Page 216 - Schaum's Outline of Theory and Problems of Applied Physics

CHAPTER 17

Fluids in Motion

FLUID FLOW

In the streamline flow of a fluid, the direction of motion of the individual particles is the same as that of the fluid
as a whole. Each particle of the fluid that passes any point follows the same path as those particles which passed
that point before. Turbulent flow, however, is characterized by the presence of irregular whirls and eddies; it
occurs at high velocities and when the fluid’s path changes direction sharply, for instance near an obstruction.
The rate (in volume/time) at which a fluid whose velocity is v flows through a pipe or channel of cross-
sectional area A is

R = v A
Rate of flow = (velocity)(cross-sectional area)

It is common for R to be expressed in such units as gallons per minute and liters per second instead of the proper
3
3
3
3
−3
units of cubic feet per second and cubic meters per second (1 U.S. gal = 0.134 ft and 1 L = 10 m = 10 cm ).
When a fluid is incompressible, which is approximately true for most liquids, its rate of flow R is constant
even though the size of the pipe or channel varies. Thus if a liquid’s velocity is v 1 when the cross-sectional area
is A 1 and v 2 when it is A 2 , then
v 1 A 1 = v 2 A 2
SOLVED PROBLEM 17.1
A garden hose has an inside diameter of 12 mm, and water flows through it at 2.5 m/s. (a) What nozzle
diameter is needed for the water to emerge at 10 m/s? (b) At what rate does water leave the nozzle?
(a) The cross-sectional areas of hose and nozzle are in the same ratio as the squares of their diameters, since
2
2
A = πr = πd /4. From v 1 A 1 = v 2 A 2 we obtain
2
v 1 d = v 2 d 2 2
1

v 1 2.5 m/s
d 2 = d 1 = (12 mm) = 6mm
v 2 10 m/s
(b) The rate of flow is
2
2
3
R = v 1 A 1 = v 1 πr = (2.5 m/s)(π)(0.006 m) = 2.83 × 10 −4 m /s
1
3
3
Because 1 m = 10 liters (L), R = 0.283 L/s. The same result, of course, would be obtained from R = v 2 A 2 .
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