MECHANICAL ENGINEERING: Fluid Mechanics and Dynamics - Conservation of Mass, mass flow rate, volume flow rate

Volume = Area * height

V = A * h


if d = h = distance travelled by the cross-sectional area

V = A * d

and the Volume flow rate or the volume per unit time

Vdot = (A * d)/t


but d/t = velocity = v

Vdot = A * v


CONSERVATION OF MASS

The amount of mass that flows past any cross-section is constant

m1 = m2

mass in = mass out


but density D = m/V

m = D * V

substituting in

m1 = m2

D1 * V1 = D2 * V2


if V1 and V2 are volume flow rates and Vdot = A * v

D1 * (A1 * v1) = D2 * (A2 * v2)


the same fluid is flowing so

D1 = D2


and it simplifies to


A1 * v1 = A2 * v2


where:

A1 = cross-sectional area at point1

A2 = cross-sectional area at point2

v1 = velocity of fluid at point1

v2 = velocity of fluid at point2


1. Problem:

Water enters a pipe with diameter of 1 in at 7 fps. Determine the exit velocity of the water if the pipe diameter is 2 in.

find:

v2 = velocity of water at exit


given:

d1 = 1 in

d2 = 2 in

v1 = 7 fps


solution:

convert v1 to in/s

v1 = 7 ft/s * 12 in/ft

v1 = 84 in/s


solving for A1 and A2

A1 = pi/4 * (d1)^2

A1 = 0.785 * (1)^2

A1 = 0.785 * 1

A1 = 0.785 sq. in


A2 = pi/4 * (d2)^2

A2 = 0.785 * (2)^2

A2 = 0.785 * 4

A2 = 3.14 sq. in


by conservation of mass

A1 * v1 = A2 * v2

0.785 * 84 = 3.14 * v2

v2 = 21 in/s


convert v2 to fps

v2 = 21 in/s * 1 ft/12 in

v2 = 21/12

v2 = 1.75 fps


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ALTERNATE SOLUTION:


A1 * v1 = A2 * v2


because (pi/4) and (sq. in) will cancel, there's no need for unit conversion

(d1)^2 * v1 = (d2)^2 * v2

(1)^2 * 7 = (2)^2 * v2

v2 = 7/4

v2 = 1.75 fps