Three reservoirs problem and Pipe networks

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. Take the flow into the network as positive. External flows: $Q_A = 20$ out, $Q_C = 30$ out, $Q_D = 100$ in; $Q_B$ is unknown (out).

Directions of the known flows

Water enters only at D (100 units). Pipe CD has $Q_3 = 40$ and $h_{f3} = 120$. If the flow in CD went from C to D, then going round the loop gives $h_{f4} + h_{f5} + h_{f3}$ with all heads adding up to a contradiction (the head at D would need to be higher than C and lower than C at the same time). So flow in CD is from D to C. Likewise $Q_1 = 30$ flows from A to B (A is fed from D).

Discharges

• Node D (100 in): $Q_3 + Q_4 = 100 \Rightarrow Q_4 = 100 - 40 = \mathbf{60}$ (D to A)
• Node A: inflow $Q_4 = 60$; outflow $= Q_A + Q_1 + Q_5 = 20 + 30 + Q_5 \Rightarrow Q_5 = 60 - 50 = \mathbf{10}$ (A to C)
• Node C: inflow $Q_3 + Q_5 + Q_2 = 40 + 10 + Q_2$; outflow $Q_C = 30 \Rightarrow Q_2 = 30 - 50 = -20$, i.e. $Q_2 = \mathbf{20}$ flowing from C to B
• Node B: inflow $Q_1 + Q_2 = 30 + 20 = 50$; the outflow $Q_B = \mathbf{50}$

Check of the whole network: inflow 100 = outflows $20 + 50 + 30 = 100$.

Head losses (pressure drops)

Let $H$ denote the head at a node.

• Path A to B to C: $H_A - H_B = h_{f1} = 60$. The flow in BC is from C to B, so $H_C - H_B = h_{f2} = 40$. Hence

$$H_A - H_C = 60 - 40 = 20 \;\Rightarrow\; h_{f5} = \mathbf{20}$$

• Path D to C: $H_D - H_C = h_{f3} = 120$. Path D to A to C: $H_D - H_C = h_{f4} + h_{f5}$:

$$h_{f4} = 120 - 20 = \mathbf{100}$$

(Check with the square law $h_f \propto Q^2$: $h_{f1}/Q_1^2 = 0.0667$, $h_{f2}/Q_2^2 = 0.1$, $h_{f3}/Q_3^2 = 0.075$, $h_{f4}/Q_4^2 = 0.0278$, $h_{f5}/Q_5^2 = 0.2$; all are positive resistances, 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: $Q_B = 50$, $Q_2 = 20$ (C to B), $Q_4 = 60$ (D to A), $Q_5 = 10$ (A to C); $h_{f4} = 100$, $h_{f5} = 20$.

Data. Reservoir levels: A = 100 m, B = 70 m, C = 30 m; junction D at elevation 50 m. Only friction losses are considered (velocity heads and minor losses neglected). For each pipe $h_f = rQ^2$ with $r = \dfrac{8fL}{\pi^2gD^5}$:

Method (trial of the piezometric head $H_D$)

Let $H_D$ be the piezometric head at the junction. Flow in each pipe: $Q_i = \sqrt{|H_i - H_D|/r_i}$, directed from the higher head to the lower. Continuity at D: $\sum Q_{in} = \sum Q_{out}$.

Since $70 < H_D < 100$ is expected (D is between the highest and the lowest reservoir), first try whether B supplies or takes water:

• Trial $H_D = 70$ m (B just balanced): $Q_A = \sqrt{30/1563.1} = 0.1385$ m³/s in, $Q_C = \sqrt{40/12729.6} = 0.0561$ m³/s out. Inflow exceeds outflow, so water must leave via B, and $H_D$ must rise above 70 m. This is case: A feeds B and C, with $Q_A = Q_B + Q_C$.

Trials (with $Q_A = Q_B + Q_C$):

The balance is reached at

$$H_D = 71.98\ \text{m}$$

Discharges

$$Q_{AD} = \sqrt{\frac{100 - 71.98}{1563.1}} = 0.1339\ \text{m}^3/\text{s} = 133.9\ \text{l/s}\quad(\text{A to D})$$ $$Q_{DB} = \sqrt{\frac{71.98 - 70}{338.1}} = 0.0765\ \text{m}^3/\text{s} = 76.5\ \text{l/s}\quad(\text{D to B})$$ $$Q_{DC} = \sqrt{\frac{71.98 - 30}{12729.6}} = 0.0574\ \text{m}^3/\text{s} = 57.4\ \text{l/s}\quad(\text{D to C})$$

Check: $76.5 + 57.4 = 133.9$ l/s $\approx 133.9$ l/s.

Piezometric head and pressure at D

Piezometric head at D $= 71.98$ m. As the junction is at elevation 50 m, the pressure head there is $71.98 - 50 = 21.98$ m of water ($p_D \approx 216$ kPa).

Answer: $Q_{AD} = 133.9$ l/s (in), $Q_{DB} = 76.5$ l/s and $Q_{DC} = 57.4$ l/s (out); $H_D = 71.98$ m.

Data. $Z_1 = 29$ m, $Z_2 = 129$ m, $Z_3 = 69$ m; $L_1 = 80$ m, $L_2 = 150$ m, $L_3 = 110$ m; $D = 0.25$ m, $k = 0.5$ mm, $\nu = 1.02\times10^{-6}$ m²/s. The friction factor is found from the Colebrook-White equation (no Moody chart), neglecting minor losses and velocity heads at the reservoirs.

$$\frac{k}{D} = \frac{0.5}{250} = 0.0020,\qquad \frac{1}{\sqrt f} = -2\log_{10}\left(\frac{k/D}{3.7} + \frac{2.51}{Re\sqrt f}\right)$$

A starting value comes from the fully rough formula $\dfrac{1}{\sqrt f} = 2\log_{10}\dfrac{D}{k} + 1.14$, giving $f_0 = 0.0234$.

Method

Let $H_J$ be the piezometric head at the junction. For pipe $i$ the head loss is $|Z_i - H_J| = f_i\dfrac{L_i}{D}\dfrac{V_i^2}{2g}$, so $V_i = \sqrt{\dfrac{2gD|Z_i - H_J|}{f_iL_i}}$, with $f_i$ updated from $Re_i = V_iD/\nu$. The correct $H_J$ satisfies continuity at the junction: $\sum Q = 0$.

The middle reservoir is the highest (129 m) and the lowest is 29 m, so the junction head lies between 29 and 129 m. If $H_J > 69$ m, reservoir 3 also feeds the junction; if $H_J < 69$ m, it receives water. Trying values (each with Colebrook for $f$):

• $H_J = 69$ m: $Q_1 \approx 0.50$ m³/s out, $Q_2 \approx 0.45$ m³/s in, $Q_3 = 0$: outflow > inflow, so raise $H_J$.
• $H_J = 68.5$ to 70: the balance is reached at $H_J \approx 68.48$ m.

Final solution ($H_J = 68.48$ m)

(Positive $Q$ = into the junction; negative = out of the junction.) Check: $450.4 + 47.9 = 498.3$ l/s in, and $498.3$ l/s out.

Answer: Reservoir 2 delivers $Q_2 = 450.4$ l/s and reservoir 3 delivers $Q_3 = 47.9$ l/s to the junction; both flow on to reservoir 1 with $Q_1 = 498.3$ l/s. Junction piezometric head $H_J = 68.48$ m (about 68.5 m).

Data. Main pipe AJ: $L = 10$ km, $D = 0.75$ m, slope $2.2\times10^{-3}$. Branch JB: $L = 5.5$ km, slope $2.75\times10^{-3}$. Branch JC: $L = 3$ km, slope $3.2\times10^{-3}$. $Q_B = Q_C = 0.24$ m³/s, so the main pipe carries $Q = 0.48$ m³/s. $f = 0.025$. Only friction loss: $h_f = \dfrac{8fLQ^2}{\pi^2gD^5}$. Each reservoir level is 3 m above the end of its pipe (A: above the start of the main pipe; B and C: above the ends of the branches).

Elevations (datum: level of junction J)

• The main pipe falls $10\,000(2.2\times10^{-3}) = 22$ m from A to J, so the level of A is $22 + 3 = 25$ m above J.
• The branch to B falls $5500(2.75\times10^{-3}) = 15.125$ m, so the level of B is $-15.125 + 3 = -12.125$ m (12.125 m below J).
• The branch to C falls $3000(3.2\times10^{-3}) = 9.6$ m, so the level of C is $-9.6 + 3 = -6.6$ m.

Head loss in the main pipe and the head at J

$$h_{f,AJ} = \frac{8(0.025)(10\,000)(0.48)^2}{\pi^2(9.81)(0.75)^5} = 20.06\ \text{m}$$

Piezometric head at J (above J's level): $H_J = 25 - 20.06 = 4.94$ m.

Head available for each branch

• To B: $h_B = H_J - (-12.125) = 17.07$ m
• To C: $h_C = H_J - (-6.6) = 11.54$ m

Branch diameters

$$D = \left(\frac{8fLQ^2}{\pi^2g\,h_f}\right)^{1/5}$$ $$D_B = \left(\frac{8(0.025)(5500)(0.24)^2}{\pi^2(9.81)(17.07)}\right)^{1/5} = 0.5209\ \text{m} \approx 521\ \text{mm}$$ $$D_C = \left(\frac{8(0.025)(3000)(0.24)^2}{\pi^2(9.81)(11.54)}\right)^{1/5} = 0.4989\ \text{m} \approx 499\ \text{mm}$$

Answer: Diameter of the branch to B $\approx 521$ mm; diameter of the branch to C $\approx 499$ mm (head at J is 4.94 m above the junction level).

Data. Main pipe AJ: $L = 10$ km, $D = 0.75$ m, slope $2.2\times10^{-3}$. Branch JB: $L = 5.5$ km, slope $2.75\times10^{-3}$. Branch JC: $L = 3$ km, slope $3.2\times10^{-3}$. $Q_B = Q_C = 0.24$ m³/s, so the main pipe carries $Q = 0.48$ m³/s. $f = 0.024$. Only friction loss: $h_f = \dfrac{8fLQ^2}{\pi^2gD^5}$. Each reservoir level is 3 m above the end of its pipe (A: above the start of the main pipe; B and C: above the ends of the branches).

Elevations (datum: level of junction J)

• The main pipe falls $10\,000(2.2\times10^{-3}) = 22$ m from A to J, so the level of A is $22 + 3 = 25$ m above J.
• The branch to B falls $5500(2.75\times10^{-3}) = 15.125$ m, so the level of B is $-15.125 + 3 = -12.125$ m (12.125 m below J).
• The branch to C falls $3000(3.2\times10^{-3}) = 9.6$ m, so the level of C is $-9.6 + 3 = -6.6$ m.

Head loss in the main pipe and the head at J

$$h_{f,AJ} = \frac{8(0.024)(10\,000)(0.48)^2}{\pi^2(9.81)(0.75)^5} = 19.25\ \text{m}$$

Piezometric head at J (above J's level): $H_J = 25 - 19.25 = 5.75$ m.

Head available for each branch

• To B: $h_B = H_J - (-12.125) = 17.87$ m
• To C: $h_C = H_J - (-6.6) = 12.35$ m

Branch diameters

$$D = \left(\frac{8fLQ^2}{\pi^2g\,h_f}\right)^{1/5}$$ $$D_B = \left(\frac{8(0.024)(5500)(0.24)^2}{\pi^2(9.81)(17.87)}\right)^{1/5} = 0.5119\ \text{m} \approx 512\ \text{mm}$$ $$D_C = \left(\frac{8(0.024)(3000)(0.24)^2}{\pi^2(9.81)(12.35)}\right)^{1/5} = 0.4883\ \text{m} \approx 488\ \text{mm}$$

Answer: Diameter of the branch to B $\approx 512$ mm; diameter of the branch to C $\approx 488$ mm (head at J is 5.75 m above the junction level).

Data missing from the question. The figure gives only the levels (A = 40 m, outlet D = 10 m) and the pipe table; no discharge is given, and two quantities are unknown ($L_{AEC}$ and $Z_B$). As the paper says "assume suitable data", the following standard conditions are assumed:

1. The two parallel pipes AEC and AFC (which together join A to C) carry equal discharges.
2. Reservoir B neither supplies nor takes water (no flow in BC), so its level equals the piezometric head at C. If B supplies a discharge $Q_B$ the level would be $Z_B = H_C + r_{BC}Q_B^2$.

Friction losses only; $h_f = rQ^2$ with $r = \dfrac{8fL}{\pi^2gD^5}$.

Length of pipe AEC

Parallel pipes have the same head loss: $\dfrac{f_1L_1Q_1^2}{D_1^5} = \dfrac{f_2L_2Q_2^2}{D_2^5}$. With $f$ equal and $Q_1 = Q_2$:

$$L_{AEC} = L_{AFC}\left(\frac{D_{AEC}}{D_{AFC}}\right)^5 = 1500\left(\frac{0.60}{0.75}\right)^5 = 1500(0.32768) = 491.5\ \text{m}$$

(General case: $L_{AEC} = L_{AFC}\left(\dfrac{D_{AEC}}{D_{AFC}}\right)^5\left(\dfrac{Q_{AFC}}{Q_{AEC}}\right)^2$.)

Discharge from A to D

Resistance of AFC: $r_{AFC} = 15.67$ s²/m⁵. Since AEC has the same resistance (equal $L/D^5$ ratio and flow), the pair in parallel has $r_{par} = r_{AFC}/4 = 3.917$ s²/m⁵. Resistance of CD: $r_{CD} = \dfrac{8(0.02)(3000)}{\pi^2(9.81)(0.4)^5} = 484.1$ s²/m⁵.

Between the levels of A and D:

$$40 - 10 = (r_{par} + r_{CD})Q^2 \;\Rightarrow\; Q = \sqrt{\frac{30}{3.917 + 484.1}} = 0.2479\ \text{m}^3/\text{s}$$

Each parallel pipe carries 0.1240 m³/s.

Elevation of reservoir B (no flow in BC)

Head loss in the parallel pair $= r_{par}Q^2 = 0.24$ m; in CD $= r_{CD}Q^2 = 29.76$ m. The piezometric head at the junction C is the head at D plus the loss in CD:

$$H_C = 10 + 29.76 = 39.76\ \text{m}$$ $$Z_B = H_C = 39.76\ \text{m}$$

Answer: $L_{AEC} \approx 491.5$ m (about 492 m) and $Z_B \approx 39.8$ m, with $Q \approx 0.248$ m³/s in CD under the stated assumptions. (For a B that supplies water, $Z_B$ is higher than this by $r_{BC}Q_B^2$, with $r_{BC} = 126.9$ s²/m⁵.)

Data. A = 110 m, B = 95 m, C = 90 m; junction D at elevation 100 m with $p_D = 39.24$ kN/m². Friction only, $h_f = rQ^2$, $r = \dfrac{8fL}{\pi^2gD^5}$ for the pipes and the valve resistances given.

Piezometric head at D

$$H_D = z_D + \frac{p_D}{\gamma} = 100 + \frac{39.24}{9.81} = 100 + 4.00 = 104.00\ \text{m}$$

Since $H_D = 104$ m is below A (110 m) and above B (95 m) and C (90 m), pipe AD supplies D, and D supplies both B and C.

Pipe resistances (s²/m⁵)

Discharges

$$Q_{AD} = \sqrt{\frac{110 - 104}{1692.2}} = 0.05955\ \text{m}^3/\text{s} = 59.55\ \text{l/s}$$

Equal discharge to B and C: $Q_B = Q_C = \dfrac{Q_{AD}}{2} = 0.02977$ m³/s ($29.77$ l/s).

Valve $V_1$ in pipe DB

Head available from D to B: $104 - 95 = 9$ m $= (r_{BD} + r_{V1})Q_B^2$:

$$r_{V1} = \frac{9}{0.02977^2} - 4131.3 = 6021.8 = \left(\frac{6000}{n_1}\right)^2 \;\Rightarrow\; n_1 = \frac{6000}{\sqrt{6021.8}} = 77.3\ \%$$

Valve $V_2$ in pipe DC

Head available from D to C: $104 - 90 = 14$ m $= (r_{CD} + r_{V2})Q_C^2$:

$$r_{V2} = \frac{14}{0.02977^2} - 6197.0 = 9596.8 = \left(\frac{8000}{n_2}\right)^2 \;\Rightarrow\; n_2 = \frac{8000}{\sqrt{9596.8}} = 81.7\ \%$$

Answer: $n_1 \approx 77.3\ \%$ and $n_2 \approx 81.7\ \%$ (each with $Q = 29.8$ l/s to B and to C).

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).

Derivation of the correction $\Delta Q$ (Hardy Cross). Head loss in a pipe is $h_f = rQ^n$ (with $n = 2$ for the Darcy-Weisbach law, $r = \dfrac{8fL}{\pi^2gD^5}$). For a correct distribution the algebraic sum of head losses around every closed loop is zero (clockwise flows positive):

$$\sum rQ|Q|^{n-1} = 0$$

Take assumed discharges $Q_0$ which satisfy continuity at the nodes but not the loop condition. Let the true discharge be $Q = Q_0 + \Delta Q$, where the same correction $\Delta Q$ applies to every pipe of the loop (so continuity is preserved). Then

$$\sum r(Q_0 + \Delta Q)^n = 0 \;\Rightarrow\; \sum rQ_0^n + n\,\Delta Q\sum rQ_0^{n-1} + \frac{n(n-1)}{2}\Delta Q^2\sum rQ_0^{n-2} + \dots = 0$$

$\Delta Q$ is small compared with $Q_0$, so the terms in $\Delta Q^2$ and higher are neglected:

$$\Delta Q = -\frac{\sum rQ_0|Q_0|^{n-1}}{n\sum r|Q_0|^{n-1}} \;\xrightarrow{n = 2}\; \boxed{\Delta Q = -\frac{\sum rQ_0|Q_0|}{2\sum r|Q_0|}}$$

The numerator is the algebraic sum of head losses in the loop (signs by direction), the denominator is the sum of absolute values. Apply the correction to every pipe (add it to flows in the clockwise direction, subtract it from anticlockwise ones, and apply both loops' corrections to a common pipe), and repeat until $\Delta Q$ is negligible.

Network discharge through FG

Reading of the figure. The scan is cut off at the bottom, so the part below B and D is assumed to be: A (inlet) joined to B and C; B-D, C-D; D-E, D-G; E-F and G-F. Nodes B, E, F, G have the outflows shown (2, 2, 4, 1) and the inlet at A must therefore be $2 + 2 + 4 + 1 = 9$. Resistances $r$ ($h_f = rQ^2$) as in the table: AB 50, AC 50, BD 100, CD 50, DE 100, EF 100, FG 50, DG 50 (the $f$ and $D$ values are already contained in $r$). Flow units are those of the figure.

Initial distribution (satisfying continuity): AB 4, AC 5, BD 2, CD 5, DE 4, EF 2, GF 2 (G to F), DG 3.

Loop 1 (A-B-D-C, clockwise: AB, BD positive; CD, AC negative) and loop 2 (D-E-F-G, clockwise: DE, EF positive; GF as F to G, DG as G to D negative).

Correction in each iteration, $\Delta Q = -\dfrac{\sum rQ|Q|}{2\sum r|Q|}$:

| Iter | $\sum rQ|Q|$ (L1) | $2\sum r|Q|$ (L1) | $\Delta Q_1$ | $\sum rQ|Q|$ (L2) | $2\sum r|Q|$ (L2) | $\Delta Q_2$ | |---|---|---|---|---|---|---| | 0 | -1300.00 | 1800.00 | 0.722 | 1350.00 | 1700.00 | -0.794 | | 1 | 26.08 | 1872.22 | -0.014 | 63.06 | 1541.18 | -0.041 | | 2 | 0.01 | 1870.83 | -0.000 | 0.17 | 1532.99 | -0.000 |

After the third iteration the corrections are below 0.001. Final discharges:

Check at the nodes: D receives $2.708 + 4.292 = 7.000$ and sends $3.165 + 3.835 = 7.000$; F receives $1.165 + 2.835 = 4.000$, equal to its outflow of 4.

Answer: The discharge in link FG is $2.84$ units (flowing from G to F), $\approx 2.8$.

Data. Reservoir levels: 100 m (pipe 1), 90 m (pipe 2, with pump), 30 m (pipe 3). Pump head $H_p = 30$ m in pipe 2 (flow from the reservoir to J). Only friction loss, $h_f = rQ^2$ with $r = \dfrac{8fL}{\pi^2gD^5}$.

Method

The pump adds 30 m of head, so on the pipe-2 side the effective head at the reservoir end is $90 + 30 = 120$ m. This is higher than the 100 m of reservoir 1 and the junction head must lie below 120 m, between the reservoir at 30 m and the others. Let $H_J$ be the energy (piezometric) head at J.

• Pipe 1 (reservoir 100 m to J): $Q_1 = \sqrt{(100 - H_J)/r_1}$ (if $H_J < 100$)
• Pipe 2 (reservoir 90 m plus pump to J): $Q_2 = \sqrt{(90 + 30 - H_J)/r_2}$
• Pipe 3 (J to reservoir 30 m): $Q_3 = \sqrt{(H_J - 30)/r_3}$
• Continuity: $Q_1 + Q_2 = Q_3$

Trial of $H_J$

Result

$$H_J = 78.65\ \text{m}$$ $$Q_1 = 0.661\ \text{m}^3/\text{s},\qquad Q_2 = 4.387\ \text{m}^3/\text{s},\qquad Q_3 = 5.048\ \text{m}^3/\text{s}$$

Check: $Q_1 + Q_2 = 0.661 + 4.387 = 5.048 = Q_3$. The directions are: reservoir 1 to J, pump-reservoir to J, and J to reservoir 3. Velocities: $V_1 = 2.34$, $V_2 = 5.59$, $V_3 = 4.46$ m/s.

Answer: $Q_1 = 0.661$, $Q_2 = 4.387$, $Q_3 = 5.048$ m³/s; energy head at J $= 78.65$ m.

Layout assumed from the figure. A feeds junction D by pipe 1; B joins D by pipe 2; from D two pipes, 3 and 4, run in parallel to a junction E; pipe 5 then runs from E to reservoir C. Minor losses are included through $\sum K$: for each pipe the head loss is $h = \left(\dfrac{fL}{D} + \sum K\right)\dfrac{V^2}{2g} = rQ^2$, with

$$r = \frac{8}{\pi^2gD^4}\left(\frac{fL}{D} + \sum K\right)$$

(i) Relation between the flows in the two parallel pipes

They have the same head loss, so $r_3Q_3^2 = r_4Q_4^2$:

$$\frac{Q_3}{Q_4} = \sqrt{\frac{r_4}{r_3}} = \sqrt{\frac{15.578}{7.536}} = 1.438 \;\Rightarrow\; Q_3 = 1.438\,Q_4$$

The equivalent resistance of the pair is $\dfrac{1}{\sqrt{r_{par}}} = \dfrac{1}{\sqrt{r_3}} + \dfrac{1}{\sqrt{r_4}}$, so $r_{par} = 2.621$ s²/m⁵, and $Q_5 = Q_3 + Q_4$.

(ii) HGL at junction D

Let $H_D$ be the HGL at D. Pipes 3, 4 and 5 are in series with the pair, so $H_D - Z_C = (r_{par} + r_5)Q_5^2$.

• Pipe 1: $Q_1 = \sqrt{(185 - H_D)/r_1}$ (A to D)
• Pipe 2: $Q_2 = \sqrt{|105 - H_D|/r_2}$, towards B if $H_D > 105$, from B if $H_D < 105$
• Pipe 5: $Q_5 = \sqrt{(H_D - 70)/(r_{par} + r_5)}$
• Continuity at D: $Q_1 = Q_2 + Q_5$ (when B is receiving), or $Q_1 + Q_2 = Q_5$ (when B is supplying).

Trials ($Q_2 > 0$ means B supplies D, $Q_2 < 0$ means B receives):

The sum changes sign between 108 and 110 m, and the balance is at

$$H_D = 108.95\ \text{m}$$

Since $H_D > 105$ m, pipe 2 carries water to reservoir B.

(iii) Flow rates

Pipe 5: $Q_5 = \sqrt{(108.95 - 70)/(2.621 + 8.539)} = 1.868$ m³/s. The HGL at the junction E is $70 + r_5Q_5^2 = 99.80$ m, and $Q_3 = \sqrt{(108.95 - 99.80)/7.536} = 1.102$, $Q_4 = \sqrt{(108.95 - 99.80)/15.578} = 0.766$ m³/s.

Continuity check at D: $Q_1 = 1.942 = Q_2 + Q_5 = 0.073 + 1.868$.

Answer: (i) $Q_3 = 1.438\,Q_4$; (ii) HGL at D $\approx 109.0$ m; (iii) $Q_1 = 1.942$, $Q_2 = 0.073$ (to B), $Q_3 = 1.102$, $Q_4 = 0.766$, $Q_5 = 1.868$ m³/s.

Data. Resistance $r$ in $h_f = rQ^2$ (discharges in l/min): AB 1, BC 2.5, CD 1.5, AD 2, DB 3. The valve (r = 1.5) sits in the diagonal, in series with DB, so the diagonal has $r_{BD} = 3 + 1.5 = 4.5$. Inflow 70 at A; outflows 30 at B and 40 at C. Sign rule: clockwise flow positive.

Initial distribution (continuity satisfied)

A to B 40, A to D 30; at B (out 30): B to C 10, B to D 0; at C (out 40): the 10 from B plus 30 from D (D to C 30).

Loop I = A-B-D-A, Loop II = B-C-D-B.

Hardy-Cross iterations

$\Delta Q = -\dfrac{\sum rQ|Q|}{2\sum r|Q|}$ for each loop, applied to all its pipes (the diagonal BD gets both corrections).

| Iter | $\sum rQ|Q|$ (I) | $2\sum r|Q|$ (I) | $\Delta Q_I$ | $\sum rQ|Q|$ (II) | $2\sum r|Q|$ (II) | $\Delta Q_{II}$ | |---|---|---|---|---|---|---| | 1 | -200.00 | 200.00 | 1.000 | -1100.00 | 140.00 | 7.857 | | 2 | -212.59 | 259.71 | 0.819 | 273.33 | 217.43 | -1.257 | | 3 | 57.52 | 239.40 | -0.240 | -29.55 | 196.23 | 0.151 | | 4 | -7.23 | 243.39 | 0.030 | 11.05 | 200.05 | -0.055 |

After about 10 iterations the corrections fall below 0.001.

Final distribution (l/min)

Checks: node A: $41.60 + 28.40 = 70.00$; node B: $41.60 + 5.10 = 30 + 16.70$ (46.70 = 46.70); node C: $16.70 + 23.30 = 40.00$.

Answer: AB = 41.6, AD = 28.4, BC = 16.7, CD = 23.3 (D to C) and the diagonal BD = 5.1 l/min (flowing D to B).

Data. Suggested discharges (units): B to A 58, B to C 42, A to C 20, A to D 18, C to D 32 (these satisfy continuity at every node with 100 in at B, 20 out at A, 30 out at C, 50 out at D). $r$: AB 2, BC 4, AC 1, AD 5, CD 1; $h_f = rQ^2$ ($n = 2$).

Loop I = A-C-B-A (AC +, CB $-$, BA +). Loop II = A-D-C-A (AD +, DC $-$, CA $-$).

Test of the suggested distribution (first iteration)

Loop I: $\sum rQ|Q| = 1(20)^2 - 4(42)^2 + 2(58)^2 = 400 - 7056 + 6728 = 72$, $2\sum r|Q| = 2(20 + 168 + 116) = 608$, so $\Delta Q_I = -\dfrac{72}{608} = -0.118$.

Loop II: $\sum rQ|Q| = 5(18)^2 - 1(32)^2 - 1(20)^2 = 1620 - 1024 - 400 = 196$, $2\sum r|Q| = 2(90 + 32 + 20) = 284$, so $\Delta Q_{II} = -\dfrac{196}{284} = -0.690$.

The loop head-loss sums are not zero (72 and 196), so strictly the suggested distribution is not exact, although the corrections are small (under 1.2% of the pipe discharges). Continuing the Hardy-Cross iterations:

| Iter | $\sum rQ|Q|$ (I) | $2\sum r|Q|$ (I) | $\Delta Q_I$ | $\sum rQ|Q|$ (II) | $2\sum r|Q|$ (II) | $\Delta Q_{II}$ | |---|---|---|---|---|---|---| | 1 | 72.00 | 608.00 | -0.118 | 196.00 | 284.00 | -0.690 | | 2 | 27.90 | 609.62 | -0.046 | 6.32 | 279.62 | -0.023 | | 3 | 0.93 | 609.75 | -0.002 | 1.88 | 279.40 | -0.007 |

Corrected distribution

(Checks: $57.83 + 42.17 = 100$; at D, $17.28 + 32.72 = 50$.) The suggested values are therefore acceptable to within about 4%; the correction mainly shifts about 0.7 unit of flow from pipe AD to the route A-C-D.

Pressure at D

Piezometric head at B: $H_B = 50 + 40 = 90$ m. Head loss from B to D via A: $h_{BA} + h_{AD} = 2(57.83)^2 + 5(17.28)^2 = 6689.5 + 1493.0 = 8182.5$ (the same through C: $7111.9 + 1070.6$).

$$H_D = 90 - 8182.5 = -8092.5, \qquad \frac{p_D}{\gamma} = H_D - z_D = -8092.5 - 40 = -8132.5$$

With $r$ and $Q$ used exactly as given the head loss ($8183$) is far larger than the 50 m available, so the resulting pressure is negative. This means the given $r$ values are on a scale in which $rQ^2$ is not in metres of head; if $h_f$ is read in the same (scaled) unit as $r$, the pressure at D is $p_D/\gamma = 50 - h_{BD}$ (m of water) and the pressure is $\gamma\,(50 - h_{BD})$, with $h_{BD}$ the actual loss in metres.

Answer: Distribution is nearly but not exactly satisfactory; proper values: BA = 57.8, BC = 42.2, AC = 20.6, AD = 17.3, CD = 32.7. Pressure head at D = $50 - h_{BD}$ with $h_{BD} = 8182.5$ in the units of the data.