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(A) In a venturi-meter, let $A_1$ be the area of the broader part and $a_2$ be the area of the narrow part (throat). Let $v_1$ be the velocity of the

Vedclass · Accepted Answer

$v_1 = a_2 \sqrt{\frac{2gh}{A_1^2 - a_2^2}}$

Vedclass · Answer

(A) In a venturi-meter, let $A_1$ be the area of the broader part and $a_2$ be the area of the narrow part (throat). Let $v_1$ be the velocity of the fluid in the broader part and $v_2$ be the velocity in the throat.
By the equation of continuity, $A_1 v_1 = a_2 v_2$, which implies $v_2 = \frac{A_1}{a_2} v_1$.
Applying Bernoulli's equation at both sections: $P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$.
The pressure difference is given by $P_1 - P_2 = \rho gh$, where $h$ is the difference in height of the liquid column in the manometer.
Substituting $v_2$: $P_1 - P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2) = \frac{1}{2} \rho [(\frac{A_1}{a_2} v_1)^2 - v_1^2]$.
Solving for $v_1$: $2gh = v_1^2 [(\frac{A_1}{a_2})^2 - 1] = v_1^2 [\frac{A_1^2 - a_2^2}{a_2^2}]$.
Thus, $v_1 = \frac{a_2}{\sqrt{A_1^2 - a_2^2}} \sqrt{2gh}$.

Give the formula for the measurement of the velocity of a fluid in the broader part of a venturi-meter.

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