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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

Solution pending in psadquestions/q368.json.

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