CONTENTS. Sl.No Lecture Notes Page No. Dynamic Action of Fluid and Concept of Velocity Triangles. 2 Water Turbines Gas and Steam Turbines 38

Size: px
Start display at page:

Download "CONTENTS. Sl.No Lecture Notes Page No. Dynamic Action of Fluid and Concept of Velocity Triangles. 2 Water Turbines Gas and Steam Turbines 38"

Transcription

1 CONTENTS Sl.No Lecture Notes Page No. 1 Dynamic Action of Fluid and Concept of Velocity Triangles 1 Water Turbines 4 3 Gas and Steam Turbines 38 4 Industrial Pumps and its Applications 79 5 Computational Fluid Dynamics 87 6 Dimensional Analysis Pnematics 119

2 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle Dynamic force DYNAMIC ACTION OF FLUID ON VANES Consider a stream of fluid entering a machine such as a hydraulic turbine or steam turbine, or a pump or a fan. The stream of fluid has a more or less defined direction. For a force to be exerted by the fluid on the machine, the stream of fluid must undergo a change in its velocity either in its magnitude or direction or both. When the fluid stream enters the machine, the machine exerts a force on the fluid bringing about a change in the velocity of fluid either in its magnitude or in its direction. According to Newton s third law of motion for every action there is an equal and opposite reaction. Hence, the fluid stream exerts an equal and opposite force upon the machine that causes the change in velocity of fluid stream. This force exerted by the virtue of fluid in motion is called the dynamic force of fluid. The dynamic force of fluid always involves a change in its velocity and thus a change in its momentum. Hence, the force exerted by the machine on the fluid is the action and the force, in turn, exerted by the fluid on the machine is the reaction. The major problem in hydraulic machinery is to determine the power developed by a particular machine or the power consumed in a particular machine. For instance, a hydraulic turbine develops power while a pump consumes power in order to run. The power consumed or developed by a machine can be determined from the dynamic force or forces exerted by the flowing fluid on the boundaries of the flow passage and which are due to the change of momentum. These are determined by applying Newton s second law of motion. Fundamental principle of dynamics The fundamental principle of dynamics is Newton s Second Law of Motion. It states that the rate of change of momentum (linear momentum) is proportional to the applied force and takes place in the direction of the force. More precisely, this statement may be written as the resultant external force F x acting on the particle of mass m along any arbitrarily chosen direction x is equal to the time rate of change of linear momentum of the particle in the same direction, i.e., x direction. Prepared by: Prof. A. MURUGAPPAN Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar

3 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle Linear momentum of a body is the product of its mass and its velocity. Let m be the mass of the fluid moving with a velocity v. Let the fluid mass undergoes a change in velocity dv in time dt. Hence, change in linear momentum of fluid mass = m.dv Time rate of change of linear momentum of fluid mass = m. According to Newton s Second Law of Motion, Dynamic force applied in x direction = time rate of change of linear momentum of fluid mass in x direction dv dt i.e., dvx Fx m. dt (1) where the suffix x, denotes the components in x direction. Equation (1) is known as linear momentum equation. Equation (1) can also be written as: F x. dt m. dv x () The term on the LHS of equation () represents the product of the dynamic force F x and the time increment dt during which it acts. This is known as the impulse of applied dynamic force. The term on the RHS of equation () represents the product of mass of fluid and the change in velocity dv x undergone by the fluid mass in x-direction in time increment dt. This term represents the change in linear momentum of fluid mass. Note: As velocity is a vector quantity, any change in magnitude or direction or both will change the velocity, and hence momentum. Equation () is known as the Impulse-Momentum Equation. It states that the impulse of the dynamic force is equal to the resulting change in linear momentum of body. Newton s Second Law of Motion is generally applied to a system. A system is a definite mass of fluid (or material) and all other matter around it is known as its surroundings. The boundaries of the system will form a closed surface and this surface may change Prepared by: Prof. A. MURUGAPPAN Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar

4 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle with time so that it contains the same mass during which the change takes place. When Newton s Second Law of Motion is applied to a system, equation (1) may be written as: dvx Fx m. dt (3) where m is the constant mass of the system. F x represents the algebraic sum (or resultant) of all body forces such as gravity as well as surface forces acting on fluid mass m in any arbitrary direction x. v x is the velocity of the centre of mass of the system in x-direction. dv x is the change in v x in time dt. Concept of Control Volume and One-dimensional form of momentum equation Control volume is a specific region in space, its size and shape being entirely arbitrary. However, the shape and size of a control volume are made to coincide with solid boundaries. Considering a control volume and a fluid entering the control volume with uniform velocity vx 1 in the arbitrary x direction and leaving the control volume after time t with a uniform velocity vx in the x direction, equation (3) can be written as Fx m t vx vx 1 (4) Since t m represents the mass of fluid flowing per unit time, we have, m t (mass density of fluid) x (rate of flow of fluid) Q where = mass density of fluid (kg m -3 ) Q = rate of flow of fluid or discharge (m 3 s -1 ) m has units of kg s -1. t Hence, equation (4) can be written as F x Q v v x x 1 (5) Prepared by: Prof. A. MURUGAPPAN Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar

5 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle Equation (5) represents the one-dimensional form of steady flow momentum equation. This equation is important in the study of hydraulic machines as it enables the determination of forces developed by the flow of fluid in the machine. Dynamic force exerted by a fluid jet on a stationary flat plate held normal to direction of jet Fluid jet moving with velocity v v Stationary flat plate held normal to fluid jet x x Nozzle v Figure 1 Fluid jet impinging on a stationary flat plate held normal to jet Let a fluid jet issued from a nozzle strikes a flat plate with a velocity v. Let the fluid jet be oriented in x direction (horizontal direction). The flat plate is held stationary, vertical and normal (perpendicular) to the direction of the jet. Let the flow rate of fluid issued from the nozzle and impinging on the plate be Q. Let be the specific weight of fluid and be the mass density of fluid. Then weight of fluid flowing per second is Q. Mass of fluid issued from nozzle and striking the plate per second, m t Q Q g Velocity of fluid issued from the nozzle = v Velocity with which the fluid jet strikes the plate in x direction = v The fluid jet after striking the plate gets divided into two equal halves, with each half deflected by 90 from the original direction of the fluid jet (at the instant of striking the plate). One half of the jet moves in the vertical upward direction and the other half of the Prepared by: Prof. A. MURUGAPPAN Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar

6 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle jet moves in the vertical downward direction as shown in Figure 1. Assuming the plate to be frictionless, the jet leaving the plate will move with the same magnitude v of the jet incident on the plate. As the direction of the velocity of fluid jet is changed after impingement on the plate, it is said that the velocity of fluid jet has changed causing a linear momentum. Final velocity of fluid jet in x - direction after striking the plate = component of velocity of fluid jet leaving the plate in x direction = v cos 90 = v x 0 = 0 ms -1 Hence change in velocity of fluid jet in x direction = final velocity of fluid jet in x direction (after impingement on plate) initial velocity of fluid jet in x - direction (before impingement) = 0 v = - v ms -1 Dynamic force exerted by plate on fluid = rate of change of momentum of fluid = (mass of fluid striking the plate per second) x (change of velocity of fluid normal to the plate) Applying Equation (5) F Q v v = Q0 v - x x x 1 F x Qv Qv (6) The negative sign on RHS of above equation represents that the velocity of fluid jet is decreasing, while the negative sign on the LHS of the equation represents that the force exerted by the plate on the fluid jet is acting in the negative direction of x axis. Equation (6) gives the force exerted by the plate on the fluid jet. Here, the plate is responsible for effecting a change in the velocity of fluid jet. Therefore, the force is exerted by the plate on the fluid jet (this is the action). Since the final velocity of fluid jet is less than the initial velocity, the force exerted by the plate on the fluid jet is a retarding force, thus it acts in a direction opposite to the direction of flow of fluid. As per Newton s Third Law of motion, for every action there is an equal and opposite reaction. Now, the reaction, in this case, is the force exerted by the fluid jet on the plate. Prepared by: Prof. A. MURUGAPPAN Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar

7 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle The magnitude of the force exerted by the fluid jet on the plate is equal to Qv. The direction of force exerted by fluid jet on the plate is diametrically opposite to that exerted by the plate on the fluid jet. As the direction of the force exerted by the plate on the fluid jet is in the negative x - direction, the direction of the force exerted by the fluid jet on the plate must be in x - direction. Hence, the dynamic force exerted by the fluid jet on the plate is given by F x Qv (7) Dynamic force exerted by a fluid jet on a stationary flat plate held inclined to direction of jet Fluid jet moving with velocity v v Stationary flat plate held inclined to fluid jet x v x Nozzle v Let a fluid jet of diameter d and cross-sectional area a issued from a nozzle and moving with a velocity v impinges on a stationary flat plate held inclined to the direction of jet. The jet is oriented in the horizontal direction (x direction). The plate makes an angle with the horizontal. After impingement on the plate, the jet gets divided into two equal parts, with one part moving upward at an angle to the horizontal and the other part moving downward at the same angle to the horizontal. Assuming the plate to be frictionless, the jet leaves the plate with a velocity whose magnitude is equal to v which is the same as that of the velocity of the jet incident on the plate. Prepared by: Prof. A. MURUGAPPAN Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar

8 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle v t = v cos 90 v n = v sin 90 - v The velocity of jet issued form nozzle and incident on the inclined plate can be resolved into two mutually perpendicular components, one component normal (pe rpendicular) to the plate and other component along the plate (parallel to the plate). Let the velocity component defined normal to the plate be denoted by v n and the component defined along the plate be defined by the symbol v t. Here, and v n v cos( 90 ) v sin v t v cos As the plate surface is frictionless, the jet leaving the plate has the same magnitude v as that of the jet incident on the plate. The jet leaving the plate is tangential to the plane surface of the plate. The component of the velocity of jet leaving the plate in a direction normal to the plate is v cos 90 v x 0 0 m s. a direction normal to the plate is given by -1 Hence, the change in velocity of jet in Change in velocity of jet in a direction normal to the plate = final velocity of jet normal to the plate initial velocity of jet normal to the plate = 0 v sin v sin Dynamic force exerted by plate on fluid in a direction normal to the plate, F n = rate of change of momentum of fluid in a direction normal to the plate = (mass of fluid striking the plate per second) x (change of velocity of fluid normal to the plate) Prepared by: Prof. A. MURUGAPPAN Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar

9 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle Replacing the direction x (horizontal) by n (normal) in Equation (5) F Q v v = Q0 v sin Qv sin - n n n 1 F n Qv sin (8) The negative sign on RHS of above equation represents that the velocity of fluid jet is decreasing in the direction normal to the jet, while the negative sign on the LHS of the equation represents that the force exerted by the plate on the fluid jet in a direction normal to the fluid jet is acting in the negative direction of x axis. The dynamic force exerted by the fluid jet on the plate in a direction normal to the plate is oriented in a direction diametrically opposite to that exerted by the plate on the fluid jet. It is given by Fn Fn ( Qv sin ) Qv sin (9) The component of F n in x direction (along the direction of jet issued from nozzle) is given by Fx Fn cos( 90 ) Fn sin ( Qv sin ) sin Qv sin (10) Dynamic Force exerted by Fluid Jet on a Moving Flat Plate The plate is held normal to the direction of fluid jet. The fluid jet moving with velocity v impinges on the plate and after the jet strikes the plate, the plate moves with a velocity u in the direction same as the direction of the fluid jet incident on the plate. Hence, the velocity of jet leaving the plate becomes (v - u). as the plate moves away progressively from the jet with velocity u, the quantity of fluid striking the plate per second is given by multiplying the cross-sectional area of fluid jet and the velocity of jet relative to the plate, w. Q aw a v u where, w = velocity of fluid jet relative to the motion of the plate v = absolute velocity of fluid jet Prepared by: Prof. A. MURUGAPPAN Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar

10 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle Fluid jet moving with velocity v v - u Moving flat plate held normal to fluid jet u x v x Nozzle v - u Figure. Fluid jet impinging on a moving flat plate held normal to jet Force exerted by the fluid on the plate (in x direction) = (mass of fluid striking the plate per second) x v u v ucos 90 Q v u Qv u F x Q 0 Here, Q av u Hence, a v uv u av u F x (change of velocity of fluid jet (in x direction)) Here, under the impact of the jet on the plate, the plate starts moving away from the jet at a velocity u. that is, the distance between the plate and the nozzle from which the jet is issued increases constantly at u m s -1. Therefore, a single moving plate is not a practical case. However, if a series of plates (refer Figure below) were so arranged that each plate appeared successively before the fluid jet in the same position and always moving with a velocity u in the direction of the jet, the entire flow issued from the nozzle will be utilized in making impact on all the plates. Thus, the mass of fluid striking the plates per second would be Q av. Hence, F u (mass of water striking the plates per second) x (change in velocity of fluid jet) cos 90 av v u = Q v u v u 0 avv u Work done by the fluid jet on the plates =. u Qv u Prepared by: Prof. A. MURUGAPPAN Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar F u u 9

11 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle This gives the output of the system under consideration v u Figure. Fluid jet impinging on a series of moving flat plates Kinetic energy of the fluid jet = 1 mv Where m = mass of fluid issued from the nozzle per second = Hence, kinetic energy of the fluid jet (input to the system under consideration) Efficiency of the system, output input Q work done by the fluid jet on the plates kinetic energy of the fluid jet 1 Q v = = Q v u u 1 Qv = v u v u For maximum efficiency, d max, 0 du Prepared by: Prof. A. MURUGAPPAN Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar

12 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle i.e., d du d v du v u u 0 v v uu 0 d du vu u 0 v u 0 v u Substituting the value of u in the expression for, the value of max can be obtained. max v v v v v v v v v (or) 50% Dynamic force exerted by fluid jet on a stationary curved plate v 1 x 1 Inlet x x Outlet v x Figure. Fluid jet impinging on a stationary curved plate with acute discharge angle The fluid jet moving with a velocity v 1 impinges on a curved plate such that it makes an angle 1 with the x axis at the point of impingement (inlet of plate). After impingement Prepared by: Prof. A. MURUGAPPAN 11 Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar

13 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle at inlet, the jet glides over the surface of the plate and leaves the plate tangentially at its outlet. The jet leaving the plate makes an angle with the x axis. The angle also indicates the angle of curvature of the plate at its outlet as the jet leaves the plate tangentially to the curvature of the plate. Let v be the velocity of fluid jet leaving the plate. If the surface of the plate along which the jet glides past the plate is frictionless, the magnitude v of the velocity of fluid jet leaving the plate will be the same as the magnitude v 1 of the velocity of the jet incident on the plate at its inlet. Velocity of fluid jet at inlet in x - direction = v 1 cos 1 Velocity of fluid jet at outlet in x - direction = v cos Therefore, force exerted by fluid jet on plate in x - direction is given by F x (mass of fluid striking the plate per second) x where = Qv cos v Q quantity of 1 1 cos fluid striking the plate per second (change of velocity of fluid jet in x direction) (cross - sectional area of fluid jet) x (velocity of fluid jet incident on the plate at its inlet) = av 1 Hence, av v cos v F x cos If the curvature of plate at its outlet is such that the discharge angle,, is more than 90 (as shown in Figure below), then cos is negative, hence, v cos is negative. Hence, in order to get more force, the curvature of the plate should be such that the discharge angle must be more than 90. Prepared by: Prof. A. MURUGAPPAN Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar

14 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle v 1 x 1 Inlet x x Outlet x v Figure. Fluid jet impinging on a stationary curved plate with obtuse discharge angle Dynamic force exerted by a fluid jet on a single moving curved plate Let a fluid jet of velocity v 1 issued from a nozzle impinges at the inlet 1 of a curved plate which moves as a result of impingement of fluid jet at its inlet. The velocity with which the fluid jet is issued from the nozzle and impinges on the plate at tits inlet is termed as the absolute velocity of jet at inlet. The absolute velocity of fluid jet makes an angle 1 with the X-direction. Prepared by: Prof. A. MURUGAPPAN Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar

15 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle 1 - Inlet v 1 w Outlet X 1 1 u 1 X u = u 1 = u X u w v X Once the jet is incident on the plate at inlet, let the plate move in X-direction with a velocity u. Let the velocity of plate at inlet be referred by u 1. It should be noted that as the entire plate moves in a unique direction (X-direction) with the same velocity, u, we have, u 1 = u. Once the jet strikes the plate at its inlet, the velocity of jet remains no more equal to v 1, but it becomes the velocity of jet relative to the motion of the plate at inlet (that is, relative velocity of fluid jet) be referred by the symbol w 1. The direction of relative velocity of jet at inlet is tangential to the curvature of the plate at inlet. The tangent to the curvature of the plate at its inlet may or may not make the same angle as that of the fluid jet with the X-direction. In Figure drawn above, the angle which the tangent to the curvature of the plate makes with the X-direction is different from that which the fluid jet at inlet makes with the X-direction. Let the angle which the relative velocity of jet at inlet, w 1, makes with the reversed direction of motion of plate at inlet, that is, - u, be 1. That is, the angle of curvature of the plate at inlet (tangent to the curvature of the plate at inlet) makes an angle 1 with the reversed direction of motion of plate at inlet. Prepared by: Prof. A. MURUGAPPAN Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar

16 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle Here, the magnitudes of v 1 and u 1 (= u) are known. The magnitude of w 1 can be determined by subtracting u 1 from v 1 vectorially by applying the Law of Parallelogram of Velocities. Vecorially speaking, we have, by Law of Parallelogram of Velocities, w 1 v1 u1 Construction of Parallelogram of Velocities at Inlet: Draw AC v 1 to a suitable scale such that v 1 makes an angle 1 with the X-direction. At A, draw AB u1 in X direction to the same scale as that of v 1. Join B and C so that BC w 1. Measure BC which gives the magnitude of w1 to the same scale as those of v 1 and u 1. A X u 1 B v 1 1 w 1 X XC Then, the jet glides past the surface of the plate and leaves the plate at its outlet with a relative velocity equal to w. The magnitude of the relative velocity of jet at outlet, w, may remain equal to the relative velocity of jet at inlet, w 1, provided the surface of the plate is perfectly smooth, that is, when there is no energy loss due to friction of the surface of the plate. Let the plate at outlet be inclined at an angle. This angle is measured such that it is the angle between relative velocity of jet at outlet, w, and reversed direction of motion of plate at outlet, i.e., - u Inlet Prepared by: Prof. A. MURUGAPPAN Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar

17 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle Construction of Parallelogram of Velocities at Outlet: u X u X w v w v Draw AB u to a suitable scale in X direction. At B set out angle representing the direction of w. Draw BC w along the set direction. Join A and C so that v Measure AC which gives the magnitude of v. Vectorially speaking, we have, AC. v u w The magnitude and direction of v can be determined by applying the Law of Parallelogram of Forces. The angle which the absolute velocity of jet at outlet, v with X direction is referred by. It is the angle between v and u., makes Force exerted by the fluid jet on the plate in X direction or in the direction of motion of plate is determined by applying the Linear-Momentum Equation: F x mass of fluid striking the plate per unit time x change in velocity of fluid jet in X - direction = Q x v cos v 1 1 cos where Q av u av u 1 1 1, since the plate progressively moves away from the jet and hence, the velocity of jet falling on the plate is reduced by u, or the velocity with which the jet falls on the plate is (v 1 u). The quantity of fluid jet issued from the nozzle (i.e., av 1 ). For <, cos > 0, while for >, cos < 0, hence, when cos < 0, the second term ( v cos ) within the parentheses will become negative. Hence, the quantity v1 cos1 v cos will be higher for the same Q, v 1, 1 and v. Hence, in Prepared by: Prof. A. MURUGAPPAN Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar

18 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle order to get more magnitude for force F, the curvature of the plate at outlet should be such that is obtuse. Velocity Diagrams for Turbine Blades At Inlet: v 1 w Inlet u 1 = A v u1 u B D w 1 v m 1 v 1 Figure. Typical velocity triangle for flow over turbine blade at Inlet u 1 = circumferential or peripheral velocity of vanes at inlet w 1 = velocity of jet relative to motion of vane at inlet v 1 = absolute velocity (i.e., relative to earth) of jet at inlet 1 = angle between v 1 and u 1, i.e., angle of jet with the direction of motion of vane at inlet 1 = angle between w 1 and u 1, i.e., angle of vane tip at inlet. This angle is measured between w 1 and u 1 reversed The absolute velocity of jet, v 1, can be resolved into two mutually perpendicular components. C Prepared by: Prof. A. MURUGAPPAN Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar

19 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle (i) Tangential component, v u1, equal to v 1 cos 1. This component is parallel to the direction of motion of vane, i.e., u 1, and hence is responsible for doing work. Therefore, this velocity component is referred to as Velocity of Whirl at inlet. (ii) Radial component, v m1, equal to v 1 sin 1. This component is perpendicular to the direction of motion of vane and hence they do not do any work on the blades (runner). This component causes the water to flow through the turbine blade and therefore, is called the Velocity of Flow at inlet. Drawing Velocity Triangle at Inlet: Step 1: Draw AB u1, in the horizontal direction (x direction) to a suitable scale. Step : At A, set out angle 1 downward, to mark the direction of v 1. It should be noted that the angle between v 1 and u 1 is 1 Step 3: Along the direction set out at A as outlined in Step, set out v1 to the same scale so that AC v1 Step 4: Join B and C so that BC w1 Step 5: Measure ABC = 1, the vane angle at inlet The relationship between the velocity vectors, v, u and 1 1 w1 is v 1 u1 w1 w1 v1 u1 The absolute velocity of water jet at inlet, v 1, is resolved into two mutually perpendicular velocity components namely, the Velocity of whirl at inlet, v u1, which is the tangential component, and the Velocity of flow at inlet, v m1, which is the normal or radial Prepared by: Prof. A. MURUGAPPAN Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar

20 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle component. It should be noted that the direction of vu 1 direction of vm is perpendicular (normal) to u 1 1 is along the direction of u 1. The The arrow heads mark the directions of the respective velocity vectors. At outlet: Outlet u w v = v u A D u B v m v w Figure. Typical velocity triangle for flow over turbine blade at Outlet C u = circumferential or peripheral velocity of vanes at outlet w = velocity of jet relative to motion of vane at outlet v = absolute velocity (i.e., relative to earth) of jet at outlet = angle between v and u, i.e., angle of jet with the direction of motion of vane at Prepared by: Prof. A. MURUGAPPAN Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar

21 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle outlet = angle between w and u, i.e., angle of vane tip at outlet. This angle is measured between w and u reversed The absolute velocity of jet, v, can be resolved into two mutually perpendicular components. (i) Tangential component, v u, equal to v cos. This component is parallel to the direction of motion of vane at outlet, i.e., u, and hence is responsible for doing work. Therefore, this velocity component is referred to as Velocity of Whirl at outlet. (ii) Radial component, v m, equal to v sin. This component is perpendicular to the direction of motion of vane and hence they do not do any work on the blades (runner). This component causes the water to flow through the turbine blade and therefore, is called the Velocity of Flow at inlet. Drawing Velocity Triangle at Outlet: Step 1: Draw AB u, in the horizontal direction (x direction) to a suitable scale. Step : At B, set out angle downward, to mark the direction of w. It should be noted that the angle between w and u reversed is Step 3: Along the direction set out at B as outlined in Step, set out w to the same scale so that BC w Step 4: Join A and C so that AC v Step 5: Measure BAC =, the angle which the absolute velocity of jet at outlet, v, makes with the circumferential or peripheral velocity of runner at outlet, u. The relationship between the velocity vectors, v, u and w is Prepared by: Prof. A. MURUGAPPAN Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar

22 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle v u w The absolute velocity of water jet at outlet, v, is resolved into two mutually perpendicular velocity components namely, the Velocity of whirl at outlet, v u, which is the tangential component, and the Velocity of flow at outlet, v m, which is the normal or radial component. It should be noted that the direction of vu is along the direction of u. The direction of vm is perpendicular (normal) to u. Note: The velocity of whirl at outlet, v u, may be positive or negative depending upon whether the angle is acute or obtuse. When is acute, cos is positive, and hence, v u, is positive. When is obtuse, cos is negative, and hence, v cos v u, is negative. v cos Fluid jet on moving curved surface of a Pelton turbine blade 1 Inlet (splitter) Outlet Relative path 1 Nozzle Absolute path Figure: Plan view of a double hemispherical bucket of a Pelton runner Prepared by: Prof. A. MURUGAPPAN Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar

23 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle Each bucket consists of two approximately hemispherical cups separated by a sharp edge (called splitter ) at the centre. The water jet impinges at the centre of the bucket (splitter) and is divided by the splitter into two equal halves, each half moving sideways in opposite direction. Thus the incident jet at inlet is deflected backward when leaving the bucket at outlet. The theoretical angle of deflection of jet is 180. But due to practical reasons, the actual angle of deflection of jet is kept less than 180, say, about = 0 1 = 180 v 1 u 1 w 1 u w v The direction of absolute velocity of jet at inlet is tangential to the curvature of blade at inlet. Further, the direction of absolute velocity of jet at inlet is along the direction of peripheral velocity of runner at inlet, i.e., v1 and u 1 are in the same direction. Hence, 1 0. Hence, the direction of relative velocity of jet at inlet is also along the same direction as those of v1 and u 1. Therefore, 1 = 180. There is no formation of velocity triangle at inlet, as all the three velocity vectors, v 1, u1 and w1 are along the same direction. Further, we have, w 1 v1 u1 As the outlet of the runner blade is located at the same radial distance (in the same plane) from the axis of runner, as that of the inlet, the peripheral velocity of blade at outlet, peripheral velocity of blade at inlet, u 1. The shape of the outlet velocity triangle is shown below. u Prepared by: Prof. A. MURUGAPPAN Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar

24 Topic: Dynamic Action of Fluid on Vanes and Concept of Velocity Triangle w 1 u w v v u u v m v w The direction of absolute velocity of jet at outlet is against the direction of motion of runner, i.e., v is against u. Further, as the angle between v and u, i.e.,, is obtuse (more than 90), cos is negative. Hence, v u = v cos, is negative. What is the impact of this condition in development of the dynamic force by the fluid jet on the runner vanes? Prepared by: Prof. A. MURUGAPPAN Professor and Head Department of Civil Engineering, Annamalai University, Annamalainagar

25 Topic: Water Turbines Introduction: WATER TURBINES Water turbines were developed in the 19th century and were widely used for industrial power prior to electrical grids. Now they are mostly used for electric power generation. Water turbines are mostly found in dams to generate electric power from water kinetic energy. Water wheels have been used for hundreds of years for industrial power. Their main shortcoming is size, which limits the flow rate and head that can be harnessed. The migration from water wheels to modern turbines took about one hundred years. Development occurred during the Industrial revolution, using scientific principles and methods. They also made extensive use of new materials and manufacturing methods developed at the time. Turbine is a device in which a mechanical energy is transferred from the flowing liquid through the machine to its operating member. The inlet energy of the liquid is greater than the outlet energy of the liquid are referred as Water turbines. It is well known from Newton s law that to change of momentum of Fluid, a force is required. Similarly, when momentum of fluid is changed, a force is generated. This principle is made use in hydraulic turbine. It converts energy in the form of falling water into rotating shaft power. The geometry of turbines is such that the fluid exerts a torque on the rotor in the direction of its rotation. The shaft power generated is available to derive generators or other devices. Classification of turbines: According to action of water on moving blades: The two basic types of hydraulic turbines are impulse and reaction turbines. In impulse turbine the entire pressure energy of water is converted into kinetic energy, also it converts the kinetic energy of a jet of water to mechanical energy. The static pressure of water at the entrance to the runner is equal to the static pressure at exit and the rotation of wheel is caused purely due to tangential force created by the impact of the jet, and hence it is called as impulse turbine. Reaction turbines convert potential in pressurized water to mechanical energy. The static pressure at inlet to the runner is higher than the static pressure at the exit, and there is a gradual conversion of static pressure into kinetic energy while water is flowing through the runner. In this type of turbine, the rotation of the runner is partly due to impulse Prepared by: Prof. N. Ashok Kumar 4 Assistant Professor Department of Civil Engineering, Annamalai University, Annamalainagar

26 Topic: Water Turbines action and partly due to change in pressure over the runner blades, hence the turbine is called reaction turbine. According to available head: Turbines can be classified as high head, medium head or low head machines. The turbines which are capable of working under high heads (i.e greater than 400m) are called high head turbines. When the gross head lies between 50m to 400m, the turbines are medium head turbines. Turbines capable of working under heads varying from 50m to.5m are called low head turbines. Type High head Medium head Low head Impulse turbines pelton turgo cross-flow multi-jet pelton cross-flow turgo Reaction turbines francis propeller kaplan According to direction of flow of water: The turbines are classified into Tangential flow, Radial flow, Axial and mixed flow turbines. Tangential flow turbines: In this type of turbines, the water strikes the runner in the direction tangent to the wheel. Example: pelton wheel Radial flow turbines: In this type of turbines, the water strikes in the radial direction. It is further classified as inward radial flow turbine and outward radial flow turbine. Inward radial flow turbine: The flow is inward from periphery to the centre. Example: old Francis turbine. Outward radial flow turbine: The flow is outward from centre to periphery. Example: Fourneyron turbine. Axial flow turbine: When flow of water is in the direction parallel to the axis of the shaft. Example: Kaplan and propeller turbine. Mixed flow turbine: The water enters the runner in the radial direction and leaves in the axial direction. Example: Modern Francis turbine. Prepared by: Prof. N. Ashok Kumar Assistant Professor Department of Civil Engineering, Annamalai University, Annamalainagar

27 Topic: Water Turbines Selection of a type of turbine: The selection of the best turbine for any particular hydro site depends on the site characteristics, the dominant ones being the head and flow available. Selection also depends on the desired running speed of the generator or other device loading the turbine. Other considerations such as whether the turbine is expected to produce power under part-flow conditions also play an important role in the selection. All turbines have a power-speed characteristic. They will tend to run most efficiently at a particular speed, head and flow combination. In large scale hydro installation Pelton turbines are normally only considered for heads above 150 m, but for micro-hydro applications Pelton turbines can be used effectively at heads down to about 0 m. Pelton turbines are not used at lower heads because their rotational speeds become very slow and the runner required is very large and heavy. If runner size and low speed do not pose a problem for a particular installation, then a Pelton turbine can be used efficiently with fairly low heads. If a higher running speed and smaller runner are required then there are two further options: Increasing the number of jets. Having two or more jets enables a smaller runner to be used for a given flow and increases the rotational speed. The required power can still be attained and the part-flow efficiency is especially good because the wheel can be run on a reduced number of jets with each jet in use still receiving the optimum flow. Twin runners. Two runners can be placed on the same shaft either side by side or on opposite sides of the generator. This configuration is unusual and would only be used if number of jets per runner had already been maximized, but it allows the use of smaller diameter and hence faster rotating runners. Impulse Turbine: A Pelton turbine consists of a set of specially shaped buckets mounted on a periphery of a circular disc. It is turned by jets of water which are discharged from one or more nozzles and strike the buckets. The buckets are split into two halves so that the central area does not act as a dead spot incapable of deflecting water away from the oncoming jet. The Prepared by: Prof. N. Ashok Kumar Assistant Professor Department of Civil Engineering, Annamalai University, Annamalainagar

28 Topic: Water Turbines cutaway on the lower lip allows the following bucket to move further before cutting off the jet propelling the bucket ahead of it and also permits a smoother entrance of the bucket into the jet. The Pelton bucket is designed to deflect the jet through 165 degrees (not 180 degrees) which is the maximum angle possible without the return jet interfering with the following bucket for the oncoming jet. Impulse turbines are generally more suitable for micro-hydro applications compared with reaction turbines because they have the following advantages: greater tolerance of sand and other particles in the water, better access to working parts, no pressure seals around the shaft, easier to fabricate and maintain, better part-flow efficiency. Runner of a Pelton wheel. Source: Components of Pelton Turbine: The main components of pelton wheel are 1. Nozzle and flow regulating arrangements,. Runner with buckets, 3. Casing, 4. Breaking jet. Working Principle of Pelton Turbine High speed water jets emerging from the nozzles (obtained by expanding high pressure water to the atmospheric pressure in the nozzle) strike a series of spoon-shaped buckets Prepared by: Prof. N. Ashok Kumar Assistant Professor Department of Civil Engineering, Annamalai University, Annamalainagar

29 Topic: Water Turbines mounted around the edge of the pelton wheel. High pressure water can be obtained from any water body situated at some height or streams of water flowing down the hills. Components of Pelton Turbine. Source: Web image As water flows into the bucket, the direction of the water velocity changes to follow the contour of the bucket. These jets flow along the inner curve of the bucket and leave it in the direction opposite to that of incoming jet. When the water-jet contacts the bucket, the water exerts pressure on the bucket and the water is decelerated as it does a "u-turn" and flows out the other side of the bucket at low velocity.the change in momentum (direction as well as speed) of water jet produces an impulse on the blades of the wheel of Pelton Turbine. This "impulse" does work on the turbine and generates the torque and rotation in the shaft of Pelton Turbine.To obtain the optimum output from the Pelton Turbine the impulse received by the blades should be maximum. For that, change in momentum of the water jet should be maximum possible. This is obtained when the water jet is deflected in the direction opposite to which it strikes the buckets and with the same speed relative to the buckets. For maximum power and efficiency, the turbine system is designed such that the water-jet velocity is twice the velocity of the bucket. A very small percentage of the water's original kinetic energy will still remain in the water. However, this allows the bucket to be emptied at the same rate at which it is filled, thus allowing the water flow to continue uninterrupted. Often two buckets are mounted side-by-side, Prepared by: Prof. N. Ashok Kumar Assistant Professor Department of Civil Engineering, Annamalai University, Annamalainagar

30 Topic: Water Turbines thus splitting the water jet in half (see photo). The high speed water jets emerging form the nozzles strike the buckets at splitters, placed at the middle of the buckets, from where jets are divided into two equal streams. This balances the side-load forces on the wheel, and helps to ensure smooth, efficient momentum transfer of the fluid jet to the turbine wheel. Because water and most liquids are nearly incompressible, almost all of the available energy is extracted in the first stage of the hydraulic turbine. Therefore, Pelton wheels have only one turbine stage, unlike gas turbines that operate with compressible fluid. Bucket shape. Source: Velocity triangle for pelton Wheel: Prepared by: Prof. N. Ashok Kumar Assistant Professor Department of Civil Engineering, Annamalai University, Annamalainagar

31 Topic: Water Turbines Velocity triangles for the jet striking the bucket. Source VTU Learning From the impulse momentum equation Force= (Mass/second) change in velocity in x direction Applications of Pelton Wheel: Pelton wheels are the preferred turbine for hydro-power, when the available water source has relatively high hydraulic head at low flow rates. Pelton wheels are made in all sizes. There exist multi-ton Pelton wheels mounted on vertical oil pad bearings in hydroelectric plants. The largest units can be up to 00 megawatts. The smallest Pelton wheels are only a few inches across, and can be used to tap power from mountain streams having flows of a few gallons per minute. Some of these systems utilize household plumbing fixtures for water delivery. These small units are recommended for use with thirty meters or more of head, in order to generate significant power levels. Depending on water flow and design, Pelton wheels operate best with heads from 15 meters to 1,800 meters, although there is no theoretical limit. Thus, more power can be extracted from a water source with high-pressure and low-flow than from a source with low-pressure and high-flow, even though the two flows Prepared by: Prof. N. Ashok Kumar Assistant Professor Department of Civil Engineering, Annamalai University, Annamalainagar

32 Topic: Water Turbines theoretically contain the same power. Also a comparable amount of pipe material is required for each of the two sources, one requiring a long thin pipe, and the other a short wide pipe. Reaction Turbines: The reaction turbines considered here are the Francis turbine and the propeller turbine. A special case of the propeller turbine is the Kaplan. In all these cases, specific speed is high, i.e. reaction turbines rotate faster than impulse turbines given the same head and flow conditions. This has the very important consequences in that a reaction turbine can often be coupled directly to an alternator without requiring a speed-increasing drive system. Some manufacturers make combined turbine-generator sets of this sort. Significant cost savings are made in eliminating the drive and the maintenance of the hydro unit is very much simpler. The Francis turbine is suitable for medium heads, while the propeller is more suitable for low heads. On the whole, reaction turbines require more sophisticated fabrication than impulse turbines because they involve the use of larger and more intricated profile blades together with carefully profiled casings. Francis turbines can either be volute-cased or open-flume machines. The spiral casing is tapered to distribute water uniformly around the entire perimeter of the runner and the guide vanes feed the water into the runner at the correct angle. The runner blades are profiled in a complex manner and direct the water so that it exits axially from the centre of the runner. In doing so, the water imparts most of its pressure energy to the runner before leaving the turbine via a draft tube. The Francis turbine is generally fitted with adjustable guide vanes. These regulates the water flow as it enters the runner and are usually coupled to a governing system which equals flow to turbine loading in the same way as a spear valve or deflector plate in a Pelton turbine. When the flow is reduced the efficiency of the turbine falls away. Francis Turbine.(Radial flow turbine) Construction and Working: Figure shows schematic diagram of a Francis turbine. Prepared by: Prof. N. Ashok Kumar Assistant Professor Department of Civil Engineering, Annamalai University, Annamalainagar

33 Topic: Water Turbines The main parts are: Penstock: It is a large size conduit which conveys water from the upstream to the dam/reservoir to the turbine runner. Spiral Casing: It constitutes a closed passage whose cross-sectional area gradually decreases along the flow direction; area is maximum at inlet and nearly zero at exit. Guide Vanes: These vanes direct the water on to the runner at an angle appropriate to the design, the motion of them is given by means of hand wheel or by a governor. Governing Mechanism: It changes the position of the guide blades/vanes to affect a variation in water flow rate, when the load conditions on the turbine change. Runner and Runner Blades: The driving force on the runner is both due to impulse and reaction effect. The number of runner blades usually varies between 16 to 4. Draft Tube: It is gradually expanding tube which discharges water, passing through the runner to the tail race. Prepared by: Prof. N. Ashok Kumar Assistant Professor Department of Civil Engineering, Annamalai University, Annamalainagar

34 Topic: Water Turbines Source:VTU Learning Working: Francis turbine has a purely radiate flow runner. Water under pressure, enters the runner from the guide vanes towards the center in radial direction and discharges out Prepared by: Prof. N. Ashok Kumar Assistant Professor Department of Civil Engineering, Annamalai University, Annamalainagar

35 Topic: Water Turbines of the runner axially. Francis turbine operates under medium heads. Water is brought down to the turbine through a penstock and directed to a number of stationary orifices fixed all around the circumference of the runner. These stationary orifices are called as guide vanes. The head acting on the turbine is transformed into kinetic energy and pressure head. Due to the difference of pressure between guide vanes and the runner (called reaction pressure), the motion of runner occurs. That is why a Francis turbine is also known as reactionturbine. The pressure at inlet is more than that at outlet. In Francis turbine runner is always full of water. The moment of runner is affected by the change of both the potential and kinetic energies of water. After doing the work the water is discharged to the tail race through a closed tube called draft tube. Draft Tubes: In radial flow turbines, as the water flows from higher pressure to lower pressure, it cannot come out of the turbine and hence a divergent tube is connected to the end of the turbine. This divergent tube, one end of which is connected to the outlet of the turbine and the other is immersed well below the tail race. Functions of Draft tube: 1. It is to increase the pressure from inlet to outlet of the draft tube as it flows through it and hence increase it more than atmospheric pressure.. To safely discharge the water that has worked on the turbine to the tail race. Velocity Triangle for a radial flow Turbines: Prepared by: Prof. N. Ashok Kumar Assistant Professor Department of Civil Engineering, Annamalai University, Annamalainagar

36 Topic: Water Turbines Source VTU Learning Component of Velocity at inlet Component of Velocity at outlet Now angular momentum per second at inlet and outlet is the product of momentum inlet and outlet with respect to their radial distance R 1 and R. Torque= Work done per second= Torque ω = ω Kaplan Turbine (axial flow turbine) Prepared by: Prof. N. Ashok Kumar Assistant Professor Department of Civil Engineering, Annamalai University, Annamalainagar

FLUID MECHANICS. TUTORIAL No.7 FLUID FORCES. When you have completed this tutorial you should be able to. Solve forces due to pressure difference.

FLUID MECHANICS. TUTORIAL No.7 FLUID FORCES. When you have completed this tutorial you should be able to. Solve forces due to pressure difference. FLUID MECHANICS TUTORIAL No.7 FLUID FORCES When you have completed this tutorial you should be able to Solve forces due to pressure difference. Solve problems due to momentum changes. Solve problems involving

More information

When the fluid velocity is zero, called the hydrostatic condition, the pressure variation is due only to the weight of the fluid.

When the fluid velocity is zero, called the hydrostatic condition, the pressure variation is due only to the weight of the fluid. Fluid Statics When the fluid velocity is zero, called the hydrostatic condition, the pressure variation is due only to the weight of the fluid. Consider a small wedge of fluid at rest of size Δx, Δz, Δs

More information

Fluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur

Fluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Fluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture - 20 Conservation Equations in Fluid Flow Part VIII Good morning. I welcome you all

More information

CENTRIFUGAL PUMP OVERVIEW Presented by Matt Prosoli Of Pumps Plus Inc.

CENTRIFUGAL PUMP OVERVIEW Presented by Matt Prosoli Of Pumps Plus Inc. CENTRIFUGAL PUMP OVERVIEW Presented by Matt Prosoli Of Pumps Plus Inc. 1 Centrifugal Pump- Definition Centrifugal Pump can be defined as a mechanical device used to transfer liquid of various types. As

More information

1. Units of a magnetic field might be: A. C m/s B. C s/m C. C/kg D. kg/c s E. N/C m ans: D

1. Units of a magnetic field might be: A. C m/s B. C s/m C. C/kg D. kg/c s E. N/C m ans: D Chapter 28: MAGNETIC FIELDS 1 Units of a magnetic field might be: A C m/s B C s/m C C/kg D kg/c s E N/C m 2 In the formula F = q v B: A F must be perpendicular to v but not necessarily to B B F must be

More information

Differential Relations for Fluid Flow. Acceleration field of a fluid. The differential equation of mass conservation

Differential Relations for Fluid Flow. Acceleration field of a fluid. The differential equation of mass conservation Differential Relations for Fluid Flow In this approach, we apply our four basic conservation laws to an infinitesimally small control volume. The differential approach provides point by point details of

More information

Copyright 2011 Casa Software Ltd. www.casaxps.com. Centre of Mass

Copyright 2011 Casa Software Ltd. www.casaxps.com. Centre of Mass Centre of Mass A central theme in mathematical modelling is that of reducing complex problems to simpler, and hopefully, equivalent problems for which mathematical analysis is possible. The concept of

More information

KEYWORDS Micro hydro turbine, Turbine testing, Cross flow turbine

KEYWORDS Micro hydro turbine, Turbine testing, Cross flow turbine DEVELOPMENT OF COST EFFECTIVE TURBINE FOR HILLY AREAS [Blank line 11 pt] A. Tamil Chandran, Senior Research Engineer Fluid Control Research Institute, Kanjikode west, Plalakkad, Kerala, India tamilchandran@fcriindia.com

More information

Unit 4 Practice Test: Rotational Motion

Unit 4 Practice Test: Rotational Motion Unit 4 Practice Test: Rotational Motion Multiple Guess Identify the letter of the choice that best completes the statement or answers the question. 1. How would an angle in radians be converted to an angle

More information

Chapter 28 Fluid Dynamics

Chapter 28 Fluid Dynamics Chapter 28 Fluid Dynamics 28.1 Ideal Fluids... 1 28.2 Velocity Vector Field... 1 28.3 Mass Continuity Equation... 3 28.4 Bernoulli s Principle... 4 28.5 Worked Examples: Bernoulli s Equation... 7 Example

More information

FLUID MECHANICS. TUTORIAL No.8A WATER TURBINES. When you have completed this tutorial you should be able to

FLUID MECHANICS. TUTORIAL No.8A WATER TURBINES. When you have completed this tutorial you should be able to FLUID MECHAICS TUTORIAL o.8a WATER TURBIES When you have completed this tutorial you should be able to Explain the significance of specific speed to turbine selection. Explain the general principles of

More information

du u U 0 U dy y b 0 b

du u U 0 U dy y b 0 b BASIC CONCEPTS/DEFINITIONS OF FLUID MECHANICS (by Marios M. Fyrillas) 1. Density (πυκνότητα) Symbol: 3 Units of measure: kg / m Equation: m ( m mass, V volume) V. Pressure (πίεση) Alternative definition:

More information

Figure 1.1 Vector A and Vector F

Figure 1.1 Vector A and Vector F CHAPTER I VECTOR QUANTITIES Quantities are anything which can be measured, and stated with number. Quantities in physics are divided into two types; scalar and vector quantities. Scalar quantities have

More information

Chapter 5 MASS, BERNOULLI AND ENERGY EQUATIONS

Chapter 5 MASS, BERNOULLI AND ENERGY EQUATIONS Fluid Mechanics: Fundamentals and Applications, 2nd Edition Yunus A. Cengel, John M. Cimbala McGraw-Hill, 2010 Chapter 5 MASS, BERNOULLI AND ENERGY EQUATIONS Lecture slides by Hasan Hacışevki Copyright

More information

Solution Derivations for Capa #11

Solution Derivations for Capa #11 Solution Derivations for Capa #11 1) A horizontal circular platform (M = 128.1 kg, r = 3.11 m) rotates about a frictionless vertical axle. A student (m = 68.3 kg) walks slowly from the rim of the platform

More information

Chapter 10 Rotational Motion. Copyright 2009 Pearson Education, Inc.

Chapter 10 Rotational Motion. Copyright 2009 Pearson Education, Inc. Chapter 10 Rotational Motion Angular Quantities Units of Chapter 10 Vector Nature of Angular Quantities Constant Angular Acceleration Torque Rotational Dynamics; Torque and Rotational Inertia Solving Problems

More information

There are four types of friction, they are 1).Static friction 2) Dynamic friction 3) Sliding friction 4) Rolling friction

There are four types of friction, they are 1).Static friction 2) Dynamic friction 3) Sliding friction 4) Rolling friction 2.3 RICTION The property by virtue of which a resisting force is created between two rough bodies that resists the sliding of one body over the other is known as friction. The force that always opposes

More information

3600 s 1 h. 24 h 1 day. 1 day

3600 s 1 h. 24 h 1 day. 1 day Week 7 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution

More information

Physics 201 Homework 8

Physics 201 Homework 8 Physics 201 Homework 8 Feb 27, 2013 1. A ceiling fan is turned on and a net torque of 1.8 N-m is applied to the blades. 8.2 rad/s 2 The blades have a total moment of inertia of 0.22 kg-m 2. What is the

More information

TURBINE SELECTION RANGE

TURBINE SELECTION RANGE GILKES GILKESTURGO TURGOIMPULSE IMPULSEHYDRO HYDROTURBINE TURBINE TURBINE SELECTION On 17th August 1856 we received our first order for a water turbine. It produced mechanical power to drive agricultural

More information

UCCS PES/ENSC 2500: Renewable Energy Spring 2014 Test 3 name:

UCCS PES/ENSC 2500: Renewable Energy Spring 2014 Test 3 name: UCCS PES/ENSC 2500: Renewable Energy Spring 2014 Test 3 name: 1. When a wind turbine is positioned between radio, television or microwave transmitter and receiver it can sometime reflect some of the in

More information

Chapter 3.8 & 6 Solutions

Chapter 3.8 & 6 Solutions Chapter 3.8 & 6 Solutions P3.37. Prepare: We are asked to find period, speed and acceleration. Period and frequency are inverses according to Equation 3.26. To find speed we need to know the distance traveled

More information

SOLID MECHANICS TUTORIAL MECHANISMS KINEMATICS - VELOCITY AND ACCELERATION DIAGRAMS

SOLID MECHANICS TUTORIAL MECHANISMS KINEMATICS - VELOCITY AND ACCELERATION DIAGRAMS SOLID MECHANICS TUTORIAL MECHANISMS KINEMATICS - VELOCITY AND ACCELERATION DIAGRAMS This work covers elements of the syllabus for the Engineering Council exams C105 Mechanical and Structural Engineering

More information

Lab #4 - Linear Impulse and Momentum

Lab #4 - Linear Impulse and Momentum Purpose: Lab #4 - Linear Impulse and Momentum The objective of this lab is to understand the linear and angular impulse/momentum relationship. Upon completion of this lab you will: Understand and know

More information

Mechanical Principles

Mechanical Principles Unit 4: Mechanical Principles Unit code: F/601/1450 QCF level: 5 Credit value: 15 OUTCOME 4 POWER TRANSMISSION TUTORIAL 2 BALANCING 4. Dynamics of rotating systems Single and multi-link mechanisms: slider

More information

6. Vectors. 1 2009-2016 Scott Surgent (surgent@asu.edu)

6. Vectors. 1 2009-2016 Scott Surgent (surgent@asu.edu) 6. Vectors For purposes of applications in calculus and physics, a vector has both a direction and a magnitude (length), and is usually represented as an arrow. The start of the arrow is the vector s foot,

More information

SOLID MECHANICS BALANCING TUTORIAL BALANCING OF ROTATING BODIES

SOLID MECHANICS BALANCING TUTORIAL BALANCING OF ROTATING BODIES SOLID MECHANICS BALANCING TUTORIAL BALANCING OF ROTATING BODIES This work covers elements of the syllabus for the Edexcel module 21722P HNC/D Mechanical Principles OUTCOME 4. On completion of this tutorial

More information

CHAPTER 6 WORK AND ENERGY

CHAPTER 6 WORK AND ENERGY CHAPTER 6 WORK AND ENERGY CONCEPTUAL QUESTIONS. REASONING AND SOLUTION The work done by F in moving the box through a displacement s is W = ( F cos 0 ) s= Fs. The work done by F is W = ( F cos θ). s From

More information

Lecture L22-2D Rigid Body Dynamics: Work and Energy

Lecture L22-2D Rigid Body Dynamics: Work and Energy J. Peraire, S. Widnall 6.07 Dynamics Fall 008 Version.0 Lecture L - D Rigid Body Dynamics: Work and Energy In this lecture, we will revisit the principle of work and energy introduced in lecture L-3 for

More information

UCCS ENSC/PES 2500: Renewable Energy Spring 2011 Test 3 name:

UCCS ENSC/PES 2500: Renewable Energy Spring 2011 Test 3 name: UCCS ENSC/PES 2500: Renewable Energy Spring 2011 Test 3 name: 1. These waves travel through the body of the Earth and are called S waves. a. Transverse b. Longitudinal c. Amplitude d. Trough 2. These waves

More information

www.mathsbox.org.uk Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx Acceleration Velocity (v) Displacement x

www.mathsbox.org.uk Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx Acceleration Velocity (v) Displacement x Mechanics 2 : Revision Notes 1. Kinematics and variable acceleration Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx differentiate a = dv = d2 x dt dt dt 2 Acceleration Velocity

More information

Weight The weight of an object is defined as the gravitational force acting on the object. Unit: Newton (N)

Weight The weight of an object is defined as the gravitational force acting on the object. Unit: Newton (N) Gravitational Field A gravitational field as a region in which an object experiences a force due to gravitational attraction Gravitational Field Strength The gravitational field strength at a point in

More information

Physics 1A Lecture 10C

Physics 1A Lecture 10C Physics 1A Lecture 10C "If you neglect to recharge a battery, it dies. And if you run full speed ahead without stopping for water, you lose momentum to finish the race. --Oprah Winfrey Static Equilibrium

More information

SURFACE TENSION. Definition

SURFACE TENSION. Definition SURFACE TENSION Definition In the fall a fisherman s boat is often surrounded by fallen leaves that are lying on the water. The boat floats, because it is partially immersed in the water and the resulting

More information

Unit 24: Applications of Pneumatics and Hydraulics

Unit 24: Applications of Pneumatics and Hydraulics Unit 24: Applications of Pneumatics and Hydraulics Unit code: J/601/1496 QCF level: 4 Credit value: 15 OUTCOME 2 TUTORIAL 3 HYDRAULIC AND PNEUMATIC MOTORS The material needed for outcome 2 is very extensive

More information

Chapter 5 Using Newton s Laws: Friction, Circular Motion, Drag Forces. Copyright 2009 Pearson Education, Inc.

Chapter 5 Using Newton s Laws: Friction, Circular Motion, Drag Forces. Copyright 2009 Pearson Education, Inc. Chapter 5 Using Newton s Laws: Friction, Circular Motion, Drag Forces Units of Chapter 5 Applications of Newton s Laws Involving Friction Uniform Circular Motion Kinematics Dynamics of Uniform Circular

More information

VELOCITY, ACCELERATION, FORCE

VELOCITY, ACCELERATION, FORCE VELOCITY, ACCELERATION, FORCE velocity Velocity v is a vector, with units of meters per second ( m s ). Velocity indicates the rate of change of the object s position ( r ); i.e., velocity tells you how

More information

HYDROPOWER SYSTEMS BY APPOINTMENT TO H.M. THE QUEEN WATER TURBINE ENGINEERS, GILBERT GILKES & GORDON LTD, KENDAL

HYDROPOWER SYSTEMS BY APPOINTMENT TO H.M. THE QUEEN WATER TURBINE ENGINEERS, GILBERT GILKES & GORDON LTD, KENDAL HYDROPOWER SYSTEMS BY APPOINTMENT TO H.M. THE QUEEN WATER TURBINE ENGINEERS, GILBERT GILKES & GORDON LTD, KENDAL 1 WWW.GILKES.COM CONTENTS THE COMPANY 3 GILKES HYDROPOWER 5 GILKES PACKAGE 7 TURBINE SELECTION

More information

v v ax v a x a v a v = = = Since F = ma, it follows that a = F/m. The mass of the arrow is unchanged, and ( )

v v ax v a x a v a v = = = Since F = ma, it follows that a = F/m. The mass of the arrow is unchanged, and ( ) Week 3 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution

More information

Chapter 4. Forces and Newton s Laws of Motion. continued

Chapter 4. Forces and Newton s Laws of Motion. continued Chapter 4 Forces and Newton s Laws of Motion continued 4.9 Static and Kinetic Frictional Forces When an object is in contact with a surface forces can act on the objects. The component of this force acting

More information

PHY231 Section 2, Form A March 22, 2012. 1. Which one of the following statements concerning kinetic energy is true?

PHY231 Section 2, Form A March 22, 2012. 1. Which one of the following statements concerning kinetic energy is true? 1. Which one of the following statements concerning kinetic energy is true? A) Kinetic energy can be measured in watts. B) Kinetic energy is always equal to the potential energy. C) Kinetic energy is always

More information

PHYSICS 111 HOMEWORK SOLUTION #9. April 5, 2013

PHYSICS 111 HOMEWORK SOLUTION #9. April 5, 2013 PHYSICS 111 HOMEWORK SOLUTION #9 April 5, 2013 0.1 A potter s wheel moves uniformly from rest to an angular speed of 0.16 rev/s in 33 s. Find its angular acceleration in radians per second per second.

More information

What are the Benefits?

What are the Benefits? Micro hydro power system introduction Not everyone is lucky enough to have a source of running water near their homes. But for those with river-side homes or live-on boats, small water generators (micro-hydro

More information

Physics Midterm Review Packet January 2010

Physics Midterm Review Packet January 2010 Physics Midterm Review Packet January 2010 This Packet is a Study Guide, not a replacement for studying from your notes, tests, quizzes, and textbook. Midterm Date: Thursday, January 28 th 8:15-10:15 Room:

More information

PHYS 211 FINAL FALL 2004 Form A

PHYS 211 FINAL FALL 2004 Form A 1. Two boys with masses of 40 kg and 60 kg are holding onto either end of a 10 m long massless pole which is initially at rest and floating in still water. They pull themselves along the pole toward each

More information

3 Work, Power and Energy

3 Work, Power and Energy 3 Work, Power and Energy At the end of this section you should be able to: a. describe potential energy as energy due to position and derive potential energy as mgh b. describe kinetic energy as energy

More information

Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam

Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam INSTRUCTIONS: Use a pencil #2 to fill your scantron. Write your code number and bubble it in under "EXAM NUMBER;" an entry

More information

Magnetism. d. gives the direction of the force on a charge moving in a magnetic field. b. results in negative charges moving. clockwise.

Magnetism. d. gives the direction of the force on a charge moving in a magnetic field. b. results in negative charges moving. clockwise. Magnetism 1. An electron which moves with a speed of 3.0 10 4 m/s parallel to a uniform magnetic field of 0.40 T experiences a force of what magnitude? (e = 1.6 10 19 C) a. 4.8 10 14 N c. 2.2 10 24 N b.

More information

AP Physics Circular Motion Practice Test B,B,B,A,D,D,C,B,D,B,E,E,E, 14. 6.6m/s, 0.4 N, 1.5 m, 6.3m/s, 15. 12.9 m/s, 22.9 m/s

AP Physics Circular Motion Practice Test B,B,B,A,D,D,C,B,D,B,E,E,E, 14. 6.6m/s, 0.4 N, 1.5 m, 6.3m/s, 15. 12.9 m/s, 22.9 m/s AP Physics Circular Motion Practice Test B,B,B,A,D,D,C,B,D,B,E,E,E, 14. 6.6m/s, 0.4 N, 1.5 m, 6.3m/s, 15. 12.9 m/s, 22.9 m/s Answer the multiple choice questions (2 Points Each) on this sheet with capital

More information

p atmospheric Statics : Pressure Hydrostatic Pressure: linear change in pressure with depth Measure depth, h, from free surface Pressure Head p gh

p atmospheric Statics : Pressure Hydrostatic Pressure: linear change in pressure with depth Measure depth, h, from free surface Pressure Head p gh IVE1400: n Introduction to Fluid Mechanics Statics : Pressure : Statics r P Sleigh: P..Sleigh@leeds.ac.uk r J Noakes:.J.Noakes@leeds.ac.uk January 008 Module web site: www.efm.leeds.ac.uk/ive/fluidslevel1

More information

PHY231 Section 1, Form B March 22, 2012

PHY231 Section 1, Form B March 22, 2012 1. A car enters a horizontal, curved roadbed of radius 50 m. The coefficient of static friction between the tires and the roadbed is 0.20. What is the maximum speed with which the car can safely negotiate

More information

Chapter 6 Work and Energy

Chapter 6 Work and Energy Chapter 6 WORK AND ENERGY PREVIEW Work is the scalar product of the force acting on an object and the displacement through which it acts. When work is done on or by a system, the energy of that system

More information

Head Loss in Pipe Flow ME 123: Mechanical Engineering Laboratory II: Fluids

Head Loss in Pipe Flow ME 123: Mechanical Engineering Laboratory II: Fluids Head Loss in Pipe Flow ME 123: Mechanical Engineering Laboratory II: Fluids Dr. J. M. Meyers Dr. D. G. Fletcher Dr. Y. Dubief 1. Introduction Last lab you investigated flow loss in a pipe due to the roughness

More information

Rotation: Moment of Inertia and Torque

Rotation: Moment of Inertia and Torque Rotation: Moment of Inertia and Torque Every time we push a door open or tighten a bolt using a wrench, we apply a force that results in a rotational motion about a fixed axis. Through experience we learn

More information

SOLID MECHANICS DYNAMICS TUTORIAL MOMENT OF INERTIA. This work covers elements of the following syllabi.

SOLID MECHANICS DYNAMICS TUTORIAL MOMENT OF INERTIA. This work covers elements of the following syllabi. SOLID MECHANICS DYNAMICS TUTOIAL MOMENT OF INETIA This work covers elements of the following syllabi. Parts of the Engineering Council Graduate Diploma Exam D5 Dynamics of Mechanical Systems Parts of the

More information

PHY121 #8 Midterm I 3.06.2013

PHY121 #8 Midterm I 3.06.2013 PHY11 #8 Midterm I 3.06.013 AP Physics- Newton s Laws AP Exam Multiple Choice Questions #1 #4 1. When the frictionless system shown above is accelerated by an applied force of magnitude F, the tension

More information

Lecture 6. Weight. Tension. Normal Force. Static Friction. Cutnell+Johnson: 4.8-4.12, second half of section 4.7

Lecture 6. Weight. Tension. Normal Force. Static Friction. Cutnell+Johnson: 4.8-4.12, second half of section 4.7 Lecture 6 Weight Tension Normal Force Static Friction Cutnell+Johnson: 4.8-4.12, second half of section 4.7 In this lecture, I m going to discuss four different kinds of forces: weight, tension, the normal

More information

Mechanics 1: Conservation of Energy and Momentum

Mechanics 1: Conservation of Energy and Momentum Mechanics : Conservation of Energy and Momentum If a certain quantity associated with a system does not change in time. We say that it is conserved, and the system possesses a conservation law. Conservation

More information

C B A T 3 T 2 T 1. 1. What is the magnitude of the force T 1? A) 37.5 N B) 75.0 N C) 113 N D) 157 N E) 192 N

C B A T 3 T 2 T 1. 1. What is the magnitude of the force T 1? A) 37.5 N B) 75.0 N C) 113 N D) 157 N E) 192 N Three boxes are connected by massless strings and are resting on a frictionless table. Each box has a mass of 15 kg, and the tension T 1 in the right string is accelerating the boxes to the right at a

More information

Problem Set 1. Ans: a = 1.74 m/s 2, t = 4.80 s

Problem Set 1. Ans: a = 1.74 m/s 2, t = 4.80 s Problem Set 1 1.1 A bicyclist starts from rest and after traveling along a straight path a distance of 20 m reaches a speed of 30 km/h. Determine her constant acceleration. How long does it take her to

More information

Unified Lecture # 4 Vectors

Unified Lecture # 4 Vectors Fall 2005 Unified Lecture # 4 Vectors These notes were written by J. Peraire as a review of vectors for Dynamics 16.07. They have been adapted for Unified Engineering by R. Radovitzky. References [1] Feynmann,

More information

Chapter 2. Derivation of the Equations of Open Channel Flow. 2.1 General Considerations

Chapter 2. Derivation of the Equations of Open Channel Flow. 2.1 General Considerations Chapter 2. Derivation of the Equations of Open Channel Flow 2.1 General Considerations Of interest is water flowing in a channel with a free surface, which is usually referred to as open channel flow.

More information

PHYSICS 111 HOMEWORK SOLUTION #10. April 8, 2013

PHYSICS 111 HOMEWORK SOLUTION #10. April 8, 2013 PHYSICS HOMEWORK SOLUTION #0 April 8, 203 0. Find the net torque on the wheel in the figure below about the axle through O, taking a = 6.0 cm and b = 30.0 cm. A torque that s produced by a force can be

More information

Force on Moving Charges in a Magnetic Field

Force on Moving Charges in a Magnetic Field [ Assignment View ] [ Eðlisfræði 2, vor 2007 27. Magnetic Field and Magnetic Forces Assignment is due at 2:00am on Wednesday, February 28, 2007 Credit for problems submitted late will decrease to 0% after

More information

Kinetic Energy (A) stays the same stays the same (B) increases increases (C) stays the same increases (D) increases stays the same.

Kinetic Energy (A) stays the same stays the same (B) increases increases (C) stays the same increases (D) increases stays the same. 1. A cart full of water travels horizontally on a frictionless track with initial velocity v. As shown in the diagram, in the back wall of the cart there is a small opening near the bottom of the wall

More information

Physical Quantities and Units

Physical Quantities and Units Physical Quantities and Units 1 Revision Objectives This chapter will explain the SI system of units used for measuring physical quantities and will distinguish between vector and scalar quantities. You

More information

Fric-3. force F k and the equation (4.2) may be used. The sense of F k is opposite

Fric-3. force F k and the equation (4.2) may be used. The sense of F k is opposite 4. FRICTION 4.1 Laws of friction. We know from experience that when two bodies tend to slide on each other a resisting force appears at their surface of contact which opposes their relative motion. The

More information

Midterm Solutions. mvr = ω f (I wheel + I bullet ) = ω f 2 MR2 + mr 2 ) ω f = v R. 1 + M 2m

Midterm Solutions. mvr = ω f (I wheel + I bullet ) = ω f 2 MR2 + mr 2 ) ω f = v R. 1 + M 2m Midterm Solutions I) A bullet of mass m moving at horizontal velocity v strikes and sticks to the rim of a wheel a solid disc) of mass M, radius R, anchored at its center but free to rotate i) Which of

More information

Solving Simultaneous Equations and Matrices

Solving Simultaneous Equations and Matrices Solving Simultaneous Equations and Matrices The following represents a systematic investigation for the steps used to solve two simultaneous linear equations in two unknowns. The motivation for considering

More information

RECYCLED MICRO HYDROPOWER GENERATION USING HYDRAULIC RAM PUMP (HYDRAM)

RECYCLED MICRO HYDROPOWER GENERATION USING HYDRAULIC RAM PUMP (HYDRAM) IMPACT: International Journal of Research in Engineering & Technology (IMPACT: IJRET) Vol., Issue 3, Aug 23, - Impact Journals RECYCLED MICRO HYDROPOWER GENERATION USING HYDRAULIC RAM PUMP (HYDRAM) C.

More information

Structural Axial, Shear and Bending Moments

Structural Axial, Shear and Bending Moments Structural Axial, Shear and Bending Moments Positive Internal Forces Acting Recall from mechanics of materials that the internal forces P (generic axial), V (shear) and M (moment) represent resultants

More information

III. Applications of Force and Motion Concepts. Concept Review. Conflicting Contentions. 1. Airplane Drop 2. Moving Ball Toss 3. Galileo s Argument

III. Applications of Force and Motion Concepts. Concept Review. Conflicting Contentions. 1. Airplane Drop 2. Moving Ball Toss 3. Galileo s Argument III. Applications of Force and Motion Concepts Concept Review Conflicting Contentions 1. Airplane Drop 2. Moving Ball Toss 3. Galileo s Argument Qualitative Reasoning 1. Dropping Balls 2. Spinning Bug

More information

3. KINEMATICS IN TWO DIMENSIONS; VECTORS.

3. KINEMATICS IN TWO DIMENSIONS; VECTORS. 3. KINEMATICS IN TWO DIMENSIONS; VECTORS. Key words: Motion in Two Dimensions, Scalars, Vectors, Addition of Vectors by Graphical Methods, Tail to Tip Method, Parallelogram Method, Negative Vector, Vector

More information

XI / PHYSICS FLUIDS IN MOTION 11/PA

XI / PHYSICS FLUIDS IN MOTION 11/PA Viscosity It is the property of a liquid due to which it flows in the form of layers and each layer opposes the motion of its adjacent layer. Cause of viscosity Consider two neighboring liquid layers A

More information

If you put the same book on a tilted surface the normal force will be less. The magnitude of the normal force will equal: N = W cos θ

If you put the same book on a tilted surface the normal force will be less. The magnitude of the normal force will equal: N = W cos θ Experiment 4 ormal and Frictional Forces Preparation Prepare for this week's quiz by reviewing last week's experiment Read this week's experiment and the section in your textbook dealing with normal forces

More information

Physics 590 Homework, Week 6 Week 6, Homework 1

Physics 590 Homework, Week 6 Week 6, Homework 1 Physics 590 Homework, Week 6 Week 6, Homework 1 Prob. 6.1.1 A descent vehicle landing on the moon has a vertical velocity toward the surface of the moon of 35 m/s. At the same time it has a horizontal

More information

Universal Law of Gravitation

Universal Law of Gravitation Universal Law of Gravitation Law: Every body exerts a force of attraction on every other body. This force called, gravity, is relatively weak and decreases rapidly with the distance separating the bodies

More information

Physics 111: Lecture 4: Chapter 4 - Forces and Newton s Laws of Motion. Physics is about forces and how the world around us reacts to these forces.

Physics 111: Lecture 4: Chapter 4 - Forces and Newton s Laws of Motion. Physics is about forces and how the world around us reacts to these forces. Physics 111: Lecture 4: Chapter 4 - Forces and Newton s Laws of Motion Physics is about forces and how the world around us reacts to these forces. Whats a force? Contact and non-contact forces. Whats a

More information

General Physics (PHY 2140)

General Physics (PHY 2140) General Physics (PHY 2140) Lecture 12 Electricity and Magnetism Magnetism Magnetic fields and force Application of magnetic forces http://www.physics.wayne.edu/~apetrov/phy2140/ Chapter 19 1 Department

More information

The Dot and Cross Products

The Dot and Cross Products The Dot and Cross Products Two common operations involving vectors are the dot product and the cross product. Let two vectors =,, and =,, be given. The Dot Product The dot product of and is written and

More information

F N A) 330 N 0.31 B) 310 N 0.33 C) 250 N 0.27 D) 290 N 0.30 E) 370 N 0.26

F N A) 330 N 0.31 B) 310 N 0.33 C) 250 N 0.27 D) 290 N 0.30 E) 370 N 0.26 Physics 23 Exam 2 Spring 2010 Dr. Alward Page 1 1. A 250-N force is directed horizontally as shown to push a 29-kg box up an inclined plane at a constant speed. Determine the magnitude of the normal force,

More information

Hydroelectric Power (3 Semester Hours)

Hydroelectric Power (3 Semester Hours) ECET 3811 Hydroelectric Power (3 Semester Hours) I. Course Overview: Hydroelectric power is a form of hydropower which exploits the movement of water to generate electricity. Hydroelectricity is a well-established

More information

Practice Problems on Boundary Layers. Answer(s): D = 107 N D = 152 N. C. Wassgren, Purdue University Page 1 of 17 Last Updated: 2010 Nov 22

Practice Problems on Boundary Layers. Answer(s): D = 107 N D = 152 N. C. Wassgren, Purdue University Page 1 of 17 Last Updated: 2010 Nov 22 BL_01 A thin flat plate 55 by 110 cm is immersed in a 6 m/s stream of SAE 10 oil at 20 C. Compute the total skin friction drag if the stream is parallel to (a) the long side and (b) the short side. D =

More information

DRAFTING MANUAL. Gears (Bevel and Hypoid) Drafting Practice

DRAFTING MANUAL. Gears (Bevel and Hypoid) Drafting Practice Page 1 1.0 General This section provides the basis for uniformity in engineering gears drawings and their technical data for gears with intersecting axes (bevel gears), and nonparallel, nonintersecting

More information

Downloaded from www.studiestoday.com

Downloaded from www.studiestoday.com Class XI Physics Ch. 4: Motion in a Plane NCERT Solutions Page 85 Question 4.1: State, for each of the following physical quantities, if it is a scalar or a vector: Volume, mass, speed, acceleration, density,

More information

Mechanical Principles

Mechanical Principles Unit 4: Mechanical Principles Unit code: F/60/450 QCF level: 5 Credit value: 5 OUTCOME 3 POWER TRANSMISSION TUTORIAL BELT DRIVES 3 Power Transmission Belt drives: flat and v-section belts; limiting coefficient

More information

Lecture L2 - Degrees of Freedom and Constraints, Rectilinear Motion

Lecture L2 - Degrees of Freedom and Constraints, Rectilinear Motion S. Widnall 6.07 Dynamics Fall 009 Version.0 Lecture L - Degrees of Freedom and Constraints, Rectilinear Motion Degrees of Freedom Degrees of freedom refers to the number of independent spatial coordinates

More information

W i f(x i ) x. i=1. f(x i ) x = i=1

W i f(x i ) x. i=1. f(x i ) x = i=1 Work Force If an object is moving in a straight line with position function s(t), then the force F on the object at time t is the product of the mass of the object times its acceleration. F = m d2 s dt

More information

Theory of turbo machinery / Turbomaskinernas teori. Chapter 4

Theory of turbo machinery / Turbomaskinernas teori. Chapter 4 Theory of turbo machinery / Turbomaskinernas teori Chapter 4 Axial-Flow Turbines: Mean-Line Analyses and Design Power is more certainly retained by wary measures than by daring counsels. (Tacitius, Annals)

More information

Characteristics of Centrifugal Blower and Its Effective Use in High Static Pressure Area

Characteristics of Centrifugal Blower and Its Effective Use in High Static Pressure Area Characteristics of Centrifugal Blower and Its Effective Use in High Static Pressure Area Masayuki TAKAHASHI With small fans, selecting the right fan that most fits the purpose is extremely important from

More information

8.2 Elastic Strain Energy

8.2 Elastic Strain Energy Section 8. 8. Elastic Strain Energy The strain energy stored in an elastic material upon deformation is calculated below for a number of different geometries and loading conditions. These expressions for

More information

Copyright 2011 Casa Software Ltd. www.casaxps.com

Copyright 2011 Casa Software Ltd. www.casaxps.com Table of Contents Variable Forces and Differential Equations... 2 Differential Equations... 3 Second Order Linear Differential Equations with Constant Coefficients... 6 Reduction of Differential Equations

More information

FRICTION, WORK, AND THE INCLINED PLANE

FRICTION, WORK, AND THE INCLINED PLANE FRICTION, WORK, AND THE INCLINED PLANE Objective: To measure the coefficient of static and inetic friction between a bloc and an inclined plane and to examine the relationship between the plane s angle

More information

Lecture L3 - Vectors, Matrices and Coordinate Transformations

Lecture L3 - Vectors, Matrices and Coordinate Transformations S. Widnall 16.07 Dynamics Fall 2009 Lecture notes based on J. Peraire Version 2.0 Lecture L3 - Vectors, Matrices and Coordinate Transformations By using vectors and defining appropriate operations between

More information

Lab 7: Rotational Motion

Lab 7: Rotational Motion Lab 7: Rotational Motion Equipment: DataStudio, rotary motion sensor mounted on 80 cm rod and heavy duty bench clamp (PASCO ME-9472), string with loop at one end and small white bead at the other end (125

More information

COMPUTATIONAL FLUID DYNAMICS (CFD) ANALYSIS OF INTERMEDIATE PRESSURE STEAM TURBINE

COMPUTATIONAL FLUID DYNAMICS (CFD) ANALYSIS OF INTERMEDIATE PRESSURE STEAM TURBINE Research Paper ISSN 2278 0149 www.ijmerr.com Vol. 3, No. 4, October, 2014 2014 IJMERR. All Rights Reserved COMPUTATIONAL FLUID DYNAMICS (CFD) ANALYSIS OF INTERMEDIATE PRESSURE STEAM TURBINE Shivakumar

More information

The purposes of this experiment are to test Faraday's Law qualitatively and to test Lenz's Law.

The purposes of this experiment are to test Faraday's Law qualitatively and to test Lenz's Law. 260 17-1 I. THEORY EXPERIMENT 17 QUALITATIVE STUDY OF INDUCED EMF Along the extended central axis of a bar magnet, the magnetic field vector B r, on the side nearer the North pole, points away from this

More information

Lab 8: Ballistic Pendulum

Lab 8: Ballistic Pendulum Lab 8: Ballistic Pendulum Equipment: Ballistic pendulum apparatus, 2 meter ruler, 30 cm ruler, blank paper, carbon paper, masking tape, scale. Caution In this experiment a steel ball is projected horizontally

More information

9. The kinetic energy of the moving object is (1) 5 J (3) 15 J (2) 10 J (4) 50 J

9. The kinetic energy of the moving object is (1) 5 J (3) 15 J (2) 10 J (4) 50 J 1. If the kinetic energy of an object is 16 joules when its speed is 4.0 meters per second, then the mass of the objects is (1) 0.5 kg (3) 8.0 kg (2) 2.0 kg (4) 19.6 kg Base your answers to questions 9

More information