A cylindrical pressure vessel made of steel has diameter of 3 m and wall thickness of 15 mm. For steel E = 210 GPa and μ = 0.3. The cylinder is designed such that the allowable normal strain at outer cylindrical surface is equal to 0.00034. The permissible pressure in tank is ______ kPa (Rounded off to one decimal place)

The correct answer is 840.

To understand why this is the correct answer, let's go through the step-by-step physical and mathematical derivation of the normal strain in a thin-walled cylindrical pressure vessel.

First, let's identify the given values from the problem statement:

Diameter of the vessel, $d=3\text{m}$
Thus, the inner radius is $r=\frac{d}{2}=1.5\text{m}=1500\text{mm}$.
Wall thickness, $t=15\text{mm}=0.015\text{m}$.
Young's modulus of steel, $E=210\text{GPa}=210\times {10}^9\text{Pa}=2.1\times {10}^8\text{kPa}$.
Poisson's ratio, $\mu =0.3$.
Allowable normal strain at the outer cylindrical surface (which is the circumferential or hoop strain), ${\epsilon }_h=0.00034$.

For a thin-walled cylindrical pressure vessel subjected to internal pressure $p$, the principal stresses developed in the walls are the circumferential (hoop) stress ${\sigma }_h$ and the longitudinal (axial) stress ${\sigma }_a$. These are given by the following formulas:

${\sigma }_h=\frac{pd}{2t}$

${\sigma }_a=\frac{pd}{4t}$

According to Hooke's Law for plane stress, the normal strain in the circumferential (hoop) direction at the outer surface is related to these stresses by:

${\epsilon }_h=\frac{1}{E}({\sigma }_h-\mu {\sigma }_a)$

Substituting the stress equations into the strain equation, we get:

${\epsilon }_h=\frac{1}{E}(\frac{pd}{2t}-\mu \frac{pd}{4t})$

Factoring out common terms:

${\epsilon }_h=\frac{pd}{4tE}(2-\mu )$

Now, we can rearrange this equation to solve for the permissible internal pressure $p$:

$p=\frac{4tE{\epsilon }_h}{d(2-\mu )}$

Let's substitute the given values into the rearranged equation:

$p=\frac{4\times 0.015\times (210\times {10}^6)\times 0.00034}{3\times (2-0.3)}$

Evaluating the numerator:

$\text{Numerator}=0.06\times 210\times {10}^6\times 0.00034=12600000\times 0.00034=4284\text{kPa}$

Evaluating the denominator:

$\text{Denominator}=3\times 1.7=5.1$

Now, calculate the value of $p$:

$p=\frac{4284}{5.1}=840\text{kPa}$

Thus, the permissible pressure in the tank is exactly 840 kPa.