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3 years ago

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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 min

ute. 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

1 answer:

Harman [31]3 years ago

4 0

Answer:

$1.302\ \text{kg/s}$

Explanation:

$\rho$ = Density of water = $1\ \text{kg/L}$

$dV$ = Change in volume = $220\times 0.355\ \text{L}$

$dt$ = Time elapsed = 1 minute = 60 seconds

Mass flow rate is given by

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

The mass flow rate is $1.302\ \text{kg/s}$ .

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