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Pvc Pipes Expanion Loops

Reference data and engineering information about pvc pipes expanion loops for fluid mechanics applications.

pvcpipesexpanionloops

Overview

Engineering reference data for Pvc Pipes Expanion Loops in fluid mechanics.

Key Formulas

Reynolds Number

Re=ρvDμRe = \frac{\rho v D}{\mu}

Ratio of inertial to viscous forces — determines flow regime.

Bernoulli's Equation

P+12ρv2+ρgh=constP + \frac{1}{2}\rho v^2 + \rho g h = \text{const}

Conservation of energy for steady, inviscid, incompressible flow.

Continuity Equation

A1v1=A2v2A_1 v_1 = A_2 v_2

Conservation of mass for incompressible flow.

Darcy-Weisbach

ΔP=fLDρv22\Delta P = f \frac{L}{D} \frac{\rho v^2}{2}

Pressure drop due to friction in a pipe.

Variables

SymbolDescriptionUnit
ReReReynolds number
ρ\rhoFluid densitykg/m³
vvFlow velocitym/s
DDCharacteristic dimensionm
μ\muDynamic viscosityPa·s
PPPressurePa
ffDarcy friction factor

Material Properties

The modulus of elasticity and allowable working stress for PVC and CPVC vary with temperature.

Modulus of Elasticity (psi)

Temperature (°F)PVCCPVC
73400,000423,000
100352,000385,000
140280,000330,000
200-241,000

Maximum Allowable Working Stress (psi)

Temperature (°F)PVCCPVC
732,0002,000
1001,2401,640
1404401,000
200-400

Thermal Expansion Coefficient

  • PVC: 28×106 in/in°F28 \times 10^{-6} \ \text{in/in} \cdot °\text{F}
  • CPVC: 44×106 in/in°F44 \times 10^{-6} \ \text{in/in} \cdot °\text{F}

Example: PVC Expansion Loop Calculation

Given:

  • 2" Schedule 40 PVC pipe, outside diameter D=2.375 inD = 2.375 \ \text{in}
  • Length L0=300 ftL_0 = 300 \ \text{ft}
  • Installed at 70°F70°F, operated at 140°F140°F

Step 1: Temperature change δt=140°F70°F=70°F\delta t = 140°F - 70°F = 70°F

Step 2: Pipe expansion (δl\delta l) Using δl=αL0δt\delta l = \alpha \cdot L_0 \cdot \delta t: δl=(28×106)(30012)70=7.1 in\delta l = (28 \times 10^{-6}) \cdot (300 \cdot 12) \cdot 70 = 7.1 \ \text{in}

Step 3: Material properties at 140°F140°F

  • E=280,000 psiE = 280,000 \ \text{psi}
  • S=440 psiS = 440 \ \text{psi}

Step 4: Required loop length (LlL_l) Ll=3EDδl2SL_l = \sqrt{\frac{3 \cdot E \cdot D \cdot \delta l}{2 \cdot S}} Ll=32800002.3757.12440126.5 inL_l = \sqrt{\frac{3 \cdot 280000 \cdot 2.375 \cdot 7.1}{2 \cdot 440}} \approx 126.5 \ \text{in}

Step 5: Loop dimensions A=Ll5=25.3 inA = \frac{L_l}{5} = 25.3 \ \text{in} B=2Ll5=50.6 inB = \frac{2L_l}{5} = 50.6 \ \text{in}

References