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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. M

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boffeemadrid

Answer:

[tex]1.302\ \text{kg/s}[/tex]

Explanation:

[tex]\rho[/tex] = Density of water = [tex]1\ \text{kg/L}[/tex]

[tex]dV[/tex] = Change in volume = [tex]220\times 0.355\ \text{L}[/tex]

[tex]dt[/tex] = Time elapsed = 1 minute = 60 seconds

Mass flow rate is given by

[tex]\dot{m}=\rho\dfrac{dV}{dt}\\ =1\times \dfrac{220\times 0.355}{60}\\ =1.302\ \text{kg/s}[/tex]

The mass flow rate is [tex]1.302\ \text{kg/s}[/tex].

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