Answered

How fast is the water traveling (in feet per second) in a 14-inch pipeline if the flow rate is 2400 gpm?



Answer :

Answer:

5.00 ft/s

Explanation:

The velocity (v) is equal to the flow rate (Q) divided by the cross-sectional area (A). For a round pipe, the cross sectional area is equal to one-fourth pi times the square of the diameter (d). We will need the following conversion factors:

7.48 gal = 1 ft³

12 in = 1 ft

1 min = 60 s

First, find the cross-sectional area.

[tex]\Large \text {$ A= $} \huge \text {$ \frac{1}{4} $} \Large \text {$ \pi d^2$}\\\Large \text {$ A= $} \huge \text {$ \frac{1}{4} $} \Large \text {$ \pi (14\ in\times$}\huge \text {$ \frac{1\ ft}{12\ in} $} \Large \text {$ )^2 $}\\\Large \text {$ A=1.069\ ft^2 $}[/tex]

Next, convert the flow rate from gpm to ft³/s.

[tex]\Large \text {$ Q=2400\ $} \huge \text {$ \frac{gal}{min} $} \Large \text {$ \ \times \ $}\huge \text {$ \frac{1\ ft^3}{7.48\ gal} $} \Large \text {$ \ \times \ $}\huge \text {$ \frac{1\ min}{60\ s} $}\\\Large \text {$ Q=5.348\ ft^3/s $}[/tex]

Finally, find the speed.

[tex]\Large \text {$ v= $} \huge \text {$ \frac{Q}{A} $}\\\Large \text {$ v= $} \huge \text {$ \frac{5.348\ ft^3/s}{1.069\ ft^2} $}\\\Large \text {$ v=5.00\ ft/s $}[/tex]