Question

A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 0.355 - L cans per minute. At point 2 in the pipe, the gauge pressure is 152kPa and the cross-sectional area is 8.00cm2. At point 1, 1.35m above point 2, the cross-sectional area is 2.00cm2.

Part a

Find the mass flow rate. kg/s

part b

Find the volume flow rate. L/s

part c

Find the flow speed at point 1.   m/s

Answer 1

Mass flow rate \(=\) density \(^{*}\) volume flow rate

now first we will find the volume flow rate

Volume rate \(=\frac{\text { Volume }}{\text { time }}\)

Volume rate \(=\frac{220 * 0.355}{60}\)

PART B)

Volume rate \(=1.30 \mathrm{~L} / \mathrm{s}\)

PART A)

Mass flow rate is given as

Mass rate \(=\) density \(*\) Volume rate

Mass rate \(=1000 * 1.30 * 10^{-3} \mathrm{~m}^{3} / \mathrm{s}\)

Mass rate \(=1.30 \mathrm{~kg} / \mathrm{s}\)

PART C)

by Equation of continuity we can say

\(A_{1} v_{1}=A_{2} v_{2}\)

\(8 * v_{2}=2 * v_{1}=\) flow rate

\(2 * 10^{-4} m^{2} * v_{1}=1.30 * 10^{-3} \mathrm{~m}^{3} / s\)

\(v_{1}=\frac{1.30 * 10^{-3}}{2 * 10^{-4}}\)

\(v_{1}=6.5 \mathrm{~m} / \mathrm{s}\)