PVC Pipes  Temperature Expansion Loops
Calculate temperature expansion and contraction in PVC piping systems.
Temperature expansion and contraction in PVC piping systems can be compensated with
 expansion loops consisting of pipes and 90^{o} elbows
 flexible bends
 bellows and rubber expansion joints
 piston type expansion joints
Expansion Loops
Expansion loops are made of standard pipes and elbows and can be produced on the site and adapted to the actual situation.
The length of the loop can be calculated as
L_{l} = ((3 E D δl) / (2 S))^{1/2} (1)
where
L_{l} = length of loop (in)
E = modulus of elasticity (psi)
D = outside diameter of pipe (in)
δl = change in pipe length L_{0} due to temperature change (in)
S = allowable working stress at maximum temperature (psi)
A = L_{l} / 5 (2)
B = 2 L_{l} / 5 (3)
Modulus of elasticity for PVC:
 73^{o}F : 400000 psi
 100^{o}F : 352000 psi
 140^{o}F : 280000 psi
Modulus of elasticity for CPVC:
 73^{o}F : 423000 psi
 100^{o}F : 385000 psi
 140^{o}F : 330000 psi
 200^{o}F : 241000 psi
Maximum working stress for PVC:
 73^{o}F : 2000 psi
 100^{o}F : 1240 psi
 140^{o}F : 440 psi
Maximum working stressfor CPVC:
 73^{o}F : 2000 psi
 100^{o}F : 1640 psi
 140^{o}F : 1000 psi
 200^{o}F : 400 psi
The temperature expansion of the pipe can be calculated as
δl = α L_{o} δt (4)
where
α = expansion coefficient (in/in ^{o}F)
L_{o} = initial length of pipe (ft)
δt = temperature change (^{o}F)
Expansion coefficient for PVC:
 28 10^{6} in/in ^{o}F
Expansion coefficient for CPVC:
 44 10^{6} in/in ^{o}F
Example  Expansion Loop
A 2" PVC Schedule 40 straight pipe with outside diameter 2.375 inches and length 300 feet is installed at 70 ^{o}F and operated at 140 ^{o}F. The expansion coefficient for PVC is 28 10^{6} in/in ^{o}F.
The expansion of the PVC pipe can from eq. 4 be calculated as
δl = α L_{o} δt
= (28 10^{6} in/in ^{o}F) (300 ft) (12 in/ft) ((140 ^{o}F)  (70 ^{o}F))
= 7.1 inches
The modulus of elasticity is 280000 psi and the maximum working stress is 440 psi at maximum temperature. The length of the loop can from eq. 1 be calculated as
L_{l} = ((3 (280000 psi) (2.375 in) (7.1 in)) / (2 (440 psi)))^{1/2}
= 126.5 in
A can be calculated from eq. 2 as
A = (126 in) / 5
= 25.3 in
B can be calculated from eq. 3 as
A = 2 (126 in) / 5
= 50.6 in
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