fluid_mechanics

5.2 Newton’s Second Law—The Linear Momentum and Moment-of-Momentum Equations 207
F l u i d s i n t h e N e w s
Motorized surfboard When Bob Montgomery, a former professional
surfer, started to design his motorized surfboard
(called a jet board), he discovered that there were many engineering
challenges to the design. The idea is to provide surfing
to anyone, no matter where they live, near or far from the ocean.
The rider stands on the device like a surfboard and steers it like
a surfboard by shifting his/her body weight. A new, sleek, compact
45-horsepower engine and pump was designed to fit within
the surfboard hull. Thrust is produced in response to the change
in linear momentum of the water stream as it enters through the
inlet passage and exits through an appropriately designed nozzle.
Some of the fluid dynamic problems associated with designing
the craft included one-way valves so that water does not
get into the engine (at both the intake or exhaust ports), buoyancy,
hydrodynamic lift, drag, thrust, and hull stability. (See
Problem 5.68.)
To further demonstrate the use of the linear momentum equation 1Eq. 5.222, we consider
another one-dimensional flow example before moving on to other facets of this important
equation.
E XAMPLE 5.12
Linear Momentum—Pressure and Change in Flow Direction
FIND Calculate the horizontal 1x and y2 components of the anchoring
2 stant at a value of through the bend. The magnitude of the
GIVEN Water flows through a horizontal, 180° pipe bend as
on the fluid and the pipe bend are resolved and combined into the
two resultant components, F Ax and F Ay . These two forces act on vrV nˆ dA F Ay p 1 A 1 p 2 A 2
cs
(2)
illustrated in Fig. E5.12a. The flow cross-sectional area is con-
force required to hold the bend in place.
2 0.1 ft
flow velocity everywhere in the bend is axial and 50 fts. The
absolute pressures at the entrance and exit of the bend are 30 psia
and 24 psia, respectively.
SOLUTION
Since we want to evaluate components of the anchoring force to At sections 112 and 122, the flow is in the y direction and therefore
hold the pipe bend in place, an appropriate control volume 1see u 0 at both cross sections. There is no x direction momentum
dashed line in Fig. E5.12a2 contains the bend and the water in the flow into or out of the control volume and we conclude from Eq. 1
bend at an instant. The horizontal forces acting on the contents of that
this control volume are identified in Fig. E5.12b. Note that the
weight of the water is vertical 1in the negative z direction2 and
F Ax 0
(Ans)
does not contribute to the x and y components of the anchoring For the y direction, we get from Eq. 5.22
force. All of the horizontal normal and tangential forces exerted
the control volume contents, and thus for the x direction, Eq. 5.22
leads to
For one-dimensional flow, the surface integral in Eq. 2 is easy to
evaluate and Eq. 2 becomes
urV nˆ dA F Ax
cs
(1) 1v 1 21m # 12 1v 2 21m # 22 F Ay p 1 A 1 p 2 A 2 (3)
V =
50 ft/s
V = 50 ft/s
x
Section (1) A = 0.1 ft 2
z
Section (2)
y
(a)
F I G U R E E5.12
Control
volume
180° pipe bend
x
z
w
u v y
p 1 A 1
v 1
Pipe bend
and water
F Az
F Ay
p 2 A 2
v 2
Control volume
(b)
F Ax