1. Jet Propulsion The propeller is one form of jet propulsion in that it creates a jet and by so doing has a thrust exerted upon it that is the propelling force. In jet engines, air (initially at rest) is taken into the engine and burned with a small amount of fuel; the gases are then ejected with a much higher velocity than in a propeller slipstream. The jet diameter is necessarily smaller than the propeller slipstream. If the mass of fuel burned is neglected, the propelling force F [Eq. (3.11.5)] is (3.11.14) V abs = ∆V (Fig. 3.29) is the absolute velocity of fluid in the jet and is the mass per unit time being discharged. The theoretical mechanical efficiency is the same expression as that for efficiency of the propeller, Eq (3.11.13). V abs /V 1 should be as small as possible. For V 1, the resistance force F is determined by the body and fluid in which it moves; hence, for V abs in Eq. (3.11.13) to be very small, ρQ must be very large.

2. Figure 3.29 Figure 3.29 Walls of flow passages through jet engines taken as inpenetrable part of control surface for plane when viewed as a steady-state problem

3. An example is the type of propulsion system to be used on a boat. If the boat requires a force of 2000 N to move it through water at 25 km/h, first a method of jet propulsion can be considered in which water is taken in at the bow and discharged our the stern by a 100 percent efficient pumping system. To analyze the propulsion system, the problem is converted to steady state by superposition of the boat speed - V 1 on boat and surroundings (Fig. 3.30). With addition enlarging of the jet pipe and the pumping of more water with less velocity head, the efficiency can be further increased. The type of pump best suited for large flows at small head is the axial- flow propeller pump. Increasing the size of pump and jet pipe would increase weight greatly and take up useful space in the boat; the logical limit is the drop the propeller down below or behind the boat and thus eliminate the jet pipe, which is the usual propeller for boats.

4. Figure 3.30 Figure 3.30 Steady-state flow around a boat

5. To take the weight of fuel into account in jet propulsion of aircraft, let m air be the mass of air unit time and r the ratio of mass of fuel burned to mass of air. Then (Fig. 3.29), the propulsive force F is The second term on the right is the mass of fuel per unit time multiplied by its change in velocity. Rearranging gives (3.11.15) Defining the mechanical efficiency again as the useful work divided by the sum of useful work and kinetic energy remaining gives and by Eq. (3.11.15) (3.11.16) The efficiency becomes unity for V 1 = V 2, as the combustion products are then brought to rest and no kinetic energy remains in the jet.

6. Example 3.12 An airplane consumes 1 kg fuel for each 20 kg air and discharges hot gases from the tailpipe at V2 = 1800 m/s. Determine the mechanical efficiency for airplane speeds of 300 and 150 m/s. Solution For 300 m/s, From Eq. (3.11.16), For 150 m/s,

7. Propulsion through air of water in each case is caused by reaction to the formation of a jet behind the body. The various means include the propeller, turbojet, turboprop, ram jet, and rocket motor. The momentum relations for a propeller determine that its theoretical efficiency increases as the speed of the aircraft increases and the absolute velocity of the slipstream decreases. As the speed of the blade tips approaches the speed of sound compressibility effects greatly increase the drag on the blades and thus decrease the overall efficiency of the propulsion system.

8. A turbojet is an engine consisting of a compressor, a combustion chamber, a turbine, and a jet pipe. Air is scooped through the front of the engine and is compressed, and fuel is added and burned with a great excess of air. The air and combustion gases then pass through a gas turbine that drives the compressor. Only a portion of the energy of the hot gases is removed by the turbine, since the only means of propulsion is the issuance of the hot gas through the jet pipe. The overall efficiency of jet engine increases with speed of the aircraft. The overall efficiencies of the turbojet and propeller systems are about the same at the speed of sound. The turboprop is a system combining thrust from a propeller with thrust from the ejection of hot gases. The gas turbine must drive both compressor and propeller. The proportion of thrust between the propeller and the jet may be selected arbitrarily by the designer.

9. The ram jet is a high-speed engine that has neither compressor nor turbine. The ram pressure of the air forces air into the front of the engine, where some of the kinetic energy is converted into pressure energy by enlarging the flow cross section. It then enters a combustion chamber, where fuel is burned, and the air and gases of combustion are ejected through a jet pipe. It is a supersonic device requiring very high speed for compression of the air. An intermittent ram jet was used by the Germans in the V-1 buzz bomb. Air is admitted through spring-closed flap valves in the nose. Fuel is ignited to build up pressure that closed the flap valves and ejects the hot gases as a jet. The ram pressure then opens the valves in the nose to repeat the cycle. The cyclic rate is around 40 s -1.

10. Rocket Mechanics The rocket motor carries with it an oxidizing agent to mix with its fuel so that it develops a thrust that is independent of the medium through which it travels. In contrast, a gas turbine can eject a mass many time the mass of fuel it carries because it take in air to mix with the fuel. To determine the acceleration of a rocket during flight, Fig. 3.31, it is convenient to take the control volume as the outer surface of the rocket, with a plane area normal to the jet across the nozzle exit. The control volume has a velocity equal to the velocity of the rocket at the instant the analysis is made.

11. Figure 3.31 Figure 3.31 Control surface for analysis of rocket acceleration. Frame of reference has the velocity V 1 of the rocket.

12. Let R be the air resistance, m R the mass of the rocket body, m f the mass of fuel, m · the rate at which fuel is being burned, and v r the exit-gas velocity relative to the rocket. V 1 is the actual velocity of the rocket (and of the frame of reference), and V is the velocity of the rocket relative to the frame of reference. V is zero, but dV/dt = dV 1 /dt is the rocket acceleration. The basic linear-momentum equation for the y direction (vertical motion ). (3.11.17) becomes (3.11.18) Since V is a function of t only, the equation can be written as a total differential equation (3.11.19) The mass of propellant reduces with time; for constant burning rate m ·,, with m f o the initial mass of fuel and oxidizer.

13. The theoretical efficiency of a rocket motor (based on available energy) is shown to increase with rocket speed. E represents the available energy in the propellant per unit mass. When the propellant is ignited, its available energy is converted into kinetic energy; E = v r 2 /2, in which v r is the jet velocity relative to the rocket. The kinetic energy being used up per unit time is due to mass loss of the unburned propellant and to the burning mE, or (3.11.20) The mechanical efficiency e is (3.11.21) When v r /V 1 = 1, the maximum efficiency e = 1 is obtained. In this case the absolute velocity of ejected gas is zero. When the thrust on a vertical rocket is greater than the total weight plus resistance, the rocket accelerates. Its mass is continuously reduced. To lift a rocket off its pad, its the thrust m · v r must exceed its total weight.

14. Example 3.13 (a) Determine the burning time for a rocket that initially weighs 4.903 MN, of which 70 percent is propellant. It consumes fuel at a constant rate, and its initial thrust is 10 percent greater than its gravity force. v r = 3300 m/s. (b) Considering constant at 9.8 m/s2 and the flight to be vertical without air resistance, find the speed of the rocket at burnout time, its height above ground, and the maximum height it will attain. Solution (a) From the thrust relation and m = 1634.3 kg/s. The available mass of propellant is 350,000 kg; hence, the burning time is

15. (b) From Eq. (3.11.19) Simplifying gives When t = 0, V 1 = 0; hence, When t = 214.2, V 1 = 1873.24 m/s. The height at t = 214.2 s is The rocket will glide V 2 1 /2g ft higher after burnout, or

16. Moving Vanes Turbomachinery utilizes the forces resulting from the motion over moving vanes. No work can be done on or by a fluid that flows over a fixed vane. When vanes can be displaced, work can be done either on the vane or on the fluid. In Fig. 3.34a a moving vane is shown with fluid flowing onto it tangentially. Forces exerted on the fluid by the vane are indicated by F x and F y. To analyze the flow, the problem is reduced to steady-state by superposition of vane velocity u to the left (Fig. 3.34b) on both vane and fluid. The control volume then encloses the fluid in contact with the vane, with the control surface normal to the flow at sections 1 and 2.

17. Figure 3.34 Figure 3.34 (a) Moving vane. (b) Vane flow viewed as stead-state problem by superposition of velocity u to the left. (c) Polar vector diagram

18. Fig. 3.34c shows the polar rector diagram for flow through the vane. The absolute-velocity vectors originate at the origin O, and the relative-velocity vector V 0 - u is turned through the angle θ of the vane as shown. V 2 is the final absolute velocity leaving the vane. The relative velocity v r = V 0 - u is unchanged in magnitude as it traverses the vane. The mass per unit time is given by ρA 0 v r and is not the mass rate being discharged from the nozzle. If a series of vanes is employed, as on the periphery of a wheel, so arranged that one or another of jets intercept all flow form the nozzle and the velocity substantially u, then mass per second is the total mass per second being discharged.

19. Application of Eq. (3.11.2) to the control volume of Fig. 3.34b (for the single vane): For a series of vanes:

20. When a vane or series of vanes moves toward a jet, work is done by the vane system on the fluid, thereby increasing the energy of the fluid. Figure 3.37 illustrates this situation; the polar vector diagram shows the exit velocity to be greater than the entering velocity. In turbulent flow, losses generally must be determined from experimental tests on the system or a geometrically similar model of the system. In the following two cases, application of the continuity, energy, and momentum equations permits the losses to be evaluated analytically.

21. Figure 3.37 Figure 3.37 Vector diagram for vane doing work on a jet

22. Example 3.14 Determine for a single moving vane of Fig. 3.37a the force components due to the water jet and the rate or work done on the vane. Solution Figure 3.35b is the steady-state reduction with a control volume shown. The polar vector diagram is shown in Fig. 3.37c. By applying Eq. (3.11.2) in the x and y directions to the control volume of Fig. 3.37b, The power exerted on the vane is

23. Figure 3.37 Figure 3.37 Jet acting on a moving vane

24. Losses Due to Sudden Expansion in a Pipe The losses due to sudden enlargement in a pipeline may be calculated with both the energy and momentum equations. Fig. 3.38: for steady, incompressible, turbulent flow through the control volume between sections 1 and 2 of the sudden expansion, the small shear force exerted on the walls between the two sections may be neglected. By assuming uniform velocity over the flow cross sections, which is approached in turbulent flow, application of Eq. (3.11.2) produces

25. At section 1 the redial acceleration of fluid particles in the eddy along the surface is small, and so generally a hydrostatic pressure variation occurs across the section. The energy equation (3.10.1) applied to sections 1 and 2, with the loss term h l, is (for α = 1) Solving for (p 1 – p 2 )/γ in each equation and equating the results give As V 1 A 1 = V 2 A 2, (3.11.22) Which indicates that the losses in turbulent flow are proportional to the square of the velocity.

26. Hydraulic Jump The hydraulic jump is the second application of the basic equations to determine losses due to a turbulent flow situation. Under proper conditions a rapidly flowing stream of liquid in an open channel suddenly changes to a slowly flowing stream with a larger cross- sectional area and a sudden rise in elevation of liquid surface - hydraulic jump (an example of steady nonuniform flow). In effect, the rapidly flowing liquid jet expands (Fig. 3.39) and converts kinetic energy into potential energy and losses or irreversibilities. A roller develops on the inclined surface of the expanding liquid jet and draws air into the liquid. The surface of the jump is very rough and turbulent, the losses being greater as the jump height is greater. For small heights, the form of the jump changes to a standing wave (Fig. 3.40).

27. Figure 3.39 Figure 3.39 Hydraulic jump in a rectangular channel Figure 3.40 Figure 3.40 Standing wave

28. The relations between the variables for the hydraulic jump in a horizontal rectangular channel: For convenience, the width of channel is taken as unity The continuity equation (Fig. 3.39) is (A 1 = y 1, A 2 = y 2 ) The momentum equation is The energy equation (for points on the liquid surface) is h j - losses due to the jump. Eliminating V 2 in the first two equations (3.11.23) the plus sign has been taken before the radical (a negative y 2 has no physical significance). The depths y 1 and y 2 : conjugate depths. Solving for h j and eliminating V 1 and V 2 give (3.11.24)

29. The hydraulic jump, which a very effective device for creating irreversibilities, is commonly used at the ends of chutes or the bottoms of spillways to destroy much of the kinetic energy in the flow. It is also an effective mixing chamber, because of the violent agitation that takes place in the roller. Experimental measurements of hydraulic jumps show that the equations yield the correct value of y 2 to within 1 percent.

30. Example 3.15 If 12 m 3 /s of water per meter of width flows down a slipway onto a horizontal floor and the velocity is 20 m/s, determine the downstream depth required to cause a hydraulic jump and the losses in power by the jump per meter of width. Solution Substituting into Eq. (3.11.23) gives With Eq. (3.11.24),

31. 3.12 The Moment-of-Momentum Equation The general unsteady lineal-momentum equation applied to a control volume, Eq. (3.11.1), is (3.12.1) The moment of a force F about a point O (Fig. 3.42) is given by which is the cross (or vector), product of F and the position vector r of a point on the line of action of the vector from O. The cross product of two vectors is a vector at right angles to the plane defined by the first two vectors and with magnitude Fr sin θ which is the product of F and the shortest distance from O to the line of action of F. The sense of the final vector follows the right-hand rule.

32. Figure 3.42 Figure 3.42 Notation for moment of a vector Right-hand rule: On figure the force tends to cause a counterclockwise rotation around O. If this were a right-hand screw thread turning in this direction, it would tend to come up, and so the vector is likewise directed up put of the paper. If one curls the fingers of the right hand in the direction the force tends to cause rotation, the thumb yields the direction, or sense, of the vector.

33. Using Eq. (3.12.1) (3.12.2) This is general moment-of-momentum equation for a control volume. The left-hand side - the torque exerted by any forces on the control volume. The right-hand side - the rate of change of moment of momentum within the control volume plus the net efflux of moment of momentum from the control volume. When applied to a case of flow in the xy plane, with r the shortest distance to the tangential component of the velocity v t (as Fig. 3.43a), v n is the normal component of velocity, and T z is the torque (3.12.3) When applied to an annular control volume, in steady flow (Fig. 3.43b) (3.12.4) For complete circular symmetry: (3.12.5)

34. Figure 3.43 Figure 3.43 Two-dimensional flow in a centrifugal pump impeller

35. Example 3.16 Sprinkler shown in Fig. 3.47 discharges water upward and outward from the horizontal plane so that it makes an angle of Θ° with the t axis when the sprinkler arm is at rest. It has a constant cross-sectional flow area of A 0 and discharges q starting with ω = 0 and t = 0. The resisting torque due to bearing and seals is the constant T 0, and the moment of inertia of the rotating empty sprinkler head is I s. Determine the equation for ω as a function of time. Solution Equation (3.12.2) can be applied. The control volume is the cylindrical area enclosing the rotating sprinkler head. The inflow is along the axis, so that it has no moment of momentum; hence, the torque – T 0 due to friction is equal to the time rate of change of moment of momentum of sprinkler head and fluid within the sprinkler head plus the net efflux of moment of momentum from the control volume. Let V r = q/2A 0 The total derivative can be used. Simplifying gives For rotation to start, ρqr 0 V r cosΘ must be greater then T 0. The equation is easily integrated to find ω as a function of t. The final value of ω is obtained by setting dω/dt = 0 in the equation.

36. Figure 3.47 Figure 3.47 Plan view of sprinkler and control surface

37. Example 3.17 A turbine discharging 10 m 3 /s is to be so designed that a torque of 10 kN · m is to be exerted on an impeller turning at 200 rpm that takes all the moment of momentum out of the fluid. At the outer periphery of the impeller, r = 1 m. What must the tangential component of velocity be at this location? Solution Equation (3.12.5) is In this case, since the outflow has v t = 0. Solving for v tin gives