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

a) 1.3 kg/s

b) 0.0013 m³/s

c) 3.25 m/s

Step-by-step explanation:

Given

Mass flow rate = (220 * 0.355) / 60

Mass flow rate = 78.1 / 60

Mass flow rate = 1.3 kg/s

Volume flow rate =

Recall, Density = mass / volume, so

Volume = mass / density, where

Density of water = 1000 kg/m³

Volume flow rate = mass flow rate / density

Volume flow rate = 1.3 / 1000

Volume flow rate = 0.0013 m³/s

Flow speed at point 1

V = volume flow rate / area

Area at point 1 = 2 cm² = 0.0004 m²

Flow speed at point 1 is

V = 0.0013 / 0.0004

V = 3.25 m/s