Question

A transit time ultrasonic flowmeter uses a pair of transmitter receiver placed at angle of ${45}^o$ to the fluid flow axis in a pipe of 0.5 m internal diameter. The velocity of ultrasound in fluid is 1000 m/s. If the minimum change in fluid velocity is found to be 0.5 m/s then the maximum clock frequency that can be used to trigger the crystal in MHz is?

• A. 2

Answer 1

The correct option is A 2
$\text{The transit time difference is given by :}$

$\Delta t=\frac{2vlcos\theta }{c^2}$

$v\text{= fluid velocity}$

$\text{d = distance between transmitter and receiver}$

$\theta \text{= wedge angle}$

$\text{c = ultrasound velocity in fluid}$

$\Delta t=\frac{2\times 0.5\times l\times cos45}{(1000{)}^2}$

$l=\frac{0.5}{sin\theta }=\frac{0.5}{sin15}$

$\Delta t=\frac{2\times 0.5\times 0.5cos45}{sin45\times (1000{)}^2}$

$\therefore \Delta t=0.5\mu s\to \text{min. transit time}$

$\text{Max. clock frequency}=\frac{1}{\Delta t}$

$f_{max}=\frac{1}{0.5\mu s}$

$\therefore f_{max}=2MHz$