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

a) 1.3 kg/s

b) 0.0013 m³/s

c) 3.25 m/s

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

answer: the direction of the wind is expressed as the direction from which the wind is blowing. for example, easterly winds blow from east to west, while westerly winds blow from west to east.

explanation:

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