Answer to Question #151055 in Mechanics | Relativity for nidhi chandra

Question

a horizontal venturimeter with inlet diameter 30 cm and throat diameter 15cm is used to measure the flow of oil of specific gravity 0.8. The discharge of oil through venturimeter is 50 litres/s. Take C=0.98. write the reading of the oil-mercury differential manometer (x)

Answer 1

Explanations & Calculations

• Refer to the figure attached
• Assume the flow is incompressible& non-viscous.
• $\small \rho ,d$ represent the densities of the oil & mercury respectively.
• Considering the central flow line & applying Bernoulli's equation for the inlet & throat sections,

$\qquad\qquad \begin{aligned} \small P_1 +\frac{1}{2}\rho v^2_1&=P_2+\frac{1}{2}\rho v^2_2\\ \small P_1-P_2&= \small \frac{1}{2}\rho(v_2^2-v_1^2)\cdots(1) \end{aligned}$

• Being incompressible, flow has a uniform rate throughout. Therefore, by considering continuity,

$\qquad\qquad \begin{aligned} \small Q&= \small CA_1v_1=CA_2v_2\cdots \end{aligned}$ without C, the ideal volumetric flow is addressed.

• Then substituting the velocities in equation (1),

$\qquad\qquad \begin{aligned} \small P_1-P_2&= \small \frac{1}{2}\rho\bigg[\frac{Q^2}{C^2A_2^2}-\frac{Q^2}{C^2A_1^2}\bigg] \\ &= \small \frac{1}{2}\rho\frac{Q^2}{C^2}\bigg[\frac{1}{A_2^2}-\frac{1}{A_1^2}\bigg]\cdots(2) \end{aligned}$

• By considering the hydrostatics of the Manometer: pressure at the down oil-mercury interface equals that of the same level within Mercury. Therefore,

$\qquad\qquad \begin{aligned} \small P_1+(h+x)\rho g&= \small P_2 +h\rho g+xdg\\ \small P_1-P_2 &= \small xg(d-\rho)\cdots(2) \end{aligned}$

• By equaling (1) & (2),

$\qquad\qquad \begin{aligned} \small \end{aligned}$ $\qquad\qquad \begin{aligned} \small xg(d-\rho) &= \small \frac{1}{2}\rho\frac{Q^2}{C^2}\bigg[\frac{1}{A_2^2}-\frac{1}{A_1^2}\bigg]\\ \small x&= \small \frac{1}{2}\rho\frac{Q^2}{C^2}\bigg[\frac{1}{A_2^2}-\frac{1}{A_1^2}\bigg]\times\frac{1}{g(d-\rho)}\cdots(3) \end{aligned}$

• Here,

$\qquad\qquad \begin{aligned} \small A_1= \small \pi\frac{D^2}{4}=\pi\frac{(30cm)^2}{4}&=\small706.858\,cm^2\\ \small A_2 &=\small 176.715cm^2\\ \small Q&= \small 50\times 10^3 cm^3/s\\ \small \rho &= \small 800\,kgcm^{-3}\\ \small d&= \small 13600\,kgcm^{-3}\\ \small g&= \small 9.8\times 10^2cms^{-2} \end{aligned}$

• Substituting these in (3),

$\qquad\qquad \begin{aligned} \small x&= \small \bold{2.492\,cm} \end{aligned}$