50 | 0.2 | ## Short answer Combine series–parallel pipe resistances to total R = 6184.1 s²/m⁵. Pump power P = γ R Q³ → Q = (500/(9810×6184.1))^(1/3) = 0.0202 m³/s (20.2 L/s). ## Answer **Reading of the figure.** The network is a closed circuit: A to D through a diamond (two parallel paths ABD and ACD), the link DE, E to H through a second diamond (paths EFH and EGH), then link HI, the pump at I, and link AI back to A. The same discharge $Q$ passes through the whole circuit, and the pump head equals the total head loss. Only friction is considered, $h_f = rQ^2$ with $r = \dfrac{8fL}{\pi^2gD^5}$. #### Resistances (s²/m⁵) - Each small link ($f = 0.06$, $L = 10$ m, $D = 0.1$ m): $r_1 = 4957.6$ - DE ($f = 0.05$, $L = 15$ m, $D = 0.2$ m): $r_{DE} = 193.7$ - HI and AI ($f = 0.04$, $L = 50$ m, $D = 0.2$ m): $r_{HI} = r_{AI} = 516.4$ each #### Equivalent resistance of one diamond Each path (e.g. A-B-D) has two links in series: $2r_1$. The two paths are identical, so the flow splits equally ($Q/2$ in each), and the head loss across the diamond is $$ h = 2r_1\left(\frac{Q}{2}\right)^2 = \frac{r_1}{2}Q^2 $$ Two diamonds in series: $2\times\dfrac{r_1}{2} = r_1 = 4957.6$. #### Total resistance of the circuit $$ r_{total} = r_1 + r_{DE} + r_{HI} + r_{AI} = 4957.6 + 193.7 + 2(516.4) = 6184.1\ \text{s}^2/\text{m}^5 $$ #### Pump power and discharge The pump supplies the head $H_p = r_{total}Q^2$; with power $P = \gamma QH_p$: $$ P = \gamma\,r_{total}\,Q^3 \;\Rightarrow\; Q = \left(\frac{P}{\gamma\,r_{total}}\right)^{1/3} = \left(\frac{500}{9810(6184.1)}\right)^{1/3} = 0.02020\ \text{m}^3/\text{s} $$ The head developed is $H_p = 6184.1(0.02020)^2 = 2.52$ m. The velocity in the 0.2 m links is $0.64$ m/s (and about 1.29 m/s in the 0.1 m links, where the flow is halved). **Answer:** Maximum discharge $Q \approx 20.2$ l/s ($0.02020$ m³/s). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the relation between the pump head and the network? A. Pump head equals the total head loss around the circuit at the same Q B. Pump head equals the largest link loss C. Pump head is zero D. Pump head equals static lift Answer: A. In a closed circulation system the same Q passes round and the pump supplies the total friction loss. (Key point: Series circuit) 2. For a diamond of two identical parallel paths each of two links r1, what is the equivalent resistance? A. r1/2 B. 2 r1 C. r1/4 D. r1 Answer: A. Head loss is 2 r1 (Q/2)^2 = r1 Q^2/2, so equivalent resistance is r1/2. (Key point: Diamond resistance) 3. What is r_total? A. Two diamonds (r1) + r_DE + r_HI + r_AI = 6184 s^2/m^5 B. Only the diamonds C. r_DE + r_HI D. Product of all resistances Answer: A. Series elements add: two diamonds r1/2 each, plus DE, HI and AI. (Key point: Total resistance) 4. How is Q obtained from the pump power? A. P = gamma r Q^3, so Q = (P/(gamma r))^(1/3) B. P = gamma r Q^2, so Q = sqrt(P/(gamma r)) C. Q = P/gamma D. Q = P/r Answer: A. P = gamma Q H_p with H_p = r Q^2 gives P = gamma r Q^3. 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