Question

Pergunta 5 You need to fill a football with air to play with it. You know that your pump expels air at speed of 8.2 ft/s. The needle of your pump has a radius of 4.5 millimeters. What is the volume flow rate of the air being pumped into the football?

Answer 1

The volume flow rate of the air being pumped into the football will be given by:
 Q=V*A
 Where
 V=Speed
 A=Area
 You know that your pump expels air at speed of 8.2 ft/s
 The area of The needle of your pump is
 A=pi*r^2=pi*((4.5/1000)*(3.28))^2=0.00068442 ft^2
 Finally
 Q=V*A=(8.2 ft/s)*(0.00068442ft^2)=0.005612243 ft^3/s

Answer 2

The volume flow rate of the air being pumped into the football is 0.0056 ft³/sec.

What is the volume flow rate?

Volume flow rate is the rate at which volume leaves a nozzle. It is the product of the area of the nozzle and the rate of flow.

Given the pump expels air at a speed of 8.2 ft/sec. Also, the radius of the air nozzle is 4.5 millimetres. Therefore, the area of the air nozzle can be written as,

$\rm \text{Area of nozzel} = \pi r^2 = \pi \times (4.5^2) = 63.6172\ mm^2$

Since 1 mm 0.00328084 foot, therefore, the area of the nozzle will be,

$\rm \text{Area of nozzel} = 63.6172\ mm^2 = 63.6172 \times 0.00328 ^2 = 0.000684\ ft^2$

Now, the volume flow rate of the air being pumped into the football can be written as,

$\rm \text{Volume flow rate} = \text{(Area of nozzel)} \times air\ speed$

                           $\rm = 0.000684\ ft^2\times 8.2\ ft/sec\\\\= 0.0056\ ft^3/sec$

Hence, the volume flow rate of the air being pumped into the football is 0.0056 ft³/sec.

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