Question

1.Water flows through a horizontal pipe at a rate of 39 gpm (1 ft3 = 7.48 gal). What is the velocity in the larger opening of the pipe that reduces from 5.4 to 2.4 inches in diameter?

Larger Pipe Velocity = ft/s

2. What must be the diameter of a hose if it is to deliver 7 liters of oil in 1.2 minutes with an exit velocity of 2 m/s?

Pipe Diameter = milimeters

Answer 1

Given

1.

the rate of flow of water is 39 gpm

1 ft^3 = 7.48 gal ===> 39 gpm = 39*1/(7.48*60 ) ft^3/s

flow rate is = 0.086898396 ft^3/s

we know that the flow rate is AV = constant

where A is area of cross section and V is velocity  

and 1 inch = 0.08333 ft ===> 5.4 inch = 0.45 ft

0.086898396 = pi*(0.45/2)^2*V

solvinf for V , V = 0.5464 ft/s

2. the flow rate is 7 liters of oil in 1.2 minutes

velocity is V = 2 m/s

A*V = constant

we know that 1 L = 0.001 m^3

so the rate of flow is 7 l in 1.2 min = 7*0.001/(1.2*60) m^3/s = 9.72222*10^-5 m^3/s

9.72222*10^-5 = pi(D/2)^2*2

solving for D diameter of the hose

D = 0.00786725 m

D = 7.86725 mm