Pump-driven circulating pipe network

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).