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13 Chapter MOMENTUM ANALYSIS OF FLOW SYSTEMS
Newton's Laws and Conservation of Momentum 13-1C Mass, energy, momentum, and electric charge are conserved, and volume and entropy are not conserved during a process. 13-2C 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 when the net force acting on it is zero ." Therefore, a body tends to preserve 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." r 13-3C Since momentum ( mV ) is the product of a vector (velocity) and a scalar (mass), momentum must be a vector that points in the same direction as the velocity vector. 13-4C The conservation of momentum principle is expressed as "the momentum of a system remains constant when the net force acting on it is zero, and thus the momentum of such systems is conserved". The momentum of a body remains constant if the net force acting on it is zero. 13-5C Newton's second law of motion, also called the angular momentum equation, is expressed as "the rate of change of the angular momentum of a body is equal to the net torque acting it ." For a non-rigid body with zero net torque, the angular momentum remains constant, but the angular velocity changes in accordance with I = constant where I is the moment of inertia of the body. 13-6C No. Two rigid bodies having the same mass and angular speed will have different angular momentums unless they also have the same moment of inertia I. Linear Momentum Equation 13-7C The relationship between the time rates of change of an extensive property for a system and for a control volume is expressed by the Reynolds transport theorem, which provides the link between the r system and control volume concepts. The linear momentum equation is obtained by setting b =V and thus r B = mV in the Reynolds transport theorem. 13-8C The forces acting on the control volume consist of body forces that act throughout the entire body of the control volume (such as gravity, electric, and magnetic forces) and surface forces that act on the control surface (such as the pressure forces and reaction forces at points of contact). The net force acting on a control volume is the sum of all body and surface forces. Fluid weight is a body force, and pressure is a surface force (acting per unit area). 13-9C All of these surface forces arise as the control volume is isolated from its surroundings for analysis, and the effect of any detached object is accounted for by a force at that location. We can minimize the number of surface forces exposed by choosing the control volume such that the forces that we are not interested in remain internal, and thus they do not complicate the analysis. A well-chosen control volume exposes only the forces that are to be determined (such as reaction forces) and a minimum number of other forces. 13-10C The momentum-flux correction factor enables us to express the momentum flux in terms of the r r r r & V ( n )dAc = mV m . The value of is unity for mass flow rate and mean flow velocity as V
Ac
uniform flow, such as a jet flow, nearly unity for turbulent flow (between 1.01 and 1.04), but about 1.3 for laminar flow. So it should be considered in laminar flow.
Chapter 13 Momentum Analysis of Flow Systems 13-11C The momentum equation for steady one-dimensional flow for the case 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 control volume, and first term on the right hand side is the incoming momentum flux and the second term is the outgoing momentum flux by mass. 13-12C In the application of the momentum equation, we can disregard the atmospheric pressure and work with gage pressures only since the atmospheric pressure acts in all directions, and its effect cancels out in every direction. 13-13C The fireman who holds the hose backwards so that the water makes a U-turn before being discharged will experience a greater reaction force since the numerical values of momentum fluxes across the nozzle are added in this case instead of being subtracted. 13-14C No, V is not the upper limit to the rocket's ultimate velocity. Without friction the rocket velocity will continue to increase as more gas exits the nozzle. 13-15C A helicopter hovers because the strong downdraft of air, caused by the overhead propeller blades, manifests a momentum in the air stream. This momentum must be countered by the helicopter lift force. 13-16C As the air density decreases, it requires more energy for a helicopter to hover, because more air must be forced into the downdraft by the helicopter blades to provide the same lift force. Therefore, it takes more power for a helicopter to hover on the top of a high mountain than it does at sea level. 13-17C In winter the air is generally colder, and thus denser. Therefore, less air must be driven by the blades to provide the same helicopter lift, requiring less power.
13-2
Chapter 13 Momentum Analysis of Flow Systems 13-18C The force required to hold the plate against the horizontal water stream will increase by a factor of 4 when the velocity is doubled since & F = mV = ( AV )V = AV 2 and thus the force is proportional to the square of the velocity. 13-19C The acceleration will not be constant since the force is not constant. The impulse force exerted by & water on the plate is F = mV = ( AV )V = AV 2 , where V is the relative velocity between the water and the plate, which is moving. The plate acceleration will be a = F/m. But as the plate begins to move, V decreases, so the acceleration must also decrease. 13-20C The maximum velocity possible for the plate is the velocity of the water jet. As long as the plate is moving slower than the jet, the water will exert a force on the plate, which will cause it to accelerate, until terminal jet velocity is reached. 13-21 It is to be shown that the force exerted by a liquid jet of velocity V on a stationary nozzle is & proportional to V2, or alternatively, to m 2 . Assumptions 1 The flow is steady, one-dimensional, and incompressible. 2 The nozzle is given to be stationary. 3 The nozzle involves a 90 turn and thus the incoming and outgoing flow streams are normal to each other. 4 The water is discharged to the atmosphere, and thus the gage pressure at the exit is zero. Analysis We take the nozzle as the control volume, and the flow direction at the exit as the x axis. Note that the nozzle makes a 90 turn, and thus the flow direction at the inlet is in the normal direction to exit flow, and thus it does not contribute to any pressure force or momentum flux term at the inlet in the x direction. & Noting that m = AV where A is the nozzle exit area and V is the mean nozzle exit velocity, the momentum equation for steady one-dimensional flow in the x direction reduces to r r r & & & & F = m e Ve - mi Vi FRx = m e Ve = mV where FRx is the reaction force on the nozzle due to liquid jet at the nozzle exit. Then, & m = AV & FRx = mV = AVV = AV 2 & & or FRx = mV = m & & m m2 = A A
Therefore, the force exerted by a liquid jet of velocity V on this & stationary nozzle is proportional to V2, or alternatively, to m 2 .
Liquid
Nozzle
V
FR
13-3
Chapter 13 Momentum Analysis of Flow Systems 13-22 A water jet of velocity V impinges on a plate moving toward the water jet with velocity V. The force required to move the plate towards the jet is to be determined in terms of F acting on the stationary plate. Assumptions 1 The flow is steady, one-dimensional, and incompressible. 2 The plate is vertical and the jet is stationary and normal to plate. 3 The pressure on both sides of the plate is atmospheric pressure (and thus its effect can be disregarded). 4 Fiction during motion is negligible. 5 There is no acceleration of the plate. 6 The water splashes off the sides of the plate in a plane normal to the jet. Analysis We take the plate as the control volume. The relative velocity between the plate and the jet is V when the plate is stationary, and 1.5V when the plate is moving with a velocity V towards the plate. Then the momentum equation for steady one-dimensional flow in the horizontal direction reduces to r r r & & & & F= mV - mV - FR = -mi Vi FR = m i Vi
out
in
Stationary plate: ( Vi = V and Moving plate: ( Vi = 1.5V and
& m i = AVi = AV ) FR = AV 2 = F & m i = AVi = A(1.5V ) ) FR = A(1.5V ) 2 = 2.25 AV 2 = 2.25 F
Therefore, the force required to hold the plate stationary against the oncoming water jet becomes 2.25 times when the jet velocity becomes 1.5 times. Discussion Note that when the plate is stationary, V is also the jet 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 off its sides) per unit time also increases by 50%.
1/2V
V
Waterjet
13-4
Chapter 13 Momentum Analysis of Flow Systems 13-23 A 90 elbow deflects water upwards and discharges it to the atmosphere at a specified rate. The gage pressure at the inlet of the elbow and the anchoring force needed to hold the elbow in place are 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 negligible. 3 The water is discharged to the atmosphere, and thus the gage pressure at the exit is zero. Properties We take the density of water to be 1000 kg/m3. Analysis (a) We take the elbow as the control volume, and designate the entrance by 1 and the exit 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 y. The continuity equation for this one-inlet one-exit steady flow system is & & & & m1 = m 2 = m = 30 kg/s. Noting that m = AV , the mean inlet and exit velocities of water are & & 25 kg/s m m = = = 3.18 m/s 2 A (D / 4) (1000 kg/m 3 )[ (0.1 m) 2 / 4] Noting that V1 = V2 and P2 = Patm, the Bernoulli equation for a streamline going through the center of the reducing elbow is expressed as V1 = V2 = V =
P V12 P V2 1 + + y1 = 2 + 2 + y2 P - P2 = g ( y2 - y1 ) P , gage = g ( y2 - y1 ) 1 1 g 2 g g 2 g
Substituting,
1 kN 2 P , gage = (1000 kg/m 3 )(9.81 m/s 2 )(0.35 m) 1 1000 kg m/s 2 = 3.434 kN/m = 3.434 kPa
r r r & & (b) The momentum equation for steady one-dimensional flow is F = m e Ve - mi Vi . We let the x- and y- components of the anchoring force of the elbow be FRx and FRy, 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 y axes become & & FRx + P1, gage A1 = 0 - m(+ V1 ) = -mV & & FRy = m(+ V2 ) = mV Solving for FRx and FRy, and substituting the given values, & FRx = -mV - P1, gage A1 1N = -(25 kg/s)(3.18 m/s) 1 kg m/s 2 = -107 N - (3434 N/m 2 )[ (0.1 m) 2 / 4] = 79.5 N = tan -1 FRy FRx = tan -1 79.5 = -36.6 = 143 - 107
Water 25 kg/s
y x FRy FRx 1 2
35 cm
1N & FRy = mV = (25 kg/s)(3.18 m/s) 1 kg m/s 2 and
2 2 FR = FRx + FRy = (-107) 2 + 79.5 2 = 133 N,
Discussion Note that the magnitude of the anchoring force is 133 N, and its line of action makes 143.4 from the positive x direction. Also, a negative value for FRx indicates the assumed direction is wrong, and should be reversed. 13-24 An 180 elbow forces the flow to make a U-turn and discharges it to the atmosphere at a specified rate. The gage pressure at the inlet of the elbow and the anchoring force needed to hold the elbow in place are 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 negligible. 3 The water is discharged to the atmosphere, and thus the gage pressure at the exit is zero.
13-5
Chapter 13 Momentum Analysis of Flow Systems Properties We take the density of water to be 1000 kg/m3. Analysis (a) We take the elbow as the control volume, and designate the entrance by 1 and the exit 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 y. The continuity equation for this one-inlet one-exit steady flow system is & & & & m1 = m 2 = m = 30 kg/s. Noting that m = AV , the mean inlet and exit velocities of water are & & 25 kg/s m m = = = 3.18 m/s A (D 2 / 4) (1000 kg/m 3 )[ (0.1 m) 2 / 4] Noting that V1 = V2 and P2 = Patm, the Bernoulli equation for a streamline going through the center of the reducing elbow is expressed as V1 = V2 = V =
P V12 P V2 1 + + y1 = 2 + 2 + y2 P - P2 = g ( y2 - y1 ) P , gage = g ( y2 - y1 ) 1 1 g 2 g g 2 g
Substituting,
1 kN 2 P , gage = (1000 kg/m 3 )(9.81 m/s 2 )(0.70 m) 1 1000 kg m/s 2 = 6.867 kN/m = 6.867 kPa
r r r & & (b) The momentum equation for steady one-dimensional flow is F = m e Ve - mi Vi . We let the x- and y- components of the anchoring force of the elbow be FRx and FRy, 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 y axes become & & & FRx + P1, gage A1 = m(-V2 ) - m(+ V1 ) = -2mV FRy = 0 Solving for FRx and substituting the given values, & FRx = -2mV - P1, gage A1 1N = -2(25 kg/s)(3.18 m/s) 1 kg m/s 2 = -213 N - (6867 N/m 2 )[ (0.1 m) 2 / 4] y x FRy FRx
Water 25 kg/s 35 cm
2
and FR = FRx = - 213 N since the y-component of the anchoring force is zero. Therefore, the anchoring force has a magnitude of 213 N and it acts in the negative x direction. Discussion Note that a negative value for FRx indicates the assumed direction is wrong, and should be reversed.
1
13-6
Chapter 13 Momentum Analysis of Flow Systems 13-25E A horizontal water jet strikes a vertical stationary plate normally at a specified velocity. For a given anchoring force needed to hold the plate in place, the flow rate of water is to be determined. Assumptions 1 The flow is steady, one-dimensional, and incompressible. 2 The water 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 splattered water is the atmospheric pressure which is disregarded since it acts on the entire control surface. 4 The vertical forces and momentum fluxes are not considered since they have no effect on the horizontal reaction force. Properties We take the density of water to be 62.4 lbm/ft 3. Analysis We take the plate as the control volume such that it contains the entire plate and cuts through the water jet and the support bar normally, and the direction of flow as the positive direction of x axis. The momentum equation for steady one-dimensional flow in the x (flow) direction reduces in this case to r r r & & & & F = m e Ve - mi Vi - FRx = -mV1 FR = mV1 We note that the reaction force acts in the opposite direction to flow, and we should not forget the negative & sign for forces and velocities in the negative x-direction. Solving for m and substituting 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 & 376 lbm/s & m V= = = 6.02 ft 3 /s 62.4 lbm/ft 3 Therefore, the volume flow rate of water under stated assumptions must be 3.45 ft 3/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.
FRx = 350 lbf
m
1
Waterjet
13-7
Chapter 13 Momentum Analysis of Flow Systems 13-26 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. 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 exit is zero. 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 entrance by 1 and the exit 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 y. The continuity equation for this one-inlet one-exit steady flow system is & & & & m1 = m 2 = m = 30 kg/s. Noting that m = AV , the inlet and exit velocities of water are & 30 kg/s m V1 = = = 2.0 m/s 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 V 2 - V12 P V12 P V2 1 + + y1 = 2 + 2 + y2 P - P2 = g 2 + y2 - y1 P , gage = g 2 + y2 1 1 2g 2g 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/m 3 )(9.81 m/s 2 ) + 0.4 1 2 1000 kg m/s 2 2(9.81 m/s )
r r r & & The momentum equation for steady one-dimensional flow is F = m e Ve - mi Vi . We let the x- and ycomponents of the anchoring force of the elbow be FRx and FRy, 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 y axes become 2 & & & FRx + P1,gage A1 = mV2 cos - mV1 and FRy - W = mV2 sin Solving for FRx and FRy, 25 cm2 and substituting the given values, & FRx = m(V2 cos - V1 ) - P1, gage A1 1 kN = (30 kg/s)[(12cos45 - 2) m/s] 1000 kg m/s 2 - (73.9 kN/m 2 )(0.0150 m 2 ) = -0.914 kN
Water 30 kg/s 45
FRy
150 m2
FRx
50 kg
1 1 kN - 0.4905 kN = 0.745 kN & FRy = mV2 sin - W = (30 kg/s)(12sin30 m/s) 1000 kg m/s 2 FRy 0.745 2 2 = tan -1 = -39.2 = 140.8 FR = FRx + FRy = (-0.914 ) 2 + 0.745 2 = 1.18 kN, = tan -1 FRx - 0.914 Discussion Note that the magnitude of the anchoring force is 1.18 kN, and its line of action makes 140.8 from the positive x direction. Also, a negative value for FRx indicates the assumed direction is wrong, and should be reversed. 13-27 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. 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 exit is zero. 13-8
Chapter 13 Momentum Analysis of Flow Systems 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 entrance by 1 and the exit 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 y. The continuity equation for this one-inlet one-exit steady flow system is & & & & m1 = m 2 = m = 30 kg/s. Noting that m = AV , the inlet and exit velocities of water are & 30 kg/s m V1 = = = 2.0 m/s 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 V 2 - V12 P V12 P V2 1 + + y1 = 2 + 2 + y2 P - P2 = g 2 + y2 - y1 P , gage = g 2 + y2 1 1 2g 2g g 2 g g 2 g (12 m/s) 2 - (2 m/s) 2 1 kN = 73.9 kN/m 2 = 73.9 kPa or, P , gage = (1000 kg/m 3 )(9.81 m/s 2 ) + 0.4 1 2 1000 kg m/s 2 2(9.81 m/s ) r r r & & The momentum equation for steady one-dimensional flow is F = m e Ve - mi Vi . We let the x- and ycomponents of the anchoring force of the elbow be FRx and FRy, 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 y axes become
& & FRx + P1,gage A1 = mV2 cos - mV1
and
& FRy - W = mV2 sin
Solving for FRx and FRy, and substituting the given values, & FRx = m(V2 cos - V1 ) - P1, gage A1 1 kN = (30 kg/s)[(12cos110 - 2) m/s] 1000 kg m/s 2 - (73.9 kN/m 2 )(0.0150 m 2 ) = -1.292 kN 1 kN & FRy = mV2 sin + W = (30 kg/s)(12sin110 m/s) 1000 kg m/s 2 + 0.4905 kN = 0.8288 kN 1 2 2 2 2 FR = FRx + FRy = (-1.292) + 0.8288 = 1.53 kN 25 cm2 = tan -1 FRy = tan -1
110
0.8288 = -32.7 FRx - 1.292 Discussion Note that the magnitude of the anchoring force is 1.53 kN, and its line of action makes 32.7 from the positive x direction. Also, a negative value for FRx indicates the assumed direction is wrong, and should be reversed.
and
FRy
FRx
150 m2
Water 1 50 kg 13-28 Water accelerated by a nozzle strikes the back surface of a cart moving horizontally at a constant 30 kg/s velocity. The braking force and the power wasted by the brakes are to be determined. .
Assumptions 1 The flow is steady, one-dimensional, and incompressible. 2 The water splatters off the sides of the plate in all directions in the plane of the back surface. 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 reaction force. 5 Fiction during motion is negligible. 6 There is no acceleration of the cart. 7 The motions of the water jet and the cart are horizontal. Analysis We take the cart as the control volume, and the direction of flow as the positive direction of x axis. The relative velocity
13-9
15 m/s
5 m/s
Chapter 13 Momentum Analysis of Flow Systems between the cart and the jet is Vr = V jet - Vcart = 15 - 10 = 10 m/s Therefore, we can assume the cart to be stationary and the jet to move 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 = m e Ve - mi Vi FRx = -mi Vi Fbrake = -mVr We note that the brake force acts in the opposite direction to flow, and we should not forget the negative sign for forces and velocities in the negative x-direction. Substituting the given values, 1N & Fbrake = - mVr = -(25 kg/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 distance and the distance traveled by the cart per unit time is the cart velocity, the power wasted by the brakes is 1 kW & W = Fbrake Vcart = (250 N)(5 m/s) = 1.25 kW 1000 N m/s Discussion Note that the power wasted is equivalent to the maximum power that can be generated as the cart velocity is maintained constant.
13-10
Chapter 13 Momentum Analysis of Flow Systems 13-29 Water accelerated by a nozzle strikes the back surface of a cart moving horizontally. The acceleration of the cart if the brakes fail is to be determined. Analysis The braking force was determined in previous problem to be 250 N. When the brakes fail, this force will propel the cart forward, and the accelerating will be a= 250 N 1 kg m/s 2 F = 1N m cart 300 kg = 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
13-11
Chapter 13 Momentum Analysis of Flow Systems 13-30E A water jet hits a stationary splitter, such that half of the flow is diverted upward at 45, and the other half is directed down. The force required to hold the splitter in place is to be determined. EES Assumptions 1 The flow is steady, one-dimensional, and incompressible. 2 The water jet is exposed to the atmosphere, and thus the pressure of the water jet before and after the split is the atmospheric pressure which is disregarded since it acts on all surfaces. 3 The gravitational effects are disregarded. Properties We take the density of water to be 62.4 lbm/ft 3. Analysis The mass flow rate of water jet is & & m = V = (62.4 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 exit 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 y. r r r & & The momentum equation for steady one-dimensional flow is F = m e Ve - mi Vi . We let the xand y- components of the anchoring force of the splitter be FRx and FRy, 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 y 2 axes become & & & FRx = 2( 1 m)V2 cos - mV1 = mV (cos - 1) 2 & & FRy = 1 m(+ V2 sin ) + 1 m(-V2 sin ) - 0 = 0 2 2 Substituting the given values, 1 lbf FRx = (6240 lbm/s)(20 ft/s)(cos45 - 1) = -1135 lbf 2 32.2 lbm ft/s FRy = 0 The negative value for FRx indicates the assumed direction is wrong, and should be reversed. Therefore, a force of 1135 lbf must be applied to the splitter in the opposite direction to flow to hold it in place. No holding force is necessary in the vertical direction. This can also be concluded from the symmetry. Discussion In reality, the gravitational effects will cause the upper stream to slow down and the lower stream to speed up after the split. But for short distances, these effects are indeed negligible.
20 ft/s
FRy
100 ft/s 45 45
FRx
13-12
Chapter 13 Momentum Analysis of Flow Systems 13-31E Problem 13-30E is reconsidered. The effect of splitter angle on the force exerted on the splitter as the half splitter angle varies from 0 to 180 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 90 100 110 120 130 140 150 160 170 180
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
4000 3000 2000 1000 0 0 20 40 60 80 100 120 140 160 180
,
13-13
Chapter 13 Momentum Analysis of Flow Systems 13-32 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, one-dimensional, 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 vertical forces and momentum fluxes are not considered since they have no effect on the horizontal motion. 5 The tract is frictionless, and thus fiction during motion is negligible. 6 The motions of the water jet and the cart are horizontal. 7 The velocity of the jet relative to the plate remains constant, Vr = Vjet = V. 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 = m e Ve - mi Vi FRx = - mi Vi FRx = -mV 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 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
= 636 N
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 = Vplate a = (9 - 0) m/s 0.636 m/s
2
FRx
= 1 4 .2 s
Frictionless track
(c) Noting that a = dV/dt and thus dV = adt and that the acceleration a is constant, the plate velocity in 20 s becomes Vplate = V0, plate + at = 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.
13-14
Chapter 13 Momentum Analysis of Flow Systems 13-33 A 90 reducer elbow deflects water downwards into a smaller diameter pipe. The resultant 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 gravitational effects are negligible. 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 exit 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 y. The continuity equation for this one-inlet one-exit steady flow system is & & & & m1 = m 2 = m = 353.4 kg/s. Noting that m = AV , the mass flow rate of water and its exit velocity are & m = V1 A1 = V1 (D12 / 4) = (1000 kg/m 3 )(5 m/s)[ (0.3 m) 2 / 4] = 353.4 kg/s & & 353.4 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 + + y1 = 2 + 2 + y2 g 2 g g 2 g
2 V 2 - V2 P2 = P + g 1 + y1 - y2 1 2g
Substituting, the gage pressure at the exit 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 )
r r r & & The momentum equation for steady one-dimensional flow is F = m e Ve - mi Vi . We let the x- and ycomponents of the anchoring force of the elbow be FRx and FRy, and assume them to be in the positive directions. Then the momentum equations along the x and y axes become & FRx + P1, gage A1 = 0 - mV1 & FRy - P2, gage A2 = m(-V2 ) - 0 Note that we should not forget the negative sign for forces and velocities in the negative x or y direction. Solving for FRx and FRy, and substituting the given values, 1 kN & FRx = -mV1 - P1, gage A1 = -(353.4 kg/s)(5 m/s) 1000 kg m/s 2 1 kN & FRy = -mV2 + P2, gage A1 = -(353.4 kg/s)(20 m/s) 1000 kg m/s 2 and FR =
2 FRx 2 - (300 kN/m 2 ) (0.3 m) = -23.0 kN 4 2 + (117.4 kN/m 2 ) (0.15 m) = -5.0 kN 4 FRy
FRx +
2 FRy
= (-23.0) + (-5.0) = 23.5 kN,
2 2
30 cm
= tan -1
FRy FRx
= tan -1
- 5.0 = 12.3 - 23.0
Water 5 m/s
Discussion Note that the magnitude of the anchoring force is 23.5 kN, and its line of action makes 12.3 from the positive x direction. Also, negative values for FRx and FRy indicate that the assumed directions are wrong, and should be reversed.
15 cm
13-34 A wind turbine with a given span diameter and efficiency is subjected to steady winds. The power generated and the horizontal force on the supporting mast of the turbine are to be determined. EES
13-15
Chapter 13 Momentum Analysis of Flow Systems Assumptions 1 The wind flow is steady, one-dimensional, and incompressible. 2 The efficiency of the turbine-generator is independent of wind speed. 3 The frictional effects are negligible, and thus none of the incoming kinetic energy is converted to thermal energy. Properties The density of air is given to be 1.25 kg/m3. Analysis (a) The power potential of the wind is its kinetic 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 D 2 4
2
Wind V1
1
D
2
V2
= (1.25 kg/m3 )(6.94 m/s)
(90 m) 2 = 55,200 kg/s 4 1 kW = 1330 kW 1 kN m/s
V (6.94 m/s) 2 1 kN & & & Wmax = mke1 = m 1 = (55,200 kg/s) 1000 kg m/s 2 2 2 Then the actual power produced becomes
& & Wact = wind turbi neW max = (0.32)(1330 kW) = 426 kW
FR
(b) The frictional effects are assumed to be negligible, and thus the portion of incoming kinetic energy not converted 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 V2 = V1 1 - wind turbine = (6.94 m/s) 1 - 0.32 = 5.72 m/s We choose the control volume around the wind turbine such that the wind is normal to the control surface at the inlet and the exit, and the entire control surface is at the atmospheric pressure. The momentum r r r & & equation for steady one-dimensional flow is F = m e Ve - mi Vi . Writing it along the x-direction (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 = mV2 - mV1 = m(V2 - 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 tower where the wind turbine is installed, and the bending moment it generates at the bottom of the tower is obtained by multiplying this force by the tower height.
13-16
Chapter 13 Momentum Analysis of Flow Systems 13-35E A horizontal water jet strikes a curved plate, which deflects the water back to its original direction. The force required to hold the plate against the water stream is to be determined. Assumptions 1 The flow is steady, one-dimensional, 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 included in the required force to hold the plate). 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. Properties We take the density of water to be 62.4 lbm/ft 3. Analysis We take the plate together with the curved water jet as the control volume, and designate the jet inlet by 1 and the exit by 2. We also designate the horizontal coordinate by x (with the direction of incoming flow as being the positive direction). The continuity equation for this one-inlet one-exit 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 = m e Ve - mi Vi . Letting the 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(-V2 ) - 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 applied 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
Waterjet
FRx
1
140 ft/s
3 in
13-17
Chapter 13 Momentum Analysis of Flow Systems 13-36E 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, one-dimensional, 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. Properties We take the density of water to be 62.4 lbm/ft 3. Analysis We take the plate together with the curved water jet as the control volume, and designate the jet inlet by 1 and the exit 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 y. The continuity equation for & & & this one-inlet one-exit 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 = m e Ve - mi Vi . We let the x- and ycomponents of the anchoring force of the plate be FRx and FRy, and assume them to be in the positive directions. Then the momentum equations along the x and y axes become & & & FRx = m(-V2 ) cos 45 - m(+ V1 ) = -mV (1 + cos 45) & & FRy = m(+ V2 ) sin 45 = mV sin 45 Substituting the given values, 1 lbf FRx = -2(428.8 lbm/s)(140 ft/s)(1 + cos45) 32.2 lbm ft/s 2 = -6365 lbf 1 lbf FRy = (428.8 lbm/s)(140 ft/s)sin45 = 1318 lbf 32.2 lbm ft/s 2 and
2 2 FRy = FRx + FRy = (-6365) 2 + 1318 2 = 6500 lbf,
2
140 ft/s Waterjet
135
FRy FRx
3 in
1
= tan -1
FRy FRx
= tan -1
1318 = -11.7 = 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.
13-18
Chapter 13 Momentum Analysis of Flow Systems 13-37 Firemen are holding a nozzle at the end of a hose while trying to extinguish a fire. The average water exit velocity and the resistance force required of the firemen to hold the nozzle are to be determined. Assumptions 1 The flow is steady, one-dimensional, 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. 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 exits horizontally (this way the pressure force and the momentum flux at the inlet are in the vertical direction, with no contribution to the force balance in the horizontal direction), and designate the entrance by 1 and the exit by 2. We also designate the horizontal coordinate by x (with the direction of flow as being the positive direction). The average exit velocity and the mass flow rate of water are determined from V= & & V V 5 m 3 /min = = = 1768 m/min = 29.5 m/s A D 2 / 4 (0.06 m) 2 / 4
& & m = V = (1000 kg/m 3 )(5 m 3 /min) = 5000 kg/min = 83.3 kg/s r r r & & (b) The momentum equation for steady one-dimensional flow is F = m e Ve - mi Vi . We let horizontal force applied 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) 2 1kg m/s 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 done by a single person. This demonstrates why several firemen are used to hold a hose with a high flow rate.
FRy FRx
5 m3/min
13-19
Chapter 13 Momentum Analysis of Flow Systems 13-38 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, one-dimensional, 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 plate. 5 The velocity of the plate, and the velocity of the water jet relative to the plate, are constant. 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 D 2
5 cm
3
10 m/s 30 m/s
FRx
Waterjet
(0.05 m) 2 = (1000 kg/m )(30 m/s) = 58.9 kg/s 4 4
The relative velocity between the plate and the jet is Vr = V jet - Vplate = 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 & & momentum equation for steady one-dimensional flow is F = m e Ve - mi Vi . We let the horizontal 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 along the x direction gives 1N & & - FRx = 0 - mVi FRx = mVr = (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 limiting 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).
13-20
Chapter 13 Momentum Analysis of Flow Systems 13-39 Problem 13-38 is reconsidered. The 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*D^2/4 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
F R, 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
13-21
Chapter 13 Momentum Analysis of Flow Systems 13-40E A fan moves air at sea level at a specified rate. The force required to hold the fan and the minimum power input required for the fan are to be determined. Assumptions 1 The flow of air is steady, one-dimensional, and incompressible. 2 Standard atmospheric conditions exist so that the pressure at sea level is 1 atm. 3 Air leaves the fan at a uniform velocity at atmospheric pressure. 4 Air approaches the fan through a large area at atmospheric pressure with negligible velocity. 5 The frictional effects are negligible, and thus the entire mechanical power input is converted to kinetic energy of air (no conversion to thermal energy through frictional effects). Properties The gas constant of air is R = 0.3704 psift3/lbmR. The standard atmospheric pressure at sea level is 1 atm = 14.7 psi. Analysis (a) We take the control volume to be a horizontal hyperbolic cylinder bounded by streamlines on the sides with air entering through the large cross-section (section 1) and the fan located at the narrow cross-section at the end (section 2), and let its centerline be the x axis. The density, mass 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 & & V V 2000 ft 3 /min V2 = = = = 636.6 ft/min = 10.6 ft/s 2 A2 D 2 / 4 (2 ft) 2 / 4 r r r & & The momentum equation for steady one-dimensional flow is F = m e Ve - mi Vi . Letting the 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(V2 ) - 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 0, the energy equation for the selected control volume reduces to
P V2 P V2 & & & & & m 1 + 1 + gz1 + W pump, u = m 2 + 2 + gz 2 + W turbine + E mech,loss 2 2
Substituting,
V2 (10.6 ft/s) 2 1 lbf & & Wfan, u = m 2 = (2.50 lbm/s) 2 2 32.2 lbm ft/s 2
V & & Wfan, u = m 2 2
2
1W = 5.91 W 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 input required for the fan. Discussion The actual power input to the fan will be larger than 5.91 W because of the fan inefficiency in converting mechanical power to kinetic energy. 1
24 in Fan
2
13-22
Chapter 13 Momentum Analysis of Flow Systems 13-41 A helicopter hovers at sea level while being loaded. The volumetric air flow rate and the required power input during unloaded hover, and the rpm and the required power input during loaded hover are to be determined. Assumptions 1 The flow of air is steady, one-dimensional, 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 conversion to thermal energy through frictional effects). 5 The change in air pressure with elevation 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. 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 bottom (section 2), and let its centerline be the z axis with upwards being the positive direction. r r r & & The momentum equation for steady one-dimensional flow is F = m e Ve - mi Vi . Noting 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 = m ( - V2 ) - 0 where A is the blade span area,
A = D 2 / 4 = (15 m) 2 / 4 = 176.7 m 2
2 & W = mV2 = ( AV2 )V2 = AV2
V2 =
W A
1
15 m
Then the discharge velocity, volume flow rate, and the mass flow rate of air in the unloaded mode become V2, unloaded = m unloaded g = A (10,000 kg)(9.81 m/s 2 ) (1.18 kg/m 3 )(176.7 m 2 ) = 21.7 m/s
2
Sea level Load 15,000 kg
& V unloaded = AV2, unloaded = (176.7 m 2 )(21.7 m/s) = 3834 m 3 /s
& & munloaded = Vunloaded = (1.18 kg/m 3 )(3834 m3/s) = 4524 kg/s
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 & & Wunloaded fan, u = m 2 = (4524 kg/s) = 1065 kW 2 1 kN m/s 1000 kg m/s 2 2 unloaded (b) We now repeat the calculations for the loaded helicopter, whose mass is 10,000+15,000 = 25,000 kg: V2,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 = Vloaded = AV2, loaded = (1.18 kg/m3 )(176.7 m 2 )(34.3 m/s) = 7152 kg/s & Wloaded
fan, u
V2 (34.3 m/s) 2 & = (7152 kg/s) = m 2 2 2 loaded
1 kW 1 kN 1000 kg m/s 2 1 kN m/s = 4207 kW
Noting that the average flow velocity is proportional to the overhead blade rotational velocity, the rpm of the loaded helicopter blades becomes
13-23
Chapter 13 Momentum Analysis of Flow Systems & V2 = kn V2,loaded V2, unloaded = & n loaded & n unloaded & n loaded = V2,loaded V2, unloaded & n unloaded = 34.3 (400 rpm) = 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.
13-24
Chapter 13 Momentum Analysis of Flow Systems 13-42 A helicopter hovers on top of a high mountain where the air density considerably lower than that at sea level. The blade rotational velocity to hover at the higher altitude 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, one-dimensional, 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 (no conversion to thermal energy through frictional effects). 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. 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 mountain. 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 r & & The momentum equation for steady one-dimensional flow is F = m e Ve - mi Vi . Noting 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 = m ( - V2 ) - 0
2 & W = mV2 = ( AV2 )V2 = AV2
V2 =
W A
where A is the blade span area. Then for a given weight W, the ratio of discharge velocities becomes V2, mountain V2,sea = W / mountain A W / sea A = sea mountain = 1.18 kg/m 3 0.79 kg/m 3 = 1.222
Noting 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 = kV2 V2, mountain & n mountain = & V2,sea n sea & n mountain = V2, mountain V2,sea & n sea = 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 2 2 2 3 1.5 1 1.5 W V2 V2 V3 & = 1 A W = W & 2 = AV2 2 = A 2 = 1 A or Wfan, u = m A 2 2 A 15 m 2 2 2 2 A Then the ratio of the required power input on top of the mountain to that at sea level becomes 2 1.5 3 & Wmountain fan, u 0.5W / mountain A 1.18 kg/m sea = = = 1.222 1.5 & Wsea fan, u mountain 0.79 kg/m3 0.5W / sea A Therefore, the required power input will increase by 22.2% on top of the mountain relative to the sea level.
Sea level Load 15,000 kg
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. 13-43 The flow rate in a channel is controlled by a sluice gate by raising or lowering a vertical plate. A relation for the force acting on a sluice gate of width w for steady and uniform flow is to be developed. Assumptions 1 The flow 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 free surfaces is the atmospheric pressure. 4 The flow is horizontal.
13-25
Chapter 13 Momentum Analysis of Flow Systems Analysis We take point 1 at the free 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
2 V2 - 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 & & & V1 = V 2 = V & A1 V1 = A2 V2 = V V1 = & & V V = A1 wy1 and V2 = & & V V = A2 wy 2 (2)
Substituting into Eq. (1), & V wy 2
2 & 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 g ( y1 - y 2 ) & V = wy 2 2 2 1 - y 2 / y1
(3)
Substituting Eq. (3) into Eqs. (2) gives the following relations for velocities, V1 = y2 y1 2 g ( y1 - y 2 ) 1-
2 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 inner surface of the sluice gate, and the bottom surface of r r r & & the channel. The momentum equation for steady one-dimensional flow is F = m e Ve - mi Vi . The force acting on the sluice gate FRx is horizontal since the wall shear at the surfaces is negligible, and it is equal and opposite to the force applied on water by the sluice gate. Noting that the pressure force acting on a vertical surface is equal to the product of the pressure at the centroid of the surface and the surface area, the momentum equation along the x direction gives y y & & & - FRx + P1 A1 - P2 A2 = mV2 - mV1 - FRx + g 1 ( wy1 ) - g 2 ( wy 2 ) = m(V2 - V1 ) 2 2 Rearranging, the force acting on the sluice gate is determined to be & FRx = m(V1 - V2 ) + w 2 2 g ( y1 - y 2 ) 2 (5) 1 FRx
V1 y1 Sluice gate y2 V2
where V1 and V2 are given in Eq. (4). Discussion Note that for y1 >> y2, Eq. (3) simplifies to & V = y w 2gy or V = 2gy which is the Toricelli equation
2 1 2 1
2
for frictionless flow from a tank through a hole a distance y1 below the free surface. 13-44 Water enters a centrifugal pump axially at a specified rate and velocity, and leaves in the normal direction along the pump casing. The force acting on the shaft in the axial direction is to be determined. We Properties take the density of water to be 1000 kg/m3. Assumptions 1 The flow is steady, one-dimensional, and incompressible. 2 The forces acting on the piping system in the horizontal direction are negligible. 3 The atmospheric pressure is disregarded since it acts on all surfaces. Analysis We take the pump as the control volume, and the inlet direction of flow as the positive direction of x axis. The momentum equation for steady one-dimensional flow in the x (flow) direction reduces in this case to
13-26
Chapter 13 Momentum Analysis of Flow Systems r r r & & F = m e Ve - mi Vi & - FRx = -mVi & & FRx = mVi = VVi
Note that the reaction force acts in the opposite direction to flow, and we should not forget the negative sign for forces and velocities in the negative x-direction. Substituting the given values, 1 kN Fbrake = (1000 kg/m 3 )(0.12 m 3 /s)(70 m/s) 1000 kg m/s 2 = 8.40 kN
Discussion To find the total force acting on the shaft, we also need to do a force balance for the vertical direction, and find the vertical component of the reaction force. y x
0.12 m3/s 7 m/s
mV
FRx
13-27
Chapter 13 Momentum Analysis of Flow Systems Angular Momentum Equation 13-45C The angular momentum equation is obtained by replacing B in the Reynolds transport theorem by r r r the total angular momentum H sys , and b by the angular momentum per unit mass r V . r r r r & 13-46C The angular momentum equation in this case is expressed as I = -r mV where is the r angular acceleration of the control volume, and r is the position vector from the axis of rotation to any r point on the line of action of F . r r r r & 13-47C The angular momentum equation in this case is expressed as I = -r mV where is the r angular acceleration of the control volume, and r is the position vector from the axis of rotation to any r point on the line of action of F .
13-28
Chapter 13 Momentum Analysis of Flow Systems 13-48 Water is pumped through a piping section. The moment acting on the elbow for the cases of downward and upward discharge is to be determined. Assumptions 1 The flow is steady and uniform. 2 The water is discharged to the atmosphere, and thus the gage pressure at the outlet is zero. 3 Effects of water falling down during upward discharge is disregarded. Properties We take the density of water to be 1000 kg/m3. Analysis We take the entire pipe as the control volume, and designate the inlet by 1 and the outlet by 2. We also take the x and y coordinates as shown. The control volume and the reference frame are fixed. The conservation of mass equation for this one-inlet one-outlet steady flow system is & & & m1 = m 2 = m , and V 1 =V 2 =V since Ac = constant. The mass flow rate and the weight of the horizontal section of the pipe are r2 = 1 m & m = AcV = (1000 kg/m 3 )[ (0.12 m) 2 / 4](4 m/s ) = 45.24 kg/s 1N W = mg = (15 kg/m )(2 m )(9.81 m/s 2 ) 1 kg m/s 2 = 294.3 N/m MA A
r1 = 2 m
r W & (a) Downward discharge: To determine the moment acting on the pipe at point A, mV 1 we need to take the moment of all forces and momentum flows about that point. This is a steady and uniform flow problem, and all forces and momentum flows are in the same plane. Therefore, the angular momentum equation in & & M = r mV - rmV where r is the moment arm, all moments in this case can be expressed as
& mV
r
2
out
in
the counterclockwise direction are positive, and all in the clockwise direction are negative. The free body diagram of the pipe section is given in the figure. Noting that the moments of all forces and momentum flows passing through point A are zero, the only force that will yield a moment about point A is the weight W of the horizontal pipe section, and the only momentum flow that will yield a moment is the exit stream (both are negative since both moments are in the clockwise direction). Then the angular momentum equation about point A becomes & M A - r1W = -r2 mV 2 Solving for MA and substituting, 1N & M A = r1W - r2 mV 2 = (1 m)(294.3 N) - (2 m)(45.54 kg/s)(4 m/s) 1 kg m/s 2 = -70.0 N m
The negative sign indicates that the assumed direction for MA is wrong, and should be reversed. Therefore, a moment of 70 Nm acts at the stem of the pipe in the clockwise direction. (b) Upward discharge: The moment due to discharge stream is positive in this case, and the moment acting on the pipe at point A is 1N & M A = r1W + r2 mV 2 = (1 m)(294.3 N) + (2 m)(45.54 kg/s)(4 m/s) 1 kg m/s 2 = 657 N m
Discussion Note direction of discharge can make a big difference in the moments applied on a piping system. This problem also shows the importance of accounting for the moments of momentums of flow streams when performing evaluating the stresses in pipe materials at critical cross-sections.
13-29
Chapter 13 Momentum Analysis of Flow Systems 13-49E A two-armed sprinkler is used to generate electric power. For a specified flow rate and rotational speed, the power produced is to be determined.
V
jet
Electric generator
& mnozzleV r
Mshaft
r = 2 ft
& mnozzleV r V jet & m total = 8 gal/s Assumptions 1 The flow is uniform and cyclically steady (i.e., steady from a frame of reference rotating with the sprinkler head). 2 The water is discharged to the atmosphere, and thus the gage pressure at the nozzle exit is zero. 3 Generator losses and air drag of rotating components are neglected.
Properties We take the density of water to be 62.4 lbm/ft 3. Analysis We take the disk that encloses the sprinkler arms as the control volume, which is a stationary & & & control volume. The conservation of mass equation for this steady flow system is m1 = m 2 = m . Noting & & & & that the two nozzles are identical, we have m = V / 2 since the density of water is = m / 2 or V
nozzle nozzle
constant. The average jet exit velocity relative to the nozzle is V
jet
=
& 1 ft 3 V nozzle 4 gal/s = 392.2 ft/s = A jet [ (0.5 / 12 ft) 2 / 4] 7.480 gal
The angular and tangential velocities of the nozzles are 1 min & = 2n = 2 (250 rev/min) = 26.18 rad/s 60 s V nozzle = r = (2 ft)(26.18 rad/s) = 52.36 ft/s The velocity of water jet relative to the control volume (or relative to a fixed location on earth) is V r =V
jet
-V nozzle = 392.2 - 52.36 = 339.8 ft/s & & M = r mV - r mV
out in
The angular momentum equation can be expressed as
where all
moments in the counterclockwise direction are positive, and all in the clockwise direction are negative. Then the angular momentum equation about the axis of rotation becomes & - M shaft = -2rm nozzleV r or & M shaft = rm totalV r
Substituting, the torque transmitted through the shaft is determined to be 1 lbf & = 1409 lbf ft M shaft = rm totalV r = (2 ft)(66.74 lbm/s)(339 .8 ft/s) 2 32.2 lbm ft/s & & since m = V = (62.4 lbm/ft 3 )(8 / 7.480 ft 3 /s ) = 66.74 lbm/s . Then the power generated becomes
total total
1 kW & & W = 2nM shaft = M shaft = (26.18 rad/s)(1409 lbf ft) = 50.0 kW 737.56 lbf ft/s Therefore, this sprinkler-type turbine has the potential to produce 50 kW of power.
13-30
Chapter 13 Momentum Analysis of Flow Systems 13-50E A two-armed sprinkler is used to generate electric power. For a specified flow rate and rotational speed, the moment acting on the rotating head when the head is stuck is to be determined.
V
jet
Electric generator
& mnozzleV r
Mshaft
r = 2 ft
& mnozzleV r V jet & m total = 8 gal/s Assumptions 1 The flow is uniform and cyclically steady (i.e., steady from a frame of reference rotating with the sprinkler head). 2 The water is discharged to the atmosphere, and thus the gage pressure at the nozzle exit is zero.
Properties We take the density of water to be 62.4 lbm/ft3. Analysis We take the disk that encloses the sprinkler arms as the control volume, which is a stationary & & & control volume. The conservation of mass equation for this steady flow system is m1 = m 2 = m . Noting & & & & that the two nozzles are identical, we have m = V / 2 since the density of water is = m / 2 or V
nozzle nozzle
constant. The average jet exit velocity relative to the nozzle is V
jet
=
& 1 ft 3 V nozzle 4 gal/s = 392.2 ft/s = A jet [ (0.5 / 12 ft) 2 / 4] 7.480 gal & & M = r mV - r mV
out in
The angular momentum equation can be expressed as
where all
moments in the counterclockwise direction are positive, and all in the clockwise direction are negative. Then the angular momentum equation about the axis of rotation becomes & - M shaft = -2rm nozzleV
jet
or
& M shaft = rm totalV
jet
Substituting, the torque transmitted through the shaft is determined to be & M shaft = rm totalV
jet
1 lbf = (2 ft)(66.74 lbm/s)(392 .2 ft/s) = 1626 lbf ft 32.2 lbm ft/s 2
& & since m total = V total = (62.4 lbm/ft 3 )(8 / 7.480 ft 3 /s ) = 66.74 lbm/s . Discussion When the sprinkler is stuck and thus the angular velocity is zero, the torque developed is maximum since V nozzle = 0 and thus V r =V jet = 392.2 ft/s , giving M shaft, max = 1626 lbf ft . But the power generated is zero in this case since the shaft does not rotate.
13-31
Chapter 13 Momentum Analysis of Flow Systems 13-51 A three-armed sprinkler is used to water a garden. For a specified flow rate and resistance torque, the angular velocity of the sprinkler head is to be determined.
V
jet
Electric generator
& mnozzleV r
& mnozzleV r
To = 50 Nm
r = 40 cm
& V total = 40 L/s
V
jet
& mnozzleV r
Assumptions 1 The flow is uniform and cyclically steady (i.e., steady from a frame of reference rotating with the sprinkler head). 2 The water is discharged to the atmosphere, and thus the gage pressure at the nozzle exit is zero. 3 Air drag of rotating components are neglected. Properties We take the density of water to be 1000 kg/m3 = 1 kg/L. Analysis We take the disk that encloses the sprinkler arms as the control volume, which is a stationary & & & control volume. The conservation of mass equation for this steady flow system is m1 = m 2 = m . Noting & & & & that the three nozzles are identical, we have m = V / 3 since the density of water is = m / 3 or V
nozzle nozzle
constant. The average jet exit velocity relative to the nozzle and the mass flow rate are V
jet
=
& 1m3 V nozzle 40 L/s = 117.9 m/s = A jet 3[ (0.012 m) 2 / 4] 1000 L
& & m total = V total = (1 kg/L )(40 L/s ) = 40 kg/s The angular momentum equation can be expressed as & & M = r mV - r mV
out in
where all moments
in the counterclockwise direction are positive, and all in the clockwise direction are negative. Then the angular momentum equation about the axis of rotation becomes & -T0 = -3rm nozzleV r or & T0 = rm totalV r
Solving for the relative velocity Vr and substituting, 1 kg m/s 2 T0 50 N m = 3.1 m/s Vr = = & 1N rm total (0.40 m)(40 kg/s) Then the tangential and angular velocity of the nozzles become V nozzle =V = V
nozzle jet
-V r = 117.9 - 3.1 = 114.8 m/s
114.8 m/s = 287 rad/s r 0.4 m 287 rad/s 60 s & n= = = 2741 rpm 2 2 1 min =
Therefore, this sprinkler will rotate at 2741 revolutions per minute.
13-32
Chapter 13 Momentum Analysis of Flow Systems 13-52 A Pelton wheel is considered for power generation in a hydroelectric power plant. A relation is to be obtained for power generation, and its numerical value is to be obtained.
Mshaft
r
Shaft
Vj - r
Vj - r
Nozzle Vj
r
Assumptions 1 The flow is uniform and cyclically steady. 2 The water is discharged to the atmosphere, and thus the gage pressure at the nozzle exit is zero. 3 Generator losses and air drag of rotating components are neglected. Properties We take the density of water to be 1000 kg/m3 = 1 kg/L. & Analysis The tangential velocity of buckets corresponding to an angular velocity of = 2n is V bucket = r . Then the relative velocity of the jet (relative to the bucket) becomes V r =V j -V
bucket
=V j - r
We take the imaginary disk that contains the Pelton wheel as the control volume. The inlet velocity of the fluid into this control volume is Vr, and the component of outlet velocity normal to the moment arm is & & Vrcos. The angular momentum equation can be expressed as M = r mV - rmV where all
out
in
moments in the counterclockwise direction are positive, and all in the clockwise direction are negative. Then the angular momentum equation about the axis of rotation becomes & & - M shaft = rmV r cos - rmV r & &V or M shaft = rmV r (1 - cos ) = rm( j - r )(1 - cos )
& & & & Noting that Wshaft = 2nM shaft = M shaft and m = V , the power output of a Pelton turbine becomes & & V Wshaft = Vr ( j - r )(1 - cos ) which is the desired relation. For given values, the power output is determined to be 1 MW & Wshaft = (1000 kg/m 3 )(10 m 3 /s )(2 m)(15.71 rad/s)(50 - 2 15.71 m/s)(1 - cos160) 6 = 11.3 MW 10 N m/s where 1 min & = 2n = 2 (150 rev/min) = 15.71 rad/s 60 s
13-33
Chapter 13 Momentum Analysis of Flow Systems 13-53 Problem 13-52 is reconsidered. The effect of on the power generation as varies from 0 to 180 is to be determined, and the fraction of power loss at 160 is to be assessed. rho=1000 "kg/m3" r=2 "m" V_dot=10 "m3/s" V_jet=50 "m/s" n_dot=150 "rpm" omega=2*pi*n_dot/60 V_r=V_jet-r*omega m_dot=rho*V_dot W_dot_shaft=m_dot*omega*r*V_r*(1-cos(Beta))/1E6 "MW" W_dot_max=m_dot*omega*r*V_r*2/1E6 "MW" Effectiveness=W_dot_shaft/W_dot_max
Angle, 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 Max power, Actual power, Effectiveness, 0.000 0.008 0.030 0.067 0.117 0.179 0.250 0.329 0.413 0.500 0.587 0.671 0.750 0.821 0.883 0.933 0.970 0.992 1.000
& W max , MW
11.7 11.7 11.7 11.7 11.7 11.7 11.7 11.7 11.7 11.7 11.7 11.7 11.7 11.7 11.7 11.7 11.7 11.7 11.7
& Wshaft , MW
0.00 0.09 0.35 0.78 1.37 2.09 2.92 3.84 4.82 5.84 6.85 7.84 8.76 9.59 10.31 10.89 11.32 11.59 11.68
12
W max
10 8
Wshaft
Wshaft
6 4 2
The effectiveness for =160 is 0.97. this angle, only 3% lost.
0 0
20
40
60
80
,
100
120
140
160
180 Therefore,
of Pelton wheel at of power is
13-34
Chapter 13 Momentum Analysis of Flow Systems 13-54 A centrifugal blower is used to deliver atmospheric air. For a given angular speed and power input, the volume flow rate of air is to be determined.
Impeller shroud b1 In r1 Shaft
b2 r2
r2 r1
r V 2,t r V 1,t
Impeller blade
Impeller
Impeller region
Assumptions 1 The flow is steady in the mean. 2 Irreversible losses are negligible. 3 The tangential components of air velocity at the inlet and the outlet are said to be equal to the impeller velocity at respective locations. Properties The gas constant of air is 0.287 kPam3/kgK. The density of air at 20C and 95 kPa is = 95 kPa P = = 1.130 kg/m 3 RT (0.287 kPa m 3 /kg K)(293 K)
Analysis In the idealized case of the tangential fluid velocity being equal to the blade angular velocity both at the inlet and the exit, we have V 1,t = r1 and V 2,t = r2 , and the torque is expressed as & & & Tshaft = m(r2V 2,t - r1 1,t ) = m (r22 - r12 ) = V (r22 - r12 ) V where the angular velocity is 1 min & = 2n = 2 (800 rev/min) = 83.78 rad/s 60 s Then the shaft power becomes & & Wshaft = Tshaft = V 2 (r22 - r12 ) & Solving for V and substituting, the volumetric flow rate of air is determined to & V= & Wshaft
2
(r22
- r12 )
=
1 kg m/s 2 1N (1.130 kg/m )(83.78 rad/s) [(0.30 m) - (0.15 m) ] 120 N m/s
2 3 2 2
= 0.224 m 3 /s
The normal velocity components at the inlet and the outlet are V 1, n = V 2, n & 0.224 m 3 /s V = = 3.90 m/s 2r1b1 2 (0.15 m)(0.061 m) & V 0.224 m 3 /s = = = 3.50 m/s 2r2 b 2 2 (0.30 m)(0.034 m)
Discussion Note that the irreversible losses are not considered in analysis. In reality, the flow rate and the normal components of velocities will be smaller.
13-35
Chapter 13 Momentum Analysis of Flow Systems 13-55 A centrifugal blower is used to deliver atmospheric air at a specified rate and angular speed. The minimum power consumption of the blower is to be determined.
2 = 50
Impeller shroud b1 In r1 Shaft
b2 r2
r2 r1
r V2 r V1
Impeller blade
Impeller
Impeller region
Assumptions 1 The flow is steady in the mean. 2 Irreversible losses are negligible. Properties The density of air is given to be 1.25 kg/m3. Analysis We take the impeller region as the control volume. The normal velocity components at the inlet and the outlet are V 1,n = V 2, n & 0.70 m 3 /s V = = 6.793 m/s 2r1b1 2 (0.20 m)(0.082 m) & 0.70 m 3 /s V = = = 4.421 m/s 2r2 b2 2 (0.45 m)(0.056 m) V 1,t =V 1,n tan 1 = 0 V 2,t =V 2,n tan 1 = (4.421 m/s ) tan 50 = 5.269 m/s
The tangential components of absolute velocity are:
1 = 0: 2 = 60:
The angular velocity of the propeller is 1 min & = 2n = 2 (700 rev/min) = 73.30 rad/s 60 s & & m = V = (1.25 kg/m 3 )(0.7 m 3 /s) = 0.875 kg/s Normal velocity components V1,n and V2,n as well pressure acting on the inner and outer circumferential areas pass through the shaft center, and thus they do not contribute to torque. Only the tangential velocity components contribute to torque, and the application of the angular momentum equation gives 1N & Tshaft = m(r2V 2,t - r1 1,t ) = (0.875 kg/s)[(0.45 m)(5.269 m/s) - 0] V 1 kg m/s 2 Then the shaft power becomes 1W & W = Tshaft = (73.30 rad/s)(2.075 N m) = 152 W 1 N m/s = 2.075 N m
13-36
Chapter 13 Momentum Analysis of Flow Systems 13-56 Problem 13-55 is reconsidered. The effect of discharge angle 2 on the minimum power input requirements as 2 varies from 0 to 85 in increments of 5 is to be investigated. rho=1.25 "kg/m3" r1=0.20 "m" b1=0.082 "m" r2=0.45 "m" b2=0.056 "m" V_dot=0.70 "m3/s" V1n=V_dot/(2*pi*r1*b1) "m/s" V2n=V_dot/(2*pi*r2*b2) "m/s" Alpha1=0 V1t=V1n*tan(Alpha1) "m/s" V2t=V2n*tan(Alpha2) "m/s" n_dot=700 "rpm" omega=2*pi*n_dot/60 "rad/s" m_dot=rho*V_dot "kg/s" T_shaft=m_dot*(r2*V2t-r1*V1t) "Nm" W_dot_shaft=omega*T_shaft "W" Angle, Torque, Shaft power, V2,t, Tshaft , Nm 2 & m/s Wshaft , W
0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85
1600 1400 1200 1000
0.00 0.39 0.78 1.18 1.61 2.06 2.55 3.10 3.71 4.42 5.27 6.31 7.66 9.48 12.15 16.50 25.07 50.53
0.00 0.15 0.31 0.47 0.63 0.81 1.01 1.22 1.46 1.74 2.07 2.49 3.02 3.73 4.78 6.50 9.87 19.90
0 11 23 34 46 60 74 89 107 128 152 182 221 274 351 476 724 1459
Wshaft, W
800 600 400 200 0 0 10 20 30 40 50 60 70 80 90
2
13-37
Chapter 13 Momentum Analysis of Flow Systems 13-57E Water enters the impeller of a centrifugal pump radially at a specified flow rate and angular speed. The torque applied to the impeller is to be determined.
Impeller shroud b1 In r1 Shaft
b2 r2
r2 r1
r V 2,t r V1
Impeller blade
Impeller
Impeller region
Assumptions 1 The flow is steady in the mean. 2 Irreversible losses are negligible. Properties We take the density of water to be 62.4 lbm/ft 3. Analysis Water enters the impeller normally, and thus V 1,t = 0 . The tangential component of fluid velocity at the outlet is given to be V 2,t = 180 ft/s . The inlet radius r1 is unknown, but the outlet radius is given to be r2 = 1 ft. The angular velocity of the propeller is 1 min & = 2n = 2 (500 rev/min) = 52.36 rad/s 60 s The mass flow rate is & & m = V = (62.4 lbm/ft 3 )(80/60 ft 3 /s) = 83.2 lbm/s Only the tangential velocity components contribute to torque, and the application of the angular momentum equation gives 1 lbf & Tshaft = m(r2V 2,t - r1 1,t ) = (83.2 lbm/s)[(1 ft)(180 ft/s) - 0] V = 465 lbf ft 32.2 lbm ft/s 2 Discussion This shaft power input corresponding to this torque is 1 kW & & W = 2nTshaft = Tshaft = (52.36 rad/s)(465 lbf ft) = 33.0 kW 737.56 lbf ft/s Therefore, the minimum power input to this pump should be 33 kW.
13-38
Chapter 13 Momentum Analysis of Flow Systems 13-58 A centrifugal pump is used to supply water at a specified rate and angular speed. The minimum power consumption of the pump is to be determined.
2 = 60
Impeller shroud b1 In r1
b2 r2
r2 r1
r V2 r V1
Shaft
Impeller blade
Impeller
Impeller region
Assumptions 1 The flow is steady in the mean. 2 Irreversible losses are negligible. Properties We take the density of water to be 1000 kg/m3. Analysis We take the impeller region as the control volume. The normal velocity components at the inlet and the outlet are V 1,n = V 2, n & 0.15 m 3 /s V = = 2.296 m/s 2r1b1 2 (0.13 m)(0.080 m) & 0.15 m 3 /s V = = = 2.274 m/s 2r2 b2 2 (0.30 m)(0.035 m) V 1,t =V 1,n tan 1 = 0 V 2,t =V 2,n tan 1 = (2.274 m/s ) tan 60 = 3.938 m/s
The tangential components of absolute velocity are:
1 = 0: 2 = 60:
The angular velocity of the propeller is 1 min & = 2n = 2 (1200 rev/min) = 125.7 rad/s 60 s & & m = V = (1000 kg/m 3 )(0.15 m 3 /s) = 150 kg/s Normal velocity components V1,n and V2,n as well pressure acting on the inner and outer circumferential areas pass through the shaft center, and thus they do not contribute to torque. Only the tangential velocity components contribute to torque, and the application of the angular momentum equation gives 1 kN & Tshaft = m(r2V 2,t - r1 1,t ) = (150 kg/s)[(0.30 m)(3.938 m/s) - 0] V 1000 kg m/s 2 Then the shaft power becomes 1 kW & W = Tshaft = (125.7 rad/s)(177.2 kN m) = 22.3 kW 1 kN m/s Discussion Note that the irreversible losses are not considered in analysis. In reality, the required power input will be larger. = 177.2 kN m
13-39
Chapter 13 Momentum Analysis of Flow Systems Review Problems 13-59 Water is flowing into and discharging from a pipe U-section with a secondary discharge section normal to return flow. Net x- and y- forces at the two flanges that connect the pipes are to be determined. Assumptions 1 The flow is steady, one-dimensional, and incompressible. 2 The weight of the U-turn and the water in it is negligible. Properties We take the density of water to be 1000 kg/m3. Analysis The flow velocities of the 3 streams are & & m m1 30 kg/s V1 = 1 = = = 0.51 m/s 2 A1 (D1 / 4) (1000 kg/m 3 )[ (0.05 m) 2 / 4] V2 = & & m2 m2 22 kg/s = = = 2.80 m/s 2 A2 (D 2 / 4) (1000 kg/m 3 )[ (0.10 m) 2 / 4]
8 kg/s
3 2
22 kg/s
FRy
& & m m3 8 kg/s V3 = 3 = = = 11.3 m/s 2 A3 (D3 / 4) (1000 kg/m 3 )[ (0.03 m) 2 / 4]
1
30 kg/s
FRx
We take the entire U-section as the control volume. We designate the horizontal coordinate by x with the direction of incoming flow as being the positive direction and the vertical coordinate by y. The momentum r r r & & equation for steady one-dimensional flow is F = m e Ve - mi Vi . We let the x- and y- components of the anchoring force of the cone be FRx and FRy, and assume them to be in the positive directions. Then the momentum equations along the x and y axes become & & FRx + P1 A1 + P2 A2 = m 2 (-V2 ) - m1 V1 & FRy + 0 = m 3 V3 - 0 Substituting the given values, (0.05 m) 2 (0.10 m) 2 - [(150 - 100) kN/m 2 ] 4 4 1 kN 1 kN - (22 kg/s)(2.80 m/s) 1000 kg m/s 2 - (30 kg/s)(0.51 m/s) 1000 kg m/s 2 = -0.666 kN = -666 N 1N FRy = (8 kg/s)(11.3 m/s) 1 kg m/s 2 = 90.4 N The negative value for FRx indicates the assumed direction is wrong, and should be reversed. Therefore, a force of 666 N acts on the flanges in the opposite direction. A vertical force of 90.4 N acts on the flange in the vertical direction. FRx = -[(200 - 100) kN/m 2 ] Discussion To assess the significance of gravity forces, we estimate the weight of the weight of water in the U-turn and compare it to the vertical force. Assuming the length of the U-turn to be 0.5 m and the average diameter to be 7.5 cm, the mass of the water becomes m = V = AL = (0.075 m) 2 D 2 L = (1000 kg/m 3 ) (0.5 m) = 2.2 kg 4 4 & & FRx = - P1 A1 - P2 A2 - m 2 V2 - m1 V1 & FRy = m3 V3
whose weight is 2.29.81 = 22 N, which is much less than 90.4 N, but still significant. Therefore, disregarding the gravitational effects is a reasonable assumption if great accuracy is not required.
13-60 A fireman was hit by a nozzle held by a tripod with a rated holding force. The accident is to be investigated by calculating the water velocity, the flow rate, and the nozzle velocity. Assumptions 1 The flow is steady, one-dimensional, 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
13-40
Chapter 13 Momentum Analysis of Flow Systems acts on all surfaces. 3 Gravitational effects and vertical forces are disregarded since the horizontal resistance force is to be determined. Properties We take the density of water to be 1000 kg/m3. Analysis We take the nozzle and the horizontal portion of the hose as the system such that water enters the control volume vertically and exits horizontally (this way the pressure force and the momentum flux at the inlet are in the vertical direction, with no contribution to the force balance in the horizontal direction, and designate the entrance by 1 and the exit by 2. We also designate the horizontal coordinate by x (with the direction of flow as being the positive direction). r r r & & The momentum equation for steady one-dimensional flow is F = m e Ve - mi Vi . We let the horizontal force applied by the tripod 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 becomes & & FRx = mVe - 0 = mV = AVV = D 2 2 V 4 1kg m/s 2 (1800 N) 1N
2 = (1000 kg/m 3 ) (0.05 m) V 2 4
Solving for the water exit velocity gives V = 30.2 m/s. Then the water flow rate becomes (0.05 m) D & (30.2 m/s) = 0.0593 m 3 /s V = AV = V= 4 4
2 2
When the nozzle was released, its acceleration must have been a nozzle = F m nozzle 1800 N 1 kg m/s 2 = 10 kg 1N = 180 m/s 2
Nozzle
FRx
D=5 cm
Assuming the reaction force acting on the nozzle and thus its acceleration to remain constant, the time it takes for the nozzle to travel 60 cm and the nozzle velocity at that moment were (note that both the distance x and the velocity V are zero at time t = 0) x = 1 at 2 2 t= 2(0.6 m) 2x = = 0.0816 s a 180 m/s 2
Tripod
V = at = (180 m/s 2 )(0.0816 s) = 14.7 m/s Thus we conclude that the nozzle hit the fireman with a velocity of 14.7 m/s. Discussion Engineering analyses such as this one are frequently used in accident reconstruction cases, and they often form the basis for judgment in courts.
13-41
Chapter 13 Momentum Analysis of Flow Systems 13-61 During landing of an airplane, the thrust reverser is lowered in the path of the exhaust jet, which deflects the exhaust and provides breaking. The thrust of the engine and the braking force produced after the thrust reverser is deployed are to be determined. EES Assumptions 1 The flow of exhaust gases is steady and one-dimensional. 2 The exhaust gas stream is exposed to the atmosphere, and thus its pressure is the atmospheric pressure. 3 The velocity of exhaust gases remains constant during reversing. Analysis (a) The thrust exerted on an airplane is simply the momentum flux of the combustion gases in the reverse direction, 1N & Thrust = m ex Vex = (18 kg/s)(250 m/s) 1 kg m/s 2 = 4500 N
(b) We take the thrust reverser as the control volume such that it cuts through both exhaust streams normally and the connecting bars to the airplane, and the direction of airplane as the positive direction of x axis. The momentum equation for steady one-dimensional flow in the x direction reduces to r r r & & & & & F = m e Ve - mi Vi FRx = m(V )cos20 - m(-V ) FRx = (1 + cos 20)mVi Substituting, the reaction force is determined to be FRx = (1 + cos 20)(18 kg/s)(250 m/s) = 8729 N The breaking force acting on the plane is equal and opposite to this force, Fbreaking = 8729 N Therefore, a braking force of 8729 N develops in the opposite direction tot flight. Discussion This problem can be solved more generally by measuring the reversing angle from the direction of exhaust gases ( = 0 when there is no reversing). When < 90, the reversed gases are discharged in the negative x direction, and the momentum equation reduces to & & FRx = m(-V )cos - m(-V ) & FRx = (1 - cos )mVi
This equation is also valid for >90 since cos(180-) = - cos. Using = 160, for example, gives & & FRx = (1 - cos 160)mVi = (1 + cos 20)mVi , which is identical to the solution above.
= 160
& mV
160 Control volume 250 m/s
& mV FRx FRx
x
13-42
Chapter 13 Momentum Analysis of Flow Systems 13-62 Problem 13-61 reconsidered. The effect of thrust reverser angle on the braking force exerted on the airplane as the reverser angle varies from 0 (no reversing) to 180 (full reversing) in increments of 10 is to be investigated. V_jet=250 "m/s" m_dot=18 "kg/s" F_Rx=(1-cos(alpha))*m_dot*V_jet "N"
Reversing angle, 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180
Braking force Fbrake, N 0 68 271 603 1053 1607 2250 2961 3719 4500 5281 6039 6750 7393 7947 8397 8729 8932 9000
9000 8000 7000 6000
Fbrake, N
5000 4000 3000 2000 1000 0 0 20 40 60 80 100 120 140 160 180
,
13-43
Chapter 13 Momentum Analysis of Flow Systems 13-63E The rocket of a spacecraft is fired in the opposite direction to motion. The acceleration, the velocity change, and the thrust are to be determined. Assumptions 1 The flow of combustion gases is steady and one-dimensional during firing period, but the flight of spacecraft is unsteady. 2 There are no external forces acting on the spacecraft, and the effect of pressure force at the nozzle exit is negligible. 3 The mass of discharged fuel is negligible relative to the mass of the spacecraft, and thus the spacecraft may be treated as a solid body with a constant mass. Analysis (a) A body moving at constant velocity can be considered to be stationary for convenience. Then the velocities of fluid steams become simply their relative velocities relative to the moving body. We take the direction of motion of the spacecraft as the positive direction along the x axis. There are no external forces acting on the spacecraft, and its mass is nearly constant. Therefore, the spacecraft can be treated as a solid body with constant mass, and the momentum equation in this case is r r r r r dVspace d (mV ) CV & & & mi Vi - m e Ve m space = = -m f V f dt dt in out
Noting that the motion is on a straight line and the discharged gases move in the negative x direction, we can write the momentum equation using magnitudes as m space dVspace dt & = mf Vf dVspace dt = & mf m space Vf
Substituting, the acceleration of the spacecraft during the first 5 seconds is determined to be & dVspace mf 150 lbm/s Vf = (5000 ft/s) = 41.7 ft/s 2 a space = = 18,000 lbm dt mspace (b) Knowing acceleration, which is constant, the velocity change of the spacecraft during the first 5 seconds is determined from the definition of acceleration a space = dVspace / dt to be dVspace = a space dt Vspace = a space t = (41.7 ft/s 2 )(5 s ) = 209 ft/s
(c) The thrust exerted on the system is simply the momentum flux of the combustion gases in the reverse direction, 1 lbf & Thrust = FR = -m f V f = -(150 lbm/s)(-50 00 ft/s) = 23,290 lbf 32.2 lbm ft/s 2 Therefore, if this spacecraft were attached somewhere, it would exert a force of 23,290 lbf (equivalent to the weight of 23,290 lbm of mass) to its support.
Combustion gases 5000 ft/s 150 lbm/s 18000 lbm 1500 ft/s
13-44
Chapter 13 Momentum Analysis of Flow Systems 13-64 A horizontal water jet strikes a vertical stationary flat plate normally at a specified velocity. For a given flow velocity, the anchoring force needed to hold the plate in place is to be determined. Assumptions 1 The flow is steady, one-dimensional, and incompressible. 2 The water 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 splattered water is the atmospheric pressure which is disregarded since it acts on the entire control surface. 4 The vertical forces and momentum fluxes are not considered since they have no effect on the horizontal reaction force. Properties We take the density of water to be 1000 kg/m3. Analysis We take the plate as the control volume such that it contains the entire plate and cuts through the water jet and the support bar normally, and the direction of flow as the positive direction of x axis. We take the reaction force to be in the negative x direction. The momentum equation for steady one-dimensional flow in the x (flow) direction reduces in this case to r r r & & & & F = m e Ve - mi Vi - FRx = -mi Vi FRx = mV We note that the reaction force acts in the opposite direction to flow, and we should not forget the negative sign for forces and velocities in the negative x-direction. The mass flow rate of water is (0.05 m) D & & m = V = AV = V = (1000 kg/m 3 ) (30 m/s) = 58.90 kg/s 4 4
2 2
Substituting, the reaction force is determined to be FRx = (58.90 kg/s)(30 m/s) = 1767 N Therefore, a force of 1767 N must be applied to the plate in the opposite direction to flow to hold it in place.
5 cm 30 m/s
FRx
Discussion In reality, some water will be scattered back, and this will add to the reaction force of water.
13-45
Chapter 13 Momentum Analysis of Flow Systems 13-65 A water jet hits a stationary cone, such that the flow is diverted equally in all directions at 45. The force required to hold the cone in place against the water stream is to be determined. Assumptions 1 The flow is steady, one-dimensional, and incompressible. 2 The water jet is exposed to the atmosphere, and thus the pressure of the water jet before and after the split is the atmospheric pressure which is disregarded since it acts on all surfaces. 3 The gravitational effects are disregarded. Properties We take the density of water to be 1000 kg/m3. Analysis The mass flow rate of water jet is (0.05 m) D & & m = V = AV = V = (1000 kg/m 3 ) (30 m/s) = 58.90 kg/s 4 4
2 2
We take the diverting section of water jet, including the cone as the control volume, and designate the entrance by 1 and the exit after divergence 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 y. r r r & & The momentum equation for steady one-dimensional flow is F = m e Ve - mi Vi . We let the xand y- components of the anchoring force of the cone be FRx and FRy, and assume them to be in the positive & & & directions. Noting that V2 = V1 = V and m 2 = m1 = m , the momentum equations along the x and y axes become & & & FRx = mV2 cos - mV1 = mV (cos - 1) FRy = 0 (because of symmetry about x axis) Substituting the given values, 1N FRx = (58.90 kg/s)(30 m/s)(cos45 - 1) 1 kg m/s 2 = -518 N FRy = 0 The negative value for FRx indicates that the assumed direction is wrong, and should be reversed. Therefore, a force of 518 N must be applied to the cone in the opposite direction to flow to hold it in place. No holding force is necessary in the vertical direction due to symmetry and neglecting gravitational effects. Discussion In reality, the gravitational effects will cause the upper part of flow to slow down and the lower part to speed up after the split. But for short distances, these effects are negligible.
5 cm 30 m/s
FRx
13-46
Chapter 13 Momentum Analysis of Flow Systems 13-66 An ice skater is holding a flexible hose (essentially weightless) which directs a stream of water horizontally at a specified velocity. The velocity and the distance traveled in 5 seconds, and the time it takes to move 5 m and the velocity at that moment are to be determined. Assumptions 1 Friction between the skates and ice is negligible. 2 The flow of water is steady and onedimensional (but the motion of skater is unsteady). 3 The ice skating arena is level, and the water jet is discharged horizontally. 4 The mass of the hose and the water in it is negligible. 5 The skater is standing still initially at t = 0. Properties We take the density of water to be 1000 kg/m3. Analysis (a) The mass flow rate of water through the hose is & m = AV = (0.02 m) 2 D 2 V = (1000 kg/m 3 ) (10 m/s) = 3.14 kg/s 4 4
The thrust exerted on the skater by the water stream is simply the momentum flux of the water stream, and it acts in the reverse direction, 1N & F = Thrust = mV = (3.14 kg/s)(10 m/s) 1 kg m/s 2 = 31.4 N (constant)
The acceleration of the skater is determined from Newton's 2nd law of motion F = ma where m is the mass of the skater, a= F 31.4 N 1 kg m/s 2 = 1N m 60 kg = 0.523 m/s 2
Note that thrust and thus the acceleration of the skater is constant. The velocity of the skater and the distance traveled in 5 s are Vskater = at = (0.523 m/s 2 )(5 s) = 2.62 m/s x = 1 at 2 = 1 (0.523 m/s 2 )(5 s) 2 = 6.54 m 2 2 (b) The time it will take to move 5 m and the velocity at that moment are x = 1 at 2 2 t= 2x = a 2(5 m) 0.523 m/s 2 = 4.4 s
Hose
F
10 m/s D=2 cm
Vskater = at = (0.523 m/s 2 )( 4.4 s) = 2.3 m/s Discussion In reality, the velocity of the skater will be lower because of friction on ice and the resistance of the hose to follow the skater.
13-47
Chapter 13 Momentum Analysis of Flow Systems 13-67 Indiana Jones is to ascend a building by building a platform, and mounting four water nozzles pointing down at each corner. The minimum water jet velocity needed to raise the system, the time it will take to rise to the top of the building and the velocity of the system at that moment, the additional rise when the water is shut off, and the time he has to jump from the platform to the roof are to be determined. Assumptions 1 The air resistance is negligible. 2 The flow of water is steady and one-dimensional (but the motion of platform is unsteady). 3 The platform is still initially at t = 0. Properties We take the density of water to be 1000 kg/m3. Analysis (a) The total mass flow rate of water through the 4 hoses and the total weight of the platform are & m = AV = 4 (0.05 m) 2 D 2 V = 4(1000 kg/m 3 ) (15 m/s) = 118 kg/s 4 4 = 1472 N
1N W = mg = (150 kg)(9.81 m/s 2 ) 1 kg m/s 2
We take the platform as the system. The momentum equation for steady one-dimensional flow is r r r & & F = m e Ve - mi Vi . The minimum water jet velocity needed to raise the platform is determined by setting the net force acting on the platform equal to zero, & - W = m(-Vmin ) - 0 & W = mVmin = AVmin Vmin = 4 D 2 2 Vmin 4
Solving for Vmin and substituting,
Vmin = W = D 2 1kg m/s 2 1472 N = 13.7 m/s 3 2 (1000 kg/m ) (0.05 m) 1N
(b) We let the vertical reaction force (assumed upwards) acting on the platform be FRy. Then the momentum equation in the vertical direction becomes 1kg m/s 2 = -298 N & & & FRy - W = m(-V ) - 0 = mV FRy = W - mV = (1472 N) - (118 kg/s)(15 m/s) 1N The upward thrust acting on the platform is equal and opposite to this reaction force, and thus F = 298 N. Then the acceleration and the ascending time to rise 10 m and the velocity at that moment become a= F 298 N 1 kg m/s 2 = 1N m 150 kg
2
= 2.0 m/s 2 y 2(10 m) 2 m/s
2
x = 1 at 2
t=
2x = a
= 3.2 s
FRy x
V = at = (2 m/s 2 )(3.2 s) = 6.4 m/s (c) When water is shut off at 10 m height (where the velocity is 6.4 m/s), the platform will decelerate under the influence of gravity, and the time it takes to come to a stop and the additional rise above 10 m become V = V0 - gt = 0 t = V0 6.4 m/s = = 0.65 s g 9.81 m/s 2
15 m/s
D=5 cm
x = V0 t - 1 gt 2 = (6.4 m/s)(0.65 s) - 1 (9.81 m/s 2 )(0.65 s) 2 = 2.1 m 2 2 Therefore, Jones has 20.65 = 1.3 s to jump off from the platform to the roof since it takes another 0.65 s for the platform to descend to the 10 m level. 13-68E A box-enclosed fan is faced down so the air blast is directed downwards, and it is to be hovered by increasing the blade rpm. The required blade rpm, air exit velocity, the volumetric flow rate, and the minimum mechanical power are to be determined.
13-48
Chapter 13 Momentum Analysis of Flow Systems Assumptions 1 The flow of air is steady, one-dimensional, 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 (no conversion to thermal energy through frictional effects). 5 The change in air pressure with elevation is negligible because of the low density of air. 6 There is no acceleration of the fan, and thus the lift generated is equal to the total weight. Properties The density of air is given to be 0.078 lbm/ft 3. 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 r & & The momentum equation for steady one-dimensional flow is F = m e Ve - mi Vi . Noting 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 = m ( - V2 ) - 0 where A is the blade span area,
A = D 2 / 4 = (3 ft) 2 / 4 = 7.069 ft 2
2 & W = mV2 = ( AV2 )V2 = AV2
V2 =
W A
1
FRy
Then the discharge velocity to produce 5 lbf of upward force becomes V2 = 32.2 lbm ft/s 2 1 lbf (0.078 lbm/ft 3 )(7.069 ft 2 ) 5 lbf = 17.1 ft/s
600 rpm
(b) The volume flow rate and the mass flow rate of air are determined from their definitions, & V = AV2 = (7.069 ft 2 )(17.1 ft/s) = 121 ft 3 /s
& & m = V = (0.078 lbm/ft 3 )(121 ft 3/s) = 9.43 lbm/s
2
(c) 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 & & & & & m 1 + 1 + gz1 + W pump, u = m 2 + 2 + gz 2 + W turbine + E mech,loss 2 2
Substituting,
V & & Wfan, u = m 2 2
2
V2 (18.0 ft/s) 2 1 lbf 1W & & Wfan, u = m 2 = (9.43 lbm/s) = 64.3 W 2 2 2 0.73756 lbf ft/s 32.2 lbm ft/s Therefore, the minimum mechanical power that must be supplied to the air stream is 64.3 W. Discussion The actual power input to the fan will be considerably larger than the calculated power input because of the fan inefficiency in converting mechanical work to kinetic energy.
13-49
Chapter 13 Momentum Analysis of Flow Systems 13-69 A parachute slows a soldier from his terminal velocity VT to his landing velocity of VF. A relation is to be developed for the soldier's velocity after he opens the parachute at time t = 0. Assumptions 1 The air resistance is proportional to the velocity squared (i.e. F = -kV2). 2 The variation of the air properties with altitude is negligible. 3 The buoyancy force applied by air to the person (and the parachute) is negligible because of the small volume occupied and the low density of air. 4 The final velocity of the soldier is equal to its terminal velocity with his parachute open. Analysis The terminal velocity of a free falling object is reached when the air resistance (or air drag) equals the weight of the object, less the buoyancy force applied by the fluid, which is negligible in this case, mg 2 Fair resistance = W kVF = mg k= 2 VF This is the desired relation for the constant of proportionality k. When the parachute is deployed and the soldier starts to decelerate, the net downward force acting on him is his weight less the ai r resistance, V2 mg Fair resistance Fnet = W - Fair resistance = mg - kV 2 = mg - 2 V 2 = mg 1 - 2 V VF F dV Substituting it into Newton's 2nd law relation Fnet = ma = m gives dt V2 dV mg 1 - 2 = m V dt F Canceling m and separating variables, and integrating from t = 0 when V = VT to t = t when V = V gives dV
2 1 - V 2 / VF
= gdt
V VT
dV
2 VF - V 2
=
g
2 VF
t
Parachute
dt
0
W = mg
Using
a
dx
2
-x
2
=
1 a+x ln from integral tables and applying the integration limits, 2a a - x
1 2V F
VF + V V F + VT gt ln V - V - ln V - V = 2 VF F F T VT + VF + (VT - V F )e -2 gt / VF VT + V F - (VT - V F )e - 2 gt / VF
Rearranging, the velocity can be expressed explicitly as a function of time as V = VF
Discussion Note that as t , the velocity approaches the landing velocity of VF, as expected.
13-50
Chapter 13 Momentum Analysis of Flow Systems 13-70 An empty cart is to be driven by a horizontal water jet that enters from a hole at the rear of the cart. A relation is to be developed for cart velocity versus time. Assumptions 1 The flow of water is steady, one-dimensional, incompressible, and horizontal. 2 All the water which enters the cart is retained. 3 The path of the cart is level and frictionless. 4 The cart is initially empty and stationary, and thus V = 0 at time t = 0. 5 Friction between water jet and air is negligible, and the entire momentum of water jet is used to drive the cart with no losses. Analysis We note that the water jet velocity VJ is constant, but the car velocity V is variable. Noting that & m = A(V J - V ) where A is the cross-sectional area of the water jet and VJ - V is the velocity of the water jet relative to the cart, the mass of water in the cart at any time t is & m w = mdt =
0
t
t 0
A(V J - V )dt = AV J t - A Vdt
0
t
(1)
Cart
Also, dm w & = m = A(V J - V ) dt We take the cart as the system. The net force acting on the cart in this case is equal to the momentum flux of the water jet. Newton's 2nd law F = ma = d(mV)/dt in this case can be expressed as F= and dm w d (m total V ) d [(m c + m w )V ] dV dV dV d ( m w V ) = = mc + = mc + mw +V dt dt dt dt dt dt dt dV = (m c + m w ) + A(V J - V )V dt Substituting, A(V J - V ) 2 = (m c + m w ) dV + A(V J - V )V dt A(V J - V )(V J - 2V ) = (m c + m w ) dV dt d (m total V ) dt where & F = m(V J - V ) = A(V J - V ) 2
Waterjet
m0
V
Noting that mw is a function of t (as given by Eq. 1) and separating variables, dV dt = A(V J - V )(V J - 2V ) m c + m w dV = A(V J - V )( V J - 2V ) dt m c + AV J t - A Vdt
0
t
Integrating from t = 0 when V = 0 to t = t when V = V gives the desired integral,
V 0
dV = A(V J - V )(V J - 2V )
t o
dt m c + AV J t - A Vdt
0
t
Discussion Note that the time integral involves the integral of velocity, which complicates the solution.
13-51
Chapter 13 Momentum Analysis of Flow Systems 13-71 A plate is maintained in a horizontal position by frictionless vertical guide rails. The underside of the & plate is subjected to a water jet. The minimum mass flow rate m min to just levitate the plate is to be determined, and a relation is to be obtained for the steady state upward velocity. Also, the integral that relates velocity to time when the water is first turned on is to be obtained. Assumptions 1 The flow of water is steady and one-dimensional. 2 The water jet splatters in the plane of he plate. 3 The vertical guide rails are frictionless. 4 Times are short, so the velocity of the rising jet can be considered to remain constant with height. 5 At time t = 0, the plate is at rest. Analysis (a) We take the plate as the system. The momentum equation for steady one-dimensional flow is r r r & & & F = m e Ve - mi Vi . Noting that m = AV J where A is the cross-sectional area of the water jet and W = mpg, the minimum mass flow rate of water needed to raise the plate is determined by setting the net force acting on the plate equal to zero, & - W = 0 - m min VJ & W = m min VJ & & m p g = m min (m min / AVJ ) & m min = Am p g
& & For m > m min , a relation for the steady state upward velocity V is obtained setting the upward impulse applied by water jet to the weight of the plate (during steady motion, the plate velocity V is constant, and the velocity of water jet relative to plate is VJ V), & W = m( V J - V ) m p g = A(V J - V ) 2 VJ - V = mpg A V= mpg & m - A A
& & (b) At time t = 0 the plate is at rest (V = 0), and it is subjected to water jet with m > m min and thus the net force acting on it is greater than the weight of the plate, and the difference between the jet impulse and the weight will accelerate the plate upwards. Therefore, Newton's 2 nd law F = ma = mdV/dt in this case can be expressed as & m(V J - V ) - W = m p a A(V J - V ) 2 - m p g = m p dV dt FRy
Guide rails mp W = mp g
Separating the variables and integrating from t = 0 when V = 0 to t = t when V = V gives the desired integral,
V 0
m p dV A(V J - V ) 2 - m p g
= dt
t
t =0
t=
V 0
m p dV A(V J - V ) 2 - m p g
m
Discussion This integral can be performed with the help of integral tables. But the relation obtained will be implicit in V.
Nozzle
13-52
Chapter 13 Momentum Analysis of Flow Systems 13-72 Water enters a centrifugal pump axially at a specified rate and velocity, and leaves at an angle from the axial direction. The force acting on the shaft in the axial direction is to be determined. y x
60
& mV & n FRx
Properties We take the density of water to be 1000 kg/m3. Assumptions 1 The flow is steady, one-dimensional, and incompressible. 2 The forces acting on the piping system in the horizontal direction are negligible. 3 The atmospheric pressure is disregarded since it acts on all surfaces. & & & & & Analysis From conservation of mass we have m1 = m 2 = m , and thus V1 = V 2 and Ac1 V1 = Ac 2 V2 . Noting that the discharge area is half the inlet area, the discharge velocity is twice the inlet velocity. That is, Ac1 V2 = Ac1 V1 = 2 V1 = 2(5 m/s ) = 10 m/s Ac 2
We take the pump as the control volume, and the inlet direction of flow as the positive direction of x axis. The momentum equation in this case in the x direction reduces to r r r & & & & & F = m e Ve - mi Vi - FRx = mV2 cos - mV1 FRx = m(V1 - V2 cos ) where the mass flow rate it & & m = V = (1000 kg/m 3 )(0.20 m 3 /s) = 200 kg/s Substituting the known quantities, the reaction force is determined to be (note that cos60 = 0.5) 1N FRx = (200 kg/s)[(5 m/s) - (10 m/s)cos60] 1 kg m/s 2 =0
Discussion Note that at this angle of discharge, the bearing is not subjected to any horizontal loading. Therefore, the loading in the system can be controlled by adjusting the discharge angle.
13-53
Chapter 13 Momentum Analysis of Flow Systems 13-73 Water enters the impeller of a turbine through its outer edge of diameter D with velocity V making & an angle with the radial direction at a mass flow rate of m , and leaves the impeller in the radial direction. & && The maximum power that can be generated is to be shown to be Wshaft = nmDV sin .
V r2 =D/2 r V1
r
Tshaft Impeller region
Assumptions 1 The flow is steady in the mean. 2 Irreversible losses are negligible. Analysis We take the impeller region as the control volume. The tangential velocity components at the inlet and the outlet are V 1,t = 0 and V 2,t =V sin . Normal velocity components as well pressure acting on the inner and outer circumferential areas pass through the shaft center, and thus they do not contribute to torque. Only the tangential velocity components contribute to torque, and the application of the angular momentum equation gives & & V & V Tshaft = m(r2V 2,t - r1 1,t ) = mr2 2,t - 0 = mD ( sin ) / 2 V & The angular velocity of the propeller is = 2n . Then the shaft power becomes & && V Wshaft = Tshaft = 2nmD ( sin ) / 2 Simplifying, the maximum power generated becomes & && Wshaft = nmDV sin which is the desired relation.
13-54
Chapter 13 Momentum Analysis of Flow Systems 13-74 A two-armed sprinkler is used to water a garden. For specified flow rate and discharge angles, the rates of rotation of the sprinkler head are to be determined. EES
r = 0.45 m
Assumptions 1 The flow is uniform and cyclically steady (i.e., steady from a frame of reference rotating with the sprinkler head). 2 The water is discharged to the atmosphere, and thus the gage pressure at the nozzle exit is zero. 3 Frictional effects and air drag of rotating components are neglected. Properties We take the density of water to be 1000 kg/m3 = 1 kg/L. Analysis We take the disk that encloses the sprinkler arms as the control volume, which is a stationary & & & control volume. The conservation of mass equation for this steady flow system is m1 = m 2 = m . Noting & & & & = V / 2 since the density of water is that the two nozzles are identical, we have m = m / 2 or V
nozzle nozzle
constant. The average jet exit velocity relative to the nozzle is V
jet
=
& 1m3 V nozzle 60 L/s = 95.49 m/s = A jet 2[ (0.02 m) 2 / 4] 1000 L & & M = r mV - r mV
out in
The angular momentum equation can be expressed as
. Noting that there are
no external moments acting, the angular momentum equation about the axis of rotation becomes & 0 = -2rm nozzleV r cos Vr =0
jet, t
V
jet
jet, t
-V
nozzle
=0
Noting that the tangential component of jet velocity isV V nozzle =V
jet
=V
cos , we have
cos = (95.49 m/s)cos
& Also noting that V nozzle = r = 2nr , and angular speed and the rate of rotation of sprinkler head become 1) = 0: = V nozzle (95.49 m/s)cos0 = = 212 rad/s r 0.45 m and & n= 212 rad/s 60 s = = 2026 rpm 2 2 1 min
2) = 30: = 3) = 60: =
V nozzle (95.49 m/s)cos30 184 rad/s 60 s & = = 184 rad/s and n = = = 1755 rpm r 0.45 m 2 2 1 min V nozzle (95.49 m/s)cos60 106 rad/s 60 s & = = 106 rad/s and n = = = 1013 rpm r 0.45 m 2 2 1 min
Discussion The rate of rotation in reality will be lower because of frictional effects and air drag.
13-55
Chapter 13 Momentum Analysis of Flow Systems & 13-75 Problem 13-74 is reconsidered. The effect of discharge angle on the rate of rotation n as varies from 0 to 90 in increments of 10 is to be investigated. D=0.02 "m" r=0.45 "m" n_nozzle=2 "number of nozzles" Ac=pi*D^2/4 V_jet=V_dot/Ac/n_nozzle V_nozzle=V_jet*cos(theta) V_dot=0.060 "m3/s" omega=V_nozzle/r n_dot=omega*60/(2*pi) rad/s 212 209 199 184 163 136 106 73 37 0 & n rpm 2026 1996 1904 1755 1552 1303 1013 693 352 0
Angle, 0 10 20 30 40 50 60 70 80 90
Vnozzle , m/s 95.5 94.0 89.7 82.7 73.2 61.4 47.7 32.7 16.6 0.0
2250
1800
n, rpm
1350
900
450
0 0
10
20
30
40
,
50
60
70
80
90
13-56
Chapter 13 Momentum Analysis of Flow Systems 13-76 A stationary water tank placed on wheels on a frictionless surface is propelled by a water jet that leaves the tank through a smooth hole. Relations are to be developed for the acceleration, the velocity, and the distance traveled by the tank as a function of time as water discharges. Assumptions 1 The orifice has a smooth entrance, and thus the frictional losses are negligible. 2 The flow is steady, incompressible, and irrotational (so that the Bernoulli equation is applicable). 3 The surface under the wheeled tank is level and frictionless. 4 The water jet is discharged horizontally and rearward. 5 The mass of the tank and wheel assembly is negligible compared to the mass of water in the tank. Analysis (a) We take point 1 at the free surface of the tank, and point 2 at the exit of the hole, which is also taken to be the reference level (z2 = 0) so that the water height above the hole at any time is z. Noting that the fluid velocity at the free surface is very low (V1 0), it is open to the atmosphere (P1 = Patm), and water discharges into the atmosphere (and thus P2 = Patm), the Bernoulli equation simplifies to
P V12 P V2 1 + + z1 = 2 + 2 + z2 g 2 g g 2 g z=
2 VJ +0 2g
VJ = 2 gz
The discharge rate of water from the tank through the hole is & m = AV J =
2 2 D0 D 0 VJ = 4 4
2 gz
r r r & & The momentum equation for steady one-dimensional flow is F = m e Ve - mi Vi . Applying it to the water tank, the horizontal force that acts on the tank is determined to be 2 2 D 0 D 0 1 & & F = mVe - 0 = mV J = 2 gz = gz 4 2 The acceleration of the water tank is determined from Newton's 2 nd law of motion F = ma where m is the mass of water in the tank, m = V tank = (D 2 / 4) z , / 2) F a= = m z (D / 4)
2 gz (D0 2
z
x
D D0 VJ
a = 2g
2 D0 2
D
2
Note that the acceleration of the tank is constant. (b) Noting that a = dV/dt and thus dV = adt and acceleration a is constant, the velocity is expressed as t D2 (c) Noting that V = dx/dt and thus dx = Vdt, the distance traveled by the water tank is determined by integration to be D2 dx = Vdt dx = 2 g 0 tdt D2 since x = 0 at t = 0. V = at V = 2g
2 D0
x=g
2 D0
D2
t2
Discussion In reality, the flow rate discharge velocity and thus the force acting on the water tank will be less because of the frictional losses at the hole. But these losses can be accounted for by incorporating a discharge coefficient. 13-77 ... Design and Essay Problems
hg
13-57

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