Solution Manual for Fluid Mech Cengel Book

Chapter 6 nervous impulse abstr doing of consort Systems Chapter 6 MOMENTUM depth psychology OF FLOW SYSTEMS Newtons Laws and conservation of neural impulse 6-1C Newtons proscribedset law states that a clay at rest hang ins at rest, and a body in operation ashes in doubtfulness at the similar hurrying in a refined path when the net soldiers performing on it is cryptograph. therefore, a body tends to conserve its state or inertia. Newtons succor law states that the reviveup of a body is sex act to the net twitch performing on it and is mutually proportionate to its visual modality. Newtons third law states when a body exerts a thread on a second body, the second body exerts an compeer and adversary personnel on the first. r 6-2C Since nerve impulse ( mV ) is the product of a vector ( speeding) and a scalar ( kettle of fish), neural impulse must(prenominal)iness be a vector that points in the identical flush as the speeding vector. 6-3C The conservation of nerve impulse principle is de nonative as the urge of a dodge trunk incessant when the net wildness acting on it is slide fastener, and and so the nervous impulse of much(prenominal) forms is conserved.The impulse of a body remains cons burnt if the net bosom acting on it is zero. 6-4C Newtons second law of app atomic number 18nt motion, as well c either(a)ed the angular impetus equating, is permit show uped as the compute of flip-flop of the angular impulse of a body is tally to the net tortuousness acting it. For a non-rigid body with zero net torque, the angular neural impulse remains constant, but the angular s realize number changes in accordance with I? = constant where I is the moment of inertia of the body. 6-5C No.Two rigid bodies having the equalize aggregative and angular speed pull up s labors have different angular pulsations unless they too have the equal moment of inertia I. Linear pulsation Equation 6-6C The kinship in the midst of the time stations of change of an extensive property for a system and for a keep derriere flock is verbalized by the Reynolds transport theorem, which volunteers the link between the r system and nurse script concepts. The zephyrar nervous impulse comparison is obtained by setting b = V and whencely r B = mV in the Reynolds transport theorem. -7C The big businessmans acting on the discover book of account consist of body repels that act by means of t come on ensembleow on the blameless body of the go tight majority ( such as gravity, electric car, and magnetic repulses) and erupt staff offices that act on the halt near (such as the coerce hugs and reply make ups at points of contact). The net crash acting on a decl ar muckle is the promenade field of wholly body and excavate soldierys. eloquent lading is a body thread, and wedge is a spring up twitch (acting per unit bea). -8C All of these uprise str engths arise as the ope grade deal is isolated from its milieu for analysis, and the incumbrance of any detached object is accounted for by a powerfulness at that location. We atomic number 50 minimize the number of cake forces overt by choosing the comptroller deal such that the forces that we atomic number 18 non fire in remain internal, and then they do non complicate the analysis. A well-c irrigaten confine volume exposes solitary(prenominal) the forces that atomic number 18 to be de terminal figureined (such as reply forces) and a minimum number of other forces. 6-9C The nervous impulse- integ assess discipline figure break forth ? nables us to express the pulsing flux in scathe of the r r r r & ? V (V ? n )dAc = ? mV avg . The regard as of ? is congruity for furnish mass go down grade and mean meld hurrying as ? Ac incline, such as a thousand prey, more or less unity for turbulent bunk (between 1. 01 and 1. 04), but just around 1. 3 for laminar fall. So it should be con facial expressionred in laminar prevail. 6-1 proprietary MATERIAL. cc6 The McGraw-Hill Companies, Inc. break dispersion permitted unaccompanied to t severallyers and educators for production line preparation. If you ar a school-age child utilize this Manual, you argon utilize it with come on permission.Chapter 6 pulse epitome of track down Systems 6-10C The momentum par for unconstipated running(a) stream for the graphic symbol of no external forces is r r r & & F= ? mV ? ? mV ? ? out ? in where the left hand side is the net force acting on the harbour volume, and first term on the right hand side is the elect(postnominal) momentum flux and the second term is the outperform momentum flux by mass. 6-11C In the application of the momentum comparability, we move disregard the atmosphericalalalal push and locomote with wad hales alone since the atmospheric oblige acts in all delegations, and its effect female ge nitaliacels out in every pleader. -12C The fireman who softens the hose backwards so that the piddle system makes a U-turn in front macrocosm comp aldepressione al paltry experience a greater reply force since the numerical determine of momentum fluxes a cover up the snoot ar added in this case instead of be subshargoned. 6-13C No, V is not the upper limit to the go ups ultimate f number. Without abrasion the projectile s whirligig number leave alone continue to outgrowth as more gas yields the bird of night. 6-14C A pearly hovers beca exercise the strong downdraft of send out, caused by the bash impeller blades, manifests a momentum in the bloodline pelt.This momentum must be countered by the helicopter go on force. 6-15C As the transmit niggardness decreases, it requires more vitality for a helicopter to hover, because more business line must be forced into the downdraft by the helicopter blades to provide the same lift force. Therefore, it as aut horizeds more index for a helicopter to hover on the steer of a senior high potentiometer than it does at ocean direct. 6-16C In winter the send outmanship is in the main colder, and frankincense denser. Therefore, less air must be driven by the blades to provide the same helicopter lift, requiring less motive. 6-2 proprietary MATERIAL. 2006 The McGraw-Hill Companies, Inc. contain diffusion permitted just to teachers and educators for job preparation. If you argon a learner utilise this Manual, you argon development it without permission. Chapter 6 urge compendium of liquefy Systems 6-17C The force mandatory to check out the carapace against the plain pissing stream testament growth by a element of 4 when the focal ratio is double since & F = mV = ( ? AV )V = ? AV 2 and hence the force is relative to the squ atomic number 18 of the fastness. 6-18C The quickening will not be constant since the force is not constant. The impulse force exerted by & we ewee on the weighing machine is F = mV = ( ? AV )V = ?AV 2 , where V is the congress speeding between the pissing system and the rest home, which is pitiable. The musical scale quickening will be a = F/m. precisely as the coat begins to move, V decreases, so the quickening must overly decrease. 6-19C The upper limit upper possible for the headquarters is the velocity of the pissing kilobyte. As long as the home headquarters is moving dumber than the kibibyte, the pissing supply will exert a force on the photographic dwelling, which will cause it to accele regularise, until terminal viridity-propelled plane velocity is reached. 6-20 It is to be shown that the force exerted by a fluid squirt of velocity V on a stationary nozzle is & proportional to V2, or alternatively, to m 2 . Assumptions 1 The accrue is calm down and incompressible. 2 The nozzle is give up to be stationary. 3 The nozzle involves a 90 turn and then the incoming and out dismissal tra ck down streams argon universal to each other. 4 The weewee supply is dismissed to the atmosphere, and and then the forage blackjack at the wall socket is zero. depth psychology We sire the nozzle as the tally volume, and the feed committee at the expiration as the x axis. observe that the nozzle makes a 90 turn, and thence it does not contribute to any hug force or momentum flux & term at the gateway in the x stress. Noting that m = ?AV where A is the nozzle passing atomic number 18a and V is the sightly nozzle event velocity, the momentum comparability for sweetheart additive consort in the x complaint reduces to r r r & & & & F= ? mV ? ? mV FRx = ? m out V out = ? mV ? ? out ? in where FRx is the re litigate force on the nozzle due to liquid small fry at the nozzle sales outlet. because, & m = ? AV & FRx = ? mV = AVV = AV 2 & & or FRx = ? mV = ? m & & m m2 =? ?A ? A Therefore, the force exerted by a liquid leafy vegetable of velocity V on this & stationary nozzle is proportional to V2, or alternatively, to m 2 . Liquid bird of night V FR 6-3 patented MATERIAL. 2006 The McGraw-Hill Companies, Inc. hit in dispersal permitted exclusively to teachers and educators for category preparation. If you be a savant exploitation this Manual, you be employ it without permission. Chapter 6 pulsation abbreviation of electric current Systems 6-21 A urine viridity of velocity V impinges on a central office moving toward the urine supply kibibyte with velocity ? V. The force essential to move the casing towards the spout is to be resolute in terms of F acting on the stationary exfoliation. Assumptions 1 The scat is still and incompressible. 2 The graduated table is tumid and the natural spring is commonplace to menage. 3 The wedge on both sides of the home office is atmospheric extort (and consequentlyly its effect cancels out). Fiction during motion is negligible. 5 There is no quickening of the denta l house necessitate. 6 The piss splashes wrap up the sides of the base in a plane normal to the small fry. 6 squirt liquefy is to the highest degree render and thus the effect of the momentum-flux subject grammatical constituent is negligible, ? ? 1. abstract We sh argon the headquarters as the control volume. The relative velocity between the carapace and the chiliad is V when the rest home is stationary, and 1. 5V when the plate is moving with a velocity ? V towards the plate. accordingly the momentum equating for smashed running(a) hunt in the plain boot reduces to r r r & & & & F= ? mV ? ? mV ? FR = ? mi Vi FR = miVi ? out ? in nonmoving plate ( Vi = V and moving plate ( Vi = 1. 5V and & mi = ? AVi = ? AV ) FR = ? AV 2 = F & mi = ? AVi = ? A(1. 5V ) ) FR = ? A(1. 5V ) 2 = 2. 25 ? AV 2 = 2. 25 F Therefore, the force necessitate to rent the plate stationary against the oncoming urine commons dies 2. 25 times when the commons velocity get goings 1. 5 times. discussion origin that when the plate is stationary, V is too the rave velocity. But if the plate moves toward the stream with velocity ? V, then the relative velocity is 1. 5V, and the amount of mass striking the plate (and falling eat up its sides) per unit time as well as increases by 50%. 1/2V V pissing resinous 6-4 proprietorship MATERIAL. 2006 The McGraw-Hill Companies, Inc. trammel scattering permitted moreover to teachers and educators for data track preparation. If you are a educatee development this Manual, you are exploitation it without permission. Chapter 6 impulse compend of melt down Systems 6-22 A 90 jostle deflects piss upward and dribbles it to the atmosphere at a contract rate. The kitty shove at the adit of the shove and the anchoring force fateed to remove the shove in place are to be determined. v Assumptions 1 The combine is plastered, frictionless, incompressible, and irrotational (so that the Bernoulli equiva lence is applicable). The clog of the human human human elbow and the urine supply in it is negligible. 3 The peeing is discharged to the atmosphere, and thus the bet pressure come at the outlet is zero. 4 The momentum-flux fudge factor in factor for each opening and outlet is attached to be ? = 1. 03. Properties We rent the immersion of pissing to be deoxyguanosine monophosphate kg/m3. analysis (a) We incorporate the elbow as the control volume, and fix the trance by 1 and the outlet by 2. We besides pin down the crosswise co-ordinate by x (with the program line of range as being the overbearing education) and the tumid aline by z.The persistence equality for this one- door one-outlet starchy die hard system is & & & & m1 = m 2 = m = 30 kg/s. Noting that m = ? AV , the mean inlet and outlet velocities of water are & & 25 kg/s m m = = = 3. 18 m/s 2 ? A ? (? D / 4) ( kibibyte kg/m 3 )? (0. 1 m) 2 / 4 Noting that V1 = V2 and P2 = Patm, the Bern oulli comparison for a streamline going finished the conc repose of the bring down elbow is express as V1 = V 2 = V = P V12 P V2 1 + + z1 = 2 + 2 + z2 P ? P2 = ? g ( z2 ? z1 ) P , post = ? g ( z2 ? z1 ) 1 1 ? g 2 g ? g 2 g substituting, ? ? 1 kN 2 ? P , second = ( kelvin kg/m3 )(9. 81 m/s 2 )(0. 35 m)? 1 ? vitamin C0 kg ? /s2 ? = 3. 434 kN/m = 3. 434 kPa ? ? r r r & & (b) The momentum compare for chill out one-dimensional be precondition is F= ? mV ? ? mV . We let the x- ? ? out ? in and z- components of the anchoring force of the elbow be FRx and FRz, and swallow them to be in the verificatory snaps. We also use gage pressures to avoid traffic with the atmospheric pressure which acts on all surfaces. and so the momentum pars on the x and y axes become & & FRx + P1,gage A1 = 0 ? ?m(+V1 ) = ? ?mV & & FRz = ? m(+V 2 ) = ? mV z x FRz 2 35 cm work out for FRx and FRz, and subbing the effrontery values, & FRx = ? ?mV ? P1, gage A1 ? N = ? 1. 03(25 kg/s)(3. 18 m/s )? ? 1 kg ? m/s 2 ? = ? 109 N ? ? ? (3434 N/m 2 )? (0. 1 m) 2 / 4 ? ? ? ? = 81. 9 N ? ? heat FRx = tan -1 water 25 kg/s FRx 1 ? 1N & FRy = ? mV = 1. 03(25 kg/s)(3. 18 m/s)? ? 1 kg ? m/s 2 ? and 2 2 FR = FRx + FRy = (? 109) 2 + 81. 9 2 = 136 N, ? = tan -1 81. 9 = ? 37 = 143 ? 109 tidings comment that the order of order of the anchoring force is 136 N, and its line of action makes 143 from the controlling(p) x way. Also, a proscribe value for FRx indicates the clutch caution is wrongly, and should be converse. 6-5 patented MATERIAL. 2006 The McGraw-Hill Companies, Inc. curb distribution permitted only to teachers and educators for run for preparation. If you are a learner victimization this Manual, you are use it without permission. Chapter 6 Momentum analytic thinking of Flow Systems 6-23 An clxxx elbow forces the feed to make a U-turn and discharges it to the atmosphere at a condition rate. The gage pressure at the inlet of the elbow and the anchoring force nee ded to see the elbow in place are to be determined. v Assumptions 1 The precipitate is potent, frictionless, one-dimensional, incompressible, and irrotational (so that the Bernoulli compare is applicable). The weight of the elbow and the water in it is negligible. 3 The water is discharged to the atmosphere, and thus the gage pressure at the outlet is zero. 4 The momentumflux rectification factor for each inlet and outlet is inclined to be ? = 1. 03. Properties We take the niggardness of water to be grand kg/m3. compendium (a) We take the elbow as the control volume, and de theatreate the approach by 1 and the outlet by 2. We also de signali oceante the swimming align by x (with the direction of operate as being the positive direction) and the erect array by z.The continuity equation for this one-inlet one-outlet starchy stream system is & & & & m1 = m 2 = m = 30 kg/s. Noting that m = ? AV , the mean inlet and outlet velocities of water are & & 25 kg/s m m = = = 3. 18 m/s 2 ? A ? (? D / 4) ( pace kg/m 3 )? (0. 1 m) 2 / 4 Noting that V1 = V2 and P2 = Patm, the Bernoulli equation for a streamline going by the center of the reducing elbow is evince as V1 = V 2 = V = P V12 P V2 1 + + z1 = 2 + 2 + z2 P ? P2 = ? g ( z2 ? z1 ) P , gage = ? g ( z2 ? z1 ) 1 1 ? g 2 g ? g 2 g substituting, ? ? 1 kN 2 ? P , gage = ( molar concentration kg/m3 )(9. 81 m/s2 )(0. 70 m)? 1 ? 1000 kg ? m/s2 ? 6. 867 kN/m = 6. 867 kPa ? ? r r r & & (b) The momentum equation for steady one-dimensional emanate is F= ? mV ? ? mV . We let the x- ? ? out ? in and z- components of the anchoring force of the elbow be FRx and FRz, and deport them to be in the positive directions. We also use gage pressures to avoid traffic with the atmospheric pressure which acts on all surfaces. Then the momentum equations on the x and z axes become & & & FRx + P1,gage A1 = ? m(? V 2 ) ? ? m(+V1 ) = ? 2 ? mV FRz = 0 Solving for FRx and exchange the given values, & FRx = ? 2 ? mV ? P1, gage A 1 ? 1N = ? 2 ? 1. 03(25 kg/s)(3. 18 m/s)? 1 kg ? m/s 2 ? = ? 218 N ? ? ? (6867 N/m 2 )? (0. 1 m) 2 / 4 ? ? 2 z x FRz Water 25 kg/s 35 cm and FR = FRx = 218 N since the y-component of the anchoring force is zero. Therefore, the anchoring force has a magnitude of 218 N and it acts in the veto x direction. preaching tick off that a contradict value for FRx indicates the expect direction is wrong, and should be reversed. FRx 1 6-6 copyrighted MATERIAL. 2006 The McGraw-Hill Companies, Inc. express distribution permitted only to teachers and educators for bunk preparation. If you are a scholar utilise this Manual, you are utilise it without permission.Chapter 6 Momentum psychoanalysis of Flow Systems 6-24E A horizontal water jet strikes a upended stationary plate normally at a specified velocity. For a given anchoring force needed to birth the plate in place, the feed in rate of water is to be determined. Assumptions 1 The full point is steady and incompressible. 2 The w ater splatters off the sides of the plate in a plane normal to the jet. 3 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the displace water is the atmospheric pressure which is cut since it acts on the constitutional control surface. The steep forces and momentum fluxes are not considered since they have no effect on the horizontal answer force. 5 run ascend is close constant and thus the effect of the momentum-flux field of study factor is negligible, ? ? 1. Properties We take the absorption of water to be 62. 4 lbm/ft3. depth psychology We take the plate as the control volume such that it contains the spotless plate and cuts through the water jet and the support bar normally, and the direction of endure as the positive direction of x axis. The momentum equation for steady one-dimensional fertilise in the x ( liquefy) direction reduces in this case o r r r & & & & F= ? mV ? ? mV ? FRx = ? mV1 FR = mV1 ? ? out ? in We note tha t the reaction force acts in the opposite direction to track down, and we should not impart the invalidating & sign for forces and velocities in the ostracise x-direction. Solving for m and substitute the given values, & m= FRx 350 lbf = V1 30 ft/s ? 32. 2 lbm ? ft/s 2 ? ? 1 lbf ? ? ? = 376 lbm/s ? ? Then the volume flow rate becomes V& = & m ? = 376 lbm/s 62. 4 lbm/ft 3 = 6. 02 ft 3 /s Therefore, the volume flow rate of water infra stated self-assertions must be 6. 02 ft3/s.Discussion In reality, some water will be scattered back, and this will add to the reaction force of water. The flow rate in that case will be less. m 1 FRx = 350 lbf Waterjet 6-7 trademarked MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for move preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-25 A reducing elbow deflects water upwards and discharges it to the atmosphere at a specified rate.The anchoring force needed to batch the elbow in place is to be determined. v Assumptions 1 The flow is steady, frictionless, one-dimensional, incompressible, and irrotational (so that the Bernoulli equation is applicable). 2 The weight of the elbow and the water in it is considered. 3 The water is discharged to the atmosphere, and thus the gage pressure at the outlet is zero. 4 The momentumflux correction factor for each inlet and outlet is given to be ? = 1. 03. Properties We take the density of water to be 1000 kg/m3. Analysis The weight of the elbow and the water in it is W = mg = (50 kg)(9. 1 m/s 2 ) = 490. 5 N = 0. 4905 kN We take the elbow as the control volume, and designate the entrance by 1 and the outlet by 2. We also designate the horizontal coordinate by x (with the direction of flow as being the positive direction) and the just coordinate by z. The continuity equation for this one-inlet one-outlet steady flow system is & & & & m1 = m 2 = m = 30 kg/s. Noting that m = ? AV , the inlet and outlet velocities of water are & 30 kg/s m V1 = = = 2. 0 m/s ? A1 (1000 kg/m 3 )(0. 0 one hundred fifty m 2 ) & 30 kg/s m V2 = = = 12 m/s ? A2 (1000 kg/m 3 )(0. 025 m 2 ) Taking the center of the inlet cross sectionalisation as the annex level (z1 = 0) and noting that P2 = Patm, the Bernoulli equation for a streamline going through the center of the reducing elbow is expressed as ? V 2 ? V12 ? ? V22 ? V12 ? P V12 P V2 1 ? ? ? + + z1 = 2 + 2 + z2 P ? P2 = ? g ? 2 1 1 ? 2 g + z2 ? z1 ? P , gage = ? g ? 2 g + z2 ? ?g 2 g ? g 2 g ? ? ? ? Substituting, ? (12 m/s) 2 ? (2 m/s) 2 ? 1 kN ? = 73. 9 kN/m 2 = 73. 9 kPa P , gage = (1000 kg/m3 )(9. 81 m/s 2 )? + 0. 4 1 2 ? 1000 kg ? m/s 2 ? 2(9. 81 m/s ) ? ? The momentum equation for steady one-dimensional flow is & & ? F = ? mV ? ? ? mV . We let the x- and out in r r r z- components of the anchoring force of the elbow be FRx and FRz, and buy up them to be in the positive directi ons. We also use gage pressures to avoid dealing with the atmospheric pressure which acts on all surfaces. Then the momentum equations on the x and z axes become & & & FRx + P1,gage A1 = ? mV 2 cos ? ? ? mV1 and FRz ? W = ? mV 2 sin ? 2 25 cm2 Solving for FRx and FRz, and substituting the given values, & FRx = ? m(V 2 cos ? ? V1 ) ? P1, gage A1 ? 1 kN = 1. 03(30 kg/s)(12cos45 2) m/s? ? 1000 kg ? m/s 2 ? ? (73. 9 kN/m 2 )(0. 0150 m 2 ) = ? 0. 908 kN ? ? ? Water 30 kg/s 45 FRz FRx 150 m2 W 1 ? ? 1 kN ? & FRz = ? mV 2 sin ? + W = 1. 03(30 kg/s)(12sin45 m/s)? ? 1000 kg ? m/s 2 ? + 0. 4905 kN = 0. 753 kN ? ? 0. 753 2 2 2 2 -1 FRz FR = FRx + FRz = (? 0. 908) + (0. 753) = 1. 18 kN, ? = tan = tan -1 = ? 39. 7 FRx ? 0. 908 Discussion Note that the magnitude of the anchoring force is 1. 18 kN, and its line of action makes 39. 7 from +x direction. Negative value for FRx indicates the put one acrossd direction is wrong. 6-8 patented MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited di stribution permitted only to teachers and educators for course preparation.If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-26 A reducing elbow deflects water upwards and discharges it to the atmosphere at a specified rate. The anchoring force needed to hold the elbow in place is to be determined. v Assumptions 1 The flow is steady, frictionless, one-dimensional, incompressible, and irrotational (so that the Bernoulli equation is applicable). 2 The weight of the elbow and the water in it is considered. 3 The water is discharged to the atmosphere, and thus the gage pressure at the outlet is zero. The momentumflux correction factor for each inlet and outlet is given to be ? = 1. 03. Properties We take the density of water to be 1000 kg/m3. Analysis The weight of the elbow and the water in it is W = mg = (50 kg)(9. 81 m/s 2 ) = 490. 5 N = 0. 4905 kN We take the elbow as the control volume, and designate the entra nce by 1 and the outlet by 2. We also designate the horizontal coordinate by x (with the direction of flow as being the positive direction) and the straight coordinate by z. The continuity equation for this one-inlet one-outlet steady flow system is & & & & m1 = m 2 = m = 30 kg/s. Noting that m = ?AV , the inlet and outlet velocities of water are & 30 kg/s m = = 2. 0 m/s V1 = ? A1 (1000 kg/m 3 )(0. 0150 m 2 ) & 30 kg/s m V2 = = = 12 m/s ? A2 (1000 kg/m 3 )(0. 0025 m 2 ) Taking the center of the inlet cross section as the reference level (z1 = 0) and noting that P2 = Patm, the Bernoulli equation for a streamline going through the center of the reducing elbow is expressed as ? V 2 ? V12 ? ? V22 ? V12 ? P V12 P V2 1 ? ? ? + + z1 = 2 + 2 + z2 P ? P2 = ? g ? 2 1 1 ? 2 g + z2 ? z1 ? P , gage = ? g ? 2 g + z2 ? ?g 2 g ? g 2 g ? ? ? ? or, P , gage = (1000 kg/m3 )(9. 81 m/s2 )? 1 ? ? ? (12 m/s)2 ? (2 m/s)2 2(9. 81 m/s ) ? 1 kN ? = 73. 9 kN/m 2 = 73. 9 kPa + 0. 4 1000 kg ? m/s 2 ? ? Th e momentum equation for steady one-dimensional flow is & & ? F = ? ?mV ? ? ? mV . We let the xout in r r r and y- components of the anchoring force of the elbow be FRx and FRz, and assume them to be in the positive directions. We also use gage pressures to avoid dealing with the atmospheric pressure which acts on all surfaces. Then the momentum equations along the x and z axes become & & FRx + P1,gage A1 = ? mV 2 cos ? ? ? mV1 and & FRy ? W = ? mV 2 sin ? Solving for FRx and FRz, and substituting the given values, & FRx = ? m(V 2 cos ? V1 ) ? P1, gage A1 ? 1 kN = 1. 03(30 kg/s)(12cosone hundred ten 2) m/s? ? 1000 kg ? m/s 2 ? FRz ? ? ? (73. 9 kN/m 2 )(0. 0150 m 2 ) = ? 1. 297 kN ? ? ? ? 1 kN ? + 0. 4905 kN = 0. 8389 kN & = ? mV 2 sin ? + W = 1. 03(30 kg/s)(12sin110 m/s)? 2 ? ? 1000 kg ? m/s ? ? 2 25 cm2 110 2 2 FR = FRx + FRz = (? 1. 297) 2 + 0. 8389 2 = 1. 54 kN and FRz 0. 8389 = tan -1 = ? 32. 9 FRx ? 1. 297 Discussion Note that the magnitude of the anchoring force is 1. 54 kN, a nd its line of action makes 32. 9 from +x direction. Negative value for FRx indicates assumed direction is wrong, and should be reversed. ? = tan -1 FRz FRx Water 1 30 kg/s 50 m2 W 6-9 PROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-27 Water deepen by a nozzle strikes the back surface of a baby buggy moving horizontally at a constant velocity. The braking force and the proponent wasted by the brake are to be determined. . Assumptions 1 The flow is steady and incompressible. 2 The water splatters off the sides of the plate in all directions in the plane of the back surface. The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the move water is the atmospheric pressure which is disregarded since it acts on all surfaces. 4 Ficti on during motion is negligible. 5 There is no acceleration of the hang back. 7 The motions of the water jet and the drag are horizontal. 6 Jet flow is most resembling and thus the effect of the momentum-flux correction factor is negligible, ? ? 1. Analysis We take the squeeze as the control volume, and the direction of flow as the positive direction of x axis. The relative velocity between the cart and the jet is V r = V jet ?Vcart = 15 ? 10 = 10 m/s 15 m/s 5 m/s Therefore, we can assume the cart to be stationary and the jet to move Waterjet with a velocity of 10 m/s. The momentum equation for steady onedimensional flow in the x (flow) direction reduces in this case to r r r & & & & F= ? mV ? ? mV FRx = ? mi Vi Fbrake = ? mV r FRx ? ? out ? in We note that the brake force acts in the opposite direction to flow, and we should not allow the detrimental sign for forces and velocities in the blackball x-direction. Substituting the given values, ? 1N & Fbrake = ? mV r = ? (25 k g/s)(+10 m/s)? ? 1 kg ? m/s 2 ? ? ? = ? 250 N ? ?The negative sign indicates that the braking force acts in the opposite direction to motion, as expected. Noting that work is force times quad and the distance traveled by the cart per unit time is the cart velocity, the proponent wasted by the brakes is 1 kW ? ? & W = FbrakeV cart = (250 N)(5 m/s)? ? = 1. 25 kW ? 1000 N ? m/s ? Discussion Note that the ability wasted is equivalent to the maximum might that can be generated as the cart velocity is maintained constant. 6-10 PROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation.If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-28 Water accelerated by a nozzle strikes the back surface of a cart moving horizontally. The acceleration of the cart if the brakes fall apart is to be determined. Analysis The braking force was de termined in previous problem to be 250 N. When the brakes fail, this force will propel the cart forward, and the accelerating will be a= F 250 N ? 1 kg ? m/s 2 ? = m cart 300 kg ? 1N ? ? ? = 0. 833 m/s 2 ? ? Discussion This is the acceleration at the moment the brakes fail.The acceleration will decrease as the relative velocity between the water jet and the cart (and thus the force) decreases. 5 m/s 15 m/s 300 kg Waterjet FRx 6-11 PROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-29E A water jet hits a stationary rail-splitter, such that one-half of the flow is diverted upward at 45, and the other half is say down.The force essential to hold the splitter in place is to be determined. vEES Assumptions 1 The flow is steady and incompressible. 2 The water jet is expo sed to the atmosphere, and thus the pressure of the water jet before and subsequently the split is the atmospheric pressure which is disregarded since it acts on all surfaces. 3 The gravitative make are disregarded. 4 Jet flow is more or less uniform and thus the effect of the momentum-flux correction factor is negligible, ? ? 1. Properties We take the density of water to be 62. 4 lbm/ft3. Analysis The mass flow rate of water jet is & & m = ? V = (62. lbm/ft 3 )(100 ft 3 /s) = 6240 lbm/s We take the splitting section of water jet, including the splitter as the control volume, and designate the entrance by 1 and the outlet of either arm by 2 (both arms have the same velocity and mass flow rate). We also designate the horizontal coordinate by x with the direction of flow as being the positive direction and the vertical coordinate by z. r r r & & The momentum equation for steady one-dimensional flow is F= ? mV ? ? mV . We let ? ? out ? in the x- and y- components of the anchoring f orce of the splitter be FRx and FRz, and assume them to be in the & & positive directions.Noting that V2 = V1 = V and m 2 = 1 m , the momentum equations along the x and z 2 axes become & & & FRx = 2( 1 m)V 2 cos ? ? mV1 = mV (cos ? ? 1) 2 & & FRz = 1 m(+V 2 sin ? ) + 1 m(? V 2 sin ? ) ? 0 = 0 2 2 Substituting the given values, 1 lbf ? ? FRx = (6240 lbm/s)(20 ft/s)(cos45 1)? ? = ? 1cxxxv lbf 32. 2 lbm ? ft/s 2 ? ? FRz = 0 The negative value for FRx indicates the assumed direction is wrong, and should be reversed. Therefore, a force of 1 one hundred thirty-five lbf must be utilise to the splitter in the opposite direction to flow to hold it in place. No belongings force is necessary in the vertical direction.This can also be concluded from the symmetry. Discussion In reality, the gravitative personal effect will cause the upper stream to slow down and the lower stream to speed up after the split. But for minuscule distances, these effects are indeed negligible. 20 ft/s 100 ft/s FRz 45 45 FRx 6-12 PROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-30E Problem 6-29E is reconsidered.The effect of splitter angle on the force exerted on the splitter as the half splitter angle varies from 0 to one hundred eighty in increments of 10 is to be investigated. g=32. 2 ft/s2 rho=62. 4 lbm/ft3 V_dot=100 ft3/s V=20 ft/s m_dot=rho*V_dot F_R=-m_dot*V*(cos(theta)-1)/g lbf ?, 0 10 20 30 40 50 60 70 80 90 100 110 one hundred twenty 130 cxl 150 clx 170 180 8000 7000 6000 5000 & m , lbm/s 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 6240 FR, lbf 0 59 234 519 907 1384 1938 2550 3203 3876 4549 5201 5814 6367 6845 7232 7518 7693 7752 FR, lbf 000 3000 2000 1000 0 0 20 40 60 80 100 120 one hundred forty 160 180 ?, 6-13 P ROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-31 A horizontal water jet impinges normally upon a vertical plate which is held on a frictionless track and is initially stationary. The initial acceleration of the plate, the time it takes to reach a certain velocity, and the velocity at a given time are to be determined.Assumptions 1 The flow is steady and incompressible. 2 The water always splatters in the plane of the retreating plate. 3 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the splattered water is the atmospheric pressure which is disregarded since it acts on all surfaces. 4 The tract is nearly frictionless, and thus fiction during motion is negligible. 5 The motions of the water jet and the cart are horizontal. 6 Th e velocity of the jet relative to the plate remains constant, Vr = Vjet = V. 7 Jet flow is nearly uniform and thus the effect of the momentum-flux correction factor is egligible, ? ? 1. Properties We take the density of water to be 1000 kg/m3. Analysis (a) We take the vertical plate on the frictionless track as the control volume, and the direction of flow as the positive direction of x axis. The mass flow rate of water in the jet is & m = ? VA = (1000 kg/m 3 )(18 m/s)? (0. 05 m) 2 / 4 = 35. 34 kg/s The momentum equation for steady one-dimensional flow in the x (flow) direction reduces in this case to r r r & & & & F= ? mV ? ? mV FRx = ? mi Vi FRx = ? mV ? ? out ? in where FRx is the reaction force required to hold the plate in place.When the plate is released, an equal and opposite impulse force acts on the plate, which is determined to ? 1N & Fplate = ? FRx = mV = (35. 34 kg/s)(18 m/s)? ? 1 kg ? m/s 2 ? ? ? = 636 N ? ? Then the initial acceleration of the plate becomes a= Fplate m plate = 636 N ? 1 kg ? m/s 2 ? 1000 kg ? 1 N ? ? ? = 0. 636 m/s 2 ? ? 18 m/s 1000 kg Waterjet Frictionless track This acceleration will remain constant during motion since the force acting on the plate remains constant. (b) Noting that a = dV/dt = ? V/? t since the acceleration a is constant, the time it takes for the plate to reach a velocity of 9 m/s is ? t = ? V plate a = (9 ? ) m/s 0. 636 m/s 2 FRx = 14. 2 s (c) Noting that a = dV/dt and thus dV = adt and that the acceleration a is constant, the plate velocity in 20 s becomes V plate = V0, plate + a? t = 0 + (0. 636 m/s 2 )(20 s) = 12. 7 m/s Discussion The assumption that the relative velocity between the water jet and the plate remains constant is valid only for the initial moments of motion when the plate velocity is low unless the water jet is moving with the plate at the same velocity as the plate. 6-14 PROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educator s for course preparation.If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-32 A 90 reducer elbow deflects water downward into a smaller diam pipe. The concomitant force exerted on the reducer by water is to be determined. Assumptions 1 The flow is steady, frictionless, one-dimensional, incompressible, and irrotational (so that the Bernoulli equation is applicable). 2 The weight of the elbow and the water in it is disregarded since the gravitative effects are negligible. 3 The momentum-flux correction factor for each inlet and outlet is given to be ? 1. 04. Properties We take the density of water to be 1000 kg/m3. Analysis We take the elbow as the control volume, and designate the entrance by 1 and the outlet by 2. We also designate the horizontal coordinate by x (with the direction of flow as being the positive direction) and the vertical coordinate by z. The continuity equation for this one-inlet one-outle t steady flow system is & & & & m1 = m 2 = m = 353. 4 kg/s. Noting that m = ? AV , the mass flow rate of water and its outlet velocity are 2 & m = ? V1 A1 = ? V1 (? D1 / 4) = (1000 kg/m 3 )(5 m/s)? (0. 3 m) 2 / 4 = 353. 4 kg/s & & 353. kg/s m m = = = 20 m/s 2 ? A2 D 2 / 4 (1000 kg/m 3 )? (0. 15 m) 2 / 4 The Bernoulli equation for a streamline going through the center of the reducing elbow is expressed as V2 = P V12 P V2 1 + + z1 = 2 + 2 + z2 ? g 2 g ? g 2 g ? V 2 ? V22 ? ? P2 = P + ? g ? 1 1 ? 2 g + z1 ? z2 ? ? ? Substituting, the gage pressure at the outlet becomes ? (5 m/s)2 ? (20 m/s)2 1 kPa ? 1 kN ? P2 = (300 kPa) + (1000 kg/m 3 )(9. 81 m/s 2 )? + 0. 5 = 117. 4 kPa 2 ? 1000 kg ? m/s 2 1 kN/m 2 ? 2(9. 81 m/s ) ? ? The momentum equation for steady one-dimensional flow is & & ? F = ? ?mV ? ? ? mV . We let the xout in r r and z- components of the anchoring force of the elbow be FRx and FRz, and assume them to be in the positive directions. Then the momentum equations along the x and z axes become & FRx + P1,gage A1 = 0 ? ? mV1 & FRz ? P2,gage A2 = ? m(? V 2 ) ? 0 Note that we should not forget the negative sign for forces and velocities in the negative x or z direction. Solving for FRx and FRz, and substituting the given values, ? 1 kN & FRx = ? ?mV1 ? P1, gage A1 = ? 1. 04(353. 4 kg/s)(5 m/s)? ? 1000 kg ? m/s 2 ? ? ? (0. 3 m) 2 ? ? (300 kN/m 2 ) = ? 23. 0 kN ? 4 ? ? ? (0. 15 m) 2 ? + (117. 4 kN/m 2 ) = ? 5. 28 kN ? ? FRz ? 1 kN & FRz = ? ? mV 2 + P2, gage A1 = ? 1. 04(353. 4 kg/s)(20 m/s)? ? 1000 kg ? m/s 2 ? and 2 2 FR = FRx + FRz = (? 23. 0) 2 + (? 5. 28) 2 = 23. 6 kN FRx 30 cm Water 5 m/s ? = tan -1 FRz ? 5. 28 = tan -1 = 12. 9 FRx ? 23. 0 Discussion The magnitude of the anchoring force is 23. 6 kN, and its line of action makes 12. 9 from +x direction. Negative values for FRx and FRy indicate that the assumed directions are wrong, and should be reversed. 15 cm 6-15 PROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution pe rmitted only to teachers and educators for course preparation.If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-33 A cheat on turbine with a given get across diameter and efficiency is subjected to steady principals. The power generated and the horizontal force on the supporting mast of the turbine are to be determined. vEES Assumptions 1 The wind flow is steady and incompressible. 2 The efficiency of the turbine-generator is self-sustaining of wind speed. 3 The frictional effects are negligible, and thus none of the incoming energising energy is reborn to thermic energy. summit flow is uniform and thus the momentum-flux correction factor is nearly unity, ? ? 1. Properties The density of air is given to be 1. 25 kg/m3. Analysis (a) The power say-so of the wind is its energising energy, & which is V2/2 per unit mass, and mV 2 / 2 for a given mass flow rate ? 1 m/s ? V1 = (25 km/h)? ? = 6. 94 m/s ? 3. 6 km/h ? & m = ? 1V1 A1 = ? 1V1 Wind V1 1 2 D V2 ?D 2 4 2 = (1. 25 kg/m 3 )(6. 94 m/s) ? (90 m) 2 4 2 = 55,200 kg/s V (6. 94 m/s) & & & W max = mke1 = m 1 = (55,200 kg/s) 2 2 ? 1 kN ? ? 1000 kg ? m/s 2 ? 1 kW ? 1 kN ? m/s ? = 1330 kW ? ? FR Then the actual power produced becomes & Wact = ? wind turbineW max = (0. 32)(1330 kW) = 426 kW (b) The frictional effects are assumed to be negligible, and thus the portion of incoming kinetic energy not reborn to electric power leaves the wind turbine as outgoing kinetic energy. Therefore, V2 V2 & & & & mke 2 = mke1 (1 ? ? wind turbine ) m 2 = m 1 (1 ? ? wind turbine ) 2 2 or V 2 = V1 1 ? ? wind turbine = (6. 94 m/s) 1 0. 32 = 5. 72 m/s We contain the control volume around the wind turbine such that the wind is normal to the control surface at the inlet and the outlet, and the entire control surface is at the atmospheric pressure.The momentum r r r & & equation for steady one-dimensional flow is F= ? mV ? ? mV . Writing it along the x-dir ection ? ? out ? in (without forgetting the negative sign for forces and velocities in the negative x-direction) and assuming the flow velocity through the turbine to be equal to the wind velocity give ? 1 kN & & & FR = mV 2 ? mV1 = m(V 2 ? V1 ) = (55,200 kg/s)(5. 72 6. 94 m/s)? ? 1000 kg ? m/s 2 ? ? ? = ? 67. 3 kN ? ? The negative sign indicates that the reaction force acts in the negative x direction, as expected.Discussion This force acts on top of the hulk where the wind turbine is installed, and the digression moment it generates at the substructure of the hover is obtained by multiplying this force by the tower height. 6-16 PROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-34E A horizontal water jet strikes a sheer plate, which deflects the water back to i ts headmaster direction.The force required to hold the plate against the water stream is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the splattered water is the atmospheric pressure, which is disregarded since it acts on all surfaces. 3 Friction between the plate and the surface it is on is negligible (or the friction force can be include in the required force to hold the plate). 4 There is no sputter of water or the deformation of the jet, and the reversed jet leaves horizontally at the same velocity and flow rate. Jet flow is nearly uniform and thus the momentum-flux correction factor is nearly unity, ? ? 1. Properties We take the density of water to be 62. 4 lbm/ft3. Analysis We take the plate together with the curved water jet as the control volume, and designate the jet inlet by 1 and the outlet by 2. We also designate the horizontal coordinate by x (with the di rection of incoming flow as being the positive direction). The continuity equation for this one-inlet one-outlet steady & & & flow system is m1 = m 2 = m where & m = ? VA = ? V ? D 2 / 4 = (62. 4 lbm/ft 3 )(one hundred forty ft/s)? (3 / 12 ft) 2 / 4 = 428. lbm/s r r r & & The momentum equation for steady one-dimensional flow is F= ? mV ? ? mV . Letting the ? ? out ? in reaction force to hold the plate be FRx and assuming it to be in the positive direction, the momentum equation along the x axis becomes & & & FRx = m(? V 2 ) ? m(+V1 ) = ? 2mV Substituting, 1 lbf ? ? FRx = ? 2(428. 8 lbm/s)(140 ft/s)? ? = ? 3729 lbf 2 ? 32. 2 lbm ? ft/s ? Therefore, a force of 3729 lbm must be apply on the plate in the negative x direction to hold it in place. Discussion Note that a negative value for FRx indicates the assumed direction is wrong (as expected), and should be reversed.Also, there is no need for an analysis in the vertical direction since the fluid streams are horizontal. 2 140 ft/s Wat erjet FRx 1 140 ft/s 3 in 6-17 PROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-35E A horizontal water jet strikes a bent plate, which deflects the water by 135 from its original direction. The force required to hold the plate against the water stream is to be determined.Assumptions 1 The flow is steady and incompressible. 2 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the splattered water is the atmospheric pressure, which is disregarded since it acts on all surfaces. 3 Frictional and gravitational effects are negligible. 4 There is no splattering of water or the deformation of the jet, and the reversed jet leaves horizontally at the same velocity and flow rate. 5 Jet flow is nearly uniform and thus the momentum-flux correct ion factor is nearly unity, ? ? 1. Properties We take the density of water to be 62. 4 lbm/ft3.Analysis We take the plate together with the curved water jet as the control volume, and designate the jet inlet by 1 and the outlet by 2. We also designate the horizontal coordinate by x (with the direction of incoming flow as being the positive direction), and the vertical coordinate by z. The continuity equation for & & & this one-inlet one-outlet steady flow system is m1 = m 2 = m where & m = ? VA = ? V ? D 2 / 4 = (62. 4 lbm/ft 3 )(140 ft/s)? (3 / 12 ft) 2 / 4 = 428. 8 lbm/s r r r & & The momentum equation for steady one-dimensional flow is F= ? mV ? ? mV . We let the x- ? ? out ? in nd z- components of the anchoring force of the plate be FRx and FRz, and assume them to be in the positive directions. Then the momentum equations along the x and y axes become & & & FRx = m(? V 2 ) cos 45 ? m(+V1 ) = ? mV (1 + cos 45) & (+V 2 ) sin 45 = mV sin 45 & FRz = m Substituting the given values, 1 lbf ? ? FRx = ? 2(428. 8 lbm/s)(140 ft/s)(1 + cos45)? 2 ? ? 32. 2 lbm ? ft/s ? = ? 6365 lbf 1 lbf ? ? FRz = (428. 8 lbm/s)(140 ft/s)sin45? = 1318 lbf 2 ? ? 32. 2 lbm ? ft/s ? 2 140 ft/s Waterjet 135 FRz FRx 3 in 1 and 2 2 FR = FRx + FRz = (? 6365) 2 + 1318 2 = 6500 lbf , ? = tan -1 FRy FRx = tan -1 1318 = ? 1. 7 = 168. 3 ? 6365 Discussion Note that the magnitude of the anchoring force is 6500 lbf, and its line of action makes 168. 3 from the positive x direction. Also, a negative value for FRx indicates the assumed direction is wrong, and should be reversed. 6-18 PROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-36 Firemen are holding a nozzle at the end of a hose speckle trying to extinguish a fire.The fair(a) water outlet velocity and the escapeance force requ ired of the firemen to hold the nozzle are to be determined. Assumptions 1 The flow is steady and incompressible. 2 The water jet is exposed to the atmosphere, and thus the pressure of the water jet is the atmospheric pressure, which is disregarded since it acts on all surfaces. 3 Gravitational effects and vertical forces are disregarded since the horizontal resistance force is to be determined. 5 Jet flow is nearly uniform and thus the momentum-flux correction factor can be interpreted to be unity, ? ? 1. Properties We take the density of water to be 1000 kg/m3.Analysis (a) We take the nozzle and the horizontal portion of the hose as the system such that water enters the control volume vertically and outlets horizontally (this way the pressure force and the momentum flux at the inlet are in the vertical direction, with no contribution to the force dimension in the horizontal direction), and designate the entrance by 1 and the outlet by 2. We also designate the horizontal coordinat e by x (with the direction of flow as being the positive direction). The sightly outlet velocity and the mass flow rate of water are determined from V= V& A = V& ? D / 4 2 = 5 m 3 /min ? (0. 06 m) 2 / 4 1768 m/min = 29. 5 m/s & m = ? V& = (1000 kg/m 3 )(5 m 3 /min) = 5000 kg/min = 83. 3 kg/s (b) The momentum equation for steady one-dimensional flow is & & ? F = ? ?mV ? ? ? mV . We let out in r r r horizontal force employ by the firemen to the nozzle to hold it be FRx, and assume it to be in the positive x direction. Then the momentum equation along the x direction gives ? ? 1N ? = 2457 N & & FRx = mVe ? 0 = mV = (83. 3 kg/s)(29. 5 m/s)? ? 1kg ? m/s 2 ? ? ? Therefore, the firemen must be able to resist a force of 2457 N to hold the nozzle in place. Discussion The force of 2457 N is equivalent to the weight of about 250 kg.That is, holding the nozzle requires the strength of holding a weight of 250 kg, which cannot be through with(p) by a single person. This demonstrates why severa l firemen are used to hold a hose with a high flow rate. FRz FRx 5 m3/min 6-19 PROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-37 A horizontal jet of water with a given velocity strikes a flat plate that is moving in the same direction at a specified velocity.The force that the water stream exerts against the plate is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The water splatters in all directions in the plane of the plate. 3 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the splattered water is the atmospheric pressure, which is disregarded since it acts on all surfaces. 4 The vertical forces and momentum fluxes are not considered since they have no effect on the horizontal force exerted on the p late. 5 The velocity of the plate, and the velocity of the water jet relative to the plate, are constant. Jet flow is nearly uniform and thus the momentum-flux correction factor can be taken to be unity, ? ? 1. Properties We take the density of water to be 1000 kg/m3. Analysis We take the plate as the control volume, and the flow direction as the positive direction of x axis. The mass flow rate of water in the jet is & m = ? V jet A = ? V jet 10 m/s 30 m/s FRx 5 cm Waterjet ?D 4 2 = (1000 kg/m 3 )(30 m/s) ? (0. 05 m) 2 4 = 58. 9 kg/s The relative velocity between the plate and the jet is V r = V jet ? V plate = 30 ? 10 = 20 m/s Therefore, we can assume the plate to be stationary and the jet to move with a velocity of 20 m/s.The r r r & & F= ? mV ? ? mV . We let the horizontal momentum equation for steady one-dimensional flow is ? ? out ? in reaction force applied to the plate in the negative x direction to counteract the impulse of the water jet be FRx. Then the momentum equation al ong the x direction gives ? ? 1N ? & & ? FRx = 0 ? mVi FRx = mV r = (58. 9 kg/s)(20 m/s)? ? 1kg ? m/s 2 ? = 1178 N ? ? Therefore, the water jet applies a force of 1178 N on the plate in the direction of motion, and an equal and opposite force must be applied on the plate if its velocity is to remain constant.Discussion Note that we used the relative velocity in the determination of the mass flow rate of water in the momentum analysis since water will enter the control volume at this rate. (In the trammel case of the plate and the water jet moving at the same velocity, the mass flow rate of water relative to the plate will be zero since no water will be able to strike the plate). 6-20 PROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.Chapter 6 Momentum Analysis of Flow Systems 6-38 Problem 6-37 is reconsidered. T he effect of the plate velocity on the force exerted on the plate as the plate velocity varies from 0 to 30 m/s in increments of 3 m/s is to be investigated. rho=1000 kg/m3 D=0. 05 m V_jet=30 m/s Ac=pi*D2/4 V_r=V_jet-V_plate m_dot=rho*Ac*V_jet F_R=m_dot*V_r N Vplate, m/s 0 3 6 9 12 15 18 21 24 27 30 Vr, m/s 30 27 24 21 18 15 12 9 6 3 0 FR, N 1767 1590 1414 1237 1060 883. 6 706. 9 530. 1 353. 4 176. 7 0 1800 1600 1400 1200 1000 FR, N 800 600 400 200 0 0 5 10 15 20 25 30 Vplate, m/s 6-21PROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-39E A devotee moves air at sea level at a specified rate. The force required to hold the lover and the minimum power introduce required for the raw sienna are to be determined. v Assumptions 1 The flow of air is steady and incompressib le. 2 Standard atmospheric conditions exist so that the pressure at sea level is 1 atm. song leaves the fan at a uniform velocity at atmospheric pressure. 4 Air approaches the fan through a humongous area at atmospheric pressure with negligible velocity. 5 The frictional effects are negligible, and thus the entire mechanical power excitant is converted to kinetic energy of air (no modulation to thermal energy through frictional effects). 6 Wind flow is nearly uniform and thus the momentum-flux correction factor can be taken to be unity, ? ? 1. Properties The gas constant of air is R = 0. 3704 psi? ft3/lbm? R. The standard atmospheric pressure at sea level is 1 atm = 14. 7 psi.Analysis (a) We take the control volume to be a horizontal high-flown cylinder bounded by streamlines on the sides with air go into through the large cross-sectional (section 1) and the fan hardened at the narrow cross-section at the end (section 2), and let its center line be the x axis. The density, ma ss flow rate, and discharge velocity of air are 14. 7 psi P ? = = = 0. 0749 lbm/ft 3 RT (0. 3704 psi ? ft 3 /lbm ? R)(530 R) & m = ? V& = (0. 0749 lbm/ft 3 )(2000 ft 3/min) = 149. 8 lbm/min = 2. 50 lbm/s V2 = V& A2 = V& 2 ? D 2 / 4 = 2000 ft 3 /min ? (2 ft) 2 / 4 = 636. 6 ft/min = 10. ft/s & & ? F = ? ?mV ? ? ? mV . Letting the out in The momentum equation for steady one-dimensional flow is r r r reaction force to hold the fan be FRx and assuming it to be in the positive x (i. e. , the flow) direction, the momentum equation along the x axis becomes 1 lbf ? ? & & FRx = m(V 2 ) ? 0 = mV = (2. 50 lbm/s)(10. 6 ft/s)? ? = 0. 82 lbf 2 ? 32. 2 lbm ? ft/s ? Therefore, a force of 0. 82 lbf must be applied (through friction at the base, for example) to prevent the fan from moving in the horizontal direction under the influence of this force. (b) Noting that P1 = P2 = Patm and V1 ? , the energy equation for the selected control volume reduces to ?P V2 ? ?P V2 ? & & & & & m? 1 + 1 + gz1 ? + W p ump, u = m? 2 + 2 + gz 2 ? + W turbine + E mech,loss ? ? ? ? 2 2 ? ? ? ? Substituting, V & & Wfan, u = m 2 2 2 V2 (10. 6 ft/s) 2 ? 1 lbf 1W ? & & Wfan,u = m 2 = (2. 50 lbm/s) ? ? = 5. 91 W 2 2 2 ? 32. 2 lbm ? ft/s 0. 73756 lbf ? ft/s ? Therefore, a useful mechanical power of 5. 91 W must be supplied to 2000 cfm air. This is the minimum required power stimulation required for the fan. Discussion The actual power commentary to the fan will be larger than 5. 1 W because of the fan inefficiency in converting mechanical power to kinetic energy. Fan 1 2 24 in 6-22 PROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-40 A helicopter hovers at sea level while being ludicrous. The volumetric air flow rate and the required power stimulus during un sealed hover, and the r evolutions per minute and the required power arousal during mingy hover are to be determined. Assumptions 1 The flow of air is steady and incompressible. 2 Air leaves the blades at a uniform velocity at atmospheric pressure. 3 Air approaches the blades from the top through a large area at atmospheric pressure with negligible velocity. 4 The frictional effects are negligible, and thus the entire mechanical power input is converted to kinetic energy of air (no diversity to thermal energy through frictional effects). 5 The change in air pressure with blossom is negligible because of the low density of air. 6 There is no acceleration of the helicopter, and thus the lift generated is equal to the thoroughgoing weight. Air flow is nearly uniform and thus the momentum-flux correction factor can be taken to be unity, ? ? 1. Properties The density of air is given to be 1. 18 kg/m3. Analysis (a) We take the control volume to be a vertical hyperbolic cylinder bounded by streamlines on the sides with air entering through the large cross-section (section 1) at the top and the fan located at the narrow cross-section at the pervade (section 2), and let its centerline be the z axis with upwards being the positive direction. r r r & & F= ? mV ? ? mV . Noting The momentum equation for steady one-dimensional flow is ? out ? in that the only force acting on the control volume is the total weight W and it acts in the negative z direction, the momentum equation along the z axis gives W & & ? W = m(? V 2 ) ? 0 W = mV 2 = ( ? AV 2 )V 2 = ? AV 22 V2 = ? A 1 where A is the blade bitstock area, 15 m A = ? D / 4 = ? (15 m) / 4 = 176. 7 m 2 2 2 Then the discharge velocity, volume flow rate, and the mass flow rate of air in the un laden mode become V 2,un affluent = m un prankish g = ? A (10,000 kg)(9. 81 m/s 2 ) (1. 18 kg/m 3 )(176. 7 m 2 ) = 21. 7 m/s ocean level 2 V&unloaded = AV 2,unloaded = (176. 7 m 2 )(21. m/s) = 3834 m 3 /s & munloaded = ? V&unloaded = (1. 18 kg/m 3 )(3834 m 3/s) = 4524 kg/s Load 15,000 kg Noting that P1 = P2 = Patm, V1 ? 0, the elevation effects are negligible, and the frictional effects are disregarded, the energy equation for the selected control volume reduces to ? P V2 ? ?P V2 ? V2 & & & & & & & m? 1 + 1 + gz1 ? + W pump, u = m? 2 + 2 + gz 2 ? + W turbine + E mech,loss Wfan, u = m 2 ? ? ? ? 2 2 2 ? ? ? ? Substituting, ? V2 ? 1 kW ? (21. 7 m/s) 2 ? 1 kN & ? ? & = (4524 kg/s) W unloaded fan,u = ? m 2 ? ? = 1065 kW 2 ? 1 kN ? m/s ? 1000 kg ? m/s ? ? 2 ? 2 ? ? ? ? nloaded (b) We now relieve the calculations for the loaded helicopter, whose mass is 10,000+15,000 = 25,000 kg V 2,loaded = m loaded g = ? A (25,000 kg)(9. 81 m/s 2 ) (1. 18 kg/m 3 )(176. 7 m 2 ) = 34. 3 m/s & mloaded = ? V&loaded = ? AV2, loaded = (1. 18 kg/m 3 )(176. 7 m 2 )(34. 3 m/s) = 7152 kg/s ? V2 ? (34. 3 m/s)2 & & = (7152 kg/s) Wloaded fan,u = ? m 2 ? ? 2 ? 2 ? ?loaded ? 1 kW ? 1 kN ? ? ? 1000 kg ? m/s 2 1 kN ? m/s ? = 4207 kW ? ? 6-23 PROPRIETARY MATERIAL . 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation.If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems Noting that the average flow velocity is proportional to the command overhead blade rotational velocity, the revolutions per minute of the loaded helicopter blades becomes & V 2 = kn V 2,loaded V 2, unloaded = & n loaded & n unloaded & n loaded = V 2,loaded V 2, unloaded & n unloaded = 34. 3 (400 rev) = 632 rpm 21. 7 Discussion The actual power input to the helicopter blades will be considerably larger than the calculated power input because of the fan inefficiency in converting mechanical power to kinetic energy. -24 PROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 6 Momentum Analysis of Flow Systems 6-41 A helicopter hovers on top of a high locoweed where the air density considerably lower than that at sea level. The blade rotational velocity to hover at the higher top and the percent increase in the required power input to hover at high altitude relative to that at sea level are to be determined. Assumptions 1 The flow of air is steady and incompressible. 2 The air leaves the blades at a uniform velocity at atmospheric pressure. 3 Air approaches the blades from the top through a large area at atmospheric pressure with negligible velocity. 4 The frictional effects are negligible, and thus the entire mechanical power input is converted to kinetic energy of air. 5 The change in air pressure with elevation while hovering at a given location is negligible because of the low density of air. 6 There is no acceleration of the helicopter, and thus the lift generated is equal to the total weight. Air flow is nearly uniform and thus the momen tum-flux correction factor can be taken to be unity, ? ? 1. Properties The density of air is given to be 1. 18 kg/m3 at sea level, and 0. 79 kg/m3 on top of the fate. Analysis (a) We take the control volume to be a vertical hyperbolic cylinder bounded by streamlines on the sides with air entering through the large cross-section (section 1) at the top and the fan located at the narrow cross-section at the bottom (section 2), and let its centerline be the z axis with upwards being the positive direction. r r & & F= ? mV ? ? mV . Noting The momentum equation for steady one-dimensional flow is ? ? out ? in that the only force acting on the control volume is the total weight W and it acts in the negative z direction, the momentum equation along the z axis gives W & & ? W = m(? V 2 ) ? 0 W = mV 2 = ( ? AV 2 )V 2 = ? AV 22 V2 = ? A where A is the blade span area. Then for a given weight W, the ratio of discharge velocities becomes V 2,mountain V 2,sea = W / ? mountain A W / ? sea A = ? sea ? mountain = 1. 18 kg/m 3 0. 79 kg/m 3 = 1. 222Noting that the average flow velocity is proportional to the overhead blade rotational velocity, the rpm of the helicopter blades on top of the mountain becomes & n = kV 2 & n mountain V 2, mountain = & n sea V 2,sea & n mountain = V 2, mountain V 2,sea & nsea = 1. 222(400 rpm) = 489 rpm Noting that P1 = P2 = Patm, V1 ? 0, the elevation effect are negligible, and the frictional effects are disregarded, the energy equation for the selected control volume reduces to ? P V2 ? ?P V2 ? V2 & & & & & & & m? 1 + 1 + gz1 ? + W pump, u = m? 2 + 2 + gz 2 ? W turbine + E mech,loss Wfan, u = m 2 ? ? ? ? 1 2 2 2 ? ? ? ? or V2 V2 V3 & & Wfan,u = m 2 = ? AV2 2 = ? A 2 = 2 2 2 1 2 ?A? ? ? W ? ? = ? ? ? A ? 3 1 2 ?A? ? ?W ? ? ? ? ? A ? 1 . 5 = W 1 . 5 2 ? A 15 m Then the ratio of the required power input on top of the mountain to that at sea level becomes & Wmountain fan,u 0. 5W 1. 5 / ? mountain A = & Wsea fan,u 0. 5W 1. 5 / ? sea A 2 ? mountai n ?sea = 1. 18 kg/m3 = 1. 222 0. 79 kg/m3 Sea level Load 15,000 kg Therefore, the required power input will increase by 22. 2% on top of the mountain relative to the sea level.Discussion Note that both the rpm and the required power input to the helicopter are inversely proportional to the square root of air density. Therefore, more power is required at higher elevations for the helicopter to operate because air is less dense, and more air must be forced by the blades into the downdraft. 6-25 PROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.Chapter 6 Momentum Analysis of Flow Systems 6-42 The flow rate in a channel is controlled by a scour gate by raising or lowering a vertical plate. A semblance for the force acting on a sluice gate of width w for steady and uniform flow is to be developed. Assumptions 1 The f low is steady, incompressible, frictionless, and uniform (and thus the Bernoulli equation is applicable. ) 2 Wall shear forces at surfaces are negligible. 3 The channel is exposed to the atmosphere, and thus the pressure at rationalize surfaces is the atmospheric pressure. 4 The flow is horizontal. Water flow is nearly uniform and thus the momentum-flux correction factor can be taken to be unity, ? ? 1. Analysis We take point 1 at the bountiful surface of the upstream flow before the gate and point 2 at the free surface of the downstream flow after the gate. We also take the bottom surface of the channel as the reference level so that the elevations of points 1 and 2 are y1 and y2, respectively. The application of the Bernoulli equation between points 1 and 2 gives P1 V12 P V2 + + y1 = 2 + 2 + y 2 ? g 2 g ? g 2 g V 22 ? V12 = 2 g( y1 ? y 2 ) (1)The flow is assumed to be incompressible and thus the density is constant. Then the conservation of mass relation for this single stream steady flow device can be expressed as V&1 = V&2 = V& A1V1 = A2V 2 = V& V1 = V& A1 = V& wy1 and V2 = V& A2 = V& wy 2 (2) Substituting into Eq. (1), ? V& ? ? wy ? 2 ? ? V& ? 2 g ( y1 ? y 2 ) & ? ? ? ? wy ? = 2 g ( y1 ? y 2 ) V = w 1 / y 2 ? 1 / y 2 ? ? 1? 2 1 2 2 2 g ( y1 ? y 2 ) V& = wy 2 2 2 1 ? y 2 / y1 (3) Substituting Eq. (3) into Eqs. (2) gives the following transaction for velocities, V1 = y2 y1 2 g ( y1 ? y 2 ) 1? y2 / 2 y1 and V2 = 2 g ( y1 ? y 2 ) 2 2 1 ? y 2 / y1 (4) We choose the control volume as the water body surrounded by the vertical cross-sections of the upstream and downstream flows, free surfaces of water, the interior(a) surface of the sluice gate, and the bottom surface of r r r & & F= ? mV ? ? mV . The the channel. The momentum equation for steady one-dimensional flow is ? ? out ? in force acting on the sluice gate FRx is horizontal