Simple pipe flow problems and solution

Data. $D = 0.1$ m, $L_{total} = 90$ m, $L_{summit} = 30$ m, level difference $H = 10$ m, $f = 0.025$, $Q = 0.05$ m³/s. All losses: entrance $0.5\dfrac{V^2}{2g}$, exit $1.0\dfrac{V^2}{2g}$ (pipe discharges below the branch canal level), friction.

(a) Number of pipelines

Energy equation between the two canal surfaces (both at atmospheric pressure, velocity zero):

$$H = \left(0.5 + f\frac{L}{D} + 1.0\right)\frac{V^2}{2g}, \qquad f\frac{L}{D} = 0.025\left(\frac{90}{0.1}\right) = 22.5$$ $$10 = 24.0\,\frac{V^2}{2g} \;\Rightarrow\; \frac{V^2}{2g} = 0.4167\ \text{m},\quad V = 2.859\ \text{m/s}$$

Discharge per pipe: $q = \dfrac{\pi}{4}(0.1)^2(2.859) = 0.02246$ m³/s $= 22.46$ l/s.

$$n = \frac{0.05}{0.02246} = 2.23 \;\Rightarrow\; \textbf{3 pipelines}$$

(b) Maximum height of the summit above the main canal

Let the summit be $z$ above the main canal surface. Atmospheric pressure head = 10 m of water (given), and the minimum absolute pressure at the summit is 20 kPa, i.e. $\dfrac{p_s}{\gamma} = \dfrac{20}{9.81} = 2.039$ m (absolute).

Energy equation from the main canal surface to the summit (losses over the first 30 m, including entrance; velocity head at the summit):

$$10 = z + 2.039 + \frac{V^2}{2g}\left(1 + 0.5 + f\frac{L_s}{D}\right), \qquad f\frac{L_s}{D} = 0.025\left(\frac{30}{0.1}\right) = 7.5$$ $$z = 10 - 2.039 - 0.4167(9.0) = 4.21\ \text{m}$$

(The velocity in each pipe is fixed by the 10 m head, so $V = 2.859$ m/s is used; the extra capacity of three pipes is controlled by a valve.)

Answer: (a) 3 pipelines; (b) the summit may be at most about 4.21 m above the water level of the main canal.

Method. Use $h_f = rQ^2$ for each pipe, with $r = \dfrac{8fL}{\pi^2gD^5}$ ($f = 0.024$). Only friction is considered. Heads: $H_A = 100$ m, $H_E = 40$ m, $H_F = 60$ m. For equal discharge let $Q_E = Q_F = q$, so the flow in $BC$ is $2q$.

Resistance coefficients (s²/m⁵)

Parallel pair A to B (same head loss in both): $Q = \sqrt{h/r_1} + \sqrt{h/r_2}$, so

$$\frac{1}{\sqrt{r_{AB}}} = \frac{1}{\sqrt{r_1}} + \frac{1}{\sqrt{r_2}} \;\Rightarrow\; r_{AB} = 788.4$$

Discharge

Energy balance from A to F (pair AB, then BC carrying $2q$, then CF carrying $q$):

$$H_A - H_F = r_{AB}(2q)^2 + r_{BC}(2q)^2 + r_{CF}q^2$$ $$100 - 60 = q^2\left[4(788.4 + 1566.9) + 5949.1\right] = 15370.2\,q^2 \;\Rightarrow\; q = 0.05101\ \text{m}^3/\text{s}$$ $$Q = 2q = 0.10203\ \text{m}^3/\text{s} = 102.0\ \text{l/s}$$

Head at C

$$H_C = H_F + r_{CF}q^2 = 60 + 5949.1(0.002602) = 75.48\ \text{m}$$

(Check from A: $H_B = H_C + r_{BC}(2q)^2 = 91.79$ m; the two parallel pipes then carry 0.03714 and 0.06489 m³/s, which add to 0.10203 m³/s.)

Valve opening

Branch C to E has pipe CD and the valve DE in series, carrying $q$ with head loss $H_C - H_E = 75.48 - 40 = 35.48$ m:

$$(r_{CD} + K_{DE})q^2 = 35.48 \;\Rightarrow\; K_{DE} = \frac{35.48}{0.002602} - 2974.6 = 10659.7$$ $$K_{DE} = \left(\frac{4000}{n}\right)^2 \;\Rightarrow\; n = \frac{4000}{\sqrt{10659.7}} = 38.7\ \%$$

Answer: Head at C $= 75.48$ m; total discharge $= 0.1020$ m³/s ($102.0$ l/s); valve opening $n \approx 38.7\ \%$.

Data. $D = 0.09$ m, $L_1 = 25$ m (up to the summit), summit 2 m above the main canal, $p_s = 20$ kN/m² absolute, level difference $H = 10$ m, $f = 0.03$, $Q = 0.06$ m³/s. Minor losses: entrance $0.5\dfrac{V^2}{2g}$ and exit $1.0\dfrac{V^2}{2g}$. Atmospheric pressure is taken as 101.3 kPa, i.e. $p_a/\gamma = 10.326$ m of water.

(i) Number of pipelines

Absolute pressure head at the summit: $\dfrac{p_s}{\gamma} = \dfrac{20}{9.81} = 2.039$ m.

Energy equation from the main canal surface to the summit (velocity head at the summit, entrance loss and friction over 25 m):

$$\frac{p_a}{\gamma} = 2 + \frac{p_s}{\gamma} + \frac{V^2}{2g}\left(1 + 0.5 + f\frac{L_1}{D}\right), \qquad f\frac{L_1}{D} = 0.03\left(\frac{25}{0.09}\right) = 8.333$$ $$10.326 - 2 - 2.039 = 6.287 = 9.833\,\frac{V^2}{2g} \;\Rightarrow\; \frac{V^2}{2g} = 0.6394\ \text{m},\quad V = 3.542\ \text{m/s}$$

Discharge per pipe: $q = \dfrac{\pi}{4}(0.09)^2(3.542) = 0.02253$ m³/s $= 22.53$ l/s.

$$n = \frac{0.06}{0.02253} = 2.66 \;\Rightarrow\; \textbf{3 pipelines}$$

(ii) Total length of pipe

For the same velocity, the total head of 10 m between the canal surfaces is used up in the entrance, friction over the full length and exit:

$$10 = \left(0.5 + f\frac{L}{D} + 1.0\right)\frac{V^2}{2g} \;\Rightarrow\; f\frac{L}{D} = \frac{10}{0.6394} - 1.5 = 14.140$$ $$L = 14.140\times\frac{0.09}{0.03} = 42.4\ \text{m}$$

So the length beyond the summit is $17.4$ m.

Answer: (i) 3 pipelines; (ii) total pipe length $\approx 42.4$ m.

Data. $D = 0.075$ m, $H = 15$ m, $L_1 = 20$ m (to summit), $L = 50$ m, $f = 0.03$, minor losses ignored, $Q \ge 50$ l/s, summit pressure $\ge 25$ kPa absolute. Atmospheric pressure is taken as 101.3 kPa, $p_a/\gamma = 10.326$ m of water.

(a) Number of pipes

Level difference = friction head loss (no minor losses, so the exit velocity head is also ignored):

$$15 = f\frac{L}{D}\frac{V^2}{2g}, \qquad f\frac{L}{D} = 0.03\left(\frac{50}{0.075}\right) = 20.0$$ $$\frac{V^2}{2g} = \frac{15}{20.0} = 0.750\ \text{m},\quad V = 3.836\ \text{m/s}$$

Discharge per pipe $q = \dfrac{\pi}{4}(0.075)^2(3.836) = 0.01695$ m³/s $= 16.95$ l/s.

$$n = \frac{50}{16.95} = 2.95 \;\Rightarrow\; \textbf{3 pipes}$$

(b) Maximum height of the summit

Minimum absolute pressure head at the summit: $\dfrac{25}{9.81} = 2.548$ m. Energy equation from the main canal surface to the summit (velocity head and friction over 20 m):

$$\frac{p_a}{\gamma} = z + \frac{p_s}{\gamma} + \frac{V^2}{2g}\left(1 + f\frac{L_1}{D}\right), \qquad f\frac{L_1}{D} = 0.03\left(\frac{20}{0.075}\right) = 8.0$$ $$z = 10.326 - 2.548 - 0.750(1 + 8.0) = 1.03\ \text{m}$$

Answer: (a) 3 pipes; (b) the summit can be at most about 1.03 m above the water level of the main canal.

Setting up. A siphon has an inlet leg $AB$ (length $l_1$) from the intake A in the upper reservoir up to the summit B, and an outlet leg $BC$ (length $l_2$) down to the outlet C. The pipe has uniform diameter, so the friction loss per unit length is the same in both legs. Neglect minor losses and the velocity head. Let $Z_A$, $Z_B$, $Z_C$ be the elevations of intake water level, summit and outlet, and $P_a$ the atmospheric pressure head (m of water).

        B (summit)
       /\
      /  \
 A __/    \
 ~~~ l1    \ l2
(upper      \___ C (outlet)
 reservoir)

Step 1: energy gradient. The whole fall is used to overcome friction over the total length:

$$H = Z_A - Z_C = h_f, \qquad S = \frac{h_f}{l_1 + l_2} = \frac{Z_A - Z_C}{l_1 + l_2}$$

Step 2: pressure at the summit. Applying Bernoulli between A and B (velocity head is the same and ignored):

$$\frac{p_A}{\gamma} + Z_A = \frac{p_B}{\gamma} + Z_B + S\,l_1 \;\Rightarrow\; \frac{p_B}{\gamma} = -\left[(Z_B - Z_A) + S\,l_1\right] \quad\text{(gauge)}$$

The pressure at B is below atmospheric. The water column will break (cavitation, air release) if the absolute pressure falls below zero, i.e. $\dfrac{p_B}{\gamma} < -P_a$. For the siphon to work:

$$(Z_B - Z_A) + S\,l_1 \le P_a \;\Rightarrow\; S\,l_1 \le P_a - (Z_B - Z_A) \quad\cdots(1)$$

Step 3: substitute the gradient.

$$\frac{(Z_A - Z_C)\,l_1}{l_1 + l_2} \le P_a - (Z_B - Z_A)$$

Let $X = P_a - (Z_B - Z_A)$ and $H = Z_A - Z_C$. Then $H\,l_1 \le X\,(l_1 + l_2)$, so

$$l_1\,(H - X) \le X\,l_2 \;\Rightarrow\; \frac{l_1}{l_2} \le \frac{X}{H - X}$$

Now

$$H - X = (Z_A - Z_C) - P_a + (Z_B - Z_A) = (Z_B - Z_C) - P_a$$ $$\boxed{\frac{l_1}{l_2} \le \frac{P_a - (Z_B - Z_A)}{(Z_B - Z_C) - P_a}}$$

Meaning. For a siphon of given height the inlet leg must be shorter than the limit given by this ratio. For the denominator to be positive the outlet must be lower than $P_a$ below the summit; otherwise the siphon works for any ratio. In practice, a smaller limiting pressure (vapour pressure, about 2 to 3 m of water absolute) is used in place of zero.

Data. $D = 0.1$ m, $L = 36$ m, $f = 0.04$. Upper water surface is 1.8 m above the inlet; lower water surface is 1.2 m above the outlet; ends differ in level by 4 m, so the difference of the water surfaces is $H = 4 + 1.8 - 1.2 = 4.6$ m. Barrier at 7.5 m horizontally from the upper reservoir. Minimum absolute pressure 40 kPa; $p_a = 101.3$ kPa. All losses: entrance $0.5\dfrac{V^2}{2g}$, exit $1.0\dfrac{V^2}{2g}$, friction.

Step 1: velocity in the pipe

$$H = \left(0.5 + f\frac{L}{D} + 1.0\right)\frac{V^2}{2g}, \qquad f\frac{L}{D} = 0.04\left(\frac{36}{0.1}\right) = 14.4$$ $$4.6 = 15.9\,\frac{V^2}{2g} \;\Rightarrow\; \frac{V^2}{2g} = 0.2893\ \text{m},\quad V = 2.382\ \text{m/s}\quad(Q = 18.71\ \text{l/s})$$

Step 2: energy equation from the upper surface to the barrier X

Take the pipe inlet level as datum. Let the barrier be at height $h_X$ above the inlet. Since the pipe is straight from the inlet to the barrier, its length is $l = \sqrt{7.5^2 + h_X^2}$.

Absolute pressure heads: $\dfrac{p_a}{\gamma} = \dfrac{101.3}{9.81} = 10.326$ m; minimum $\dfrac{p_X}{\gamma} = \dfrac{40}{9.81} = 4.077$ m.

$$\frac{p_a}{\gamma} + 1.8 = h_X + \frac{p_X}{\gamma} + \frac{V^2}{2g}\left(1 + 0.5 + f\frac{l}{D}\right)$$ $$h_X = 10.326 + 1.8 - 4.077 - 0.2893\left(1.5 + 0.4\,l\right)$$

Iterating ($h_X$ then $l$): $l = 9.90$ m, so the bracket is $5.46$ and

$$h_X = 6.47\ \text{m above the pipe inlet}$$

Answer: The pipe may rise to about 6.47 m above its inlet at the barrier (about 4.67 m above the upper reservoir water level).

Data. $L = 1000$ m, $D = 0.3$ m, $f = 0.032$, level difference $H = 13$ m, summit 5 m above the upper reservoir level, minimum absolute pressure 1.2 m of water. Atmospheric pressure $= 760$ mm Hg $= 0.76(13.6) = 10.336$ m of water. Head losses are taken as pipe friction only (bend losses neglected; entry and exit losses would change the result by less than 1%).

Discharge

$$H = f\frac{L}{D}\frac{V^2}{2g}, \qquad f\frac{L}{D} = 0.032\left(\frac{1000}{0.3}\right) = 106.67$$ $$\frac{V^2}{2g} = \frac{13}{106.67} = 0.12187\ \text{m} \;\Rightarrow\; V = 1.546\ \text{m/s}$$ $$Q = \frac{\pi}{4}(0.3)^2(1.546) = 0.1093\ \text{m}^3/\text{s} = 109.3\ \text{l/s}$$

Length of pipe from the upper reservoir to the summit

The critical point is the summit S, where absolute pressure head is 1.2 m. Energy equation from the upper reservoir surface (datum, atmospheric) to the summit:

$$\frac{p_a}{\gamma} = 5 + 1.2 + \frac{V^2}{2g} + f\frac{l_1}{D}\frac{V^2}{2g}$$ $$10.336 = 6.2 + 0.12187 + f\frac{l_1}{D}(0.12187) \;\Rightarrow\; f\frac{l_1}{D}\,(0.12187) = 4.0141$$ $$l_1 = \frac{4.0141\times 0.3}{0.032\times 0.12187} = 308.8\ \text{m}$$

Answer: The summit may be at most about 309 m of pipe from the upper reservoir; the discharge is $0.1093$ m³/s ($109.3$ l/s).

Data. s.g. 0.6 so $\rho = 600$ kg/m³, $\nu = 5.0\times10^{-7}$ m²/s, $D = 0.025$ m, $k/D = 0.0004$, total hose length $L = 9$ m, length to the highest point A $= 3.25$ m. From the figure: A is 1.5 m above the liquid surface; the outlet is 5 m below A, so it is $5 - 1.5 = 3.5$ m below the liquid surface. The hose discharges freely to the atmosphere.

Volumetric flow

Energy equation from the tank surface to the outlet (entrance loss neglected, velocity head at the outlet kept):

$$3.5 = \frac{V^2}{2g}\left(1 + f\frac{L}{D}\right) = \frac{V^2}{2g}\left(1 + f\frac{9}{0.025}\right) = \frac{V^2}{2g}\left(1 + 360f\right)$$

$f$ depends on $Re = VD/\nu$ and $k/D = 0.0004$ (Colebrook-White), so iterate:

1. Assume $f = 0.02$: $\dfrac{V^2}{2g} = \dfrac{3.5}{8.2} = 0.427$ m, $V = 2.89$ m/s, $Re = 1.4\times10^5$.
2. Colebrook: $f \approx 0.0189$ and the result converges (3 iterations) to

$$f = 0.01891,\qquad \frac{V^2}{2g} = \frac{3.5}{1 + 360(0.01891)} = 0.4483\ \text{m},\quad V = 2.966\ \text{m/s},\quad Re = 148284$$ $$Q = \frac{\pi}{4}(0.025)^2(2.966) = 0.001456\ \text{m}^3/\text{s} = 1.456\ \text{l/s} \;(87.3\ \text{l/min})$$

Minimum pressure (at the top point A)

Energy equation from the tank surface (gauge pressure 0) to A, 1.5 m above the surface, with friction over 3.25 m:

$$0 = 1.5 + \frac{p_A}{\gamma} + \frac{V^2}{2g}\left(1 + f\frac{3.25}{0.025}\right), \qquad f\frac{3.25}{0.025} = 2.458$$ $$\frac{p_A}{\gamma} = -1.5 - 0.4483(3.458) = -3.050\ \text{m of liquid}$$ $$p_A = \rho g\left(\frac{p_A}{\gamma}\right) = 600(9.81)(-3.050) = -17954\ \text{Pa} \approx -17.95\ \text{kPa (gauge)}$$

This is about 83.3 kPa absolute, above the vapour pressure of the liquid unless it is very volatile.

Answer: $Q = 1.46$ l/s ($0.001456$ m³/s, $V = 2.97$ m/s); minimum pressure at A $= -17.95$ kPa gauge.

Data. $H = 15$ m, $D_1 = D_{AB} = 0.2$ m, $D_2 = D_{BC} = 0.15$ m, $L_1 + L_2 = 3000$ m, $f = 0.012$. Summit B is 2 m below the upper water level; pressure head at B $\ge -2$ m (gauge). Secondary (minor) losses are neglected.

Let $h_1 = \dfrac{V_1^2}{2g}$. By continuity $V_2 = V_1\left(\dfrac{D_1}{D_2}\right)^2 = 1.7778V_1$, so $\dfrac{V_2^2}{2g} = \left(\dfrac{D_1}{D_2}\right)^4h_1 = 3.1605\,h_1$.

Condition at B (limiting pressure)

Energy equation from the upper reservoir surface (datum, $p = 0$) to B, which is 2 m below the surface ($z_B = -2$), with $\dfrac{p_B}{\gamma} = -2$ m:

$$0 = -2 + (-2) + h_1\left(1 + f\frac{L_1}{D_1}\right) \;\Rightarrow\; h_1\left(1 + 0.060L_1\right) = 4 \quad\cdots(1)$$

(here $f/D_1 = 0.012/0.2 = 0.060$ per metre).

Condition for the total head

Level difference = friction in AB + friction in BC (no minor losses):

$$15 = h_1\,(0.060)L_1 + 3.1605\,h_1\left(\frac{0.012}{0.15}\right)(3000 - L_1)$$ $$15 = h_1\left[0.060L_1 + 0.080(3.1605)(3000 - L_1)\right] \quad\cdots(2)$$

Solve

Substitute $h_1 = \dfrac{4}{1 + 0.060L_1}$ from (1) into (2):

$$15\left(1 + 0.060L_1\right) = 4\left[0.060L_1 + 0.080(3.1605)(3000 - L_1)\right]$$ $$L_1 = 1806\ \text{m}$$

Check: $h_1 = \dfrac{4}{1 + 0.06(1806)} = 0.03657$ m, so $V_1 = 0.847$ m/s, $V_2 = 1.506$ m/s, $Q = 0.0266$ m³/s; friction in AB $= 3.963$ m and in BC $= 11.037$ m, which add to 15 m.

Answer: The maximum allowable length of portion AB $\approx 1806$ m (so BC is about 1194 m).

Data. $Q = 0.25$ m³/s, pump power $P = 54\,642$ W, $D = 0.25$ m, $e = 0.26$ mm, $\mu = 0.0021$ N s/m², water $\rho = 1000$ kg/m³. Reservoir levels 12 m and 19 m, so the static lift is $19 - 12 = 7$ m. All losses are considered: entrance $0.5\dfrac{V^2}{2g}$, exit $1.0\dfrac{V^2}{2g}$ and friction.

(i) Length of the pipe

Pump head: $P = \gamma Q H_p$

$$H_p = \frac{54\,642}{9810(0.25)} = 22.280\ \text{m}$$

Energy equation between the reservoir surfaces: $H_p = 7 + h_L$, so the head loss is $h_L = 15.280$ m.

Velocity and Reynolds number:

$$V = \frac{0.25}{\pi(0.25)^2/4} = 5.093\ \text{m/s},\quad \frac{V^2}{2g} = 1.3220\ \text{m},\quad Re = \frac{\rho VD}{\mu} = \frac{1000(5.093)(0.25)}{0.0021} = 606305$$

Relative roughness $e/D = 0.00104$. Colebrook-White (solved by iteration):

$$\frac{1}{\sqrt f} = -2\log_{10}\left(\frac{e/D}{3.7} + \frac{2.51}{Re\sqrt f}\right) \;\Rightarrow\; f = 0.02031$$

Head loss: $h_L = \left(0.5 + f\dfrac{L}{D} + 1.0\right)\dfrac{V^2}{2g}$

$$15.280 = \left(1.5 + f\frac{L}{D}\right)(1.3220) \;\Rightarrow\; f\frac{L}{D} = 10.058$$ $$L = \frac{10.058\times 0.25}{0.02031} = 123.8\ \text{m}$$

(ii) Same power, loss of head doubled

New loss $h_L' = 2(15.280) = 30.560$ m, so the new pump head is $H_p' = 7 + 30.560 = 37.560$ m. For the same power:

$$Q' = \frac{P}{\gamma H_p'} = \frac{54\,642}{9810(37.560)} = 0.1483\ \text{m}^3/\text{s} = 148.3\ \text{l/s}$$

Comment. Doubling the head loss (for example a longer or rougher pipe) cuts the discharge from 250 l/s to 148 l/s, a drop of 41%, for the same power. The static lift (7 m) is a fixed cost, but the pump must also supply the friction head; when friction rises, a larger share of the power is wasted as loss and the flow falls. (In practice the friction loss would fall as the flow falls, because $h_f \propto Q^2$, so the actual new equilibrium would be at a somewhat higher flow.)

Answer: (i) $L \approx 124$ m ($f = 0.0203$); (ii) $Q' = 148$ l/s.

Data. $D = 0.6$ m, $L = 1500$ m, $f = 0.01$, head $H = 0.3$ m (as given). Minor losses are neglected. The head is used up in friction only.

Before: single pipe of length 1500 m

$$H = \frac{fLV_1^2}{2gD} \;\Rightarrow\; \frac{V_1^2}{2g} = \frac{HD}{fL} = \frac{0.3(0.6)}{0.01(1500)} = 0.01200\ \text{m}$$ $$V_1 = 0.4852\ \text{m/s},\qquad Q_1 = \frac{\pi}{4}(0.6)^2V_1 = 0.13719\ \text{m}^3/\text{s}$$

After: first 750 m single, second 750 m has two identical pipes in parallel

Let $V$ be the velocity in the first half. In the second half the flow divides equally between two pipes of the same size, so the velocity in each is $V/2$.

$$H = \frac{f(750)V^2}{2gD} + \frac{f(750)(V/2)^2}{2gD} = \frac{f(750)}{D}\,\frac{V^2}{2g}\,(1 + 0.25)$$ $$\frac{V^2}{2g} = \frac{HD}{f(750)(1.25)} = \frac{0.3(0.6)}{0.01(750)(1.25)} = 0.01920\ \text{m} \;\Rightarrow\; V_2 = 0.6138\ \text{m/s}$$ $$Q_2 = \frac{\pi}{4}(0.6)^2V_2 = 0.17354\ \text{m}^3/\text{s}$$

(each parallel pipe carries 0.08677 m³/s).

Increase in discharge

$$\Delta Q = Q_2 - Q_1 = 0.17354 - 0.13719 = 0.03634\ \text{m}^3/\text{s} = 26.5\%\ \text{increase}$$

(Generally $\dfrac{Q_2}{Q_1} = \sqrt{\dfrac{1500}{937.5}} = \sqrt{1.6} = 1.265$.)

Answer: The discharge increases by $0.0363$ m³/s, i.e. by about 26.5%.

Setting up. Let the single pipe have length $L$, diameter $D$, friction factor $f$, joining two reservoirs with a level difference $H$. Minor losses are neglected, so the head is lost in friction only. Use $h_f = \dfrac{fLV^2}{2gD}$.

Before

$$H = \frac{fLV_1^2}{2gD} \;\Rightarrow\; V_1 = \sqrt{\frac{2gDH}{fL}}, \qquad Q_1 = AV_1$$

After

From the mid-point, a second pipe of the same diameter and length $L/2$ runs parallel to the second half. The first half carries all the flow ($V$); the second half has two identical pipes, so the flow divides equally and each carries velocity $V/2$.

$$H = \frac{f(L/2)V^2}{2gD} + \frac{f(L/2)(V/2)^2}{2gD} = \frac{fLV^2}{2gD}\left(\frac12 + \frac18\right) = \frac{fLV^2}{2gD}\left(\frac58\right)$$ $$V_2 = \sqrt{\frac{8}{5}}\sqrt{\frac{2gDH}{fL}} = \sqrt{1.6}\,V_1 = 1.2649\,V_1$$

The discharge in the first half is $Q = AV$, so

$$\frac{Q_2}{Q_1} = \sqrt{1.6} = 1.2649$$ $$\boxed{\text{Increase in flow} = 26.5\%}$$

(This value does not depend on $D$, $L$, $f$ or $H$.)

Method. Use $h_f = \dfrac{8fLQ^2}{\pi^2gD^5}$, i.e. $h_f = rQ^2$ with $r \propto D^{-5}$ for the same $L$ and $f$. Let $D_2 = D_1/2$.

Parallel arrangement (head loss $H_1$)

Both pipes have the same head loss $H_1$, so $Q_i = \sqrt{H_1/r_i}$, which gives $Q_i \propto D_i^{5/2}$:

$$Q = Q_1' + Q_2' = \sqrt{\frac{\pi^2gH_1}{8fL}}\left(D_1^{5/2} + D_2^{5/2}\right) \;\Rightarrow\; H_1 = \frac{8fLQ^2}{\pi^2g\left(D_1^{5/2} + D_2^{5/2}\right)^2}$$

With $D_2 = D_1/2$: $D_2^{5/2} = (0.5)^{2.5}D_1^{5/2} = 0.17678\,D_1^{5/2}$, so $\left(D_1^{5/2} + D_2^{5/2}\right)^2 = (1 + 0.17678)^2D_1^5 = 1.3848D_1^5$:

$$H_1 = \frac{8fLQ^2}{\pi^2g}\cdot\frac{1}{1.3848D_1^5}$$

Series arrangement (head loss $H_2$)

The same $Q$ passes through both pipes and the losses add:

$$H_2 = \frac{8fLQ^2}{\pi^2g}\left(\frac{1}{D_1^5} + \frac{1}{D_2^5}\right) = \frac{8fLQ^2}{\pi^2g}\cdot\frac{1 + 32}{D_1^5} = \frac{8fLQ^2}{\pi^2g}\cdot\frac{33}{D_1^5}$$

Ratio

$$\frac{H_1}{H_2} = \frac{1/1.3848}{33} = 0.02188 \approx \frac{1}{45.7}$$

Answer: $H_1 : H_2 = 1 : 45.7$, i.e. $H_1/H_2 \approx 0.0219$ (the parallel arrangement loses only about 2% of the head lost in series).

Data. $Q = 12$ l/s $= 0.012$ m³/s, $D_1 = 0.12$ m, $D_2 = 0.10$ m, $L = 1000$ m each, s.g. 0.85 ($\rho = 850$ kg/m³), $k = 0.26$ mm, $\nu = 9$ cm²/s $= 9\times10^{-4}$ m²/s.

The kinematic viscosity is very high, so the Reynolds numbers are small. Assume laminar flow in both pipes and verify afterwards.

Flow in each pipe

In parallel, both pipes have the same head loss. For laminar flow $h_f = \dfrac{128\nu LQ}{\pi gD^4}$, so $Q \propto D^4$:

$$\frac{Q_1}{Q_2} = \left(\frac{0.12}{0.10}\right)^4 = 2.0736$$ $$Q_1 = 0.012\left(\frac{2.0736}{1 + 2.0736}\right) = 0.008096\ \text{m}^3/\text{s} = 8.096\ \text{l/s},\qquad Q_2 = 0.003904\ \text{m}^3/\text{s} = 3.904\ \text{l/s}$$

Check for laminar flow

$$V_1 = \frac{Q_1}{\pi(0.12)^2/4} = 0.7158\ \text{m/s},\ Re_1 = \frac{V_1D_1}{\nu} = 95.4 \qquad V_2 = 0.4971\ \text{m/s},\ Re_2 = 55.2$$

Both are well below 2000, so the flow is laminar and the roughness (0.26 mm) has no effect.

Head loss and pump power

$$h_f = \frac{128\nu LQ_1}{\pi gD_1^4} = \frac{128(9\times10^{-4})(1000)(0.008096)}{\pi(9.81)(0.12)^4} = 145.94\ \text{m of oil}$$

The same value is obtained from pipe 2 ($145.94$ m). Pump head $= 145.94$ m (equal to the friction head, with no static lift).

Power delivered to the oil:

$$P = \rho g Q h_f = 850(9.81)(0.012)(145.94) = 14603\ \text{W} \approx 14.60\ \text{kW}$$

Answer: $Q_1 = 8.10$ l/s (12 cm pipe), $Q_2 = 3.90$ l/s (10 cm pipe); pump power $\approx 14.6$ kW (head 145.9 m of oil).

Data. $V = 1.2$ m/s, $D = 0.02$ m, $L = 30$ m, $e = 0.2$ mm, entrance loss $K = 0.5$, water $\nu = 1.0\times10^{-6}$ m²/s. The hose outlet (open to the atmosphere) is 3 m below the hose inlet level. Let $d$ be the depth of water above the hose inlet.

Friction factor

$$Re = \frac{VD}{\nu} = \frac{1.2(0.02)}{10^{-6}} = 24000, \qquad \frac{e}{D} = 0.010$$

Colebrook-White (iteration) gives $f = 0.0403$.

Energy equation

Between the pool surface (atmospheric, velocity zero, elevation $d + 3$ above the outlet) and the free outlet (velocity $V$):

$$d + 3 = \frac{V^2}{2g}\left(1 + K + f\frac{L}{D}\right), \qquad \frac{V^2}{2g} = \frac{1.2^2}{2(9.81)} = 0.0734\ \text{m}$$ $$f\frac{L}{D} = 0.0403\left(\frac{30}{0.02}\right) = 60.40$$ $$d + 3 = 0.0734\left(1 + 0.5 + 60.40\right) = 0.0734(61.90) = 4.543\ \text{m}$$ $$d = 4.543 - 3 = 1.54\ \text{m}$$

Answer: The water depth above the hose inlet is about $1.54$ m.

Idea. Express every loss in terms of the velocity head in the 200 mm pipe, $\dfrac{V_1^2}{2g}$, and find the length of 200 mm pipe ($f = 0.02$) that has the same total loss: $f\dfrac{L_e}{D_1} = \sum K$.

Areas: $\dfrac{A_1}{A_2} = \left(\dfrac{200}{250}\right)^2 = 0.64$, so $V_2 = 0.64V_1$.

Equivalent length of 200 mm pipe:

$$f\frac{L_e}{D_1} = 51.0392 \;\Rightarrow\; L_e = \frac{51.0392\times 0.2}{0.02} = 510.39\ \text{m}$$

Answer: Equivalent length $\approx 510.4$ m of 200 mm pipe (with $f = 0.02$). The minor losses (entrance, expansion, exit) alone are equivalent to about 10.4 m of pipe.

Data. $H = 100$ m, $L = 10$ km, $f = 0.03$, $Q = 200$ l/s $= 0.2$ m³/s. Only friction loss is considered.

Diameter of the single pipe

$$H = \frac{8fLQ^2}{\pi^2gD^5} \;\Rightarrow\; D^5 = \frac{8(0.03)(10\,000)(0.2)^2}{\pi^2(9.81)(100)} = 0.00992 \;\Rightarrow\; D = 0.397\ \text{m}$$

($D \approx 397$ mm; check $V = 1.61$ m/s.) For this pipe the head loss per metre at discharge $Q$ is $h_f = kLQ^2$ with

$$k = \frac{8f}{\pi^2gD^5} = 0.2500\ \text{s}^2/\text{m}^6 \text{ per metre of pipe}$$

(so that $100 = k(10\,000)(0.2)^2$).

Second pipe

The extra flow is $5\times10^6$ l/day $= 5000$ m³/day $= 0.05787$ m³/s, so the new total discharge is

$$Q' = 0.2 + 0.05787 = 0.25787\ \text{m}^3/\text{s}$$

A second pipe of the same diameter ($D$, same $f$) is laid in parallel along a length $x$, measured from one end. In this length the flow divides equally ($Q'/2$ in each pipe); in the remaining length $(L - x)$ the single pipe carries $Q'$.

$$H = k(L - x)Q'^2 + k\,x\left(\frac{Q'}{2}\right)^2 = kQ'^2\left(L - 0.75x\right)$$ $$100 = 0.2500(0.25787)^2(10\,000 - 0.75x) \;\Rightarrow\; 10\,000 - 0.75x = 6015.3$$ $$x = \frac{10\,000 - 6015.3}{0.75} = 5313\ \text{m}$$

Answer: Diameter of the pipe $D = 0.397$ m (about 397 mm). The second pipe, of the same diameter, must be laid in parallel for a length of about $5313$ m ($5.31$ km, about 53% of the line).

Data. $Q = 0.4$ m³/s, $L = 1000$ m, $h_f = 2$ m, $e = 0.26$ mm, $\mu = 0.0014$ Pa s, so $\nu = \mu/\rho = 1.40\times10^{-6}$ m²/s. This is a "find the diameter" problem, so iterate.

Working equation

From Darcy-Weisbach with $V = \dfrac{4Q}{\pi D^2}$:

$$h_f = \frac{8fLQ^2}{\pi^2gD^5} \;\Rightarrow\; D = \left(\frac{8fLQ^2}{\pi^2g\,h_f}\right)^{1/5} = \left(\frac{8f(1000)(0.4)^2}{\pi^2(9.81)(2)}\right)^{1/5}$$

Trials (f from Colebrook-White at each new D)

The diameter converges to $D = 0.6452$ m. Check: $V = 1.223$ m/s, $Re = 563815$, $e/D = 0.00040$, $f = 0.01692$, and

$$h_f = f\frac{L}{D}\frac{V^2}{2g} = 0.01692\left(\frac{1000}{0.6452}\right)\frac{(1.223)^2}{19.62} = 2.000\ \text{m}$$

Answer: $D \approx 645$ mm; a 650 mm cast iron pipe would be used.