Pvc Pipes Expanion Loops

Overview

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

Key Formulas

Reynolds Number

$Re = \frac{\rho v D}{\mu}$

Ratio of inertial to viscous forces — determines flow regime.

Bernoulli's Equation

$P + \frac{1}{2}\rho v^2 + \rho g h = \text{const}$

Conservation of energy for steady, inviscid, incompressible flow.

Continuity Equation

$A_1 v_1 = A_2 v_2$

Conservation of mass for incompressible flow.

Darcy-Weisbach

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

Pressure drop due to friction in a pipe.

Variables

Symbol	Description	Unit
$Re$	Reynolds number	—
$\rho$	Fluid density	kg/m³
$v$	Flow velocity	m/s
$D$	Characteristic dimension	m
$\mu$	Dynamic viscosity	Pa·s
$P$	Pressure	Pa
$f$	Darcy 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)	PVC	CPVC
73	400,000	423,000
100	352,000	385,000
140	280,000	330,000
200	-	241,000

Maximum Allowable Working Stress (psi)

Temperature (°F)	PVC	CPVC
73	2,000	2,000
100	1,240	1,640
140	440	1,000
200	-	400

Thermal Expansion Coefficient

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

Example: PVC Expansion Loop Calculation

Given:

• 2" Schedule 40 PVC pipe, outside diameter $D = 2.375 \ \text{in}$
• Length $L_0 = 300 \ \text{ft}$
• Installed at $70°F$, operated at $140°F$

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

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

Step 3: Material properties at $140°F$

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

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

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

References

• Original Source