# Estimate the differential pressure limit for a 0.5-mm-thick, 25-mmdiameter steel diaphragm...

## Question:

Estimate the differential pressure limit for a 0.5-mm-thick, 25-mm diameter steel diaphragm pressure transducer. {eq}v_p {/eq} = 0.32, {eq}E_m {/eq} = 200 GPa.

## Poisson's ratio

Poisson's ratio may be defined as a term that is used for finding a strain. Poisson's ratio is a unit less term, and the symbol of (mu) denotes Poisson's ratio.

## Answer and Explanation:

**Given Data**

Modulus of elasticity {eq}{E_m} = 200\;{\rm{GPa = }}200 \times {10^9}\;{\rm{Pa}} {/eq}

Poisson's ratio {eq}{v_p} = 0.32 {/eq}

Thickness of diaphragm {eq}t = 0.5\;{\rm{mm = 5}} \times {\rm{1}}{{\rm{0}}^{ - 04}} {/eq}

Diameter of diaphragm {eq}D = 25\;{\rm{mm = 0}}{\rm{.25}}\;{\rm{m}} {/eq}

Radius of diaphragm {eq}{\rm{a = r = 0}}{\rm{.125}}\;{\rm{m}} {/eq}

The deflection in the diaphragm must be less than {eq}1/3 {/eq} of thickness of diaphragm

Hence

{eq}\begin{align*} \delta & < \dfrac{1}{3} \times \left( {0.5} \right)\\ \delta &< 1.67\\ \delta &< 0.17\;{\rm{mm}} \end{align*} {/eq}

Now we are finding differential pressure limit

The expression for the differential pressure limit is;

{eq}p = \dfrac{{16{E_m}{t^4}}}{{3{a^4}\pi \left( {1 - v_p^2} \right)}} \times \dfrac{\delta }{t} {/eq}

Substitute the values in above expression

{eq}\begin{align*} p &= \dfrac{{16 \times 200 \times {{10}^9} \times {{\left( {5 \times {{10}^{ - 04}}} \right)}^4} \times 0.17}}{{3 \times {{\left( {0.25} \right)}^4} \times \pi \left( {1 - {{0.32}^2}} \right) \times 0.5}}\\ p &= \dfrac{{0.034}}{{0.01652}}\\ p &= 2.058\;{\rm{Pa}} \end{align*} {/eq}

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