W-6
Chapter 4 Basic Equation in Integral Form for a Control Volume
4.6 Momentum Equation for Control Volume with Arbitrary Acceleration
In Section 4.5 we obtained a momentum equation for a control volume with rectilinear acceleration. The purpose of this section is to extend this for completeness to include rotation and angular acceleration of the control volume, in addition to translation and rectilinear acceleration. First, we develop an expression for Newton’s second law in an arbitrary, noninertial coordinate system. Then we use Eq. 4.25 to complete the formulation for a control volume. Newton’s second law for a system moving relative to an inertial coordinate system is given by ~ ~ 5 dP XYZ F ð4:27Þ dt
!
system
where, as in the previous section, XYZ denotes the inertial (e.g., stationary) reference frame. Since
~ XYZ P XYZ Þsystem
5
Z
M ðsystemÞ
~ XYZ V XYZ dm
and M (system) (system) is constant,
~5 F
Z
d dt
~ XYZ dm V
5
M ðsystemÞ
~ XYZ dV dm dt M ðsystemÞ
Z
or
~5 F
Z
~ a XYZ dm
ð4:35Þ
M ðsystemÞ
a XYZ to the acceleration ~ a xyz , measured relative to a The basic problem is to relate ~ noninertial coordinate system. For this purpose, consider the noninertial reference frame, xyz, shown in Fig. 4.5. ~ relative to the fixed The noninertial frame, xyz, itself is located by position vector R frame XYZ . The noninertial frame is assumed to rotate with angular velocity ~ ω. A parti particle cle is insta instantane ntaneously ously located located relat relative ive to the moving frame by position vector ~ r 5 ix 1 jy 1 kz. Relative to inertial reference frame XYZ , the position of the particle r ~. From the geometry of the figure, X ~5R ~ 1~ r . r is deno denoted ted by posit position ion vector X The velocity of the particle relative to an observer in the XYZ system is ~ ~ r r d~ r r ~ XYZ 5 d X 5 dR 1 d~ ~rf V ð4:36Þ 5 V rf 1 dt dt dt dt ^
^
^
Particle Y ω
X
R
r
y x
z X
Z
Fig. 4.5
Location of a particle in inertial ( XYZ ) and noninert noninertial ial ( xyz ) reference frames.
4.6 Momentum Equation for Control Volume with Arbitrary Acceleration ~rf is the instantaneous velocity of the control where, as in the previous section, V volume frame itself relative to the inertial XYZ reference frame. r =dt because both the magnitude, j~ r j, and the We must be careful in evaluating d~ orientation of the unit vectors, i; j; and k, are functions of time. Thus ^
d~ r dt
5
d ð xi 1 y j 1 zkÞ dt ^
^
^
^
^
dx di dy d j dz dk 1 x 1 j 1 y 1k 1z dt dt dt dt dt dt ^
5
i
^
^
^
^
^
ð4:37aÞ
The terms dx/dt , dy/dt , and dz/dt are the velocity components of the particle relative to xyz. Thus ~ xyz 5 i dx 1 j dy 1 k dz V ð4:37bÞ dt dt dt ^
^
^
You may recall from dynamics (and as we will see in Example 4.13), for a rotating coordinate system di d j dk 1 y 1z dt dt dt ^
~ ω 3~ r 5 x
^
^
ð4:37cÞ
Combining Eqs. 4.37a, 4.37b, and 4.37c, we obtain d~ r dt
5
~ xyz 1 ~ V r ω 3~
ð4:37dÞ
Substituting into Eq. 4.36 gives
~ XYZ 5 V ~rf 1 V ~ xyz 1 ~ V r ω 3~
ð4:38Þ
The acceleration of the particle relative to an observer in the inertial XYZ system is then
~ a XYZ 5
~ XYZ dV dt
5
~rf dV dt
~ xyz dV 1 dt
!
1
XYZ
d r Þ ð~ ω 3~ dt
or
~ xyz dV ~ a XYZ 5 ~ arf 1 dt
!
1
XYZ
d r Þ ð~ ω 3~ dt
ð4:39Þ
~ xyz and ~ Both V r are measured relative to xyz, so the same caution observed in developing Eq. 4.37d applies. Thus ~ xyz dV dt
!
5
XYZ
i
du dv dw 1 j 1k dt dt dt
^
^
^
~
~
~
~
~
1 ω 3 V xyz 5 a xyz 1 ω 3 V xyz
ð4:40aÞ
and d r Þ ð~ ω 3~ dt
5
5
ω d~ dt
~ ~
3 r 1 ω 3
d~ r dt
~ xyz 1 ~ ~ r 1 ~ r Þ ω 3~ ω 3 ðV ω 3~ _
or d r Þ ð~ ω 3~ dt
5
~ xyz 1 ~ ~ r 1 ~ r Þ ω 3~ ω 3 V ω 3 ð~ ω 3~ _
ð4:40bÞ
Substituting Eqs. 4.40a and 4.40b into Eq. 4.39, we obtain
~ xyz 1 ~ ~ a XYZ 5 ~ arf 1 ~ a xyz 1 2~ r Þ 1 ~ r ω 3 V ω 3 ð~ ω 3~ ω 3~ _
ð4:41Þ
W-7
W-8
Chapter 4 Basic Equation in Integral Form for a Control Volume Equation 4.41 relates the acceleration of a fluid particle as measured in the two frames (the inertial frame XYZ and the noninertial frame xyz). From your study of dynamics you will be familiar with each of the terms in the equation. They are
~ a XYZ
: Absolute rectilinear acceleration of a particle relative to fixed reference frame XYZ . ~ : Absolute rectilinear acceleration of origin of moving reference frame xyz arf relative to fixed frame XYZ . ~ a xyz : Rectilinear acceleration of a particle relative to moving reference frame xyz (this acceleration would be that “seen” by an observer on moving ~ xyz =dt Þ ). a xyz 5 dV frame xyz; ~ xyz ~ xyz : Coriolis acceleration due to motion of the particle within moving frame 2~ ω 3 V xyz. ~ r Þ: Centripetal acceleration due to rotation of moving frame xyz. ω 3 ð~ ω 3~ ~ ~ : Tangential acceleration due to angular acceleration of moving reference ω 3 r frame xyz. _
Substituting ~ a XYZ , as given by Eq. 4.41, into Eq. 4.35, we obtain
~system F
5
Z h M ðsystemÞ
i
~ xyz 1 ~ ~ arf 1 ~ a xyz 1 2~ r Þ 1 ~ r dm ω 3 V ω 3 ð~ ω 3~ ω 3~ _
or
~2 F
Z h
i Z
~ xyz 1 ~ ~ ω 3 V ω 3 ð~ ω 3~ ω 3~ arf 1 2~ r Þ 1 ~ r dm _
M ðsystemÞ
~ a xyz dm ð4:42aÞ
5
M ðsystemÞ
But
Z
~ a xyz dm
M ðsystemÞ
5
~ xyz dV M ðsystemÞ dt
Z
!
dm
5
xyz
Z
d dt
!
~ xyz dm V
M ðsystemÞ
5
xyz
~ xyz dP dt
!
system
ð4:42bÞ Combining Eqs. 4.42a and 4.42b, we obtain
~2 F
Z h
i
~ xyz 1 ~ ~ arf 1 2~ r Þ 1 ~ r dm ω 3 V ω 3 ð~ ω 3~ ω 3~ _
M ðsystemÞ
5
~ xyz dP dt
!
~ xyz dP dt
!
system
or
~S 1 F ~B 2 F
Z h
i
~ xyz 1 ~ --- 5 ~ ω 3 V ω 3 ð~ ω 3~ ω 3~ arf 1 2~ r Þ 1 ~ r ρ dV _
V ðsystemÞ
ð4:43Þ
system
Equation 4.43 is a statement of Newton’s second law for a system. The system ~ xyz =dt , represents the rate of change of momentum, P ~ xyz , of the derivative, dP system measured relative to xyz, as seen by an observer in xyz. This system derivative can be related to control volume variables through Eq. 4.25,
dN dt
5
system
@ @ t
Z
--- 1 η ρ dV
CV
Z
~ ~ xyz Á d A η ρ V
~ xyz ; and η To obtain the control volume formulation, we set N 5 P 4.25 and 4.43 may be combined to give ~S 1 F ~B 2 F
Z
ð4:25Þ
CS 5
~ xyz . Then Eqs. V
~ xyz 1 ~ --arf 1 2~ r Þ 1 ~ r ρ dV ½~ ω 3 V ω 3 ð~ ω 3~ ω 3~ _
CV 5
@ @ t
Z
CV
~ xyz ρ dV --- 1 V
Z
CS
~ ~ xyz ρ V ~ xyz Á d A V
ð4:44Þ
4.6 Momentum Equation for Control Volume with Arbitrary Acceleration
W-9
Equation 4.44 is the most general control volume form of Newton’s second law. Comparing the momentum equation for a control volume moving with arbitrary acceleration, Eq. 4.44, with that for a control volume moving with rectilinear acceleration, Eq. 4.33, we see that the only difference is the presence of three additional terms on the left side of Eq. 4.44. These terms result from the angular motion of noninertial reference frame xyz. In dynamics these terms are often referred to as “fictitious” forces that arise due to inertia effects present when we use a noninertial xyz coordinate system: the Coriolis force due to particle motion within the xyz coordinates, and centripetal and tangential forces due to the xyz coordinate system’s rotational motion, respectively. As we should expect, the general form, Eq. 4.44, reduces to the rectilinear acceleration form, Eq. 4.33, when the angular terms are zero, and to the inertial control volume form, Eq. 4.26, when all of the terms for the control a rf ; ~ volume motion (~ ω ; and ~ ω ) are zero. The precautions concerning the use of Eqs. 4.26 and 4.33 also apply to the use of Eq. 4.44. Before attempting to apply this equation, one must draw the boundaries of the control volume and label appropriate coordinate directions. For a control volume moving with arbitrary acceleration, one must label a coordinate system xyz on the control volume and an inertial reference frame XYZ . _
E
xample
4.13
VELOCITY IN FIXED AND NONINERTIAL REFERENCE FRAMES
A reference frame, xyz, moves arbitrarily with respect to a fixed frame, XYZ . A particle moves with velocity ~ xyz 5 ðdx=dt Þi 1 ðdy=dt Þ j 1 ðdz=dt Þk, relative to frame xyz. Show that the absolute velocity of the particle is given by V ^
^
^
~ XYZ 5 V ~rf 1 V ~ xyz 1 ~ V r ω 3~ Given: Find:
Fixed and noninertial frames as shown. ~ XYZ in terms of V ~ xyz ; ~ ~rf . ω; ~ V r ; and V
ω
y
~ XYZ 5 V
~ d X dt
5
~ dR dt
Particle
Y
Solution: ~5R ~ 1~ From the geometry of the sketch, X r , so
r X
x
R 1
d~ r dt
~rf 1 V
5
d~ r
z X
dt Z
Since
~ r 5 xi 1 y j 1 zk ^
^
^
we have d~ r dt
dx dy dz di d j dk i 1 j 1 k 1 x 1 y 1 z dt dt dt dt dt dt ^
5
^
^
^
^
^
or d~ r dt
5
~ xyz 1 x V
di d j dk 1 y 1z dt dt dt ^
^
^
The problem now is to evaluate di=dt ; d j=dt , and dk=dt that result from the angular motion of frame xyz. To evaluate these derivatives, we must consider the rotation of each unit vector caused by the three components of the angular velocity, ~ ω , of frame xyz. Consider the unit vector i. It will rotate in the xy plane because of ωz, as follows: ^
^
^
^
(t + Δ t )
y(t ) ^
i (t + Δ t ) ω z ^
i (t )
x(t + Δ t ) x(t )
^
i (t + Δ t ) ^
^
i (t + Δ t ) – i (t )
Δθ ^
i (t )
W-10
Chapter 4 Basic Equation in Integral Form for a Control Volume
Now from the diagram iðt 1 Δt Þ 2 iðt Þ ^
Δθ %
But for small angles cos
^
iðt 1 Δt Þ 2 iðt Þ Δt -
0, since
#
di dt
5
^
^
Δθ % Δθ ,
^
so
ðΔθ Þ2 ð1ÞΔθ j 1 ð1Þ ð2iÞ 2 ^
^
ð1Þ Δθ j 2
5
Δθ
^
i
^
2
Δθ 5 ωz Δt ,
lim
^
di dt
ð1Þsin Δθ j 1 ð1Þð1 2 cosΔθ Þð2iÞ
1 2 [(Δθ )2/2] and sin
^
In the limit as
5
Δt -0 5
due to ωz
iðt 1 Δt Þ 2 iðt Þ Δt ^
^
2 66 4
ð1Þ ωz Δt j 2
lim
5 Δt -0
^
5 jωz
^
ωz Δt i
3 77 5
^
2
Δt
^
due to ωz
Similarly, i will rotate in the xz plane because of ω y. ^
^
i (t + Δ t ) ω y
x(t + Δ t )
Enlarged sketch
x(t )
^
i (t )
^
i (t + Δ t ) ^
^
i (t + Δ t ) – i (t )
Δθ ^
i (t )
z(t + Δ t )
z(t )
Then from the diagram iðt 1 Δt Þ 2 iðt Þ ^
^
5
ð1Þ sin Δθ ð2kÞ 1 ð1Þð1 2 cos Δθ Þð2iÞ ^
^
For small angles iðt 1 Δt Þ 2 iðt Þ ^
In the limit as
Δt -
^
0, since
di dt
5
lim
5 Δt -0
dueto ω y
di dt
^
5
ð1Þ Δθ
2k 2
Δθ
^
i
^
2
Δθ 5 ω y Δt ,
^
ðΔθ Þ2 ð1Þ Δθ ð2kÞ 1 ð1Þ ð2iÞ 2 ^
iðt 1 Δt Þ 2 iðt Þ Δt ^
^
lim
5 Δt -0
"
ð1Þ ω y Δt
2k2 ^
Δt
^
5 2kω y ^
dueto ω y
Rotation in the yz plane because of ω x does not affect i. Combining terms, ^
di dt
^
5
ωz j 2 ω y k ^
^
By similar reasoning, d j dt
dk dt
^
^
5
ω x k 2 ωz i ^
^
and
5
ω y i 2 ω x j ^
^
ω y Δt i
#
^
2
4.7 The Angular-Momentum Principle (Continued) Thus di d j dk 1 y 1z dt dt dt ^
x
^
^
5
ðzω y 2 yωz Þi 1 ð xωz 2 zω x Þ j 1 ð yω x 2 xω y Þk ^
^
^
But
~ r 5 ω 3~
i j k ω x ω y ωz x y z
^
^
Combining these results, we obtain
^
5
ðzω y 2 yωz Þi 1 ð xωz 2 zω x Þ j 1 ð yω x 2 xω y Þk ^
~ XYZ 5 V ~rf 1 V ~ xyz 1 ~ V ω 3 r
^
^
~ XYZ V
ß
The Angular-Momentum Principle (Continued) 4.7 Equation for Rotating Control Volume In problems involving rotating components, such as the rotating sprinkler of Example 4.14, it is often convenient to express all fluid velocities relative to the rotating component. The most convenient control volume is a noninertial one that rotates with the component. In this section we develop a form of the angularmomentum principle for a noninertial control volume rotating about an axis fixed in space. Inertial and noninertial reference frames were related in Section 4.6. Figure 4.5 showed the notation used. For a system in an inertial frame,
~system T
5
!
~ dH dt
ð4:3aÞ system
The angular momentum of a system in general motion must be specified relative to an inertial reference frame. Using the notation of Fig. 4.5,
~system H
5
Z
~ 1~ ~ XYZ dm r Þ 3 V ðR
5
M ðsystemÞ
Z
--V ðsystemÞ
~1~ ~ XYZ ρ dV --r Þ 3 V ðR
~ 5 0 the xyz frame is restricted to rotation within XYZ , and the equation With R becomes ~system H
5
Z
~ XYZ dm ~ r 3 V
5
M ðsystemÞ
Z
~ XYZ ρ dV --~ r 3 V
---ðsystemÞ V
so that
~system T
5
d dt
Z
~ XYZ dm ~ r 3 V
M ðsystemÞ
Since the mass of a system is constant,
~system T
5
d ~ XYZ Þ dm r 3 V ð~ M ðsystemÞ dt
Z
W-11
W-12
Chapter 4 Basic Equation in Integral Form for a Control Volume or
~system T
5
Z M ðsystemÞ
!
~ XYZ d~ r ~ dV ~ r dm 3 V 1 3 XYZ dt dt
ð4:47Þ
From the analysis of Section 4.6,
~ XYZ 5 V ~rf 1 V ~rf With xyz restricted to pure rotation, V right side of Eq. 4.47 is then
5
d~ r dt
ð4:36Þ
0. The first term under the integral on the
d~ r d~ r 3 dt dt
0
5
Thus Eq. 4.47 reduces to
~system T
5
Z
~ r 3
M ðsystemÞ
From Eq. 4.41 with ~ a rf
5
~ XYZ dV dm dt
5
Z
~ r 3 ~ a XYZ dm
ð4:48Þ
M ðsystemÞ
0 (since xyz does not translate),
~ xyz 1 ~ ~ a XYZ 5 ~ a xyz 1 2~ r Þ 1 ~ r ω 3 V ω 3 ð~ ω 3~ ω 3~ _
Substituting into Eq. 4.48, we obtain
~system T
5
Z
~ r 3
M ðsystemÞ
h
i
~ xyz 1 ~ ~ a xyz 1 2~ r Þ 1 ~ r dm ω 3 V ω 3 ð~ ω 3~ ω 3~ _
or
Z Z
~system 2 T 5
M ðsystemÞ
h
_
~ r 3 ~ a xyz dm
5
M ðsystemÞ
~ xyz dV ~ r 3 dt M ðsystemÞ
Z
We can write the last term as
Z
~ r 3
M ðsystemÞ
~ xyz dV dt
!
dm
5
xyz
0Z @
d dt
~ xyz dm ~ r 3 V
M ðsystemÞ
~ s 1 r 3 F 5~
Z
1 A
ð4:49Þ
dm
xyz
1 A
The torque on the system is given by
~system T
i
~ xyz 1 ~ ~ r 3 2~ r Þ 1 ~ r dm ω 3 V ω 3 ð~ ω 3~ ω 3~
xyz 5
~ xyz dH dt
~shaft ~ ~ dm 1 T r 3 g
!
ð4:50Þ
system
ð4:3cÞ
M ðsystemÞ
The relation between the system and control volume formulations is
dN dt
system
5
@ @ t
Z
--- 1 η ρ d V
CV
Z
~ ~ xyz Á d A η ρ V
CS
where N system
5
Z
M ðsystemÞ
η dm
ð4:25Þ
4.7 The Angular-Momentum Principle (Continued) ~ xyz Þ Setting N equal to H system and η ~ xyz dH dt
!
5
system
@ @ t
Z
5
W-13
~ xyz yields ~ r 3 V
~ xyz ρ dV --- 1 ~ r 3 V
CV
Z
~ ~ xyz ρ V ~ xyz Á d A ~ r 3 V
ð4:51Þ
CS
Combining Eqs. 4.49, 4.50, 4.51, and 4.3c, we obtain
~ s 1 ~ r 3 F
Z Z
M ðsystemÞ
~shaft ~ ~dm 1 T r 3 g
h
MðsystemÞ 5
i
~ xyz 1 ~ ~ ω 3 V ω 3 ð~ ω 3~ ω 3~ r 3 2~ r Þ 1 ~ r dm
2
@ @ t
Z
_
~ xyz ρ dV --- 1 ~ r 3 V
CV
Z
~ ~ xyz ρ V ~ xyz Á d A ~ r 3 V
CS
Since the system and control volume coincided at t 0,
~ s 1 ~ r 3 F
Z Z h Z CV
~shaft --- 1 T ~ ~ ρ dV r 3 g
2
CV
i
~ xyz 1 ~ --~ r 3 2~ r Þ 1 ~ r ρ dV ω 3 V ω 3 ð~ ω 3~ ω 3~
5
@ @ t
_
~ xyz ρ dV --- 1 ~ r 3 V
CV
Z
ð4:52Þ
~ ~ xyz ρ V ~ xyz Á d A ~ r 3 V
CS
Equation 4.52 is the formulation of the angular-momentum principle for a (noninertial) control volume rotating about an axis fixed in space. All fluid velocities in Eq. 4.52 are evaluated relative to the control volume. Comparing Eq. 4.52 with Eq. 4.46 (for inertial XYZ cooordinates) we see that the noninertial (rotating) xyz coordinates have an extra “moment” term on the left side that includes three components. As we discussed following Eq. 4.44, these components arise because of “fictitious” forces: the Coriolis force because of fluid particle motion within the xyz coordinates, and centripetal and tangential forces because of the xyz coordinates’ rotational motion, respectively. Equation 4.52 reduces to Eq. 4.46 when the control volume is not in motion (when ~ ω and ~ ω are zero). Even though we have the extra term to evaluate, Eq. 4.52 is sometimes simpler to use than Eq. 4.44 because a problem that is unsteady in XYZ coordinates becomes steady state in xyz coordinates, as we will see in Example 4.15. _
E
xample
4.15
LAWN SPRINKLER: ANALYSIS USING ROTATING CONTROL VOLUME
A small lawn sprinkler is shown in the sketch at right. At an inlet gage pressure of 20 kPa, the total volume flow rate of water through the sprinkler is 7.5 liters per minute and it rotates at 30 rpm. The diameter of each jet is 4 mm. Calculate the jet speed relative to each sprinkler nozzle. Evaluate the friction torque at the sprinkler pivot. Given: Find:
Small lawn sprinkler as shown. (a) Jet speed relative to each nozzle. (b) Friction torque at pivot.
V rel α
V rel ω
= 30°
Q = 7.5 L/min ω = 30 rpm
psupply = 20 kPa (gage) R = 150 mm
Chapter 4 Basic Equation in Integral Form for a Control Volume
W-14
Solution: Apply continuity and angular momentum equations using rotating A ˆ control volume enclosing sprinkler arms.
0(1)
Governing equations:
t
r F S
dV
CV
V xyz
· dA 0
CS
r g dV T shaft
CV
r ⎡2
⎣
CV
0(3)
V xyz
(
r )
r ⎤ dV
⎦
ð4:52Þ
0(1)
t
r V xyz V xyz · dA
r V xyz dV
CV
CV
Q = 7.5 L/min = 30 rpm
ω
z
r
V rel
CV
V rel
ω
Assumptions:
α
(1) Steady flow relative to the rotating CV. (2) Uniform flow at each section. (3) ω 5 constant.
V rel
5
Q 2 A jet
5
1 2
5
4:97 m=s
5
T f
(Control volume rotates with sprinkler arm)
From continuity V rel
= 30°
psupply = 20 kPa (gage)
R = 150 mm
Q 4 2 2 πD jet
L 3 7 :5 min
3
4
1
π ð4Þ2 mm2
3
m3 1000 L
mm2 3 10 m2 6
3
min 60 s V rel
ß
Consider terms in the angular-momentum equation separately. As in Example 4.14, the only external torque acting on the CV is friction in the pivot. It opposes the motion, so
~shaft T
ð 1Þ
5 2T f k
^
The second integral on the left of Eq. 4.52 is evaluated for flow within the CV. Let the velocity and area within the sprinkler tubes be V CV and ACV, respectively. Then, for one side, the first term (a Coriolis effect) is
Z h ~ r 3
i
~ xyz ρ dV --- 5 2~ ω 3 V
CV
R
Z h Z Z
5
ˆ
r er 3 2ωV CV eθ ρ ACV dr ˆ
0
5
^
ˆ
0 R
i
r er 3 2ωk 3 V CV er ρ ACV dr ˆ
R
0
2ωV CV ρ ACV r dr k
^
5
V A CV k ωR2 ρ VC
fone sideg
^
(The flow in the bent portion of the tube has no r component of velocity, so it does not contribute to the integral.) From continuity, Q 5 2 V CV ACV, so for both sides the integral becomes
Z h ~ r 3
i Z h Z h
~ xyz ρ dV --- 5 ωR2 ρ Qk 2~ ω 3 V
CV
ð 2Þ
^
The second term in the integral (a moment generated by centripetal acceleration) is evaluated as
Z
--- 5 ~ r 3 ½~ r Þ ρ dV ω 3 ð~ ω 3~
CV
^
^
CV
5
i
--r er 3 ωk 3 ðωk 3 r e r Þ ρ dV ^
--- 5 r er 3 ωk 3 ωr e θ ρ dV ˆ
CV
i
ˆ
^
ˆ
Z
CV
--- 5 0 r er 3 ω2 r ð2er Þ ρ dV ^
^
so it contributes no torque. (The force generated by centripetal acceleration is radial, so it generates no moment.)
4.7 The Angular-Momentum Principle (Continued)
W-15
The integral on the right side of Eq. 4.52 is evaluated for flow crossing the control surface. For the right arm of the sprinkler,
Z
CS
h
i
~ 5 R er 3 V rel cos αð2 eθ Þ 1 sin αk f1ρ Vr el A jet g ~ xyz ρ V ~ xyz Á d A ~ r 3 V ˆ
5
^
ˆ
Q RV rel cos αð2kÞ 1 sin αð2 eθ Þ ρ 2
h
^
ˆ
i
The velocity and radius vectors for flow in the left arm must be described in terms of the same unit vectors used for the right arm. In the left sprinkler arm, the θ component has the same magnitude but opposite sign, so it cancels. For the complete CV,
Z
~ 5 2RV rel cos α ρ Q k ~ xyz ρ V ~ xyz Á d A ~ r 3 V
ð3Þ
^
CV
Combining terms (1), (2), and (3), we obtain 2
2T f k 2 ωR ^
ρ Q k
^
5 2RV rel
cos α ρ Q k
^
or T f
5
RðV rel cos α
2
ωRÞρ Q
From the data given, rev min Substituting gives ωR
5
T f
30
5
3
150 mm
150 mm 3 2π
5
3
min 60 s
3
m 1000 mm
5
0:471m=s
~
0 @
4:97
3
T f
rad rev
T hi s p ro b le m i llu st r ang ular mo me nt u at es use o f t he m p ri n no ni ne rt i c ip l e f or a al ( ro t a t i ng ) c on N ot e t ha t ro l v ol um t i n t hi s a e p pr o i ne rt i al c a c h, unli ke . on t ro l t v h ol um e 4.14, t he fl ui d p ar e o f E x amp le r and v e lo ci t y v t ic l e p os i t io v ec t o r r V ar e n d e pe n d e nt . A ec t o n ot t im s w e e s r es ult s a g re e usi n ho uld e x p ec t , t he g e it he r a no ni ne rt i al c on t ro l v ol um n i ne rt i a l o r e .
m s
3
m3 1000 L
0:0718 N Á m
cos 30 2 0:471
3
min 60 s
3
N Á s2 kg Á m
1 A
m kg 999 3 s m 3
~
3
7 :5
m 1000 mm T f
ß
L min
