Question

Q4: Bellows, diaphragm, and Bourdon tube pressure sensors all exhibit second-order time response. This means that a sudden change in pressure will cause an oscillation in the displacement and, therefore, in sensor output. Because they are like springs, they have an effective spring constant and mass, so the frequency can be estimated by the following Equation: ??=12?√??
Where fN= natural frequency in Hz
k= spring constant in N/m
m= seismic mass in kg
Consider a bellows with an effective spring constant 3500 N/m and mass of 50 g. The effective area against which the pressure acts is 0.5 in2. Calculate
(a) The bellows deflection for a pressure of 20 psi
(b) The natural frequency of oscillation

Answer 1

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(a) Pressure = 20psi on an area of 0.5 inch²

Thus, total force F = P*A = 20*0.5 = 10N

Deflection = F/k = 10/3500 = 0.0029m = 0.29cm

(b) Using ??=12?√??, we have Fn = 498.71 Hz.

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