Discover the increase in pressure:
A copper tube of 50 mm diameter and 1200 mm length contains a thickness of 1.2 mm along closed ends. This is filled with water at atmospheric pressure. Discover the increase in pressure when an additional volume of 32 cc of water is pumped into the tube. Take E for copper = 100 GPa, Poisson's ratio = 0.3 and K for water = 2000 N/mm2.
Solution
The added quantity of water pumped in accounts for the change in volume of the shell in addition to the compression of the water in it.
Therefore, if p is the enhanced in pressure in water, after that,
Hoop stress = pd /2t = p × 50/(2 × 1.2) = 20.83 p
Longitudinal stress = pd /4t = p × 50 /(4 × 1.2 )= 10.42 p
Hoop strain = (1/ E) [σ_{ h} - v σ _{l} ] = (1/ E) [20.83 p - 0.3 × 10.42] = (17.7 p )/ E
Longitudinal strain = (1/ E) [σ _{l }- v σ _{h}] = (1/ E) [10.42 p - 0.3 × 20.83] = 4.17 p/E
Volumetric strain of water = 2ε _{h} + ε _{l} =( 2 × 17.7 )/ E p + (4.17 p / E )= 39.57 p/ E (increase)
Because of compression in water, its volumetric strain p /K (increase).
∴ Addition volume pumped
= Increase in volume of cylinder + Decrease in volume of water
i.e.
32 × 10^{3} = (39.57 p / E )× V + (p/K) V
= 39.57/(1 × 10 ^{5}) + (1/2000)) × p × (π d ^{2}/ 4 )× l
= 87.57 × 10^{- 5} × p ×( π/4) × 502 × 1200
∴ p = 15.51 N/mm^{2}