Athin-walled cylindrical pressure vessel is subjected to an internal gauge pressure, p=75 psip=75 psi. it had a wall thickness of 0.25 inches and an inner diameter of 8 inches. use mohr’s circle to determine the absolute maximum shear stress in the pressure vessel when it is subjected to this pressure.

To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as

Here,

P = Pressure

d = Diameter

t = Thickness

At the same time the longitudinal stress is given as,

The letters have the same meaning as before.

Then he hoop stress would be,

And the longitudinal stress would be

The Mohr's circle is attached in a image to find the maximum shear stress, which is given as

Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi

a)  Q₀ = Q = 4.27 10² pC,  b) C = 1.26 10⁻¹⁰ F,  ΔV = 3.375 V, c) ΔE = 56.88  nJ

Explanation:

a) the capacitance of a parallel plate capacitor is

        C = ε₀ A / d

let's reduce the magnitudes to the SI system

        d = 1.40 cm = 0.0140 m

        A = 25.0 cm² (1 m / 100 cm) ² = 25.0 10⁻⁴ m²

         C = 8.85 10⁻¹² 25.0 10⁻⁴ / 0.0140

         C = 1.58 10⁻¹² F

the capacitance is also

          C = Q / ΔV

          Q = V ΔV

          Q = 270  1.58 10⁻¹²

          Q = 4.27 10⁻¹⁰ C

When the battery is removed and the capacitor is inserted into the dielectric, the charge should remain the same,

           Q₀ = Q = 4.27 10⁻¹⁰ C

let's reduce to pC

           Q₀ = 4.27 10⁻¹⁰ C (10¹² pC / 1C)

           Q₀ = Q = 4.27 10² pC

b) After being immersed in the fluid of constant k = 80.0

capacitance is

            C = k Co

            C = 80 1.58 10⁻¹²

            C = 1.26 10⁻¹⁰ F

the voltage difference is

            ΔV = ΔV₀ / k

            ΔV = 270/80

            ΔV = 3.375 V

c) We stop the energy before the dive

             E₀ = ½ C₀ ΔV₀²

after the dive

              E = ½ (k C₀) (ΔV₀/k)²

              E = ½ C₀ ΔV₀² / k

              E = E₀ / k

the change in energy is

            ΔE = E -E₀

            ΔE = E₀ / k - E₀

            ΔE = E₀ ( )

we calculate

             E₀ = ½ 1.58 10⁻¹² 270²

             E₀ = 5.76 10⁻⁸ J

             ΔE = 5.76 10⁻⁸ ()

             ΔE = 5.688 10⁻⁸ J

we reduce to nJ

             ΔE = 5.688 10⁻⁸ J (10⁹ nJ / 1J)

             ΔE = 56.88  nJ