Hydraulics 3

Hydraulics 3 - Fluid Flow in Pipes and Reservoirs

Pipes

• Closed conduits for fluid transport; flow can be full or partial.
• Fluids flow full in pipes, while open channels have partial flow.
• Steady Flow Types:
  1. Laminar Flow: Fluid particles move without intersecting paths.
     • Reynold's Number (NR) indicates flow type. Laminar flow occurs when NR \le 2000.
  2. Turbulent Flow: Fluid particle paths intersect continuously.
     • Turbulent flow occurs when N_R \ge 4000.
  3. Transitional Flow: Occurs when 2000 < N_R < 4000.

Reynold's Number

• Dimensionless parameter: Ratio of inertia force to viscous force.
• Formula: N_R = \frac{\rho vD}{\mu} = \frac{vD}{\nu}
  • Where:
    • v = mean velocity
    • D = diameter
    • \mu = dynamic viscosity
    • \rho = density
    • \nu = kinematic viscosity
    • g = gravitational acceleration
• Non-circular Pipes:
  • D = 4R, where R is the hydraulic radius.
  • Hydraulic Radius, R = \frac{Area, A}{Wetted Perimeter, P}.

Velocity Distribution in Pipes

Laminar Flow

• Velocity distribution follows a parabolic law; zero velocity at the walls.
• Centerline velocity: Vc = 2V{ave}, where V_{ave} is the mean velocity.
• Velocity (u) at distance "r" from the center (pipe radius ro): u = vc(1-\frac{r^2}{r_o^2}).

Turbulent Flow

• Velocity distribution varies with Reynold's Number; zero velocity at the walls, increasing rapidly.
• Centerline velocity: v_c = v(1 + 1.33\sqrt{f})
• Mean velocity related to centerline velocity: v = vc - 3.75\sqrt{\frac{To}{\rho}}
  • Where:
    • v_c = centerline velocity
    • f = friction factor
    • T_o = maximum shearing stress in the pipe
    • v = mean velocity

Shear Stress at Pipe Walls

• Shear Stress: T = \frac{fpv^2}{8}

Head Losses in Pipe Flow

Major Head Losses

1. Darcy-Weisbach Formula:
   • h_L = f \frac{Lv^2}{2gD}
   • For circular pipes: h_L = \frac{0.0826fLQ^2}{D^5}.
2. Manning's Formula:
   • h_L = \frac{10.29n^2LQ^2}{D^{16/3}}, where n is the roughness coefficient.
3. Hazen-Williams Formula:
   • h_L = \frac{10.67LQ^{1.85}}{C^{1.85}D^{4.87}}, where C is the Hazen-Williams constant.

Values of Friction Factor, f

• Laminar Flow: f = \frac{64}{N_R}
• Turbulent Flow:
  • Smooth Pipe (3000 \le NR \le 10000): f = \frac{0.316}{NR^{0.25}}
  • All Pipes: \frac{1}{\sqrt{f}} = -2 \log(\frac{E}{3.7D} + \frac{2.51}{N_R\sqrt{f}})
    • (Colebrook and White Formula)

Pipes Connected in Series

• Flow is equal in all pipes: Q = Q1 = Q2 = Q3 = … = Qn
• Total head loss is the sum of individual head losses: HL = h{L1} + h{L2} + h{L3} + … + h_{Ln}

Pipes Connected in Parallel

• Head loss in individual pipes is equal: HL = h{L1} = h{L2} = h{L3} = … = h_{Ln}
• Total discharge is the sum of individual discharges: QT = Q1 + Q2 + Q3 + … + Q_n

Equivalent Pipe

• Must have the same discharge and head loss as the original pipe system.
• Discharge, total head loss, and total length from source to supply point are conserved.

Sample Problems

Situation 1

A 60-mm diameter pipe contains glycerin (\rho = 1250 kg/m^3) flowing at 2.36 L/sec. Viscosity of glycerin is 0.0556 Pa-s. Assume steady laminar flow:

1. Reynold's Number:
   • N_R = 2002.57
2. Direction of Flow:
   • A to B
3. Head Loss:
   • 18.56 m
4. Equivalent Friction Factor (L = 1.2 km):
   • 0.0378

Situation 2

A cast iron pipe carries a flow of 6.20 cfs, length is 30 ft, and diameter is 6 inches.

5. Friction Factor = 0.014:
   • Head Loss = 11.70 ft
6. Roughness Coefficient = 0.015:
   • Head Loss = 48.76 ft
7. Hazen-Williams Constant = 110:
   • Head Loss = 33.37 ft

Situation 3

Pipes 1, 2, and 4 are in series; pipes 2 and 3 are parallel.

• Pipeline 2 Discharge = 40 liters/sec

8. Discharge in Pipeline 3:
   • 26.4 liters/sec
9. Discharge in Pipeline 4:
   • 66.4 liters/sec

Situation 4

Two pipes are connected in parallel between two reservoirs.

10. Rate of Flow at Pipe A:
    • 0.14652 m^3/s
11. Rate of Flow at Pipe B:
    • 0.49918 m^3/s
12. Equivalent length of 1000mm Diameter Pipe (C=120):
    • 627,826 m

Flow in Three Reservoirs

Three Reservoir Problems

Step 1: Identify the Case (One Supplier or Two Suppliers)

• Compare discharges of the highest and lowest reservoirs.
• Place the top of the piezometer at the same elevation as the middle reservoir (hB = 0).
• Get the difference in elevations between the highest and middle reservoir (hA) and between the middle and lowest reservoir (hC).

Step 2: Solve for Head Losses

• Use major head loss equations (Darcy, Manning, Hazen-Williams).
• Express head losses in terms of QA, QB, and Q_C.

Step 3: Solve for QA and QC and Compare

• If QA > QC, CASE I: Reservoir A supplies both B and C.
• If QA < QC, CASE II: Reservoirs A and B supply C.

Case I: Reservoir A Supplies both B and C

• ElevA - ElevB = h{fA} + h{fB}
• ElevA - ElevC = h{fA} + h{fC}
• QA = QB + QC Calculator Technique: x = \frac{AElev{AB}}{A'} + \frac{AElev_{AC}}{C'}
  • Where: A' and C' are pipe properties, x – head loss of reservoir A

Case II: Reservoirs A and B Supply C

• ElevA - ElevC = h{fA} + h{fC}
• ElevB - ElevC = h{fB} + h{fC}
• QA + QB = QC Calculator Technique: x = \frac{AElev{AC}}{A'} + \frac{AElev_{BC}}{B'}
  • Where: A' and B' are pipe properties, x = head loss of reservoir C

Sample Problem: Three Reservoirs

Problem: Determine the flow in each pipe connecting the three reservoirs.

Given:

• Pipe 1:
  • Length = 2000 m
  • Diameter = 500 mm
  • f = 0.02
• Pipe 2:
  • Length = 2500 m
  • Diameter = 600 mm
  • f = 0.02
• Pipe 3:
  • Length = 4500 m
  • Diameter = 800 mm
  • f = 0.02
• Reservoir Elevations:
  • A = 110 m
  • B = 70 m
  • C = 20 m

Solution:

• Headloss Formula: h = \frac{0.0826fLQ^2}{D^5}
  Headloss Calculations:
  *hA = 105.728 QA^2
  *hB = 53.11 QB^2
  *hC = 22.69 QC^2
  *Case:* QA + QB = QC Discharge:
  *QA = 0.4515 m^3/s
  *QB = 1.2853 m^3/s *QC = 1.7368 m^3/s

Practice Problems

Problems 1 - 3

Three reservoirs A, B, and C with elevations 346 m, 301 m, and 241 m, respectively. Flow rate at 1 is 200 L/sec.

1. Head Loss of Pipeline 2:
   • 42.73 m
2. Discharge of Pipeline 3:
   • 94 L/sec
3. Diameter of Pipeline 3:
   • 180 mm

Problems 4 - 6

Pressure heads at points A and E are 70 m and 46 m, respectively. Assume C = 120 for all pipes.

4. Flow Rate of Water Through B:
   • 0.103 m^3/s
5. Flow Rate of Water Through C:
   • 0.039 m^3/s
6. Flow Rate of Water Through D:
   • 0.07 m^3/s

Problems 7 - 9

A 6-km cast-iron (new) pipeline conveys 320 L/sec of water at 30°C. The pipe diameter is 30 cm.

7. Head Loss (m) - Darcy-Weisbach (f = 0.0195):
   • 408
8. Head Loss (m) - Hazen-Williams (C = 130):
   • 314
9. Head Loss (m) - Manning Equation (n = 0.011):
   • 496

Problem 10

A buried, horizontal concrete pipe (n = 0.012) of unknown length needs replacement. Pressure head drop is 29.9 ft, flow rate is 30.0 cfs.

10. Length (ft) of New Pipe Needed:
    • 2000