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Chapter 6 Momentum Analysis of Flow Systems

Solutions Manual for

Essentials of Fluid Mechanics:

Fundamentals and Applications

by Cimbala & Çengel

CHAPTER 6MOMENTUM ANA!S"S O# #O$S!STEMS

PROPRIETARY AND CONFIDENTIAL

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*-+




Ne%tons a%s an' Conser(ation of Momentum

6"C Solution e are to epress ewton/s three laws.

Analysis Newton’s first law states that “a body at rest remains at rest, and a body in motion remains in motion at

the same velocity in a straight path hen the net force acting on it is zero.” Therefore, a !oy tens to preser0e its state

or inertia. Newton’s second law states that “the acceleration of a body is proportional to the net force acting on it and is

inversely proportional to its mass.” Newton’s third law states “when a body exerts a force on a second body, the second

body exerts an equal and opposite force on the first .”

Discussion 1s we shall see in later chapters, the ifferential e2uation of flui motion is !ase on ewton/s secon law.

6"-C Solution e are to iscuss if momentum is a 0ector, an its irection.

Analysis 3ince momentum ( V m ) is the prouct of a 0ector (0elocity) an a scalar (mass), momentum must be a

vector that points in the same direction as the velocity vector.

Discussion In the "eneral case, we must sol0e three components of the linear momentum e2uation, since it is a 0ector e2uation.

6"C

Solution e are to iscuss the conser0ation of momentum principle.

Analysis The conservation of momentum principle is epresse as “the momentum of a system remains constant

hen the net force acting on it is zero, and thus the momentum of such systems is conserved ”. The momentum of a

!oy remains constant if the net force actin" on it is 4ero.

Discussion Momentum is not conser0e in "eneral, !ecause when we apply a force, the momentum chan"es.

6"/C

Solution e are to iscuss ewton/s secon law for rotatin" !oies.

Analysis ewton/s secon law of motion, also calle the angular momentum equation, is epresse as “the rate of

change of the angular momentum of a body is e0ual to the net tor0ue acting it .” 5or a non-ri"i !oy with 4ero net

tor2ue, the an"ular momentum remains constant, !ut the angular velocity changes in accordance ith I

1 constant

where I is the moment of inertia of the !oy.

Discussion 1n"ular momentum is analo"ous to linear momentum in this way$ inear momentum oes not chan"e

unless a force acts on it. 1n"ular momentum oes not chan"e unless a tor2ue acts on it.

6"2C Solution e are to compare the an"ular momentum of two rotatin" !oies

Analysis No. The to bodies do not necessarily have the same angular momentum . Two ri"i !oies ha0in" thesame mass an an"ular spee may ha0e ifferent an"ular momentums unless they also ha0e the same moment of inertia I .

Discussion The reason why flywheels ha0e most of their mass at the outermost raius, is to maimi4e the an"ular

momentum.


*-&




inear Momentum E)uation

6"6C

Solution e are to iscuss the importance of the 6TT, an its relationship to the linear momentum e2uation.

Analysis The relationship beteen the time rates of change of an extensive property for a system and for acontrol volume is epresse !y the Reynolds transport theorem (6TT), which pro0ies the lin7 !etween the system an

control 0olume concepts. The linear momentum e2uation is o!taine !y setting V b = and thus V m B = in the

%eynolds transport theorem.

Discussion ewton/s secon law applies irectly to a system of fie mass, !ut we use the 6TT to transform from the

system formulation to the control 0olume formulation.

6"3C

Solution e are to escri!e an iscuss !oy forces an surface forces.

Analysis The forces actin" on the control 0olume consist of body forces that act throughout the entire body of the

control volume (such as "ra0ity, electric, an ma"netic forces) an surface forces that act on the control surface (such asthe pressure forces an reaction forces at points of contact). The net force actin" on a control 0olume is the sum of all body

and surface forces. 5lui eight is a body force, an pressure is a surface force (actin" per unit area).

Discussion In a "eneral flui flow, the flow is influence !y !oth !oy an surface forces.

6"4C Solution e are to iscuss surface forces in a control 0olume analysis.

Analysis 1ll surface forces arise as the control volume is isolated from its surroundings for analysis, and the

effect of any detached ob5ect is accounted for by a force at that location. e can minimi4e the num!er of surface forces

epose !y choosing the control volume isely7 such that the forces that e are not interested in remain internal,and thus they do not complicate the analysis. 1 well-chosen control 0olume eposes only the forces that are to !eetermine (such as reaction forces) an a minimum num!er of other forces.

Discussion There are many choices of control 0olume for a "i0en pro!lem. 1lthou"h there are not really “wron"” an

“ri"ht” choices of control 0olume, there certainly are “wise” an “unwise” choices of control 0olume.

6"8C

Solution e are to iscuss the momentum flu correction factor, an its si"nificance.

Analysis The momentum-flux correction factor β ena!les us to epress the momentum flu in terms of the mass flow

rate an mean flow 0elocity as avg c

V mdnV V c

β ρ =⋅

∫ )(

. The 0alue of β is unity for uniform flow, such as a 8etflow, nearly unity for fully e0elope tur!ulent pipe flow (!etween +.'+ an +.'9), !ut a!out +.: for fully e0elopelaminar pipe flow. 9o it is significant and should be considered in laminar flo; it is often i"nore in tur!ulent flow.

Discussion <0en thou"h β is nearly unity for many tur!ulent flows, it is wise not to i"nore it.


*-:




6":C

Solution e are to iscuss the momentum e2uation for steay one-= flow with no eternal forces.

Analysis The momentum e2uation for steay one-imensional flow for the case of no eternal forces is

out in

' ! mV mV β β = = −∑ ∑ ∑& &

where the left han sie is the net force acting on the control volume (which is 4ero here), the first term on the ri"ht han

sie is the incoming momentum flux, an the secon term on the ri"ht is the outgoing momentum flux !y mass.

Discussion This is a special simplifie case of the more "eneral momentum e2uation, since there are no forces. In this

case we can say that momentum is conser0e.

6"C Solution e are to eplain why we can usually wor7 with "a"e pressure rather than a!solute pressure.

Analysis In the application of the momentum e2uation, we can isre"ar the atmospheric pressure an wor7 with

"a"e pressures only since the atmospheric pressure acts in all directions, an its effect cancels out in e0ery irection.

Discussion In some applications, it is !etter to use a!solute pressure e0erywhere, !ut for most practical en"ineerin"

pro!lems, it is simpler to use "a"e pressure e0erywhere, with no loss of accuracy.

6"-C

Solution e are to compare the reaction force on two fire hoses.

Analysis The fireman ho holds the hose backards so that the ater makes a ;"turn before being discharged

ill experience a greater reaction force. This is !ecause of the 0ector nature of the momentum e2uation. 3pecifically, the

inflow an outflow terms en up with the same si"n (they a to"ether) for the case with the >-turn, whereas they ha0eopposite si"ns (one partially cancels out the other) for the case without the >-turn.

Discussion =irection is not an issue with the conser0ation of mass or ener"y e2uations, since they are scalar e2uations.

6"C Solution e are to iscuss if the upper limit of a roc7et/s 0elocity is limite to V , its ischar"e 0elocity.

Analysis No, V is not the upper limit to the rocket<s ultimate velocity. ithout friction the roc7et 0elocity will

continue to increase (i.e., it will continue to accelerate) as more "as is epelle out the no44le.

Discussion This is a simple application of ewton/s secon law. 1s lon" as there is a force actin" on the roc7et, it willcontinue to accelerate, re"arless of how that force is "enerate.

6"/C

Solution e are to escri!e how a helicopter can ho0er.

Analysis 1 helicopter ho0ers !ecause the strong dondraft of air, caused by the overhead propeller blades,

manifests a momentum in the air stream. This momentum must !e countere !y the helicopter lift force.

Discussion In essence, the helicopter stays aloft !y pushin" own on the air with a net force e2ual to its wei"ht.


*-9




6"2C

Solution e are to iscuss the power re2uire for a helicopter to ho0er at 0arious altitues.

Analysis 3ince the air ensity ecreases, it re0uires more energy for a helicopter to hover at higher altitudes,

!ecause more air must !e force into the ownraft !y the helicopter !laes to pro0ie the same lift force. Therefore, it

ta7es more power for a helicopter to ho0er on the top of a hi"h mountain than it oes at sea le0el.

Discussion This is consistent with the limitin" case ? if there were no air, the helicopter woul not !e a!le to ho0er at

all. There woul !e no air to push own.

6"6C

Solution e are to iscuss helicopter performance in summer 0ersus winter.

Analysis In winter the air is "enerally coler, an thus enser. Therefore, less air must !e ri0en !y the !laes to

pro0ie the same helicopter lift, re2uirin" less power. +ess energy is re0uired in the inter.

Discussion Howe0er, it is also harer for the !laes to mo0e throu"h the enser col air, so there is more tor2uere2uire of the en"ine in col weather.

6"3C

Solution e are to iscuss if the force re2uire to hol a plate stationary ou!les when the 8et 0elocity ou!les.

Analysis No, the force ill not double. In fact, the force re2uire to hol the plate a"ainst the hori4ontal water

stream ill increase by a factor of / when the 0elocity is ou!le since

&)( V V V V m ! ρ ρ ===

an thus the force is proportional to the square of the velocity.

Discussion @ou can thin7 of it this way$ 3ince momentum flu is mass flow rate times 0elocity, a ou!lin" of the0elocity ou!les !oth the mass flow rate and the 0elocity, increasin" the momentum flu !y a factor of four.

6"4C Solution e are to iscuss the acceleration of a cart hit !y a water 8et.

Analysis The acceleration is not be constant since the force is not constant. The impulse force eerte !y the

water on the plate is&)( V V V V m ! ρ ρ === , where V is the relati0e 0elocity !etween the water an the plate,

which is mo0in". The ma"nitue of the plate acceleration is thus a A ! Bm. #ut as the plate !e"ins to mo0e, V ecreases, sothe acceleration must also ecrease.

Discussion It is the relative 0elocity of the water 8et on the cart that contri!utes to the cart/s acceleration.

6"8C

Solution e are to iscuss the maimum possi!le 0elocity of a cart hit !y a water 8et.

Analysis The maximum possible velocity for the plate is the velocity of the ater 5et. 1s lon" as the plate is

mo0in" slower than the 8et, the water eerts a force on the plate, which causes it to accelerate, until terminal 8et 0elocity is

reache.

Discussion nce the relati0e 0elocity is 4ero, the 8et supplies no force to the cart, an thus it cannot accelerate further.


*-D




6"-:

Solution It is to !e shown that the force eerte !y a li2ui 8et of 0elocity V on a stationary no44le is proportional to

V &, or alternati0ely, to &m .

Assumptions The flow is steay an incompressi!le. - The no44le is "i0en to !e stationary. The no44le in0ol0es a

E'° turn an thus the incomin" an out"oin" flow streams are normal to each other. / The water is ischar"e to the

atmosphere, an thus the "a"e pressure at the outlet is 4ero.

Analysis e ta7e the no44le as the control 0olume, an the flow

irection at the outlet as the x ais. ote that the no44le ma7es a E'°turn, an thus it oes not contri!ute to any pressure force or momentum

flu term at the inlet in the x irection. otin" that V m ρ = where

is the no44le outlet area an V is the a0era"e no44le outlet 0elocity,the momentum e2uation for steay one-imensional flow in the xirection reuces to

∑∑∑ −=inout

V mV m ! β β → V mV m ! out out Rx

β β ==

where ! Rx is the reaction force on the no44le ue to li2ui 8et at the no44le outlet. Then,

V m ρ = → & V VV V m ! Rx βρ βρ β === or

m

mmV m ! Rx

ρ β

ρ β β

&

===

Therefore, the force exerted by a li0uid 5et of velocity V on this stationary nozzle is proportional to V -, or

alternatively, to &m .

Discussion If there were not a E'o turn, we woul nee to ta7e into account the momentum flu an pressure

contri!utions at the inlet.

6"-

Solution 1 water 8et of 0elocity V impin"es on a plate mo0in" towar the water 8et with 0elocity FV . The force

re2uire to mo0e the plate towars the 8et is to !e etermine in terms of ! actin" on the stationary plate.

Assumptions The flow is steay an incompressi!le. - The plate is 0ertical an the 8et is normal to plate. The pressure

on !oth sies of the plate is atmospheric pressure (an thus its effect cancels out). / 5iction urin" motion is ne"li"i!le. 2

There is no acceleration of the plate. 6 The water splashes off the sies of the plate in a plane normal to the 8et. 6 et flow is

nearly uniform an thus the effect of the momentum-flu correction factor is ne"li"i!le, β ≅ +.

Analysis e ta7e the plate as the control 0olume. The relati0e 0elocity !etween the plate an the 8et is V when the

plate is stationary, an +."V when the plate is mo0in" with a 0elocity FV towars the plate. Then the momentum e2uation

for steay one-imensional flow in the hori4ontal irection reuces to

∑∑∑ −=inout

V mV m ! β β → ii Rii R V m ! V m ! =→−=−

#tationary plate$ ( V V mV V iii ρ ρ === an ) → ! V ! R == & ρ

$oving plate$ ( )D.+(anD.+ V V mV V iii ρ ρ === )

→ ! V V ! R &D.&&D.&)D.+(&& === ρ ρ

Therefore, the force re2uire to hol the plate stationary a"ainst the oncomin" water 8et !ecomes *+*, times greater when the 8et 0elocity !ecomes +.D times "reater.

Discussion ote that when the plate is stationary, V is also the 8et 0elocity. #ut if the plate mo0es towar the stream

with 0elocity FV , then the relative 0elocity is +.DV , an the amount of mass stri7in" the plate (an fallin" off its sies) per unit time also increases !y D'.


*-*

+B&V

V

ater8et

! Rx

i2ui

o44le

V y

x




6"-

Solution 1 +'° el!ow forces the flow to ma7e a >-turn an ischar"es it to the atmosphere at a specifie rate. The

"a"e pressure at the inlet of the el!ow an the anchorin" force neee to hol the el!ow in place are to !e etermine.

Assumptions The flow is steay, frictionless, one-imensional, incompressi!le, an irrotational (so that the #ernoulli

e2uation is applica!le). - The wei"ht of the el!ow an the water in it is ne"li"i!le. The water is ischar"e to the

atmosphere, an thus the "a"e pressure at the outlet is 4ero. / The momentum-flu correction factor for each inlet an

outlet is "i0en to !e β A +.':.

Properties e ta7e the ensity of water to !e +''' 7"Bm:.

Analysis (a) e ta7e the el!ow as the control 0olume, an esi"nate the entrance !y + an the outlet !y &. e alsoesi"nate the hori4ontal coorinate !y x (with the irection of flow as !ein" the positi0e irection) an the 0ertical

coorinate !y % . The continuity e2uation for this one-inlet one-outlet steay flow system is 7"Bs.:'&+ === mmm

otin" that V m ρ = , the mean inlet an outlet 0elocities of water are

mBs+.:I9Bm)+.'()J 7"Bm(+'''

7"Bs&D

)9B( &:&&+ ======π π ρ ρ &

m

mV V V

otin" that V + A V & an ' & A ' atm, the #ernoulli e2uation for a streamline "oin" throu"h the center of the reucin" el!ow is

epresse as

( ) )( &&

+&"a"e,++&&+&

&&&

+

&++ % % g ' % % g ' ' %

g

V

g

' %

g

V

g

' −=→−=−→++=++ ρ ρ

ρ ρ

3u!stitutin",

( ) ( ) ( ): & &

+ "a"e &

+ 7+''' 7"Bm E + mBs ' L' m * *L 7Bm * *L 7Ka

+''' 7" mBs ( ' ) ) ) )

= = = ≅ ÷×

6+12 /Pa

(b) The momentum e2uation for steay one-imensional flow is ∑∑∑ −=inout

V mV m ! β β . e let the x- an %-

components of the anchorin" force of the el!ow !e ! Rx an ! R% , an assume them to !e in the positi0e irections. e also

use "a"e pressures to a0oi ealin" with the atmospheric pressure which acts on all surfaces. Then the momentum

e2uations alon" the x an % aes !ecome

'

&)()( +&+"a"e+,

=

−=+−−=+

R%

Rx

!

V mV mV m ' ! β β β

3ol0in" for ! Rx an su!stitutin" the "i0en 0alues,

&+

I9Bm)+.'()J Bm**L(mBs7"+

+mBs)7"Bs)(:.+&D(':.+&

&

&&

&

+"a"e,+

−=

−

⋅×−=

−−=

π

β ' V m ! Rx

an ! R A ! Rx 1 3 *01 N since the y-component of the anchorin" force is 4ero.Therefore, the anchorin" force has a ma"nitue of &+ an it acts in the ne"ati0e

x irection.

Discussion ote that a ne"ati0e 0alue for ! Rx inicates the assume irection is wron", an shoul !e re0erse.


*-

ater

&D 7"Bs

:D cm

! R%

! Rx

%

x

+

&




6"-/(

Solution 1 hori4ontal water 8et stri7es a 0ertical stationary plate normally at a specifie 0elocity. 5or a "i0enanchorin" force neee to hol the plate in place, the flow rate of water is to !e etermine.

Assumptions The flow is steay an incompressi!le. - The water splatters off the sies of the plate in a plane normal to

the 8et. The water 8et is epose to the atmosphere, an thus the pressure of the water 8et an the splattere water is the

atmospheric pressure which is isre"are since it acts on the entire control surface. / The 0ertical forces an momentum

flues are not consiere since they ha0e no effect on the hori4ontal reaction force. 2 et flow is nearly uniform an thus the

effect of the momentum-flu correction factor is ne"li"i!le, β ≅ +.

Properties e ta7e the ensity of water to !e *&.9 l!mBft:.

Analysis e ta7e the plate as the control 0olume such that it contains the entire plate an cuts throu"h the water 8et

an the support !ar normally, an the irection of flow as the positi0e irection of x ais. The momentum e2uation for steay one-imensional flow in the x (flow) irection reuces in this case to

∑∑∑ −=inout

V mV m ! β β → ++ V m ! V m ! R Rx =→−=−

e note that the reaction force acts in the opposite irection to flow, an we shoul not for"et the ne"ati0e si"n for forces

an 0elocities in the ne"ati0e x-irection. 3ol0in" for m an su!stitutin" the "i0en 0alues,

l!mBs:L*l!f +

ftBsl!m:&.&

ftBs:'

l!f :D' &

+

=

⋅==

V

! m Rx

Then the 0olume flow rate !ecomes

=s-

ft6+4*===:l!mBft*&.9

l!mBs:L*

ρ

mV

Therefore, the 0olume flow rate of water uner state assumptions must !e *.'& ft:Bs.

Discussion In reality, some water will !e scattere !ac7, an this will a to the reaction force of water. The flow rate inthat case will !e less.


*-E

! Rx

A :D' l!f

m

ater8et

+




6"-2

Solution 1 reucin" el!ow eflects water upwars an ischar"es it to the atmosphere at a specifie rate. Theanchorin" force neee to hol the el!ow in place is to !e etermine.


e2uation is applica!le). - The wei"ht of the el!ow an the water in it is consiere. The water is ischar"e to the




Analysis The wei"ht of the el!ow an the water in it is

7 '.9E'D 9E'.D)mBs7")(E.+D'( & ==== mg *

e ta7e the el!ow as the control 0olume, an esi"nate theentrance !y + an the outlet !y &. e also esi"nate the

hori4ontal coorinate !y x (with the irection of flow as !ein"

the positi0e irection) an the 0ertical coorinate !y % . The

continuity e2uation for this one-inlet one-outlet steay flow

system is 7"Bs.:'&+ === mmm otin" that V m ρ = ,

the inlet an outlet 0elocities of water are

mBs+&)m''&D.')( 7"Bm(+'''

7"Bs:'

mBs'.&

)m'+D'.')( 7"Bm(+'''

7"Bs:'

&:&

&

&:

+

+

===

===

mV

mV

ρ

ρ

Ta7in" the center of the inlet cross section as the reference le0el ( % + A ') an notin" that ' & A ' atm, the #ernoulli e2uation for

a streamline "oin" throu"h the center of the reucin" el!ow is epresse as

+

−=→

−+

−=−→++=++ &

&+

&&

"a"e,++&

&+

&&

&+&

&&&

+

&++

&

&

&& %

g

V V g ' % %

g

V V g ' ' %

g

V

g

' %

g

V

g

' ρ ρ

ρ ρ

3u!stitutin",

7L:.E7BmE.L:

mBs7"+'''

7+9.'

)mBs+.E(&

mBs)&(mBs)+&()mBs+.E)(7"Bm+'''(

&

&&

&&&:

"a"e,+ ==

⋅

+

−= '

The momentum e2uation for steay one-imensional flow is ∑∑∑ −=inout



use "a"e pressures to a0oi ealin" with the atmospheric pressure which acts on all surfaces. Then the momentume2uations alon" the x an % aes !ecome

θ β β θ β sinancos &+&+"a"e+, V m* ! V mV m ' ! R% Rx =−−=+

3ol0in" for ! Rx an ! R% , an su!stitutin" the "i0en 0alues,

( ) ( ) ( )& &

& + "a"e + &+

+ 7+ ': :' 7"Bs)J(+&cos9D -&) mBs L: E 7Bm ' '+D' m

+''' 7" mBs

'.E' 7

Rx ( ! m V cos V ' ) + ) )β θ

= − − = ° − ÷×

= −

&

7'.LD:79E'D.'mBs7"+'''

7+mBs)in9D7"Bs)(+&s:'(':.+sin

&& =+

⋅°=+= * V m ! R% θ β

°−=−

===+−=+= -5+2/N0+01E'.'

LD:.'tantan ,)LD:.'()E'.'(

+-+-&&&&

Rx

R% R% Rx R

!

! ! ! ! θ

Discussion ote that the ma"nitue of the anchorin" force is +.+ 7, an its line of action ma7es ?:E.L° from x

irection. e"ati0e 0alue for ! Rx inicates the assume irection is wron".


*-+'

ater

:' 7"Bs

+D' m&

&D cm&

9D°

* +

&

! R%

! Rx




6"-6

Solution 1 reucin" el!ow eflects water upwars an ischar"es it to the atmosphere at a specifie rate. Theanchorin" force neee to hol the el!ow in place is to !e etermine.


e2uation is applica!le). - The wei"ht of the el!ow an the water in it is consiere. The water is ischar"e to the




Analysis The wei"ht of the el!ow an the water in it is

7 '.9E'D 9E'.D)mBs7")(E.+D'( & ==== mg *

e ta7e the el!ow as the control 0olume, an esi"nate the entrance !y + an the outlet !y &. e also esi"nate the hori4ontal coorinate

!y x (with the irection of flow as !ein" the positi0e irection) an

the 0ertical coorinate !y % . The continuity e2uation for this one-inlet

one-outlet steay flow system is 7"Bs.:'&+ === mmm otin"

that V m ρ = , the inlet an outlet 0elocities of water are

mBs+&)m''&D.')( 7"Bm(+'''

7"Bs:'

mBs'.&)m'+D'.')( 7"Bm(+'''

7"Bs:'

&:&

&

&:+

+

===

===

mV

mV

ρ

ρ

Ta7in" the center of the inlet cross section as the reference le0el ( % + A ') an notin" that ' & A ' atm, the #ernoulli e2uation for

a streamline "oin" throu"h the center of the reucin" el!ow is epresse as

+

−=→

−+

−=−→++=++ &

&+

&&

"a"e,++&

&+

&&

&+&

&&&

+

&++

&

&

&& %

g

V V g ' % %

g

V V g ' ' %

g

V

g

' %

g

V

g

' ρ ρ

ρ ρ

or,

7L:.E7BmE.L:mBs7"+'''

7+9.'

)mBs+.E(&

mBs)&(mBs)+&()mBs+.E)(7"Bm+'''(

&

&&

&&&:

"a"e,+ ==

⋅

+

−= '

The momentum e2uation for steay one-imensional flow is

∑∑∑ −=

inout

V mV m ! β β . e let the x- an y-


use "a"e pressures to a0oi ealin" with the atmospheric pressure which acts on all surfaces. Then the momentume2uations alon" the x an % aes !ecome

θ β β θ β sinancos &+&+"a"e+, V m* ! V mV m ' ! Ry Rx =−−=+

3ol0in" for ! Rx an ! R% , an su!stitutin" the "i0en 0alues,

7&EL.+)m'+D'.')(7BmE.L:(

mBs7"+'''

7+mBsI&)-os++'7"Bs)J(+&c:'(':.+

)cos(

&&

&

+"a"e,++&

−=−

⋅°=

−−= ' V V m ! Rx θ β

7'.:E79E'D.'mBs7"+'''

7+mBs)n++'7"Bs)(+&si:'(':.+sin

&&

=+

⋅°=+= * V m !

R%

θ β

an°−=

−==

=+−=+=

-*+5

/N0+,.

&EL.+

:E.'tantan

:E.')&EL.+(

+-+-

&&&&

Rx

R%

R% Rx R

!

!

! ! !

θ

Discussion ote that the ma"nitue of the anchorin" force is +.D9 7, an its line of action ma7es ?:&.E° from x

irection. e"ati0e 0alue for ! Rx inicates assume irection is wron", an shoul !e re0erse.


*-++

ater :' 7"Bs

+D' m&

&D cm&

++'°

*

! R%

! Rx

&

+




6"-3

Solution ater accelerate !y a no44le stri7es the !ac7 surface of a cart mo0in" hori4ontally at a constant 0elocity.The !ra7in" force an the power waste !y the !ra7es are to !e etermine.

Assumptions The flow is steay an incompressi!le. - The water splatters off the sies of the plate in all irections in

the plane of the !ac7 surface. The water 8et is epose to the atmosphere, an thus the pressure of the water 8et an the

splattere water is the atmospheric pressure which is isre"are since it acts on all surfaces. / 5iction urin" motion is

ne"li"i!le. 2 There is no acceleration of the cart. 3 The motions of the water 8et an the cart are hori4ontal. 6 et flow is

nearly uniform an thus the effect of the momentum-flu correction factor is ne"li"i!le, β ≅ +.

Analysis e ta7e the cart as the control 0olume, an theirection of flow as the positi0e irection of x ais. The relati0e

0elocity !etween the cart an the 8et is

mBs+'+'+Dcart 8et =−=−= V V V r

Therefore, we can 0iew the cart as !ein" stationary an the 8et mo0in"

with a 0elocity of +' mBs. otin" that water lea0es the no44le at +D mBsan the corresponin" mass flow rate relati0e to no44le eit is &D 7"Bs,

the mass flow rate of water stri7in" the cart corresponin" to a water 8et

0elocity of +' mBs relati0e to the cart is

7"Bs*L.+*7"Bs)&D(mBs+D

mBs+' 8et

8et

=== mV

V m r

r

The momentum e2uation for steay one-imensional flow in the x (flow) irection reuces in this case to

∑∑∑ −=inout

V mV m ! β β → !ra7e r r ii Rx V m ! V m ! −=→−=

e note that the !ra7e force acts in the opposite irection to flow, an we shoul not for"et the ne"ati0e si"n for forces an0elocities in the ne"ati0e x-irection. 3u!stitutin" the "i0en 0alues,

N062−≅−=

⋅+−=−= +**.L

mBs7"+

+mBs)'+7"Bs)(*L.+*(

& !ra7e r r V m !

The ne"ati0e si"n inicates that the !ra7in" force acts in the opposite irection to motion, as epecte. otin" that wor7 is

force times istance an the istance tra0ele !y the cart per unit time is the cart 0elocity, the power waste !y the !ra7es is

$1--=

⋅==

mBs +

+mBs) )(DL.+**(cart !ra7eV ! *

Discussion ote that the power waste is e2ui0alent to the maimum power that can !e "enerate as the cart 0elocity is

maintaine constant.

6"-4

Solution ater accelerate !y a no44le stri7es the !ac7 surface of a cart mo0in" hori4ontally. The acceleration of the

cart if the !ra7es fail is to !e etermine.

Analysis The !ra7in" force was etermine in pre0ious pro!lem

to !e +*L . hen the !ra7es fail, this force will propel the cart

forwar, an the acceleration will !e

*ms4+,,6=

⋅==

+

mBs7"+

7":''

+*L&

cartm

! a

Discussion This is the acceleration at the moment the !ra7es fail. The acceleration will ecrease as the relati0e 0elocity

!etween the water 8et an the cart (an thus the force) ecreases.


*-+&

+D mBs

ater8et

D mBs

! Rx

+D mBs

ater8et

D mBs

:'' 7" ! Rx




6"-8(

Solution 1 water 8et hits a stationary splitter, such that half of the flow is i0erte upwar at 9D°, an the other half is

irecte own. The force re2uire to hol the splitter in place is to !e etermine.

Assumptions The flow is steay an incompressi!le. - The

water 8et is epose to the atmosphere, an thus the pressure

of the water 8et !efore an after the split is the atmospheric

pressure which is isre"are since it acts on all surfaces.

The "ra0itational effects are isre"are. / et flow is nearly

uniform an thus the effect of the momentum-flu correctionfactor is ne"li"i!le, β ≅ +.


Analysis The mass flow rate of water 8et is

l!mBs*&9'Bs)ft)(+''l!mBft9.*&( :: === V m ρ

e ta7e the splittin" section of water 8et, incluin" the splitter as the control 0olume, an esi"nate the entrance !y + an

the outlet of either arm !y & (!oth arms ha0e the same 0elocity an mass flow rate). e also esi"nate the hori4ontalcoorinate !y x with the irection of flow as !ein" the positi0e irection an the 0ertical coorinate !y % .


V mV m ! β β . e let the x- an

y- components of the anchorin" force of the splitter !e ! Rx an ! R% , an assume them to !e in the positi0e irections. otin"

that V & A V + A V an mm &+

& = , the momentum e2uations alon" the x an % aes !ecome

'')sin()sin(

)+(coscos)(&

&&+

&&+

+&&+

=−−++=

−=−=

θ θ

θ θ

V mV m !

V mV mV m !

R%

Rx

3u!stitutin" the "i0en 0alues,

( )&

+ l!f *&9' l!mBs (&' ftBs)(cos9D +) ++:D l!f

:&.& l!m ftBs Rx

R%

! -

!

= ° = − − ≅ ÷× = :

300.4 lbf

The ne"ati0e 0alue for ! Rx inicates the assume irection is wron", an shoul !e re0erse. Therefore, a force of ++9' l!f

must !e applie to the splitter in the opposite irection to flow to hol it in place. o holin" force is necessary in the

0ertical irection. This can also !e conclue from the symmetry.

Discussion In reality, the "ra0itational effects will cause the upper stream to slow own an the lower stream to speeup after the split. #ut for short istances, these effects are inee ne"li"i!le.


*-+:

+'' ft:Bs

&' ftBs

9D°9D°

! R%

! Rx




6":(

Solution The pre0ious pro!lem is reconsiere. The effect of splitter an"le on the force eerte on the splitter as the

half splitter an"le 0aries from ' to +'° in increments of +'° is to !e in0esti"ate.

Analysis The <<3 ,quations winow is printe !elow, followe !y the ta!ulate an plotte results.

g=32.2 "ft/s2"rho=62.4 "lbm/ft3"V_dot=100 "ft3/s"V=20 "ft/s"m_dot=rho*V_dotF_R=-m_dot*V*(cos(tht!-1/g "lbf"

θ , ° m , l!mBs ! R, l!f

'

+'&'

:'

9'D'*'

L'

'

E'+''

++'

+&'

+:'+9'

+D'

+*'

+L'+'

*&9'

*&9'*&9'

*&9'

*&9'*&9'*&9'

*&9'

*&9'

*&9'*&9'

*&9'

*&9'

*&9'*&9'

*&9'

*&9'

*&9'*&9'

'

DE&:9

D+E

E'L+:9+E:

&DD'

:&':

:L*9D9E

D&'+

D+9

*:*L*9D

L&:&

LD+

L*E:LLD&

Discussion The force rises from 4ero at θ A 'o to a maimum at θ A +'o, as epecte, !ut the relationship is not linear.


*-+9

0 20 40 60 80 100 120 140 160 180

0

1000

2000

3000

4000

5000

6000

7000

8000

θ7 8

# R 7

l b f



+ mBs

ater8et

+''' 7"

5rictionless trac7


6"

Solution 1 hori4ontal water 8et impin"es normally upon a 0ertical plate which is hel on a frictionless trac7 an isinitially stationary. The initial acceleration of the plate, the time it ta7es to reach a certain 0elocity, an the 0elocity at a

"i0en time are to !e etermine.

Assumptions The flow is steay an incompressi!le. - The water always splatters in the plane of the retreatin" plate.

The water 8et is epose to the atmosphere, an thus the pressure of the water 8et an the splattere water is the atmospheric

pressure which is isre"are since it acts on all surfaces. / The tract is nearly frictionless, an thus fiction urin" motion isne"li"i!le. 2 The motions of the water 8et an the cart are hori4ontal. 6 The 0elocity of the 8et relati0e to the plate remains

constant, V r A V 8et A V . 3 et flow is nearly uniform an thus the effect of the momentum-flu correction factor is ne"li"i!le,

β ≅ +.


Analysis (a) e ta7e the 0ertical plate on the frictionless trac7 as the control 0olume, an the irection of flow as the

positi0e irection of x ais. The mass flow rate of water in the 8et is

7"Bs:D.:9I9Bm) ('.'DmBs)J)(+7"Bm+'''( &: === π ρ Vm

The momentum e2uation for steay one-imensional flow in the x (flow) irection reuces in this case to

∑∑∑ −=inout

V mV m ! β β → V m ! V m ! Rxii Rx −=→−=

where ! Rx is the reaction force re2uire to hol the plate in place. hen the plate is release, an e2ual an opposite impulse

force acts on the plate, which is etermine to

*:*mBs7"+

+mBs)7"Bs)(+:9.:D(

& plate =

⋅==−= V m ! ! Rx

Then the initial acceleration of the plate !ecomes

*ms4+6-6=

⋅==

+

mBs7"+

7"+'''

*:*&

plate

plate

m

! a

This acceleration will remain constant urin" motion since the force

actin" on the plate remains constant.

(b) otin" that a A dV Bdt A ∆V B∆t since the acceleration a is constant, the time it ta7es for the plate to reach a 0elocity of E

mBs is

s0.+*=−

=∆

=∆&

plate

mBs*:*.'

mBs)'E(

a

V t

(c) otin" that a A dV Bdt an thus dV A adt an that the acceleration a is constant, the plate 0elocity in &' s !ecomes

ms0*+2=+=∆+= s)&')(mBs*:*.'(' & plate', plate t aV V

Discussion The assumption that the relati0e 0elocity !etween the water 8et an the plate remains constant is 0ali only

for the initial moments of motion when the plate 0elocity is low unless the water 8et is mo0in" with the plate at the same0elocity as the plate.


*-+D

! Rx





*-+*




6"-

Solution 1 E'° reucer el!ow eflects water ownwars into a smaller iameter pipe. The resultant force eerte on

the reucer !y water is to !e etermine.


e2uation is applica!le). - The wei"ht of the el!ow an the water in it is isre"are since the "ra0itational effects are

ne"li"i!le. The momentum-flu correction factor for each inlet an outlet is "i0en to !e β A +.'9.


Analysis e ta7e the el!ow as the control 0olume, an esi"nate theentrance !y + an the outlet !y &. e also esi"nate the hori4ontal coorinate !y

x (with the irection of flow as !ein" the positi0e irection) an the 0ertical

coorinate !y % . The continuity e2uation for this one-inlet one-outlet steay flow

system is 7"Bs.9.:D:&+ === mmm otin" that V m ρ = , the mass

flow rate of water an its outlet 0elocity are

7"Bs9.:D:I9Bm) ('.:mBs)J)(D7"Bm+'''()9B(&:&

++++ ==== π π ρ ρ &V V m

mBs&'I9Bm)+D.'()J 7"Bm(+'''

7"Bs9.:D:

9B &:&&&

& ====π ρπ ρ &

m

mV

The #ernoulli e2uation for a streamline "oin" throu"h the center of the reucin" el!ow is epresse as

−+−+=→++=++ &+

&&

&+

+&&

&&&

+

&++

&

&& % %

g

V V g ' ' %

g

V

g

' %

g

V

g

' ρ

ρ ρ

3u!stitutin", the "a"e pressure at the outlet !ecomes

+7Bm+

7Ka+

mBs7"+'''

7+D.'

)mBs+.E(&

mBs)&'(mBs)D()mBs+.E)(7"Bm+'''(7Ka):''(

&&&

&&&:

& =

⋅

+

−+= '



components of the anchorin" force of the el!ow !e ! Rx an ! R% , an assume them to !e in the positi0e irections. Then the

momentum e2uations alon" the x an % aes !ecome

')(

'

&&"a"e&,

++"a"e+,

−−=−

−=+

V m ' !

V m ' !

R%

Rx

β

β

ote that we shoul not for"et the ne"ati0e si"n for forces an 0elocities in the ne"ati0e x or % irection. 3ol0in" for ! Rx

an ! R% , an su!stitutin" the "i0en 0alues,

9

m):.'()7Bm:''(

mBs7"+'''

7+mBs)7"Bs)(D9.:D:('9.+

&&

&+"a"e,++ −=−

⋅−=−−=

π β ' V m ! Rx

9

m)+D.'()7Bm9.++L(

mBs7"+'''

7+mBs)7"Bs)(&'9.:D:('9.+

&&

&+"a"e,&& +

⋅−=+−=

π β ' V m ! R%

an

°=−−

==

=−+−=+=

0*+5

/N*-+6

'.&:

&.Dtantan

)&.D()'.&:(

+-+-

&&&&

Rx

R%

R% Rx R

!

!

! ! !

θ

Discussion The ma"nitue of the anchorin" force is &:.* 7, an its line of action ma7es +&.E° from x irection.

e"ati0e 0alues for ! Rx an ! Ry inicate that the assume irections are wron", an shoul !e re0erse.


*-+L

ater D mBs

:' cm

+D cm

! R%

! Rx





*-+




6" > lso solved using ,,# on enclosed &V&?

Solution 1 win tur!ine with a "i0en span iameter an efficiency is su!8ecte to steay wins. The power "enerate

an the hori4ontal force on the supportin" mast of the tur!ine are to !e etermine.

Assumptions The win flow is steay an incompressi!le. - The efficiency of the tur!ine-"enerator is inepenent of

win spee. The frictional effects are ne"li"i!le, an thus none of the incomin" 7inetic ener"y is con0erte to thermal

ener"y. / in flow is uniform an thus the momentum-flu correction factor is nearly unity, β ≅ +.

Properties The ensity of air is "i0en to !e +.&D 7"Bm:.

Analysis (a) The power potential of the win is its 7inetic ener"y,

which is V &B& per unit mass, an &B&

V m for a "i0en mass flow rate$

mBsE9.* 7mBh*.:

mBs+ 7mBh)&D(+ =

=V

7"Bs&'',DD9

m)(E'mBs)E9.*)(7"Bm&D.+(

9

&:

&

+++++ ==== π π

ρ ρ &

V V m

7+::'mBs7+

7+

mBs7"+'''

7+

&

mBs)(*.E9 7"Bs)&'',DD(

& &

&&+

+ma =

⋅

⋅===

V m-em*

Then the actual power prouce !ecomes

/$.*6=== 7)+::')(:&.'(manewin tur!iact * * η

(b) The frictional effects are assume to !e ne"li"i!le, an thus the portion of incomin" 7inetic ener"y not con0erte to

electric power lea0es the win tur!ine as out"oin" 7inetic ener"y. Therefore,

)+(&

&

)+( newin tur!i

&+

&&

newin tur!i+& η η −=→−= V

mV

m-em-em

or

mBsL&.D'.:&-+mBs)E9.*(+ newin tur!i+& ==−= η V V

e choose the control 0olume aroun the win tur!ine such that the win is normal to the control surface at the inlet an

the outlet, an the entire control surface is at the atmospheric pressure. The momentum e2uation for steay one-imensional

flow is ∑∑∑ −=inout

V mV m ! β β . ritin" it alon" the x-irection (without for"ettin" the ne"ati0e si"n for forces

an 0elocities in the ne"ati0e x-irection) an assumin" the flow 0elocity throu"h the tur!ine to !e e2ual to the win

0elocity "i0e

/N62+-−=

⋅=−=−=

&+&+&mBs7"+'''

7+mBs)*.E9-&7"Bs)(D.L&'',DD()( V V mV mV m ! R

The ne"ati0e si"n inicates that the reaction force acts in the ne"ati0e x irection, as epecte.

Discussion This force acts on top of the tower where the win tur!ine is installe, an the !enin" moment it "enerates

at the !ottom of the tower is o!taine !y multiplyin" this force !y the tower hei"ht.


*-+E

in

V +

! R

V &D

+&




6"/(

Solution 1 hori4ontal water 8et stri7es a cur0e plate, which eflects the water !ac7 to its ori"inal irection. Theforce re2uire to hol the plate a"ainst the water stream is to !e etermine.

Assumptions The flow is steay an incompressi!le. - The water 8et

is epose to the atmosphere, an thus the pressure of the water 8et an

the splattere water is the atmospheric pressure, which is isre"are

since it acts on all surfaces. 5riction !etween the plate an the surface itis on is ne"li"i!le (or the friction force can !e inclue in the re2uire

force to hol the plate). / There is no splashin" of water or the

eformation of the 8et, an the re0erse 8et lea0es hori4ontally at the same

0elocity an flow rate. 2 et flow is nearly uniform an thus the

momentum-flu correction factor is nearly unity, β ≅ +.


Analysis e ta7e the plate to"ether with the cur0e water 8et as the control 0olume, an esi"nate the 8et inlet !y +

an the outlet !y &. e also esi"nate the hori4ontal coorinate !y x (with the irection of incomin" flow as !ein" the

positi0e irection). The continuity e2uation for this one-inlet one-outlet steay flow system is mmm == &+ where

l!mBs.9&I9Bft)+&B:(JftBs)+9')(l!mBft(*&.9I9BJ &:& ==== π π ρ ρ &V Vm


V mV m ! β β . ettin" the reaction force

to hol the plate !e ! Rx an assumin" it to !e in the positi0e irection, the momentum e2uation alon" the x ais !ecomes

V mV mV m ! Rx &)()( +& −=+−−=

3u!stitutin",

( ) &

+ l!f & 9& l!mBs (+9' ftBs) :L&E l!f

:&.& l!m ftBs Rx ! )

= − = − ≅ ÷× -2-4 lbf

Therefore, a force of :L:' l!f must !e applie on the plate in the ne"ati0e x irection to hol it in place.

Discussion ote that a ne"ati0e 0alue for ! Rx inicates the assume irection is wron" (as epecte), an shoul !e

re0erse. 1lso, there is no nee for an analysis in the 0ertical irection since the flui streams are hori4ontal.


*-&'

ater8et

+9' ftBs

+9' ftBs

: in+

&

! Rx




6"2(

Solution 1 hori4ontal water 8et stri7es a !ent plate, which eflects the water !y +:D° from its ori"inal irection. The

force re2uire to hol the plate a"ainst the water stream is to !e etermine.

Assumptions The flow is steay an incompressi!le. - The water 8et is epose to the atmosphere, an thus the pressure

of the water 8et an the splattere water is the atmospheric pressure, which is isre"are since it acts on all surfaces.

5rictional an "ra0itational effects are ne"li"i!le. / There is no splatterin" of water or the eformation of the 8et, an the

re0erse 8et lea0es hori4ontally at the same 0elocity an flow rate. 2 et flow is nearly uniform an thus the momentum-flu

correction factor is nearly unity, β ≅ +.


Analysis e ta7e the plate to"ether with the cur0e water 8et as the control 0olume, an esi"nate the 8et inlet !y +

an the outlet !y &. e also esi"nate the hori4ontal coorinate !y x (with the irection of incomin" flow as !ein" the

positi0e irection), an the 0ertical coorinate !y % . The continuity e2uation for this one-inlet one-outlet steay flow system

is mmm == &+ where

l!mBs.9&I9Bft)+&B:(JftBs)+9')(l!mBft(*&.9I9BJ &:& ==== π π ρ ρ &V Vm



components of the anchorin" force of the plate !e ! Rx an ! R% , an assume them to !e in the positi0e irections. Then themomentum e2uations alon" the x an y aes !ecome

°=°+=°+−=+−°−=

9Dsin9Dsin)(

)9Dcos+()(9Dcos)(

&

+&

V mV m !

V mV mV m !

R%

Rx


l!f *:*D

ftBsl!m:&.&

l!f +)cos9DftBs)(+l!mBs)(+9'.9&(&

&

−=

⋅°+−= Rx !

l!f +:+ftBsl!m:&.&

l!f +ftBs)sin9Dl!mBs)(+9'.9&(

& =

⋅°= R% !

an( )

&& & & -+ + +:+*:*D +:+ tan tan ++ L +* :

*:*D

Ry -

R Rx R% Rx

! ! ! ! ) )

! θ = + = − + = = = = − ° = ° ≅ °

−,6,44 lbf 061

Discussion ote that the ma"nitue of the anchorin" force is *D'' l!f, an its line of action is +*° from the positi0e x

irection. 1lso, a ne"ati0e 0alue for ! Rx inicates the assume irection is wron", an shoul !e re0erse.


*-&+

+9' ftBs

ater8et

+:D°

: in

&

+

! R%

! Rx




6"6

Solution 5iremen are holin" a no44le at the en of a hose while tryin" to etin"uish a fire. The a0era"e water outlet0elocity an the resistance force re2uire of the firemen to hol the no44le are to !e etermine.


of the water 8et is the atmospheric pressure, which is isre"are since it acts on all surfaces. Gra0itational effects an

0ertical forces are isre"are since the hori4ontal resistance force is to !e etermine. 2 et flow is nearly uniform an

thus the momentum-flu correction factor can !e ta7en to !e unity, β ≅ +.


Analysis (a) e ta7e the no44le an the hori4ontal portion of the hose as the system such that water enters the

control 0olume 0ertically an outlets hori4ontally (this way

the pressure force an the momentum flu at the inlet are in

the 0ertical irection, with no contri!ution to the force !alancein the hori4ontal irection), an esi"nate the entrance !y +

an the outlet !y &. e also esi"nate the hori4ontal

coorinate !y x (with the irection of flow as !ein" the

positi0e irection). The a0era"e outlet 0elocity an the massflow rate of water are etermine from

ms*5+,===== mBmin+L*9Bm)'*.'(

BminmD

9B &

:

& π π & V

V V

7"Bs:.:7"BminD'''Bmin)m)(D7"Bm+'''( :: ==== V ρ m

(b) The momentum e2uation for steay one-imensional flow is ∑∑∑ −=inout

V mV m ! β β . e let hori4ontal force

applie !y the firemen to the no44le to hol it !e ! Rx, an assume it to !e in the positi0e x irection. Then the momentum

e2uation alon" the x irection "i0es

( )&

+ ' : : 7"Bs (&E.D mBs) &9DL

+7" mBs Rx e ! mV mV )

= − = = = ≅ ÷×

& & *.64 N

Therefore, the firemen must !e a!le to resist a force of &9*' to hol the no44le in place.

Discussion The force of &9*' is e2ui0alent to the wei"ht of a!out &D' 7". That is, holin" the no44le re2uires the

stren"th of holin" a wei"ht of &D' 7", which cannot !e one !y a sin"le person. This emonstrates why se0eral firemenare use to hol a hose with a hi"h flow rate.


*-&&

D m:Bmin

! R%

! Rx




6"3

Solution 1 hori4ontal 8et of water with a "i0en 0elocity stri7es a flat plate that is mo0in" in the same irection at aspecifie 0elocity. The force that the water stream eerts a"ainst the plate is to !e etermine.

Assumptions The flow is steay an incompressi!le. - The water splatters in all irections in the plane of the plate.

The water 8et is epose to the atmosphere, an thus the pressure of the water 8et an the splattere water is the atmospheric

pressure, which is isre"are since it acts on all surfaces. / The 0ertical forces an momentum flues are not consiere

since they ha0e no effect on the hori4ontal force eerte on the plate. 2 The 0elocity of the plate, an the 0elocity of thewater 8et relati0e to the plate, are constant. 6 et flow is nearly uniform an thus the momentum-flu correction factor can

!e ta7en to !e unity, β ≅ +. Properties e ta7e the ensity of water to !e +''' 7"Bm:.

Analysis e ta7e the plate as the control 0olume, an the flow

irection as the positi0e irection of x ais. The relati0e 0elocity

!etween the plate an the 8et is

mBs&'+':' plate 8et =−=−= V V V r

Therefore, we can 0iew the plate as !ein" stationary an the 8et to !emo0in" with a 0elocity of &' mBs. The mass flow rate of water relati0e

to the plate Ji.e., the flow rate at which water stri7es the plate is

7"Bs&L.:E

9

m)('.'DmBs)&')(7"Bm+'''(

9

&:

&

==== π π

ρ ρ &

V V m r r r


V mV m ! β β . e let the hori4ontal

reaction force applie to the plate in the ne"ati0e x irection to counteract the impulse of the water 8et !e ! Rx. Then the

momentum e2uation alon" the x irection "i0es

N21,=

⋅==→−=−

&mBs+7"

+mBs)7"Bs)(&'&L.:E( ' r r Rxi Rx V m ! V m !

Therefore, the water 8et applies a force of LD on the plate in the irection of motion, an an e2ual an opposite force

must !e applie on the plate if its 0elocity is to remain constant.

Discussion ote that we use the relati0e 0elocity in the etermination of the mass flow rate of water in the momentumanalysis since water will enter the control 0olume at this rate. (In the limitin" case of the plate an the water 8et mo0in" at

the same 0elocity, the mass flow rate of water relati0e to the plate will !e 4ero since no water will !e a!le to stri7e the plate).


*-&:

:' mBs

ater8et

+' mBs

D cm

! Rx




6"4

Solution The pre0ious pro!lem is reconsiere. The effect of the plate 0elocity on the force eerte on the plate as

the plate 0elocity 0aries from ' to :' mBs in increments of : mBs is to !e in0esti"ate.


rho=1000 "#g/m3"$=0.0% "m"V_t=30 "m/s"

'c=)*$2/4V_r=V_t-V_l!tm_dot=rho*'c*V_rF_R=m_dot*V_r "+"

V plate, mBs V r , mBs ! R,

'

:

*

E+&

+D

+

&+&9

&L

:'

:'

&L

&9

&++

+D

+&

E*

:

'

+L*L

+9:+

++:+

***:*

99&

&:

+DEL'.L

+L.L

'

Discussion hen the plate 0elocity reaches :' mBs, there is no relati0e motion !etween the 8et an the plate; hence,

there can !e no force actin".


*-&9

4 , 04 0, *4 *, -4

4

*44

.44

644

144

0444

0*44

0.44

0644

0144

9:late7 ms

# R 7

N




6"8(

Solution 1 fan mo0es air at sea le0el at a specifie rate. The force re2uire to hol the fan an the minimum power input re2uire for the fan are to !e etermine.

Assumptions The flow of air is steay an incompressi!le. - 3tanar atmospheric conitions eist so that the pressure

at sea le0el is + atm. 1ir lea0es the fan at a uniform 0elocity at atmospheric pressure. / 1ir approaches the fan throu"h a

lar"e area at atmospheric pressure with ne"li"i!le 0elocity. 2 The frictional effects are ne"li"i!le, an thus the entire

mechanical power input is con0erte to 7inetic ener"y of air (no con0ersion to thermal ener"y throu"h frictional effects). 6

in flow is nearly uniform an thus the momentum-flu correction factor can !e ta7en to !e unity, β ≅ +.

Properties The "as constant of air is R A '.:L'9 psi⋅ft:Bl!m⋅6. The stanar atmospheric pressure at sea le0el is

+ atm A +9.L psi.

Analysis (a) e ta7e the control 0olume to !e ahori4ontal hyper!olic cyliner !oune !y streamlines on the

sies with air enterin" throu"h the lar"e cross-section (section

+) an the fan locate at the narrow cross-section at the en

(section &), an let its centerline !e the x ais. The ensity,mass flow rate, an ischar"e 0elocity of air are

:

:l!mBft'L9E.'

6)6)(D:'Bl!mft psi('.:L'9

psiL.+9=

⋅⋅==

R.

' ρ

l!mBs&.D'l!mBmin.+9EBmin)ft&''')(l!mBft'L9E.'( :: ==== V ρ m

ftBs+'.*ftBmin*.*:*9Bft)&(

Bminft&'''

9B &

:

&&&

& =====π π &

V V V


V mV m ! β β . ettin" the reaction force

to hol the fan !e ! Rx an assumin" it to !e in the positi0e x (i.e., the flow) irection, the momentum e2uation alon" the x

ais !ecomes

lbf 4+1*=

⋅==−=

&&ftBsl!m:&.&

l!f +ftBs)*l!mBs)(+'.D'.&(')( V mV m ! Rx

Therefore, a force of '.& l!f must !e applie (throu"h friction at the !ase, for eample) to pre0ent the fan from mo0in" in

the hori4ontal irection uner the influence of this force.

(b) otin" that ' + A ' & A ' atm an V + ≅ ', the ener"y e2uation for the selecte control 0olume reuces to

lossmech,tur!ine&

&&&

u pump,+

&++

&& , * g%

V ' m* g%

V ' m ++

++=+

++

ρ ρ →

&

&&

ufan,

V m* =

3u!stitutin",

$,+50=

⋅

⋅

==ftBsl!f '.L:LD*

+

ftBsl!m&.:&

l!f +

&

ftBs)(+'.*l!mBs)D'.&(

& &

&&&

ufan,

V m*

Therefore, a useful mechanical power of D.E+ must !e supplie to air. This is the minimum re2uire power input re2uire

for the fan.

Discussion The actual power input to the fan will !e lar"er than D.E+ !ecause of the fan inefficiency in con0ertin"mechanical power to 7inetic ener"y.


*-&D

&''' cfm

24 in

Fan

+&




6"/:

Solution 1 helicopter ho0ers at sea le0el while !ein" loae. The 0olumetric air flow rate an the re2uire power input urin" unloae ho0er, an the rpm an the re2uire power input urin" loae ho0er are to !e etermine.

Assumptions The flow of air is steay an incompressi!le. - 1ir lea0es the !laes at a uniform 0elocity at atmospheric

pressure. 1ir approaches the !laes from the top throu"h a lar"e area at atmospheric pressure with ne"li"i!le 0elocity. /

The frictional effects are ne"li"i!le, an thus the entire mechanical power input is con0erte to 7inetic ener"y of air (no

con0ersion to thermal ener"y throu"h frictional effects). 2 The chan"e in air pressure with ele0ation is ne"li"i!le !ecauseof the low ensity of air. 6 There is no acceleration of the helicopter, an thus the lift "enerate is e2ual to the total wei"ht.

3 1ir flow is nearly uniform an thus the momentum-flu correction factor can !e ta7en to !e unity, β ≅ +. Properties The ensity of air is "i0en to !e +.+ 7"Bm:.

Analysis (a) e ta7e the control 0olume to !e a 0ertical hyper!olic cyliner !oune !y streamlines on the sies

with air enterin" throu"h the lar"e cross-section (section +) at the top an the fan locate at the narrow cross-section at the

!ottom (section &), an let its centerline !e the % ais with upwars !ein" the positi0e irection.


V mV m ! β β . otin" that the

only force actin" on the control 0olume is the total wei"ht * an it acts in the ne"ati0e % irection, the momentum e2uation

alon" the % ais "i0es

)( ')( &&

&&&&&

* V V V V V m* V m*

ρ ρ ρ =→===→−−=−

where is the !lae span area,&&& mL.+L*9Bm)+D(9B === π π &

Then the ischar"e 0elocity, 0olume flow rate, an the mass flow rate

of air in the unloae moe !ecome

mBsL.&+)m)(+L*.L7"Bm(+.+

)mBs7")(E.+''',+'(&:

&unloae

unloae,& ===

g mV

ρ

( ) ( )& :

unloae & unloae +L* L m &+ L mBs ::9 m Bs ( V ) )= = = ≅V & --1-4 m s

7"Bs9D&9Bs)m::9)(7"Bm+.+(::

unloaeunloae === V ρ m

otin" that ' + A ' & A ' atm, V + ≅ ', the ele0ation effects are ne"li"i!le, an the frictional effects are isre"are, the ener"y

e2uation for the selecte control 0olume reuces to

lossmech,tur!ine&

&&&

u pump,+

&++

&& , * g%

V '

m* g%

V '

m

++

++=+

++ ρ ρ → &

&& ufan,

V m* =

3u!stitutin",

( ) ( )

&&

&

unloae fan,u &

unloae

&+.L mBs + 7 + 79D&9 7"Bs +'*D 7

& & + 7 mBs+''' 7" mBs

V * m

= = = ≅ ÷ ÷ ÷×× & & 0424 /$

(b) e now repeat the calculations for the loaded helicopter, whose mass is +','''+D,''' A &D,''' 7"$

mBs:.:9)m)(+L*.L7"Bm(+.+

)mBs7")(E.+''',&D(&:

&loae

loae,& ===

g mV

ρ

7"BsL+D&mBs):.:9)(mL.+L*)(7"Bm+.+(&:

loae&,loaeloae ==== V m ρ ρ V

( )& &

&

loae fan,u &

loae

(:9.: mBs) + 7 + 7L+D& 7"Bs 9&'L 7

& & + 7 mBs+''' 7" mBs

V * m = = = ≅ ÷ ÷ ÷××

& & .*04 /$

otin" that the a0era"e flow 0elocity is proportional to the o0erhea !lae rotational 0elocity, the rpm of the loae

helicopter !laes !ecomes

r:m6-*===→=→= rpm)9''(L.&+

:.:9 unloae

unloae&,

loae&,

loae

unloae

loae

unloae&,

loae&,

& nV

V n

n

n

V

V n- V

Discussion The actual power input to the helicopter !laes will !e consiera!ly lar"er than the calculate power input


*-&*

+D m

oa

+D,''' 7"

3ea le0el

+

&




!ecause of the fan inefficiency in con0ertin" mechanical power to 7inetic ener"y.


*-&L




6"/

Solution 1 helicopter ho0ers on top of a hi"h mountain where the air ensity consiera!ly lower than that at seale0el. The !lae rotational 0elocity to ho0er at the hi"her altitue an the percent increase in the re2uire power input to

ho0er at hi"h altitue relati0e to that at sea le0el are to !e etermine.

Assumptions The flow of air is steay an incompressi!le. - The air lea0es the !laes at a uniform 0elocity at

atmospheric pressure. 1ir approaches the !laes from the top throu"h a lar"e area at atmospheric pressure with ne"li"i!le

0elocity. / The frictional effects are ne"li"i!le, an thus the entire mechanical power input is con0erte to 7inetic ener"y of air. 2 The chan"e in air pressure with ele0ation while ho0erin" at a "i0en location is ne"li"i!le !ecause of the low ensity

of air. 6 There is no acceleration of the helicopter, an thus the lift "enerate is e2ual to the total wei"ht. 3 1ir flow is

nearly uniform an thus the momentum-flu correction factor can !e ta7en to !e unity, β ≅ +.

Properties The ensity of air is "i0en to !e +.+ 7"Bm: at sea le0el, an '.LE 7"Bm: on top of the mountain.


with air enterin" throu"h the lar"e cross-section (section +) at the top an the fan locate at the narrow cross-section at the !ottom (section &), an let its centerline !e the % ais with upwars !ein" the positi0e irection.


V mV m ! β β . otin" that the only

force actin" on the control 0olume is the total wei"ht * an it acts in the ne"ati0e % irection, the momentum e2uationalon" the % ais "i0es

)( ')( &&

&&&&&

* V V V V V m* V m*

ρ ρ ρ =→===→−−=−

where is the !lae span area. Then for a "i0en wei"ht *( the ratio of ischar"e 0elocities !ecomes

&&&.+ 7"Bm'.LE

7"Bm+.+

B

B

:

:

mountain

sea

sea

mountain

sea,&

mountain,& ==== ρ

ρ

ρ

ρ

*

*

V

V

otin" that the a0era"e flow 0elocity is proportional to the o0erhea !lae rotational 0elocity, the rpm of the helicopter

!laes on top of the mountain !ecomes

r:.15===→=→= rpm)9''(&&&.+ sea

sea&,

mountain&,

mountain

sea&,

mountain&,

sea

mountain& n

V

V n

V

V

n

n-V n

otin" that ' + A ' & A ' atm, V + ≅ ', the ele0ation effect are ne"li"i!le, an the frictional effects are isre"are, the ener"y

e2uation for the selecte control 0olume reuces to

lossmech,tur!ine&

&

&& u pump,+

&

++&&

, * g% V '

m* g% V '

m ++

++=+

++ ρ ρ

→ &

&

& ufan, V m*

=

or

*

*

*

V

V V

V m*

ρ ρ ρ

ρ ρ ρ ρ

&&&&

D.+D.+

&+

:

&+

:&

&&

&

&&

ufan, =

=

====

Then the ratio of the re2uire power input on top of the mountain tothat at sea le0el !ecomes

&&&.+7"Bm'.LE

7"Bm+.+

BD.'

BD.':

:

mountain

sea

seaD.+

mountainD.+

ufan,sea

ufan,mountain === ρ

ρ

ρ

ρ

*

*

*

*

Therefore, the re2uire power input will increase !y **+*; on top of

the mountain relati0e to the sea le0el.

Discussion ote that !oth the rpm an the re2uire power input to the helicopter are in0ersely proportional to the

s2uare root of air ensity. Therefore, more power is re2uire at hi"her ele0ations for the helicopter to operate !ecause air is

less ense, an more air must !e force !y the !laes into the ownraft.


*-&

+D m

oa

+D,''' 7"

3ea le0el

+

&




6"/-

Solution The flow rate in a channel is controlle !y a sluice "ate !y raisin" or lowerin" a 0ertical plate. 1 relation for the force actin" on a sluice "ate of with w for steay an uniform flow is to !e e0elope.

Assumptions The flow is steay, incompressi!le, frictionless, an uniform (an thus the #ernoulli e2uation is

applica!le.) - all shear forces at surfaces are ne"li"i!le. The channel is epose to the atmosphere, an thus the

pressure at free surfaces is the atmospheric pressure. / The flow is hori4ontal. 2 ater flow is nearly uniform an thus the

momentum-flu correction factor can !e ta7en to !e unity, β ≅ +.

Analysis e ta7e point + at the free surface of the upstream flow !efore the "ate an point & at the free surface of the

ownstream flow after the "ate. e also ta7e the !ottom surface of the channel as the reference le0el so that the ele0ationsof points + an & are y+ an y&, respecti0ely. The application of the #ernoulli e2uation !etween points + an & "i0es

)"(& &&

&+&

+&

&&

&&&

+

&++ y yV V y

g

V

g

' y

g

V

g

' −=−→++=++

ρ ρ (+)

The flow is assume to !e incompressi!le an thus the ensity is constant. Then the conser0ation of mass relation for this

sin"le stream steay flow e0ice can !e epresse as

an&&

&++

+&&++&+wy

V wy

V V V V V V V

V V V V

====→==→== (&)

3u!stitutin" into <2. (+),

&+

&&

&+&&

+&&

&+&+

&

+

&

& B+

)(&

B+B+

)(& )(&

y y

y y g wy

y y

y y g w y y g

wywy −

−=→

−

−=→−=

−

V V

V V

(:)

3u!stitutin" <2. (:) into <2s. (&) "i0es the followin" relations for 0elocities,

B+

)(&an

B+

)(&&+

&&

&+&&

+&&

&+

+

&+

y y

y y g V

y y

y y g

y

yV

−

−=

−

−= (9)

e choose the control 0olume as the water !oy surroune !y the 0ertical cross-sections of the upstream an ownstream

flows, free surfaces of water, the inner surface of the sluice "ate, an the !ottom surface of the channel. The momentum

e2uation for steay one-imensional flow is ∑∑∑ −=inout

V mV m ! β β . The force actin" on the sluice "ate ! Rx is

hori4ontal since the wall shear at the surfaces is ne"li"i!le, an it is e2ual an opposite to the force applie on water !y thesluice "ate. otin" that the pressure force actin" on a 0ertical surface is e2ual to the prouct of the pressure at the centroi

of the surface an the surface area, the momentum e2uation alon" the x irection "i0es

)()(&

)(&

+&&&

++

+&&&++ V V mwy y

g wy y

g ! V mV m ' ' ! Rx Rx −=

−

+−→−=−+− ρ ρ

6earran"in", the force actin" on the sluice "ate is etermine to !e

)(&

)( &&

&+&+ y y g

wV V m ! Rx −+−= ρ (D)

where V + an V & are "i0en in <2. (9).

Discussion ote that for y+ NN y&, <2. (:) simplifies to

+& & gyw y=V or +& & gyV = which is the Toricelli e2uation

for frictionless flow from a tan7 throu"h a hole a istance y+ !elow the

free surface.


*-&E

3luice "ate y+

y&

V +

V &

+

&

! Rx




6"/

Solution ater enters a centrifu"al pump aially at a specifie rate an 0elocity, an lea0es in the normal irectionalon" the pump casin". The force actin" on the shaft in the aial irection is to !e etermine.


Assumptions The flow is steay an incompressi!le. - The forces

actin" on the pipin" system in the hori4ontal irection are ne"li"i!le.

The atmospheric pressure is isre"are since it acts on all surfaces.

Analysis e ta7e the pump as the control 0olume, an the inletirection of flow as the positi0e irection of x ais. The momentume2uation for steay one-imensional flow in the x (flow) irection

reuces in this case to

∑∑∑ −=inout

V mV m ! β β → ii Rxi Rx V V m ! V m ! V ρ ==→−=−

ote that the reaction force acts in the opposite irection to flow, an we shoul not for"et the ne"ati0e si"n for forces an0elocities in the ne"ati0e x-irection. 3u!stitutin" the "i0en 0alues,

N1.4=

⋅=

&

:: !ra7e

mBs7"+

+mBs) Bs)(Lm)('.+&7"Bm+'''( !

Discussion To fin the total force actin" on the shaft, we also nee to o a force !alance for the 0ertical irection, an

fin the 0ertical component of the reaction force.

Angular Momentum E)uation

6"//C

Solution e are to iscuss how the an"ular momentum e2uation is o!taine from the 6TT.

Analysis The angular momentum equation is o!taine !y replacing in the %eynolds transport theorem by the

total angular momentum sys / , and b by the angular momentum per unit mass V r × )

Discussion The 6TT is a "eneral e2uation that hols for any property B, either scalar or (as in this case) 0ector.

6"/2C

Solution e are to epress the an"ular momentum e2uation for a specific (restricte) control 0olume.

Analysis The an"ular momentum e2uation in this case is epresse as V mr I ×−=α where α is the an"ular

acceleration of the control 0olume, an r is the 0ector from the ais of rotation to any point on the line of action of ! .

Discussion This is a simplification of the more "eneral an"ular momentum e2uation (many terms ha0e roppe out).

6"/6C Solution e are to epress the an"ular momentum e2uation in scalar form a!out a specifie ais.

Analysis The an"ular momentum e2uation a!out a "i0en fie ais in this case can !e epresse in scalar form as

∑∑∑ −=inout

V mr V mr $ where r )s th mom,t !rm V )s th m!g,)td of th r!d)!l

loc)t !,d m )s th m!ss o r!t.

Discussion This is a simplification of the more "eneral an"ular momentum e2uation (many terms ha0e roppe out).


*-:'

'.+& m:BsL mBs

! Rx

mV

%

x




6"/3

Solution ater is pumpe throu"h a pipin" section. The moment actin" on the el!ow for the cases of ownwar anupwar ischar"e is to !e etermine.

Assumptions The flow is steay an incompressi!le. - The water is ischar"e to the atmosphere, an thus the "a"e

pressure at the outlet is 4ero. <ffects of water fallin" own urin" upwar ischar"e is isre"are. / Kipe outlet iameter

is small compare to the moment arm, an thus we use a0era"e 0alues of raius an 0elocity at the outlet.


Analysis e ta7e the entire pipe as the control 0olume, an esi"nate theinlet !y + an the outlet !y &. e also ta7e the x an y coorinates as shown.The control 0olume an the reference frame are fie. The conser0ation of mass

e2uation for this one-inlet one-outlet steay flow system is mmm == &+ , an

V V V == &+ since c A constant. The mass flow rate an the wei"ht of the

hori4ontal section of the pipe are

7"Bs&9.9D)mBs9I(9Bm)+&.'()J 7"Bm(+'''&: === π ρ V m c

Bm:.&E9mBs7"+

+)mBs+.E)(m&)( 7"Bm(+D

&

& =

⋅== mg *

(a) @onard discharge$ To etermine the moment actin" on the pipe at point , we nee to ta7e the moment of all forces

an momentum flows a!out that point. This is a steay an uniform flow pro!lem, an all forces an momentum flows arein the same plane. Therefore, the an"ular momentum e2uation in this case can !e epresse as

∑∑∑ −=inout

V mr V mr $ where r is the moment arm, all moments in the countercloc7wise irection are positi0e,

an all in the cloc7wise irection are ne"ati0e.The free !oy ia"ram of the pipe section is "i0en in the fi"ure. otin" that the moments of all forces an

momentum flows passin" throu"h point are 4ero, the only force that will yiel a moment a!out point is the wei"ht * of

the hori4ontal pipe section, an the only momentum flow that will yiel a moment is the outlet stream (!oth are ne"ati0e

since !oth moments are in the cloc7wise irection). Then the an"ular momentum e2uation a!out point !ecomes

&&+ V mr * r $ −=−

3ol0in" for $ an su!stitutin",

mN24+4 ⋅−=

⋅=−=

&&&+mBs7"+

+mBs)7"Bs)(9m)(9D.D9(&- )m)(&E9.:+(V mr * r $

The ne"ati0e si"n inicates that the assume irection for $ is wron", an shoul !e re0erse. Therefore, a moment of L'

⋅m acts at the stem of the pipe in the cloc7wise irection.

(b) ;pard discharge$ The moment ue to ischar"e stream is positi0e in this case, an the moment actin" on the pipe at

point is

mN6,5 ⋅=

⋅+=+=

&&&+mBs7"+

+mBs)7"Bs)(9m)(9D.D9(& )m)(&E9.:+(V mr * r $

Discussion ote irection of ischar"e can ma7e a !i" ifference in the moments applie on a pipin" system. This

pro!lem also shows the importance of accountin" for the moments of momentums of flow streams when performin"

e0aluatin" the stresses in pipe materials at critical cross-sections.


*-:+

&V m*

•

$

+V m

r + A & mr

&

A + m




6"/4(

Solution 1 two-arme sprin7ler is use to "enerate electric power. 5or a specifie flow rate an rotational spee, the power prouce is to !e etermine.

Assumptions The flow is cyclically steay (i.e., steay from a

frame of reference rotatin" with the sprin7ler hea). - The water is

ischar"e to the atmosphere, an thus the "a"e pressure at the no44le

outlet is 4ero. Generator losses an air ra" of rotatin" componentsare ne"lecte. / The no44le iameter is small compare to the moment

arm, an thus we use a0era"e 0alues of raius an 0elocity at the outlet.


Analysis e ta7e the is7 that encloses the sprin7ler arms as the

control 0olume, which is a stationary control 0olume. The conser0ation

of mass e2uation for this steay flow system is mmm == &+ . otin"

that the two no44les are ientical, we ha0e &Bno44le mm = or

&Bno44le totalV V = since the ensity of water is constant. The a0era"e

8et outlet 0elocity relati0e to the no44le is

ftBs&.:E&"al9'.L

ft+

I9Bft)+&BD.'(J

"alBs9 :

& 8et

no44le 8et =

==

π V

V

The an"ular an tan"ential 0elocities of the no44les are

ftBs D&.:*raBs)ft)(&*.+&(

raBs&*.+s*'

min+ re0Bmin)&D'(&&

no44le ===

=

==

ω

π π ω

r V

n

The 0elocity of water 8et relati0e to the control 0olume (or relati0e to afie location on earth) is

ftBs.::E:*.D&&.:E&no44le 8et =−=−= V V V r

The an"ular momentum e2uation can !e epresse as

∑∑∑ −=inout

V mr V mr $ where all moments in the

countercloc7wise irection are positi0e, an all in the cloc7wise

irection are ne"ati0e. Then the an"ular momentum e2uation a!out theais of rotation !ecomes

r V mr $ no44leshaft & −=− or r V mr $ totalshaft =

3u!stitutin", the tor2ue transmitte throu"h the shaft is etermine to !e

ftl!f +9'EftBsl!m:&.&

l!f +ftBs).l!mBs)(::Eft)(**.L9&(

&totalshaft ⋅=

⋅== r V mr $

since l!mBsL9.**)Bsft9'.LB)(l!mBft (*&.9::

totaltotal === V ρ m . Then the power "enerate !ecomes

/$,4+4=

⋅⋅===

ftBsl!f L:L.D*

7+ft)l!f 'EraBs)(+9+.&*(& shaftshaft $ $ n* ω π

Therefore, this sprin7ler-type tur!ine has the potential to prouce D' 7 of power.

Discussion This is, of course, the maimum possi!le power. The actual power "enerate woul !e much smaller than

this ue to all the irre0ersi!le losses that we ha0e i"nore in this analysis.


*-:&

"alBstotal =m

<lectric"enerator

ω

8etV

8etV

r V mno44le

r V mno44le

$ shaft

•

r A & ft




6"/8(

Solution 1 two-arme sprin7ler is use to "enerate electric power. 5or a specifie flow rate an rotational spee, themoment actin" on the rotatin" hea when the hea is stuc7 is to !e etermine.

Assumptions The flow is uniform an steay. - The water is ischar"e to the atmosphere, an thus the "a"e pressure at

the no44le outlet is 4ero. The no44le iameter is small compare to the moment arm, an thus we use a0era"e 0alues of

raius an 0elocity at the outlet.


Analysis e ta7e the is7 that encloses the sprin7ler arms as the control 0olume, which is a stationary control

0olume. The conser0ation of mass e2uation for this steay flow system is mmm == &+ . otin" that the two no44les are

ientical, we ha0e &Bno44le mm = or no44le total & 0 =V V & & since the ensity of water is constant. The a0era"e 8et outlet 0elocity

relati0e to the no44le is

ftBs&.:E&"al9'.L

ft+

I9Bft)+&BD.'(J

"alBs9 :

& 8et

no44le 8et =

==

π V

V

The an"ular momentum e2uation can !e epresse as ∑∑∑ −=inout

V mr V mr $ where all moments in the

countercloc7wise irection are positi0e, an all in the cloc7wise irection are ne"ati0e. Then the an"ular momentume2uation a!out the ais of rotation !ecomes

1et V mr $ no44leshaft & −=− or 1et V mr $ totalshaft =

3u!stitutin", the tor2ue transmitte throu"h the shaft is etermine to !e

( ) ( ) ( )shaft total &

+ l!f & ft **.L9 l!mBs :E&.& ftBs +*&* l!f ft

:&.& l!m ftBs 1et $ rm V

= = = × ≅ × ÷× & 06-4 lbf ft

since l!mBsL9.**)Bsft9'.LB)(l!mBft (*&.9::

totaltotal === V ρ m .

Discussion hen the sprin7ler is stuc7 an thus the an"ular 0elocity is 4ero, the tor2ue e0elope is maimum since

'no44le =V an thus ftBs &.:E& 8etr ==V V , "i0in" shaft, ma +*:' l!f ft $ = × . #ut the power "enerate is 4ero in this

case since the shaft oes not rotate.


*-::

"alBstotal =m

<lectric

"enerator

ω

8etV

8etV

r V mno44le

r V mno44le

$ shaft

•

r A & ft




6"2:

Solution 1 three-arme sprin7ler is use to water a "aren. 5or a specifie flow rate an resistance tor2ue, thean"ular 0elocity of the sprin7ler hea is to !e etermine.

Assumptions The flow is uniform an cyclically steay (i.e., steay from a frame of reference rotatin" with the

sprin7ler hea). - The water is ischar"e to the atmosphere, an thus the "a"e pressure at the no44le outlet is 4ero. 1ir

ra" of rotatin" components are ne"lecte. / The no44le iameter is small compare to the moment arm, an thus we use

a0era"e 0alues of raius an 0elocity at the outlet.

Properties e ta7e the ensity of water to !e +''' 7"Bm: A + 7"B.

Analysis e ta7e the is7 that encloses the sprin7ler arms as thecontrol 0olume, which is a stationary control 0olume. The conser0ation

of mass e2uation for this steay flow system is mmm == &+ . otin"

that the three no44les are ientical, we ha0e :Bno44le mm = or

:Bno44le total V V = since the ensity of water is constant. The a0era"e

8et outlet 0elocity relati0e to the no44le an the mass flow rate are

mBs++L.E+'''

m+

I9Bm)'+&.'(J:

Bs9':

& 8et

no44le 8et =

==

π V

V

7"Bs9')Bs9')( 7"B(+totaltotal === V ρ m

The an"ular momentum e2uation can !e epresse as

∑∑∑ −=inout

V mr V mr $

where all moments in the countercloc7wise irection are positi0e, an

all in the cloc7wise irection are ne"ati0e. Then the an"ular momentum e2uation a!out the ais of rotation !ecomes

r V mr no44le' :T −=− or r V mr total'T =

3ol0in" for the relati0e 0elocity V r an su!stitutin",

mBs+.: +

mBs7"+

7"Bs)m)(9'9'.'(

m D'T &

total

' =

⋅⋅==

mr V r

Then the tan"ential an an"ular 0elocity of the no44les !ecomemBs.++9+.:E.++L 8etno44le =−=−= r V V V

no44le ++9 mBs

' 9 m

&L raBs *' s&L9+ rpm

& & + min

V )

r )

n

ω

ω

π π

= = =

= = = ≅ ÷

&

*12 ra's

*2.4 r:m

Therefore, this sprin7ler will rotate at &L9' re0olutions per minute (to three si"nificant i"its).

Discussion The actual rotation rate will !e somewhat lower than this ue to air friction as the arms rotate.


*-:9

Bs9'total =V

<lectric"enerator

ω

8etV

8etV

r V mno44ler V mno44le

•

r A 9' cm

r V mno44le

To A D' ⋅m




6"2

Solution 1 Kelton wheel is consiere for power "eneration in a hyroelectric power plant. 1 relation is to !eo!taine for power "eneration, an its numerical 0alue is to !e o!taine.

Assumptions The flow is uniform an cyclically steay. - The water is ischar"e to the atmosphere, an thus the "a"e

pressure at the no44le outlet is 4ero. 5riction an losses ue to air ra" of rotatin" components are ne"lecte. / The no44le

iameter is small compare to the moment arm, an thus we use a0era"e 0alues of raius an 0elocity at the outlet.


Analysis The tan"ential 0elocity of !uc7ets corresponin" to an an"ular 0elocity of nπ ω &= is ω r V = !uc7et .

Then the relati0e 0elocity of the 8et (relati0e to the !uc7et) !ecomes

ω r V V V V 1 1r −=−= !uc7et

e ta7e the ima"inary is7 that contains the Kelton wheel as the control 0olume. The inlet 0elocity of the flui into this

control 0olume is V r , an the component of outlet 0elocity normal to the moment arm is V r cosβ . The an"ular momentum

e2uation can !e epresse as ∑∑∑ −=inout

V mr V mr $ where all moments in the countercloc7wise irection are

positi0e, an all in the cloc7wise irection are ne"ati0e. Then the an"ular momentum e2uation a!out the ais of rotation

!ecomes

r r V mr V mr $ −=− β cosshaft or )cos+)(()cos+(shaft β ω β −−=−= r V mr V mr $ 1r

otin" that shaftshaftshaft & $ $ n* ω π == an V ρ =m , the shaft power output of a Kelton tur!ine !ecomes

)cos+)((shaft β ω ω ρ −−= r V r * 1V

which is the esire relation. 5or "i0en 0alues, the shaft power output is etermine to !e

00+-=

⋅°×=

mBs +'

M+)cos+*'-mBs)(++D.L+&-raBs)(D'L+.+D(m)&)(Bsm+')(7"Bm+'''(

*

::shaft*

where raBsL+.+Ds*'

min+

re0Bmin)+D'(&& =

== π π ω n

Discussion The actual power will !e somewhat lower than this ue to air ra" an friction. ote that this is the shaft power; the electrical power "enerate !y the "enerator connecte to the shaft is !e lower ue to "enerator inefficiencies.


*-:D

β

V 1 - r ω

V 1 - r ω

ω 3haft

r

o44le

r ω V 1

$ shaft




6"2-

Solution The pre0ious pro!lem is reconsiere. The effect of β on the power "eneration as β 0aries from '° to +'°

is to !e etermine, an the fraction of power loss at +*'° is to !e assesse.


rho=1000 "#g/m3"r=2 "m"V_dot=10 "m3/s"V_t=%0 "m/s",_dot=1%0 "rm"omg!=2*)*,_dot/60V_r=V_t-r*omg!m_dot=rho*V_dot_dot_sh!ft=m_dot*omg!*r*V_r*(1-cos(t!/156 "" _dot_m!7=m_dot*omg!*r*V_r*2/156 "" 58c),c=_dot_sh!ft/_dot_m!7

1n"le,

β °Ma power,

ma* , M

1ctual power,

shaft* , M

<fficiency,

η

'

+'

&':'

9'

D'*'

L'

'E'

+''

++'+&'

+:'+9'

+D'+*'+L'

+'

++.L

++.L

++.L++.L

++.L

++.L++.L

++.L

++.L++.L

++.L

++.L++.L

++.L++.L

++.L++.L++.L

++.L

'.''

'.'E

'.:D'.L

+.:L

&.'E&.E&

:.9

9.&D.9

*.D

L.9.L*

E.DE+'.:+

+'.E++.:&++.DE

++.*

'.'''

'.''

'.':''.'*L

'.++L

'.+LE'.&D'

'.:&E

'.9+:'.D''

'.DL

'.*L+'.LD'

'.&+'.:

'.E::'.EL''.EE&

+.'''

Discussion The efficiency of a Kelton wheel for β A+*'° is '.EL. Therefore, at this an"le, only : of the power is lost.


*-:*

0 20 40 60 80 100 120 140 160 180

0

2

4

6

8

10

12

β, °

$ s < a f t

Wmax

Wshaft




6"2

Solution 1 centrifu"al !lower is use to eli0er atmospheric air. 5or a "i0en an"ular spee an power input, the0olume flow rate of air is to !e etermine.

Assumptions The flow is steay in the mean. - Irre0ersi!le losses are ne"li"i!le. The tan"ential components of air

0elocity at the inlet an the outlet are sai to !e e2ual to the impeller 0elocity at respecti0e locations.

Properties The "as constant of air is '.&L 7Ka⋅m:B7"⋅O. The ensity of air at &'°C an ED 7Ka is

:

:7"Bm+.+:'

O)O)(&E:B7"m7Ka('.&L

7KaED=

⋅⋅==

R.

' ρ

Analysis In the ieali4e case of the tan"ential flui 0elocity !ein" e2ual to the !lae an"ular 0elocity !oth at the

inlet an the outlet, we ha0e +,+ r V t ω = an &,& r V t ω = , an the tor2ue is epresse as

)()()(T&

+&

&&

+&

&,++,&&shaft r r r r mV r V r m t t −=−=−= ω ρ ω V

where the an"ular 0elocity is

raBsL.:

s*'

min+ re0Bmin)''(&& =

== π π ω n

Then the shaft power !ecomes

)(T &+

&&

&shaftshaft r r * −== ω ρ ω V

3ol0in" for V an su!stitutin", the 0olume flow rate of air is etermine to

sm4+**.-=

⋅⋅=

−=

+

mBs7"+

Im)('.+D-m)J('.:'raBs))(:.L7"Bm(+.+:'

mBs +&'

)(

&

&&&:&+

&&

&

shaft

r r

*

ρω

V

The normal 0elocity components at the inlet an the outlet are

ms-+,4

ms-+54

===

===

m)m)('.':9:'.'(&

Bsm&&9.'

&

m)m)('.'*++D.'(&

Bsm&&9.'

&

:

&&

,&

:

++,+

π π

π π

br V

br V

n

n

V

V

Discussion ote that the irre0ersi!le losses are not consiere in this analysis. In reality, the flow rate an the normal

components of 0elocities will !e smaller.


*-:L

r +

r &

ω

t V ,&

Impeller re"ion

t V ,+

In

Impeller

!lae

3haft ω

r +

b&

b+ r &

Impeller

shrou

Impeller




6"2/

Solution 1 centrifu"al !lower is use to eli0er atmospheric air at a specifie rate an an"ular spee. The minimum power consumption of the !lower is to !e etermine.

Assumptions The flow is steay in the mean. - Irre0ersi!le losses are ne"li"i!le.

Properties The ensity of air is "i0en to !e +.&D 7"Bm:

. Analysis e ta7e the impeller re"ion as the control 0olume. The normal 0elocity components at the inlet an the

outlet are

mBs9.9&+m)m)('.'D*9D.'(&

BsmL'.'

&

mBs*.LE:m)m)('.'&&'.'(&

BsmL'.'

&

:

&&,&

:

++,+

===

===

π π

π π

br V

br V

n

n

V

V

The tan"ential components of a!solute 0elocity are$

α1 = 0°: 'tan +,+,+ == α nt V V

α2 = 60°: mBs&*E.DD'tan)mBs9&+.9(tan +,&,& =°== α nt V V

The an"ular 0elocity of the propeller is

raBs:'.L:s*'

min+ re0Bmin)L''(&& =

== π π ω n

7"BsLD.'Bs)m)('.L7"Bm&D.+( :: === V ρ m

ormal 0elocity components V +,n an V &,n as well pressure actin" on the inner an outer circumferential areas pass throu"h

the shaft center, an thus they o not contri!ute to tor2ue. nly the tan"ential 0elocity components contri!ute to tor2ue, an

the application of the an"ular momentum e2uation "i0es

m 'LD.&mBs7"+

+'I-mBs)m)(D.&*E9D7"Bs)J('.LD.'()(T

&,++,&&shaft ⋅=

⋅=−= t t V r V r m


$0,*=

⋅⋅==

mBs +

+m) 'LDraBs)(&.:'.L:(Tshaftω *

Discussion The actual re2uire shaft power is "reater than this, ue to the friction an other irre0ersi!ilities that we

ha0e ne"lecte in our analysis. e0ertheless, this is a "oo first approimation.


*-:

In

Impeller

!lae

3haft ω

r +

b&

b+ r &

Impeller

shrou

Impeller

r +

r &

ω

Impeller re"ion

α2= 50°

+V

&V




6"22

Solution The pre0ious pro!lem is reconsiere. The effect of ischar"e an"le α & on the minimum power input

re2uirements as α & 0aries from '° to D° in increments of D° is to !e in0esti"ate.


rho=1.2% "#g/m3"r1=0.20 "m"b1=0.092 "m"r2=0.4% "m"b2=0.0%6 "m"V_dot=0.:0 "m3/s"V1,=V_dot/(2*)*r1*b1 "m/s"V2,=V_dot/(2*)*r2*b2 "m/s"'lh!1=0V1t=V1,*t!,('lh!1 "m/s"V2t=V2,*t!,('lh!2 "m/s",_dot=:00 "rm"omg!=2*)*,_dot/60 "r!d/s"m_dot=rho*V_dot "#g/s"

;_sh!ft=m_dot*(r2*V2t-r1*V1t "+m"_dot_sh!ft=omg!*;_sh!ft ""

1n"le,

α 2° V &,t,

mBs

Tor2ue,

Tshaft, m

3haft power,

shaft* ,

'

D

+'+D

&'&D

:':D9'

9D

D'DD

*'

*DL'

LD

'D

'.''

'.:E

'.L+.+

+.*+&.'*

&.DD:.+':.L+

9.9&

D.&L*.:+

L.**

E.9+&.+D

+*.D'

&D.'LD'.D:

'.''

'.+D

'.:+'.9L

'.*:'.+

+.'++.&&+.9*

+.L9

&.'L&.9E

:.'&

:.L:9.L

*.D'

E.L+E.E'

'

++

&::9

9**'

L9E+'L

+&

+D&+&

&&+

&L9:D+

9L*

L&9+9DE

Discussion hen α & A ', the shaft power is also 4ero as epecte, since there is no turnin" at all. 1s α & approaches E'o,

the re2uire shaft power rises rapily towars infinity. e can ne0er reach α & A E'o

!ecause this woul mean 4ero flownormal to the outlet, which is impossi!le.


*-:E

0 10 20 30 40 50 60 70 80 0

0

200

400

600

800

1000

1200

1400

1600

α*

$ s < a f t 7 $




6"26(

Solution ater enters the impeller of a centrifu"al pump raially at a specifie flow rate an an"ular spee. Thetor2ue applie to the impeller is to !e etermine.



Analysis ater enters the impeller normally, an thus ',+ =t V . The tan"ential component of flui 0elocity at the

outlet is "i0en to !e ftBs+',& =t V . The inlet raius r + is un7nown, !ut the outlet raius is "i0en to !e r & A + ft. The

an"ular 0elocity of the propeller is

raBs:*.D&s*'

min+ re0Bmin)D''(&& =

== π π ω n

The mass flow rate is

l!mBs&.:Bs)ft)('B*'l!mBft9.*&( :: === V ρ m

nly the tan"ential 0elocity components contri!ute to tor2ue, an the application of the an"ular momentum e2uation "i0es

ftlbf .6, ⋅=

⋅=−= &,++,&&shaftftBsl!m:&.&

l!f +

'I-ftBs)ft)(+'l!mBs)J(+&.:()(T t t V r V r m

Discussion This shaft power input corresponin" to this tor2ue is

/$--+4=

⋅⋅===

ftBsl!f L:L.D*

7+ft)l!f DraBs)(9*:*.D&(& shaftshaft . . n* ω π

Therefore, the minimum power input to this pump shoul !e :: 7.


*-9'

In

Impeller

!lae

3haft ω

r +

b&

b+ r &

Impeller

shrou

Impeller

r +

r &

ω

t V ,&

Impeller re"ion

+V




6"23

Solution 1 centrifu"al pump is use to supply water at a specifie rate an an"ular spee. The minimum power consumption of the pump is to !e etermine.


Properties e ta7e the ensity of water to !e +''' 7"Bm:

. Analysis e ta7e the impeller re"ion as the control 0olume. The normal 0elocity components at the inlet an the

outlet are

mBs&.&L9m)m)('.':D:'.'(&

Bsm+D.'

&

mBs&.&E*m)m)('.''+:.'(&

Bsm+D.'

&

:

&&,&

:

++,+

===

===

π π

π π

br V

br V

n

n

V

V

The tan"ential components of a!solute 0elocity are$

α1 = 0°: 'tan +,+,+ == α nt V V

α2 = 60°: mBsE:.:*'tan)mBs&L9.&(tan +,&,& =°== α nt V V

The an"ular 0elocity of the propeller is

raBsL.+&Ds*'

min+ re0Bmin)+&''(&& =

== π π ω n

7"Bs+D'Bs)m)('.+D7"Bm+'''( :: === V ρ m

ormal 0elocity components V +,n an V &,n as well pressure actin" on the inner an outer circumferential areas pass throu"h

the shaft center, an thus they o not contri!ute to tor2ue. nly the tan"ential 0elocity components contri!ute to tor2ue, an

the application of the an"ular momentum e2uation "i0es

m7&.+LLmBs7"+'''

7+'I-mBs)m)(:.E::'7"Bs)J('.+D'()(T

&,++,&&shaft ⋅=

⋅=−= t t V r V r m


/$**+-=

⋅⋅==

mBs7+

7+m)7L.&raBs)(+LL.+&D(Tshaftω *

Discussion ote that the irre0ersi!le losses are not consiere in analysis. In reality, the re2uire power input will !elar"er.


*-9+

In

Impeller

!lae

3haft ω

r +

b&

b+ r &

Impeller

shrou

Impeller

r +

r &

ω

&V

Impeller re"ion

α2= 60°

+V




Re(ie% Problems

6"24 Solution ater is flowin" into an ischar"in" from a pipe >-section with a seconary ischar"e section normal toreturn flow. et x- an % - forces at the two flan"es that connect the pipes are to !e etermine.

Assumptions The flow is steay an incompressi!le. - The wei"ht of the >-turn an the water in it is ne"li"i!le. / Themomentum-flu correction factor for each inlet an outlet is "i0en to !e β A +.':.


Analysis The flow 0elocities of the : streams are

mBs:.+DI9Bm)'D.'()J7"Bm(+'''

7"Bs:'

)9B( &:&+

+

+

++ ====

π π ρ ρ &

m

mV

mBs'.&I9Bm)+'.'()J7"Bm(+'''

7"Bs&&

)9B( &:&&

&

&

&& ====

π π ρ ρ &

m

mV

mBs:.++I9Bm)':.'()J7"Bm(+'''

7"Bs

)9B( &:&:

:

:

:: ==== π π ρ ρ &

m

mV

e ta7e the entire >-section as the control 0olume. e esi"nate the hori4ontal coorinate !y x with the irection of incomin" flow as !ein" the positi0e irection an the 0ertical coorinate !y % . The momentum e2uation for steay one-

imensional flow is ∑∑∑ −=inout

V mV m ! β β . e let the x- an %- components of the anchorin" force of the cone

!e ! Rx an ! R% , an assume them to !e in the positi0e irections. Then the momentum e2uations alon" the x an % aes !ecome

'-'

)( )(

::::

++&&&&++++&&&&++

V m ! V m !

V mV m ' ' ! V mV m ' ' !

R% R%

Rx Rx

β

β β β

=→=++−−−=→−−=++


N2--−=−=

⋅+

⋅−

−−−−=

7 '.L::

mBs7"+'''

7+mBs)7"Bs)(+D.::'(

mBs7"+'''

7+mBs)7"Bs)(&.'&&(':.+

9

m)('.+'I7Bm)+''+D'J(

9

m)('.'DI7Bm)+''&''J(

&&

&&

&& π π

Rx !

N5-+0=

⋅=

&mBs7"+

+mBs):7"Bs)(++.(':.+ R% !

The ne"ati0e 0alue for ! Rx inicates the assume irection is wron", an shoul !e re0erse. Therefore, a force of L::

acts on the flan"es in the opposite irection. 1 0ertical force of E:.+ acts on the flan"e in the 0ertical irection.

Discussion To assess the si"nificance of "ra0ity forces, we estimate the wei"ht of the wei"ht of water in the >-turn ancompare it to the 0ertical force. 1ssumin" the len"th of the >-turn to !e '.D m an the a0era"e iameter to !e L.D cm, the

mass of the water !ecomes

7"&.&m)('.D9

m)('.'LD)7"Bm+'''(

9

&:

&

===== π π

ρ ρ ρ 3 &

3m V

whose wei"ht is &.&×E.+ A && , which is much less than E:.+, !ut still si"nificant. Therefore, isre"arin" the

"ra0itational effects is a reasona!le assumption if "reat accuracy is not re2uire.


*-9&

7"Bs

+

&:

&& 7"Bs

:' 7"Bs

! R%

! Rx





*-9:




6"28

Solution 1 fireman was hit !y a no44le hel !y a tripo with a rate holin" force. The accient is to !e in0esti"ate !y calculatin" the water 0elocity, the flow rate, an the no44le 0elocity.


of the water 8et is the atmospheric pressure, which is isre"are since it acts on all surfaces. Gra0itational effects an

0ertical forces are isre"are since the hori4ontal resistance force is to !e etermine. / et flow is nearly uniform an



Analysis e ta7e the no44le an the hori4ontal portion of the hose as the system such that water enters the control0olume 0ertically an outlets hori4ontally (this way the pressure force an the momentum flu at the inlet are in the 0ertical

irection, with no contri!ution to the force !alance in the hori4ontal irection, an esi"nate the entrance !y + an the

outlet !y &. e also esi"nate the hori4ontal coorinate !y x (with the irection of flow as !ein" the positi0e irection).


V mV m ! β β . e let the

hori4ontal force applie !y the tripo to the no44le to hol it !e ! Rx, an assume it to !e in the positi0e x irection. Then the

momentum e2uation alon" the x irection !ecomes

&&

:&

&&

9

m)('.'D)7"Bm+'''(

+

mBs7"+ )(+''

9' V V

& VV V mV m ! e Rx

π π ρ ρ =

⋅→===−=

3ol0in" for the water outlet 0elocity "i0es V A :. m=s. Then the water flow rate !ecomes

sm4+4,5,-==== mBs)(:'.:

9

m)('.'D

9

&& π π V

& V V

hen the no44le was release, its acceleration must ha0e !een

mBs+' +

mBs7"+

7"+'

+'' &

&

no44leno44le =

⋅==

m

! a

1ssumin" the reaction force actin" on the no44le an thus its acceleration to remain constant, the time it ta7es for the

no44le to tra0el *' cm an the no44le 0elocity at that moment were (note that !oth the istance x an the 0elocity V are 4ero

at time t A ')

sa

xt at x '+*.'mBs+'

m)*.'(&& &

&

&+ ===→=

ms0.+2=== s)'+*.')(mBs+'( &at V

Thus we conclue that the no44le hit the fireman with a 0elocity of +9.L mBs.

Discussion <n"ineerin" analyses such as this one are fre2uently use in accient reconstruction cases, an they oftenform the !asis for 8u"ment in courts.


*-99

Tripo

o44le

& A D cm

! Rx




6"6:

Solution =urin" lanin" of an airplane, the thrust re0erser is lowere in the path of the ehaust 8et, which eflects theehaust an pro0ies !ra7in". The thrust of the en"ine an the !ra7in" force prouce after the thrust re0erser is eploye

are to !e etermine.

Assumptions The flow of ehaust "ases is steay an one-imensional. - The ehaust "as stream is epose to the

atmosphere, an thus its pressure is the atmospheric pressure. The 0elocity of ehaust "ases remains constant urin"

re0ersin". / et flow is nearly uniform an thus the momentum-flu correction factor can !e ta7en to !e unity, β ≅ +.

Analysis (a) The thrust eerte on an airplane is simply the momentum flu of the com!ustion "ases in the re0erse

irection,

N.,44=

⋅==

&mBs7"+

+mBs)7"Bs)(&D'+(Thrust exexV m

(b) e ta7e the thrust re0erser as the control 0olume such that it cuts throu"h !oth ehaust streams normally an the

connectin" !ars to the airplane, an the irection of airplane as the positi0e irection of x ais. The momentum e2uation for

steay one-imensional flow in the x irection reuces to

∑∑∑ −=inout

V mV m ! β β →

)&'cos+( )()cos&'( i Rx Rx V m ! V mV m ! °+=→−−°=

3u!stitutin", the reaction force is etermine to !e

L&EmBs)7"Bs)(&D'+)(&'cos+( =°+= Rx !

The !rea7in" force actin" on the plane is e2ual an opposite to this force,

!rea7in" L&E ! = ≅ 12-4 N

Therefore, a !ra7in" force of L:' e0elops in the opposite irection to fli"ht.

Discussion This pro!lem can !e sol0e more "enerally !y measurin" the re0ersin" an"le from the irection of ehaust

"ases (α A ' when there is no re0ersin"). hen α P E'°, the re0erse "ases are ischar"e in the ne"ati0e x irection, an

the momentum e2uation reuces to

)cos+( )()cos( i Rx Rx V m ! V mV m ! α α −=→−−−=

This e2uation is also 0ali for α NE'° since cos(+'°-α) A - cosα. >sin" α A +*'°, for eample, "i0es

)&'cos+()+*'cos+( ii Rx V mV m ! +=−= , which is ientical to the solution a!o0e.


*-9D

+*'°&D' mBs

! Rx

α A +*'°

! Rx

V m

V m

x

Control0olume




6"6

Solution The pre0ious pro!lem is reconsiere. The effect of thrust re0erser an"le on the !ra7in" force eerte on

the airplane as the re0erser an"le 0aries from ' (no re0ersin") to +'° (full re0ersin") in increments of +'° is to !e

in0esti"ate.


V_t=2%0 "m/s"

m_dot=19 "#g/s"F_R7=(1-cos(!lh!*m_dot*V_t "+"

6e0ersin"

an"le,

α °

#ra7in" force

! !ra7e,

'

+'

&'

:'9'

D'

*'L'

'

E'

+''++'

+&'

+:'

+9'+D'

+*'

+L'

+'

'

*

&L+

*':+'D:

+*'L

&&D'&E*+

:L+E

9D''

D&+*':E

*LD'

L:E:

LE9L:EL

L&E

E:&

E'''

Discussion 1s epecte, the !ra7in" force is 4ero when the an"le is 4ero (no eflection), an maimum when the an"le

is +'o (completely re0erse). f course, it is impossi!le to completely re0erse the flow, since the 8et ehaust cannot !eirecte !ac7 into the en"ine.


*-9*

0 20 40 60 80 100 120 140 160 180

0

1000

2000

3000

4000

5000

6000

7000

8000

000

α, °

# b r a / e 7

N




6"6-(

Solution The roc7et of a spacecraft is fire in the opposite irection to motion. The eceleration, the 0elocity chan"e,an the thrust are to !e etermine.

Assumptions The flow of com!ustion "ases is steay an one-imensional urin" the firin" perio, !ut the fli"ht of the

spacecraft is unsteay. - There are no eternal forces actin" on the spacecraft, an the effect of pressure force at the no44le

outlet is ne"li"i!le. The mass of ischar"e fuel is ne"li"i!le relati0e to the mass of the spacecraft, an thus the spacecraft

may !e treate as a soli !oy with a constant mass. / The no44le is well-esi"ne such that the effect of the momentum-

flu correction factor is ne"li"i!le, an thus β ≅ +.

Analysis (a) e choose a reference frame in which the control0olume mo0es with the spacecraft. Then the 0elocities of flui steams

!ecome simply their relati0e 0elocities (relati0e to the mo0in" !oy).

e ta7e the irection of motion of the spacecraft as the positi0eirection alon" the x ais. There are no eternal forces actin" on the

spacecraft, an its mass is nearly constant. Therefore, the spacecraft can

!e treate as a soli !oy with constant mass, an the momentum

e2uation in this case is

f f 4V

V mdt

V d mV mV m

dt

V md

−=→−+= ∑∑ space

space

inout

)(

' β β

otin" that the motion is on a strai"ht line an the ischar"e "ases

mo0e in the positi0e x irection (to slow own the spacecraft), we write

the momentum e2uation usin" ma"nitues as

f

f

f f V m

m

dt

dV V m

dt

dV m

space

spacespace

space

−=→−=

3u!stitutin", the eceleration of the spacecraft urin" the first D secons is etermine to !e

fts.0+2*-ftBs)(D'''

l!m+,'''

l!mBs+D'

space

space

space =−=−== f

f V

m

m

dt

dV a

(b) Onowin" the eceleration, which is constant, the 0elocity chan"e of the spacecraft urin" the first D secons is

etermine from the efinition of acceleration dt dV a Bspacespace = to !e

fts*45−==∆=∆→= )sD)(ftBs(-9+.L &spacespacespacespace t aV dt adV

(c) The thrust eerte on the system is simply the momentum flu of the com!ustion "ases in the re0erse irection,

( ) ( )&

+ l!f Thrust +D' l!mBs D''' ftBs &: &E' l!f

:&.& l!m ftBs R f f ! m V (

= = − = − = − ≅ − ÷× & *-7-44 lbf

Therefore, if this spacecraft were attache somewhere, it woul eert a force of &:,:'' l!f (e2ui0alent to the wei"ht of

&:,:'' l!m of mass on earth) to its support in the ne"ati0e x irection.

Discussion In Kart (b) we approimate the eceleration as constant. Howe0er, since mass is lost from the spacecrafturin" the time in which the 8et is on, a more accurate solution woul in0ol0e sol0in" a ifferential e2uation. Here, the time

span is short, an the lost mass is li7ely ne"li"i!le compare to the total mass of the spacecraft, so the more complicate

analysis is not necessary.


*-9L

+D'' ftBs

+''' l!m

+D' l!mBsD''' ftBs

x




6"6

Solution 1 hori4ontal water 8et stri7es a 0ertical stationary flat plate normally at a specifie 0elocity. 5or a "i0enflow 0elocity, the anchorin" force neee to hol the plate in place is to !e etermine.

Assumptions The flow is steay an incompressi!le. - The water splatters off the sies of the plate in a plane normal to

the 8et. The water 8et is epose to the atmosphere, an thus the pressure of the water 8et an the splattere water is the

atmospheric pressure which is isre"are since it acts on the entire control surface. / The 0ertical forces an momentum

flues are not consiere since they ha0e no effect on the hori4ontal reaction force. 2 et flow is nearly uniform an thus the

momentum-flu correction factor can !e ta7en to !e unity, β ≅ +.


Analysis e ta7e the plate as the control 0olume such that it contains the entire plate an cuts throu"h the water 8et

an the support !ar normally, an the irection of flow as the positi0e irection of x ais. e ta7e the reaction force to !e in

the ne"ati0e x irection. The momentum e2uation for steay one-imensional flow in the x (flow) irection reuces in this

case to

∑∑∑ −=inout

V mV m ! β β → V m ! V m ! Rxii Rx =→−=−

e note that the reaction force acts in the opposite irection to flow, an we shoul not for"et the ne"ati0e si"n for forces

an 0elocities in the ne"ati0e x-irection. The mass flow rate of water is

7"BsE'.DmBs)(:'

9

m)('.'D)7"Bm+'''(

9

&:

&

===== π π

ρ ρ ρ V &

V m V

3u!stitutin", the reaction force is etermine to !e

( ) ( )D E' 7"Bs :' mBs +L*L Rx ! )= = ≅ 0224 N

Therefore, a force of +LL' must !e applie to the plate in the opposite irection

to the flow to hol it in place.

Discussion In reality, some water may !e scattere !ac7, an this woul a to the reaction force of water.


*-9

:' mBs

D cm

! Rx




6"6/

Solution 1 water 8et hits a stationary cone, such that the flow is i0erte e2ually in all irections at 9D °. The force

re2uire to hol the cone in place a"ainst the water stream is to !e etermine.


of the water 8et !efore an after the split is the atmospheric pressure which is isre"are since it acts on all surfaces. The

"ra0itational effects are isre"are. / et flow is nearly uniform an thus the momentum-flu correction factor can !e

ta7en to !e unity, β ≅ +.


Analysis The mass flow rate of water 8et is

7"BsE'.DmBs)(:'9

m)('.'D)7"Bm+'''(

9

&:

&

===== π π

ρ ρ ρ V &

V m V

e ta7e the i0ertin" section of water 8et, incluin" the cone as the control 0olume, an esi"nate the entrance !y + an the

outlet after i0er"ence !y &. e also esi"nate the hori4ontal coorinate !y x with the irection of flow as !ein" the

positi0e irection an the 0ertical coorinate !y y.


V mV m ! β β . e let the x- an y-

components of the anchorin" force of the cone !e ! Rx an ! Ry, an assume them to !e in the positi0e irections. otin" that

V & A V + A V an mmm == +& , the momentum e2uations alon" the x an y aes !ecome

ais)a!outsymmetryof (!ecause '

)+(coscos +&

=−=−=

Ry

Rx

!

V mV mV m ! θ θ


4

N,01

=−=

⋅°=

Ry

Rx

!

! &mBs7"+

++)-mBs)(cos9D7"Bs)(:'E'.D(

The ne"ati0e 0alue for ! Rx inicates that the assume irection is wron", an shoul !e re0erse. Therefore, a force of D+

must !e applie to the cone in the opposite irection to flow to hol it in place. o holin" force is necessary in the

0ertical irection ue to symmetry an ne"lectin" "ra0itational effects.

Discussion In reality, the "ra0itational effects will cause the upper part of flow to slow own an the lower part tospee up after the split. #ut for short istances, these effects are ne"li"i!le.


*-9E

:' mBs

D cm

! Rx




6"62

Solution 1n ice s7ater is holin" a flei!le hose (essentially wei"htless) which irects a stream of water hori4ontallyat a specifie 0elocity. The 0elocity an the istance tra0ele in D secons, an the time it ta7es to mo0e D m an the

0elocity at that moment are to !e etermine.

Assumptions 5riction !etween the s7ates an ice is ne"li"i!le. - The flow of water is steay an one-imensional (!ut

the motion of s7ater is unsteay). The ice s7atin" arena is le0el, an the water 8et is ischar"e hori4ontally. / The mass

of the hose an the water in it is ne"li"i!le. 2 The s7ater is stanin" still initially at t A '. 6 et flow is nearly uniform an



Analysis (a) The mass flow rate of water throu"h the hose is

7"Bs+9.:mBs)(+'9

m)('.'&)7"Bm+'''(

9

&:

&

==== π π

ρ ρ V &

V m

The thrust eerte on the s7ater !y the water stream is simply the momentum flu of the water stream, an it acts in the

re0erse irection,

(constant) 9.:+mBs7"+

+mBs)7"Bs)(+'+9.:(Thrust

& =

⋅=== V m !

The acceleration of the s7ater is etermine from ewton/s &n law of motion ! A ma where m is the mass of the s7ater,

&&

mBs'.D&: +

mBs7"+

7"*'

9.:+=

⋅==

m

! a

ote that thrust an thus the acceleration of the s7ater is constant. The 0elocity of the s7ater an the istance tra0ele in D s

are

m6+,.

ms*+6*

===

===&&

&+&

&+

&s7ater

s)D)(mBs'.D&:(

s)D)(mBs'.D&:(

at x

at V

(b) The time it will ta7e to mo0e D m an the 0elocity at that moment are

ms*+-

s.+.

======→=

s)9.9)(mBs'.D&:(

mBs'.D&:

m)D(&&

&

s7ater

&

&

&

+

at V

a

x

t at x

Discussion In reality, the 0elocity of the s7ater will !e lower !ecause of friction on ice an the resistance of the hose to

follow the s7ater. 1lso, in the V mβ epressions, V is the flui stream spee relati0e to a fie point. Therefore, the

correct epression for thrust is )( s-ater 1et V V m ! −= , an the analysis a!o0e is 0ali only when the s7ater spee is low

relati0e to the 8et spee. 1n eact analysis woul result in a ifferential e2uation.


*-D'

+' mBs

Hose

=A& cm

!




6"66

Solution Iniana ones is to ascen a !uilin" !y !uilin" a platform, an mountin" four water no44les pointin"own at each corner. The minimum water 8et 0elocity neee to raise the system, the time it will ta7e to rise to the top of

the !uilin" an the 0elocity of the system at that moment, the aitional rise when the water is shut off, an the time he

has to 8ump from the platform to the roof are to !e etermine.

Assumptions The air resistance is ne"li"i!le. - The flow of water is steay an one-imensional (!ut the motion of

platform is unsteay). The platform is still initially at t A '. / et flow is nearly uniform an thus the momentum-flu

correction factor can !e ta7en to !e unity, β ≅ +.


Analysis (a) The total mass flow rate of water throu"h the 9 hoses an the total wei"ht of the platform are

7"Bs++mBs)(+D9

m)('.'D)7"Bm+'''(9

99

&:

&

==== π π

ρ ρ V &

V m

+9L&mBs7"+

+)mBs7")(E.++D'(

&

&

=

⋅== mg *

e ta7e the platform as the system. The momentum e2uation for steay one-imensional flow is

∑∑∑ −=inout

V mV m ! β β . The minimum water 8et 0elocity neee to raise the platform is etermine !y settin"

the net force actin" on the platform e2ual to 4ero,

&min

&

minminminmin99 ')( V

&V V V m* V m*

π ρ ρ ===→−−=−

3ol0in" for V min an su!stitutin",

ms0-+2=

⋅==

+

mBs+7"

m)('.'D)7"Bm+'''(

+9L&&

&:&minπ ρπ &

* V

(b) e let the 0ertical reaction force (assume upwars) actin" on the platform !e ! R% . Then the momentum e2uation in the0ertical irection !ecomes

&E+

mBs+7"mBs)7"Bs)(+D(++- )(+9L& ')(

&

−=

⋅=−=→=−−=− V m* ! V mV m* ! R% R%

The upwar thrust actin" on the platform is e2ual an opposite to this reaction force, an thus ! A &E . Then theacceleration an the ascenin" time to rise +' m an the 0elocity at that moment !ecome

&&

mBs'.& +

mBs7"+

7"+D'

&E=

⋅==

m

! a

s-+*===→=&

&&+

mBs&

m)+'(&&

a

xt at x

ms6+.=== s)&.:)(mBs&( &at V

(c) hen water is shut off at +' m hei"ht (where the 0elocity is *.9 mBs), the

platform will ecelerate uner the influence of "ra0ity, an the time it ta7es tocome to a stop an the aitional rise a!o0e +' m !ecome

s4+6,===→=−=&

''

mBsE.+

mBs9.* '

g

V t gt V V

m*+0==−= &&

&+&

&+

' s))('.*DmBs(E.+-s)mBs)('.*D9.*( gt t V %

Therefore, ones has &×'.*D A +.: s to 8ump off from the platform to the roof since it ta7es another '.*D s for the platform to

escen to the +' m le0el.

Discussion i7e most stunts in the Iniana ones mo0ies, this woul not !e practical in reality.


*-D+

+D mBs

&AD cm

%

x

! Ry




6"63(

Solution 1 !o-enclose fan is face own so the air !last is irecte ownwars, an it is to !e ho0ere !yincreasin" the !lae rpm. The re2uire !lae rpm, air outlet 0elocity, the 0olumetric flow rate, an the minimum

mechanical power are to !e etermine.

Assumptions The flow of air is steay an incompressi!le. - The air lea0es the !laes at a uniform 0elocity at

atmospheric pressure, an thus the momentum-flu correction factor can !e ta7en to !e unity, β ≅ +. 1ir approaches the

!laes from the top throu"h a lar"e area at atmospheric pressure with ne"li"i!le 0elocity. / The frictional effects arene"li"i!le, an thus the entire mechanical power input is con0erte to 7inetic ener"y of air (no con0ersion to thermal ener"y

throu"h frictional effects). 2 The chan"e in air pressure with ele0ation is ne"li"i!le !ecause of the low ensity of air. 6There is no acceleration of the fan, an thus the lift "enerate is e2ual to the total wei"ht.

Properties The ensity of air is "i0en to !e '.'L l!mBft:.


with air enterin" throu"h the lar"e cross-section (section +) at the top an the fan locate at the narrow cross-section at the !ottom (section &), an let its centerline !e the % ais with upwars !ein" the positi0e irection.


V mV m ! β β . otin" that the

only force actin" on the control 0olume is the total wei"ht * an it acts in the ne"ati0e % irection, the momentum e2uationalon" the % ais "i0es

)( ')( &&

&&&&&

* V V V V V m* V m*

ρ ρ ρ =→===→−−=−

where is the !lae span area,

&&&ft'*E.L9Bft):(9B === π π &

Then the ischar"e 0elocity to prouce D l!f of upwar force !ecomes

fts02+0=

⋅=

l!f +

ftBsl!m:&.&

)ft)(L.'*El!mBft('.'L

l!f D&

&:&V

(b) The 0olume flow rate an the mass flow rate of air are etermine from their

efinitions,

sft 0*0-=== ftBs)+.+L)(ft'*E.L(

&& V V

l!mBs9:.EBs)ft+&+)(l!mBft'L.'( :: === V ρ m

(c) otin" that ' + A ' & A ' atm, V + ≅ ', the ele0ation effects are ne"li"i!le, an the frictional effects are isre"are, the

ener"y e2uation for the selecte control 0olume reuces to

lossmech,tur!ine&

&&&

u pump,+

&++

&& , * g%

V ' m* g%

V ' m ++

++=+

++

ρ ρ →

&

&&

ufan,

V m* =

3u!stitutin",

$6.+-=

⋅

⋅==

ftBsl!f '.L:LD*

+

ftBsl!m&.:&

l!f +

&

ftBs)(+.'l!mBs)9:.E(

& &

&&&

ufan,

V m*

Therefore, the minimum mechanical power that must !e supplie to the air stream is *9.: .

Discussion The actual power input to the fan will !e consiera!ly lar"er than the calculate power input !ecause of the

fan inefficiency in con0ertin" mechanical wor7 to 7inetic ener"y.


*-D&

*'' rpm

! Ry

+

&




6"64

Solution 1 parachute slows a solier from his terminal 0elocity V T to his lanin" 0elocity of V 5. 1 relation is to !ee0elope for the solier/s 0elocity after he opens the parachute at time t A '.

Assumptions The air resistance is proportional to the 0elocity s2uare (i.e. ! A -V &). - The 0ariation of the air

properties with altitue is ne"li"i!le. The !uoyancy force applie !y air to the person (an the parachute) is ne"li"i!le

!ecause of the small 0olume occupie an the low ensity of air. / The final 0elocity of the solier is e2ual to its terminal

0elocity with his parachute open.

Analysis The terminal 0elocity of a free fallin" o!8ect is reache when the air resistance (or air ra") e2uals the

wei"ht of the o!8ect, less the !uoyancy force applie !y the flui, which is ne"li"i!le in this case,

&

&anceair resist

!

! V

mg - mg -V * ! =→=→=

This is the esire relation for the constant of proportionality . hen the parachute is eploye an the solier starts to

ecelerate, the net ownwar force actin" on him is his wei"ht less the air resistance,

−=−=−=−=

&

&&

&

&anceair resistnet +

! ! V

V mg V

V

mg mg -V mg ! * !

3u!stitutin" it into ewton/s &n law relationdt

dV mma ! ==net "i0es

dt

dV m

V

V mg

! =

− &

&

+

Cancelin" m an separatin" 0aria!les, an inte"ratin" from t A ' when V A V . tot A t when V A V "i0es

gdt V V

dV

!

=− && B+

→ dt V

g

V V

dV t

! ! . ∫ ∫ =

−

'&

Q

Q &&

>sin" xa

xa

a xa

dx

−+=

−∫ ln&

+

&& from inte"ral ta!les an applyin" the

inte"ration limits,

&lnln

&

+

! . !

. !

!

!

! V

gt

V V

V V

V V

V V

V =

−

+

−−

+

6earran"in", the 0elocity can !e epresse eplicitly as a function of time as

!

!

V gt ! . ! .

V gt ! . ! .

! eV V V V

eV V V V V V

B&

B&

)(

)(−

−

−−+

−++=

Discussion ote that as t → ∞, the 0elocity approaches the lanin" 0elocity of V ! , as epecte.


*-D:

Karachute

! air resistance

* A m"




6"68

Solution 1n empty cart is to !e ri0en !y a hori4ontal water 8et that enters from a hole at the rear of the cart. 1relation is to !e e0elope for cart 0elocity 0ersus time.

Assumptions The flow of water is steay, one-imensional, incompressi!le, an hori4ontal. - 1ll the water which enters

the cart is retaine. The path of the cart is le0el an frictionless. / The cart is initially empty an stationary, an thus V A

' at time t A '. 2 5riction !etween water 8et an air is ne"li"i!le, an the entire momentum of water 8et is use to ri0e the

cart with no losses. 6 et flow is nearly uniform an thus the momentum-flu correction factor can !e ta7en to !e unity, β ≅

+.

Analysis e note that the water 8et 0elocity V is constant, !ut the car 0elocity V is 0aria!le. otin" that)( V V m 5 −= ρ where is the cross-sectional area of the water 8et an V - V is the 0elocity of the water 8et relati0e to

the cart, the mass of water in the cart at any time t is

∫ ∫ ∫ −=−==t

5

t

5

t

w Vdt t V dt V V dt mm

'

'

')( ρ ρ ρ (+)

1lso,

)( V V mdt

dm 5

w −== ρ

e ta7e the cart as the mo0in" control 0olume. The net force actin" on the cart in this case is e2ual to the momentum flu

of the water 8et. ewton/s &n law ! A ma A d (mV 7=dt in this case can !e epresse as

dt

V md !

)( total= where 5 5 5 in V V V V mV mV mV m ! )()(

outin

−===−= ∑∑ ρ β β

an

V V V dt

dV mm

dt

dmV

dt

dV m

dt

dV m

dt

V md

dt

dV m

dt

V mmd

dt

V md

5 wc

wwc

wc

w

)()(

)(I)J()( ctotal

−++=

++=+=+

=

ρ

ote that in V mβ epressions, we use the flui stream 0elocity relati0e to a fie point. 3u!stitutin",

V V V dt

dV mmV V V 5 wc 5 5 )()()( −++=− ρ ρ → dt

dV mmV V V V wc 5 5 )())(( +=−− ρ

otin" that mw is a function of t (as "i0en !y <2. +) an separatin" 0aria!les,

wc 5 mm

dt

V V

dV

+=

− &)( ρ →

∫ −+=

− t

5 c 5 Vdt t V m

dt

V V

dV

'

&)( ρ ρ ρ

Inte"ratin" from t A ' when V A ' to t A t when V A V "i0es the esire inte"ral,

∫ ∫

∫ −+

=−

t

ot

5 c

V

5 Vdt t V m

dt

V V

dV

'

' &)( ρ ρ ρ

Discussion ote that the time inte"ral in0ol0es the inte"ral of 0elocity, which complicates the solution.


*-D9

V

Cartm

6

ater8et




6"3:

Solution 1 plate is maintaine in a hori4ontal position !y frictionless 0ertical "uie rails. The unersie of the plate

is su!8ecte to a water 8et. The minimum mass flow rate minm to 8ust le0itate the plate is to !e etermine, an a relation is

to !e o!taine for the steay state upwar 0elocity. 1lso, the inte"ral that relates 0elocity to time when the water is firstturne on is to !e o!taine.

Assumptions The flow of water is steay an one-imensional. - The water 8et splatters in the plane of he plate. The

0ertical "uie rails are frictionless. / Times are short, so the 0elocity of the risin" 8et can !e consiere to remain constant

with hei"ht. 2 1t time t A ', the plate is at rest. 6 et flow is nearly uniform an thus the momentum-flu correction factor

can !e ta7en to !e unity, β ≅ +.

Analysis (a) e ta7e the plate as the system. The momentum e2uation for steay one-imensional flow is

∑∑∑ −=inout

V mV m ! β β . otin" that 5 V m ρ = where is the cross-sectional area of the water 8et an * A

m p g , the minimum mass flow rate of water neee to raise the plate is etermine !y settin" the net force actin" on the platee2ual to 4ero,

g mm V mm g mV m* V m* p p ρ =→=→=→−=− minGminminGminGmin )B( '

5or minmm > , a relation for the steay state upwar 0elocity V is o!taine settin" the upwar impulse applie !y water 8et

to the wei"ht of the plate (urin" steay motion, the plate 0elocity V is constant, an the 0elocity of water 8et relati0e to

plate is V ? V ),

g m

mV

g mV V V V g mV V m*

p p

5 5 p 5 ρ ρ ρ

ρ −=→=−→−=→−= )( )(

&

(b) 1t time t A ' the plate is at rest (V A '), an it is su!8ecte to water

8et with minmm > an thus the net force actin" on it is "reater than the

wei"ht of the plate, an the ifference !etween the 8et impulse an the

wei"ht will accelerate the plate upwars. Therefore, ewton/s &n law ! A ma A mdV0dt in this case can !e epresse as

dt

dV m g mV V am* V V m p p 5 p 5 =−−→=−− &

)( )( ρ

3eparatin" the 0aria!les an inte"ratin" from t A ' when V A ' to t A t

when V A V "i0es the esire inte"ral,

Q

& ' '

t p

t 5 p

m dV dt

+V V 7 m g ρ == →

− −∫ ∫ Q

& '

p

5 p

m dV t

+V V 7 m g ρ =

− −∫

Discussion This inte"ral can !e performe with the help of inte"ral ta!les. #ut therelation o!taine will !e implicit in V .


*-DD

m p

Guie

rails

m

o44le

! R%

* A m p g

)




6"3

Solution ater enters a centrifu"al pump aially at a specifie rate an 0elocity, an lea0es at an an"le from the aialirection. The force actin" on the shaft in the aial irection is to !e etermine.

Assumptions The flow is steay an incompressi!le. - The forces actin" on the pipin" system in the hori4ontal

irection are ne"li"i!le. The atmospheric pressure is isre"are since it acts on all surfaces. / ater flow is nearly

uniform at the outlet an thus the momentum-flu correction factor can !e ta7en to !e unity, β ≅ +.


Analysis 5rom conser0ation of mass we ha0e mmm

== &+ , anthus &+ V V = an &&++ V V cc = . otin" that the ischar"e area is

half the inlet area, the ischar"e 0elocity is twice the inlet 0elocity. That is,

mBs+')mBsD(&& ++&

+&+ ==== V V

V

c

cc

e ta7e the pump as the control 0olume, an the inlet irection of flow as

the positi0e irection of x ais. The linear momentum e2uation in this case

in the x irection reuces to

∑∑∑ −=inout

V mV m ! β β → )cos( cos &++& θ θ V V m ! V mV m ! Rx Rx −=→−=−

where the mass flow rate it

7"Bs&''Bs)m)('.&'7"Bm+'''( :: === V ρ m

3u!stitutin" the 7nown 2uantities, the reaction force is etermine to !e (note that cos*'° A '.D)

4=

⋅=

&6mBs7"+

+mBs)cos*'I(+'-mBs)7"Bs)J(D&''( !

Discussion ote that at this an"le of ischar"e, the !earin" is not su!8ecte to any hori4ontal loain". Therefore, theloain" in the system can !e controlle !y a8ustin" the ischar"e an"le.

6"3- Solution ater enters the impeller of a tur!ine throu"h its outer e"e of iameter & with 0elocity V ma7in" an an"le

α with the raial irection at a mass flow rate of m

, an lea0es the impeller in the raial irection. The maimum power that can !e "enerate is to !e shown to !e shaft sin* nm&V π α =& && .

Assumptions The flow is steay in the mean. - Irre0ersi!le losses arene"li"i!le.

Analysis e ta7e the impeller re"ion as the control 0olume. The

tan"ential 0elocity components at the inlet an the outlet are ',+ =t V an

α sin,& V V t = .

ormal 0elocity components as well pressure actin" on the inner an

outer circumferential areas pass throu"h the shaft center, an thus they o not

contri!ute to tor2ue. nly the tan"ential 0elocity components contri!ute to

tor2ue, an the application of the an"ular momentum e2uation "i0es

&B)sin(')(T ,&&,++,&&shaft α V &mV r mV r V r m t t t =−=−=

The an"ular 0elocity of the propeller is nπ ω &= . Then the shaft power

!ecomes

&B)sin(&Tshaftshaft α π ω V &mn* ==

3implifyin", the maimum power "enerate !ecomes α π sinshaft &V mn* = which is the esire relation.

Discussion The actual power is less than this ue to irre0ersi!le losses that are not ta7en into account in our analysis.


*-D*

n

*' °

! Rx

%

x

V m

r & 8&02

ω

V

Impeller re"ion

α

V

Tshaft




6"3

Solution 1 two-arme sprin7ler is use to water a "aren. 5or specifie flow rate an ischar"e an"les, the rates of rotation of the sprin7ler hea are to !e etermine.

Assumptions The flow is uniform an cyclically steay (i.e., steay from a frame of reference rotatin" with the

sprin7ler hea). - The water is ischar"e to the atmosphere, an thus the "a"e pressure at the no44le outlet is 4ero.

5rictional effects an air ra" of rotatin" components are ne"lecte. / The no44le iameter is small compare to themoment arm, an thus we use a0era"e 0alues of raius an 0elocity at the outlet.


Analysis e ta7e the is7 that encloses the sprin7ler arms as the control 0olume, which is a stationary control

0olume. The conser0ation of mass e2uation for this steay flow system is mmm == &+ . otin" that the two no44les are

ientical, we ha0e &Bno44le mm = or no44le total & 0 =V V & & since the ensity of water is constant. The a0era"e 8et outlet 0elocity

relati0e to the no44le is

mBs9E.ED+'''

m+

I9Bm)'&.'(J&

Bs*':

& 8et

no44le 8et =

==

π V

V

The an"ular momentum e2uation can !e epresse as ∑∑∑ −=inout

V mr V mr $ . otin" that there are no eternal

moments actin", the an"ular momentum e2uation a!out the ais of rotation !ecomesθ cos&' no44le r V mr −= → '=r V → 'no44let 8et, =−V V

otin" that the tan"ential component of 8et 0elocity is θ cos 8ett 8et, V V = , we ha0e

θ θ mBs)cos9E.ED(cos 8etno44le ==V V

1lso notin" that r nr V π ω &no44le == , an an"ular spee an the rate of rotation of sprin7ler hea !ecome

+) θ A '°$( )no44leED 9E mBs cos' &+& raBs *' s

an &'&* rpm' 9D m & & + min

)V n

r )

ω ω

π π

= = = = = = ≅ ÷

&*0* ra's *4-4 r:m

&) θ A :'°$( )no44leED 9E mBs cos:' +9 raBs *' s

an +LDD rpm' 9D m & & + min

)V n

r )

ω ω

π π

° = = = = = = ≅ ÷

&01. ra's 0264 r:m

:) θ A *'°$( )no44leED 9E mBs cos*' +'* raBs *' s

an +'+: rpm' 9D m & & + min

)V n

r )

ω ω

π π

° = = = = = = ≅ ÷

&046 ra's 0404 r:m

Discussion 5inal results are "i0en to three si"nificant i"its, as usual. The rate of rotation in reality will !e lower

!ecause of frictional effects an air ra".


*-DL

•

r A '.9D m

θ

θ




6"3/

Solution The pre0ious pro!lem is reconsiere. The effect of ischar"e an"le θ on the rate of rotation n as θ 0aries

from ' to E'° in increments of +'° is to !e in0esti"ate.


$=0.02 "m"r=0.4% "m",_,o<<l=2 ",mbr of ,o<<ls"'c=)*$2/4V_t=V_dot/'c/,_,o<<lV_,o<<l=V_t*cos(tht!V_dot=0.060 "m3/s"omg!=V_,o<<l/r,_dot=omg!*60/(2*)

1n"le,

θ °V no44le ,

mBsω

raBs

n

rpm

'

+'&'

:'

9'D'

*'

L'

'E'

ED.D

E9.'E.L

&.L

L:.&*+.9

9L.L

:&.L

+*.*'.'

&+&

&'E+EE

+9

+*:+:*

+'*

L:

:L'

&'&*

+EE*+E'9

+LDD

+DD&+:':

+'+:

*E:

:D&'

Discussion The maimum rpm occurs when θ A 'o, as epecte, since this represents purely tan"ential outflow. hen θ

A E'o, the rpm rops to 4ero, as also epecte, since the outflow is purely raial an therefore there is no tor2ue to spin the

sprin7ler.


*-D

0 10 20 30 40 50 60 70 80 0

0

450

00

1350

1800

2250

θ, °

n 7 r : m




6"32

Solution 1 stationary water tan7 place on wheels on a frictionless surface is propelle !y a water 8et that lea0es thetan7 throu"h a smooth hole. 6elations are to !e e0elope for the acceleration, the 0elocity, an the istance tra0ele !y

the tan7 as a function of time as water ischar"es.

Assumptions The orifice has a smooth entrance, an thus the frictional losses are ne"li"i!le. - The flow is steay,

incompressi!le, an irrotational (so that the #ernoulli e2uation is applica!le). The surface uner the wheele tan7 is le0el

an frictionless. / The water 8et is ischar"e hori4ontally an rearwar. 2 The mass of the tan7 an wheel assem!ly isne"li"i!le compare to the mass of water in the tan7. / et flow is nearly uniform an thus the momentum-flu correction

factor can !e ta7en to !e unity, β ≅ +. Analysis (a) e ta7e point + at the free surface of the tan7, an point & at

the outlet of the hole, which is also ta7en to !e the reference le0el ( % & A ') so that

the water hei"ht a!o0e the hole at any time is % . otin" that the flui 0elocity at

the free surface is 0ery low (V + ≅ '), it is open to the atmosphere ( ' + A ' atm), an

water ischar"es into the atmosphere (an thus ' & A ' atm), the #ernoulli e2uation

simplifies to

g% V g

V % %

g

V

g

' %

g

V

g

' 5

5 & '&

&&

&

&

&&&

+

&++ =→+=→++=++

ρ ρ

The ischar"e rate of water from the tan7 throu"h the hole is

g% &V & V m 5 5 &99

&

'

&

' π ρ π ρ ρ ===


V mV m ! β β . 1pplyin" it to the water

tan7, the hori4ontal force that acts on the tan7 is etermine to !e

&&

9'

&'

&' &

g% g% &

V mV m ! 5 e

π ρ

π ρ ===−=

The acceleration of the water tan7 is etermine from ewton/s &n law of motion ! A ma where m is the mass of water in

the tan7, % &m )9B( &tan7 π ρ ρ == V ,

( )

( )

&

'

&

&

9

g% & 0 ! a

m % & 0

ρ π

ρ π

= = →&

'

&

& &

a g &

=

ote that the acceleration of the tan7 is constant.

(b) otin" that a A dV Bdt an thus dV A adt an acceleration a is constant, the 0elocity is epresse as

V at = →&

'

&&

&V g t

&=

(c) otin" that V A dxBdt an thus dx 8 Vdt , the istance tra0ele !y the water tan7 is etermine !y inte"ration to !e

&

'

& &

&dx Vdt dx g tdt

&= → = →

&

&'

&

& x g t

&=

since x A ' at t A '.

Discussion In reality, the flow rate ischar"e 0elocity an thus the force actin" on the water tan7 will !e less !ecause of

the frictional losses at the hole. #ut these losses can !e accounte for !y incorporatin" a ischar"e coefficient.

=esign an' Essay Problems

6"36

Solution 3tuents/ essays an esi"ns shoul !e uni2ue an will iffer from each other.


*-DE

D &6

%

x

V

+

&

ch 6 solution of cvg 2116

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