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. M

Answer 1

`1.302\ \text{kg/s}`

Step-by-step explanation:

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

`dV` = Change in volume =
`220* 0.355\ \text{L}`

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

Mass flow rate is given by

`\dot{m}=\rho(dV)/(dt)\\ =1* (220* 0.355)/(60)\\ =1.302\ \text{kg/s}`

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