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Series Pipe Rules

For pipes in series, the same discharge passes through each pipe and the total head loss is the sum of the individual losses.

$$Q_1=Q_2=Q_3$$$$h_L=h_{L1}+h_{L2}+h_{L3}$$

Parallel Pipe Rules

For parallel pipes between two common junctions, head loss is the same in each branch and total flow is the sum of branch flows.

$$h_{L1}=h_{L2}=h_{L3}$$$$Q=Q_1+Q_2+Q_3$$

Darcy-Weisbach Head Loss

$$h_f=f\frac{L}{D}\frac{V^2}{2g}$$$$V=\frac{4Q}{\pi D^2}$$

Use Darcy-Weisbach when $f$ is given; use Hazen-Williams only when the coefficient $C$ is specified.

Parallel Flow Split Check

Two parallel pipes carry 0.10 m3/s total. If branch A carries 0.065 m3/s, find branch B and the head-loss condition.

$$Q_B=Q-Q_A=0.10-0.065=0.035\text{ m}^3/\text{s}$$$$h_{fA}=h_{fB}$$

Answer: Branch B carries 0.035 m3/s, and both branches must have equal head loss between the same junctions.

Series Pipes Total Head Loss

Water flows at 0.050 m3/s through two pipes in series. Pipe 1 is 200 m long and 200 mm in diameter. Pipe 2 is 150 m long and 150 mm in diameter. Take $f=0.020$ for both pipes and neglect minor losses. Find the total head loss.

$$V_1=\frac{Q}{A_1}=\frac{0.050}{\pi(0.20)^2/4}=1.59\text{ m/s}$$$$V_2=\frac{0.050}{\pi(0.15)^2/4}=2.83\text{ m/s}$$$$h_{f1}=0.020\frac{200}{0.20}\frac{1.59^2}{2(9.81)}=2.58\text{ m}$$$$h_{f2}=0.020\frac{150}{0.15}\frac{2.83^2}{2(9.81)}=8.12\text{ m}$$$$h_L=2.58+8.12=10.70\text{ m}$$

Answer: The total head loss is 10.70 m.

Parallel Pipes Same Diameter Flow Division

Two parallel pipes have the same diameter and friction factor. Branch A is 100 m long and branch B is 400 m long. The total flow is 0.120 m3/s. Determine the flow in each branch.

$$h_{fA}=h_{fB},\quad h_f\propto LQ^2\text{ for same }D\text{ and }f$$$$L_AQ_A^2=L_BQ_B^2$$$$100Q_A^2=400Q_B^2\Rightarrow Q_A=2Q_B$$$$Q_A+Q_B=0.120\Rightarrow 3Q_B=0.120$$$$Q_B=0.040\text{ m}^3/\text{s},\quad Q_A=0.080\text{ m}^3/\text{s}$$

Answer: Shorter branch A carries 0.080 m3/s; branch B carries 0.040 m3/s.

Problem: Equivalent Pipe for Three Pipes in Series

Three pipes are connected in series between two reservoirs with a total head difference of 25 m. Pipe 1: L = 500 m, D = 300 mm, f = 0.020. Pipe 2: L = 400 m, D = 250 mm, f = 0.022. Pipe 3: L = 300 m, D = 200 mm, f = 0.018. Neglect minor losses. Find the discharge through the system.

Express friction head loss for each pipe in terms of Q using $h_f = \frac{8fLQ^2}{\pi^2 g D^5}$:

$$K_i = \frac{8f_i L_i}{\pi^2 g D_i^5}$$ $$K_1 = \frac{8(0.020)(500)}{\pi^2(9.81)(0.30)^5} = \frac{80}{9.81\pi^2(0.00243)} = 337.9$$ $$K_2 = \frac{8(0.022)(400)}{\pi^2(9.81)(0.25)^5} = \frac{70.4}{9.81\pi^2(0.000977)} = 744.5$$ $$K_3 = \frac{8(0.018)(300)}{\pi^2(9.81)(0.20)^5} = \frac{43.2}{9.81\pi^2(0.000320)} = 1399.5$$ $$\sum h_f = (K_1 + K_2 + K_3)Q^2 = (337.9 + 744.5 + 1399.5)Q^2 = 2481.9Q^2$$ $$25 = 2481.9Q^2 \Rightarrow Q = \sqrt{\frac{25}{2481.9}} = 0.1004 \text{ m}^3/\text{s}$$

Answer: Discharge is 0.100 m³/s (100 L/s). The smallest pipe (200 mm) contributes the most friction loss, controlling the system capacity.

Problem: Three Unequal Parallel Pipes — Flow Distribution

Three parallel pipes connect junction J1 to junction J2 with a head loss of 8.0 m between them. All have f = 0.020. Branch 1: L = 600 m, D = 250 mm. Branch 2: L = 400 m, D = 200 mm. Branch 3: L = 300 m, D = 150 mm. Find the discharge in each branch and the total discharge.

For each branch: $h_f = K_i Q_i^2 = 8.0$ m, so $Q_i = \sqrt{8.0/K_i}$.

$$K_1 = \frac{8(0.020)(600)}{\pi^2(9.81)(0.25)^5} = \frac{96}{9.81\pi^2(0.000977)} = 1014.8$$ $$K_2 = \frac{8(0.020)(400)}{\pi^2(9.81)(0.20)^5} = \frac{64}{9.81\pi^2(0.000320)} = 2065.2$$ $$K_3 = \frac{8(0.020)(300)}{\pi^2(9.81)(0.15)^5} = \frac{48}{9.81\pi^2(0.0000759)} = 6539.6$$
$$Q_1 = \sqrt{\frac{8.0}{1014.8}} = 0.0888 \text{ m}^3/\text{s}$$ $$Q_2 = \sqrt{\frac{8.0}{2065.2}} = 0.0622 \text{ m}^3/\text{s}$$ $$Q_3 = \sqrt{\frac{8.0}{6539.6}} = 0.0350 \text{ m}^3/\text{s}$$ $$Q_{total} = 0.0888 + 0.0622 + 0.0350 = 0.1860 \text{ m}^3/\text{s}$$

Answer: Q1 = 88.8 L/s, Q2 = 62.2 L/s, Q3 = 35.0 L/s. Total = 186 L/s. The 250 mm pipe carries the most flow even though it is the longest, because its large diameter (raised to the 5th power) dominates.

Problem: Hazen-Williams Formula — Pipe Sizing

A water main must carry 0.085 m³/s over a length of 800 m with a maximum allowable head loss of 12 m. Using the Hazen-Williams formula $V = 0.8492 C R^{0.63} S^{0.54}$ with C = 120 and $R = D/4$ for a full circular pipe, determine the required pipe diameter in mm. Round up to the next standard size.

$$S = \frac{h_f}{L} = \frac{12}{800} = 0.015$$ $$Q = VA = 0.8492C\left(\frac{D}{4}\right)^{0.63} S^{0.54} \cdot \frac{\pi D^2}{4}$$ $$0.085 = 0.8492(120)\left(\frac{D}{4}\right)^{0.63}(0.015)^{0.54}\frac{\pi D^2}{4}$$ $$0.085 = 101.9(0.01488)(D/4)^{0.63}\frac{\pi D^2}{4}$$

Solve by trial: try D = 0.25 m:

$$Q_{trial} = 0.8492(120)\left(\frac{0.25}{4}\right)^{0.63}(0.015)^{0.54}\frac{\pi(0.25)^2}{4}$$ $$= 101.9(0.0625)^{0.63}(0.01488)(0.04909)$$ $$= 101.9(0.1633)(0.01488)(0.04909) = 0.1215 \text{ m}^3/\text{s}$$

Too large. Try D = 0.20 m: Q ≈ 0.068 m³/s (too small). Interpolate: required D ≈ 0.225 m.

Answer: Required diameter is approximately 225 mm. Use the next standard size of 250 mm to ensure adequate capacity with margin.

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q347

HGE - Hydraulics / Pipe Network / Engr. Janclyde Espinosa (Clidez)

A 400 mmø pipeline discharges water and branches into 3 pipes at junction A. The first pipe has a diameter of 300 mm, and length of 3000 m, the 2nd pipe has a diameter of 200mm and length of 1300m, and the third pipe has a diameter of 250mm and length of 2600 m. These 3 pipes then merge together at junction B to form a single pipeline having a diameter of 400 mm. Headloss between junction A and B = 24 m. Assume C = 120.

q347

Determine the rate of flow of the first pipeline in m3/s.

  1. 0.103
  2. 0.056
  3. 0.069
  4. 0.467

Determine the rate of flow of the second pipeline in m3/s.

  1. 0.056
  2. 0.103
  3. 0.069
  4. 0.467

Determine the rate of flow of the third pipeline in m3/s.

  1. 0.069
  2. 0.103
  3. 0.056
  4. 0.467

Part 1.

For each parallel branch, the head loss is 24 m. Using Hazen-Williams in SI form:
$h_f=\frac{10.67LQ^{1.852}}{C^{1.852}D^{4.87}}$
For pipeline 1, $L=3000$ m, $D=0.300$ m, $C=120$:
$Q=\left(\frac{h_fC^{1.852}D^{4.87}}{10.67L}\right)^{1/1.852}$
$Q=\left(\frac{24(120)^{1.852}(0.300)^{4.87}}{10.67(3000)}\right)^{1/1.852}$
$\boxed{Q\approx0.103\text{ m}^3\!/s}$

Part 2.

For branch 2, use Hazen-Williams with $h_f=24$ m, $D=0.200$ m, $L=1300$ m, and $C=120$:
$Q=\left(\frac{h_fC^{1.852}D^{4.87}}{10.67L}\right)^{1/1.852}$
$Q=\left(\frac{24(120)^{1.852}(0.200)^{4.87}}{10.67(1300)}\right)^{1/1.852}$
$\boxed{Q\approx0.056\text{ m}^3\!/s}$

Part 3.

For branch 3, use Hazen-Williams with $h_f=24$ m, $D=0.250$ m, $L=2600$ m, and $C=120$:
$Q=\left(\frac{h_fC^{1.852}D^{4.87}}{10.67L}\right)^{1/1.852}$
$Q=\left(\frac{24(120)^{1.852}(0.250)^{4.87}}{10.67(2600)}\right)^{1/1.852}$
$\boxed{Q\approx0.069\text{ m}^3\!/s}$

Question Bank: q368

HGE - Hydraulics / Pipes in Series and Parallel / Engr. Janclyde Espinosa (Clidez)

Three pipes A, B and C are connected in parallel. If the combined discharged of the 3 pipes is equal to 0.61 m3/s, and assuming they have equal values of friction factor "f", compute the following using the tabulated data shown:

q368

Compute the rate of flow of pipeline A in liters/sec.

  1. 170
  2. 390
  3. 50
  4. 230

Compute the rate of flow of pipeline B in liters/sec.

  1. 390
  2. 170
  3. 50
  4. 230

Compute the rate of flow of pipeline C in liters/sec.

  1. 50
  2. 390
  3. 170
  4. 230
### Parallel-pipe discharges The head loss is the same in every parallel branch. With equal friction factors, Darcy-Weisbach gives $$h_f\propto\frac{LQ^2}{D^5},$$ so $$Q\propto\frac{D^{5/2}}{\sqrt L}.$$ Using the tabulated lengths and diameters, form the three relative-flow factors and scale them so that $$Q_A+Q_B+Q_C=0.61\ \text{m}^3/\text{s}.$$ The branch discharges are $$Q_A=\boxed{170\ \text{L/s}},\qquad Q_B=\boxed{390\ \text{L/s}},\qquad Q_C=\boxed{50\ \text{L/s}}.$$

Question Bank: v27

HGE - Hydraulics / Pipe Flow / HGE May 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

Three pipes connect the same two points in parallel and carry a total discharge of 0.56 m3/s with equal friction factors. Line 1: $D_1$=0.15 m, $L_1$=650 m. Line 2: $D_2$=0.2 m, $L_2$=550 m. Line 3: $D_3$=0.1 m, $L_3$=720 m.

Compute the flow rate in line 1, in L/s.

  1. 349.4 L/sec
  2. 156.6 L/sec
  3. 54.0 L/sec
  4. 560.0 L/sec

Compute the flow rate in line 2, in L/s.

  1. 156.6 L/sec
  2. 349.4 L/sec
  3. 560.0 L/sec
  4. 54.0 L/sec

Compute the flow rate in line 3, in L/s.

  1. 156.6 L/sec
  2. 62.6 L/sec
  3. 81.5 L/sec
  4. 54.0 L/sec

Parallel pipes share the same head loss. With equal $f$, $h_f=\dfrac{8fLQ^{2}}{\pi^{2}gD^{5}}$ equal for all branches means $\dfrac{L_iQ_i^{2}}{D_i^{5}}$ is the same. Taking line 1 as reference,

$$Q_i=r_i\,Q_1,\qquad r_i=\sqrt{\frac{L_1/D_1^{5}}{L_i/D_i^{5}}}.$$

Continuity gives $Q_1(1+r_2+r_3)=Q_{total}$, hence

$$Q_1=\frac{Q_{total}}{1+r_2+r_3},\quad Q_2=r_2Q_1,\quad Q_3=r_3Q_1\ \ (\times1000\ \text{for L/s}).$$

Computed answers:
1. 156.6 L/sec
2. 349.4 L/sec
3. 54.0 L/sec

Question Bank: v66

HGE - Hydraulics / Pipe Flow / HGE November 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

Find the capacity in m3/min of a 0.9-m-diameter pipe with head loss 5 m per 900 m and Darcy friction factor 0.018.

  1. 89.1 m³/min
  2. 65.0 m³/min
  3. 47.2 m³/min
  4. 45.4 m³/min
Darcy-Weisbach gives $$h_L=\frac{8fLQ_s^2}{\pi^2gD^5}.$$ Thus $Q_s=\sqrt{h_L\pi^2gD^5/(8fL)}$ and the capacity is $Q=60Q_s=89.1095810369$ m3/min.
Computed answer: 89.1 m³/min

Question Bank: v68

HGE - Hydraulics / Pipe Flow / HGE November 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

Water flows at 0.18 m3/s through a horizontal pipe of diameter 180 mm. Mercury gages 140 m apart differ by 1.2 m. Evaluate the Darcy friction factor.

  1. 0.011
  2. 0.006
  3. 0.008
  4. 0.014
For the uniform horizontal pipe, the gage difference represents head loss: $h_L=SG_{Hg}\Delta h$. From Darcy-Weisbach, $$f=\frac{h_L\pi^2gD^5}{8LQ^2}=0.00822788853965.$$
Computed answer: 0.008

Question Bank: v103

HGE - Hydraulics / Pipes in Series and Parallel / HGE November 2021

Formula-mode item rendered with fixed values for lecture/PDF export.

A water pipeline consists of a larger pipe of diameter 500 mm and length 3250 m joined in series to a smaller pipe of diameter 325 mm and length 6250 m. The elevation difference is 28 m. Neglect velocity head and minor losses and use friction factor f = 0.022 for both pipes. Determine the velocity in the larger pipe, velocity in the smaller pipe, and flow rate.

The velocity of water in the larger pipeline, in m/s.

  1. 0.34 m/s
  2. 0.63 m/s
  3. 0.75 m/s
  4. 0.47 m/s

The velocity of water in the smaller pipeline, in m/s.

  1. 1.11 m/s
  2. 0.89 m/s
  3. 1.33 m/s
  4. 1.66 m/s

The rate of flow in the pipes, in m^3/s.

  1. 0.072 m^3/s
  2. 0.119 m^3/s
  3. 0.092 m^3/s
  4. 0.147 m^3/s
For each pipe, Darcy-Weisbach gives $h_f=KQ^2$, where $$K= rac{8fL}{\pi^2gD^5}.$$ Thus $$Q=\sqrt{ rac{H}{K_1+K_2}}=0.091802367281 ext{ m}^3/ ext{s}.$$ The velocities are $$v_1= rac{Q}{A_1}=0.46754561729 ext{ m/s},\qquad v_2= rac{Q}{A_2}=1.10661684566 ext{ m/s}.$$


Computed answers:
1. 0.47 m/s
2. 1.11 m/s
3. 0.092 m^3/s

Question Bank: v104

HGE - Hydraulics / Pipes in Series and Parallel / HGE November 2021

Formula-mode item rendered with fixed values for lecture/PDF export.

A main carries total flow Q = 2.8 m^3/s and divides into two parallel pipes. Pipe 1 has diameter 1.3 m, length 300 m, and friction factor 0.025; pipe 2 has diameter 0.6 m, length 880 m, and friction factor 0.029. The branches have equal head loss. Determine the flow in the 0.6-m pipe, the flow in the 1.3-m pipe, and the velocity in pipe 2.

The flow in the 600-mm-diameter pipeline, in m^3/s.

  1. 0.153 m^3/s
  2. 0.265 m^3/s
  3. 0.326 m^3/s
  4. 0.204 m^3/s

The flow in the larger-diameter pipeline, in m^3/s.

  1. 2.596 m^3/s
  2. 2.285 m^3/s
  3. 2.908 m^3/s
  4. 3.635 m^3/s

The velocity of flow in the smaller pipeline, in m/s.

  1. 0.97 m/s
  2. 0.72 m/s
  3. 0.36 m/s
  4. 0.58 m/s
Continuity gives $Q=Q_1+Q_2$. Equal Darcy-Weisbach losses give $K_1Q_1^2=K_2Q_2^2$, where $K=fL/D^5$. Therefore $$Q_1= rac{Q\sqrt{K_2}}{\sqrt{K_1}+\sqrt{K_2}}=2.59631135697,\quad Q_2=Q-Q_1=0.203688643033 ext{ m}^3/ ext{s},$$ and $$v_2=Q_2/A_2=0.720401208674 ext{ m/s}.$$


Computed answers:
1. 0.204 m^3/s
2. 2.596 m^3/s
3. 0.72 m/s

Question Bank: v105

HGE - Hydraulics / Pipes in Series and Parallel / HGE November 2021

Formula-mode item rendered with fixed values for lecture/PDF export.

Three pipes are parallel between junctions J and K. Their diameters and lengths are D1 = 0.35 m, L1 = 2400 m; D2 = 0.2 m, L2 = 1000 m; and D3 = 0.25 m, L3 = 3200 m. The pressure-head difference is 33 m and Hazen-Williams C = 125 for all pipes. Determine the flow rate in each pipe.

The flow rate in pipeline 1, in m^3/s.

  1. 0.173 m^3/s
  2. 0.216 m^3/s
  3. 0.281 m^3/s
  4. 0.346 m^3/s

The flow rate in pipeline 2, in m^3/s.

  1. 0.095 m^3/s
  2. 0.056 m^3/s
  3. 0.119 m^3/s
  4. 0.079 m^3/s

The flow rate in pipeline 3, in m^3/s.

  1. 0.061 m^3/s
  2. 0.076 m^3/s
  3. 0.095 m^3/s
  4. 0.114 m^3/s
For each parallel pipe, the head loss is the common pressure-head difference. Hazen-Williams gives $$Q_i=\left( rac{HC^{1.85}D_i^{4.87}}{10.67L_i} ight)^{1/1.85}.$$ Thus $Q_1=0.21607898397$, $Q_2=0.0794970251945$, and $Q_3=0.0762791590478 ext{ m}^3/ ext{s}$.

Computed answers:
1. 0.216 m^3/s
2. 0.079 m^3/s
3. 0.076 m^3/s

Question Bank: v165

HGE - Hydraulics / Pipes in Series and Parallel / HGE Refresher Series

Minor losses through valves, bends, and fittings are conventionally expressed as a coefficient multiplied by which head?

  1. Total head
  2. Static head
  3. Velocity head
  4. Pressure drop
Minor head loss is hL = K V2/(2g).

Question Bank: v181

HGE - Hydraulics / Pipes in Series and Parallel / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

Two parallel pipes connect the same reservoirs. Their lengths are 2500 m and 2600 m, diameters 0.22 m and 0.29 m, Darcy friction factors 0.021 and 0.019, and common head loss 35 m.

Find the discharge in the smaller pipe.

  1. 0.064 m^3/s
  2. 0.050 m^3/s
  3. 0.073 m^3/s
  4. 0.081 m^3/s

Find the discharge in the larger pipe.

  1. 0.104 m^3/s
  2. 0.151 m^3/s
  3. 0.133 m^3/s
  4. 0.167 m^3/s

Find the diameter of one 3000 m replacement pipe with f = 0.018 carrying the combined flow.

  1. 0.270 m
  2. 0.346 m
  3. 0.393 m
  4. 0.436 m
Apply Darcy-Weisbach to each branch using the common head loss, add the discharges, and size the equivalent single pipe.
Computed answers: 0.0644942577419, 0.132640307124, 0.345912506291.

Question Bank: v184

HGE - Hydraulics / Pipes in Series and Parallel / HGE Refresher Series

The continuity equation for fluid flow follows directly from which conservation law?

  1. Conservation of mass
  2. Conservation of momentum
  3. Conservation of energy
  4. Conservation of force
Continuity is the mathematical statement of mass conservation.

Question Bank: v214

HGE - Hydraulics / Pipes in Series and Parallel / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

Oil of specific gravity 0.93 flows through a horizontal pipe of length 45 m and diameter 0.14 m at mean velocity 3.75 m/s with Darcy factor 0.025.

Find the pressure drop.

  1. 41.04 kPa
  2. 52.55 kPa
  3. 59.74 kPa
  4. 66.26 kPa
Convert Darcy-Weisbach head loss to pressure loss with Δp = γhf.
Computed answers: 52.5460379464.

Question Bank: v226

HGE - Hydraulics / Pipes in Series and Parallel / HGE Refresher Series

For a very long pipe, which local losses are commonly neglected compared with friction loss?

  1. Entrance only
  2. Outlet only
  3. Both entrance and outlet
  4. Pipe friction
Major friction loss dominates the relatively small entrance and exit losses.

Question Bank: v254

HGE - Hydraulics / Pipes in Series and Parallel / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A pipeline length 900 m and diameter 0.33 m carries 0.19 m3/s with Darcy factor 0.024.

Find head loss.

  1. 12.85 m
  2. 18.71 m
  3. 16.46 m
  4. 20.75 m
Apply Darcy-Weisbach in discharge form.
Computed answers: 16.4577818345.

Question Bank: v264

HGE - Hydraulics / Pipes in Series and Parallel / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

Water and oil enter a mixing chamber through two pipes. Water flows at 0.17 m3/s, oil of specific gravity 0.77 at 0.05 m3/s, and the outlet diameter is 0.3 m.

Find the outlet mixture velocity.

  1. 3.11 m/s
  2. 2.43 m/s
  3. 3.54 m/s
  4. 3.92 m/s

Find the mixture density.

  1. 947.7 kg/m^3
  2. 740.2 kg/m^3
  3. 1077.6 kg/m^3
  4. 1195.1 kg/m^3
Apply volumetric continuity and mass conservation.
Computed answers: 3.11236333157, 947.727272727.

Question Bank: v278

HGE - Hydraulics / Pipes in Series and Parallel / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A 0.31 m pipeline has Darcy factor 0.022. Two pressure taps 500 m apart show a pressure difference 125 kPa; a leak between them causes unequal upstream and downstream flows.

Estimate the leak rate.

  1. 16.40 L/s
  2. 12.81 L/s
  3. 18.64 L/s
  4. 20.68 L/s
Apply the energy equation to each side of the leak and subtract downstream flow from upstream flow.
Computed answers: 16.3960710648.