2 Review of Fundamentals
2.5 Steady Flow Momentum Equation
Newton's second law of motion for a control volume o~ is
~ - ' ~ F ~ = 1 ( dM~ )
g-~ T + n o u t - Win (2.20) where M~ = f~ V dm is the momentum of the mass within the control volume o- and Mout = lout Vdrh is the momentum flux leaving the control volume. In words, Eq. (2.20) says that the net force acting on a fixed control volume o-is equal to the time rate of increase of momentum within o-plus the net flux of momentum from o-.
This very important momentum equation is in fact a vector equation, which implies that it must be applied in a specified direction to solve for an unknown quantity.
Applying control volume equations to a steady flow problem gives useful results with only a knowledge of conditions at the control surface. Nothing needs to be known about the state of the fluid interior to the control volume.
The following examples illustrate the use of the steady one-dimensional flow condition and the momentum equation. We suggest that the procedure of sketch- ing a control surface (and showing the applicable fluxes through the surface and the applicable forces acting on the surface) be followed whenever a control volume equation is used. This situation is similar to the use of free body diagrams for the analysis of forces on solid bodies. We illustrate this procedure in the solutions that follow.
Example 2.2
Water (p = 1000 k g / m 3) is flowing at a steady rate through a convergent duct as illustrated in Fig. 2.6a. For the data given in the figure, find the force of the fluid F~o acting on the convergent duct D between stations 1 and 2.
Solution: We first select the control volume o" such that the force of interest is acting at the control surface. Because we want the force interaction between D and the flowing water, we choose a control surface coincident with the inner wall surface of D bounded by the permeable surfaces 1 and 2, as illustrated in Fig. 2.6a. By applying the steady one-dimensional continuity equation [Eq.
(2.11)], as depicted in Fig. 2.6b, we find V1 as follows:
plAI VI = P2A2V2 A2 1/2 V1 =All
= 3 m / s
(Pl = P 2 )
REVIEW OF FUNDAMENTALS 77
v~ II , ~ v2
(1) (2)
P1 = 137,900 Pa P2 = 101,325 Pa A1 = 0 . 2 m 2 A2 = 0.1 m 2 Pl = P2 = 1000 kg/m 3 V 2 = 6 m/s
O
PlA1VI I , I -- I P2A2V2 .I . . . .2 Mass flux in = mass flux out
P1 A1
a) Given data 0
, I I t
I _ _l P2 A2
b) Continuity equation
VI(PlA1V1)' ~ -- ~ I , "~1 V2(y 2A2V2)
x forces on c Net x momentum from G
c) Momentum equation
Fig. 2.6 Flow through a convergent duct.
W i t h V 1 d e t e r m i n e d , w e can a p p l y the m o m e n t u m Eq. (2.20) to o- a n d find the f o r c e o f t h e d u c t walls on or, d e n o t e d b y FDo- (Fo-D = --Foo.). B y s y m m e t r y , Foo- is a h o r i z o n t a l force, and so the h o r i z o n t a l x c o m p o n e n t s o f f o r c e s and m o m e n t u m fluxes w i l l be c o n s i d e r e d . T h e x f o r c e s acting on o- are d e p i c t e d in Fig. 2.6c a l o n g w i t h the x m o m e n t u m fluxes t h r o u g h o-.
F r o m Fig. 2.6c, w e h a v e m o m e n t u m :
P1AI - FDo. -- P2A2 = 1 [(p2A2V2)V 2 _ (PlA 1Vl)Vl]
A n d b y Fig. 2.6b, c o n t i n u i t y :
plA1V1 = p2A2V2 =
C o m b i n i n g t h e c o n t i n u i t y and m o m e n t u m e q u a t i o n s , w e o b t a i n
i n
P1A1 - Foo- -- P2A2 = ~ ( V 2 - V1) gc
78
o r
ELEMENTS OF PROPULSION
rh V Foa = P1A1 - P2A2 - - - ( 2 - - V l )
gc
With rh = plA1V! -- 1000 k g / m 3 x 0.2 m 2 x 3 m / s = 600 k g / s , we have FD~ = 137,000 N / m 2 x 0.2 m 2 - 101,325 N / m z x 0.1 m 2
- 600 k g / s x (6 - 3) m / s
o r
and
Fo~ = 2 7 , 5 8 0 N - 10, 132 N - 1800 N
---- 15,648 N acts to left in assumed position Finally, the force of the water on the duct is
F~rD = --FDcr = --15,648 N
Example 2.3
acts to fight
Figure 2.7 shows the steady flow conditions at sections 1 and 2 about an airfoil mounted in a wind tunnel where the frictional effects at the wall are negligible.
Determine the section drag coefficient Cd of this airfoil.
Solution: Since the fluid is incompressible and the flow is steady, the continuity equation may be used to find the u n k n o w n velocity VB as follows:
o r
but
thus
( p A g ) l = ( p A V ) 2 that is, rhL = rh2
(2 1 )
p l A I V a = P 2 ~VB-q-~Vc A2
P l = P2 and
2 1V
Va ~- "~ VB "~ 5 C
A1 = A 2
_ 3 V 1V
VB - - "~ A -- ~ C = 31.5 m / s
REVIEW OF FUNDAMENTALS 79
/ / / / / l -V-£ Y 3O m,
0 . 3 m .I
I I
"1
/ / / / /
(1) P1 = 74,730 Pa Tunnel area = 0.1 m 2 Chord = 0.15 m
G
3 0 m / s I VB=7
I I
/ ~ . ~ Wake I
p = 0.618 kg/m3 = constant ~ ~ I
I
/ / / / / / , ( / ? ,
I (2) P2 - 74,700 Pa] / / " 0.1 m
O . l m VC = 27 m/s
VB= ? t_
O . l mvB VA
Fig. 2.7 W i n d t u n n e l d r a g d e t e r m i n a t i o n f o r a n airfoil section.
The momentum equation may now be used to find the drag on the airfoil. This drag force will include both the skin friction and pressure drag. We sketch the control volume o- with the terms of the momentum equation as shown in Fig. 2.8.
Taking forces to right as positive, we have from the sketch
Z
F~ = PIA1 - P2A2 + FDo-a//1 = ( P l A 1 V A ) V A = plalV2a
2 2
M 2 =
p2(-~A2)V~ +p2(1Az)V~
=p2A2( 2V2 ' 1B-t-~V2"~c)
P1 AI
~tl
I P2 A2
I " ~t2
[
Momentum equation sketch Z F o = ~/2 - ~'/1
F i g . 2.8 S k e t c h for m o m e n t u m e q u a t i o n f o r airfoil section.
80 ELEMENTS OF PROPULSION F o r p : Pl = P2 and A : A1 = A2,
or 74 00)01+0618 01i 0
= 3.0 N - 0.278 N
- F o a = 2.722 N .'. F o a acts to left Fan = drag force for section
: --Foa and FaD acts to left FaD 2.722 N
/ ~ . . . 8.174 N / m b 0 . 3 3 3 m
' p V 2 where
Cd ---- F b and q = 2gc qc
F b 8.174
Cd = [(pVZ)/(2gc)]C -- (0.618 × 302/2)0.15
V~=Va
= 0.196
Example 2.4
Figure 2.9 shows a test stand for determining the thrust o f a liquid-fuel rocket.
The propellants enter at section 1 at a mass flow rate o f 15 k g / s , a velocity o f 30 m / s , and a pressure of 0.7 MPa. The inlet pipe for the propellants is very flexible, and the force it exerts on the rocket is negligible. At the nozzle exit, section 2, the area is 0.064 m 2, and the pressure is 110 kPa. The force read by the scales is 2700 N, atmospheric pressure is 82.7 kPa, and the flow is steady.
~0~ ~
//;,
, 0 . 8 m ,
!
~///////~
Propellant n] = 15 kg/s
- , ~ Combustion =
~ / chamber Nozzle~
0 / ( ) ()
U////////////////////
Fig. 2.9 Liquid-fuel rocket test setup.
REVIEW OF FUNDAMENTALS 81 Determine the exhaust velocity at section 2, assuming one-dimensional flow exists. Mechanical frictional effects m a y be neglected.
Solution: First, determine the force on the lever by the rocket to develop a 2700 N scale reading. This m a y be done by summing moments about the fulcrum point 0 (see Fig. 2.10a). Sum the horizontal forces on the rocket engine as shown in Fig. 2.10b. W e note that the unbalanced pressure force on the exterior o f the rocket engine is PaA2, and the interior forces (pressure and fric- tion) are contained within the force Fc~. Next, draw an internal volume ~r around
a)
2700 N
° 0 . 2 m
Y_,Mo = 2700 x 0.8-FR× 0.2 = 0 ... FR = 2700 x 0.8 = 10,800 N
0.2
b) P~2
FR
1
_J
x
Fc~
c)
I I f °
y \ S
Fcc
¢ "%..
• X
PzA2
£'outx = k2v2
F i g . 2 . 1 0 M o m e n t u m e q u a t i o n s k e t c h ~ F ~ = (~/2 -
h41)/gc.
82 ELEMENTS OF PROPULSION
the fluid within the rocket engine as shown, and indicate the horizontal forces and momentum flux (see Fig. 2.10c).
Summing the forces on the rocket engine as shown in Fig. 2.10b, we obtain FR + PaA2 = Fc~r
Applying the momentum equation to the control volume o-shown in Fig. 2.10c, we obtain
m2V2
F o c - P 2 A 2 = gc
Combining these two equations to remove Foc gives k2V2 FR -- (/)2 - - Pa)A2 =
gc
which is the same as Eq. (1.52). Because the flow is steady, the continuity equation yields rhl = th2 = 15 kg/s. Therefore,
v2 = F R - ( P 2 -- Pa)A2 tn2/gc
10,800 - [(110 - 82.7) x 103 N/m2](0.064 m 2) 15kg/s
= 603.5 m / s