# Fuel oil density 820" kg/m"^3 flows through a venturi meter having a throat diameter of 4.0 cm and an entrance diameter of 8.0 cm. The pressure drop between the entrance and the throat is 16 cm of Mercury. The density of mercury is 13600" kg/m"^3?

Apr 22, 2018

The flow rate is $= 0.0094 {m}^{3} {s}^{-} 1$

#### Explanation:

Apply Bernouilli's Equation between points $A$ and $B$

${p}_{A} + \frac{1}{2} \rho {v}_{A}^{2} + \rho g {h}_{A} = {p}_{B} + \frac{1}{2} \rho {v}_{B}^{2} + \rho g {h}_{B}$

The flow rate is constant in the pipe

$Q = {v}_{A} \times A = {v}_{B} \times B$

where

$A = \pi {d}_{A}^{2} / 4 = \pi \times {0.08}^{2} / 4$

and

$B = \pi {d}_{B}^{2} / 4 = \pi \times {0.04}^{2} / 4$

${v}_{A} \times \pi \times {0.08}^{2} / 4 = {v}_{B} \times \pi \times {0.04}^{2} / 4$

${v}_{A} = {v}_{B} \cdot {\left(\frac{0.04}{0.08}\right)}^{2} = 0.25 {v}_{B}$

${v}_{B} = \frac{1}{0.25} {v}_{A} = 4 {v}_{A}$

But

${h}_{A} = {h}_{B}$

Therefore,

${p}_{A} + \frac{1}{2} \rho {v}_{A}^{2} = {p}_{B} + \frac{1}{2} \rho {v}_{B}^{2}$

The pressure difference is $\Delta p = \rho g h = 13600 \cdot 9.8 \cdot 0.16 = 21324.8 P a$

$\Delta p = 21324.8 = \frac{1}{2} \rho \left({v}_{B}^{2} - {v}_{A}^{2}\right)$

$\frac{1}{2} \cdot 820 \cdot \left(16 {v}_{A}^{2} - {v}_{A}^{2}\right) = 21324.8$

${v}_{A} = \sqrt{\frac{52.01}{15}} = 1.86 m {s}^{-} 1$

The flow rate is

$q = 1.86 \cdot \pi \cdot {0.08}^{2} / 4 = 0.0094 {m}^{3} {s}^{-} 1$