Propellers and propulsion 537 Calculate the forward speed at which it will absorb 750 kW at 1250 rpm at 3660 m (c = 0.693) and the thrust under these conditions. Compare the efficiency of the airscrew with that of the ideal actuator disc of the same area, giving the same thrust under the same conditions. Power = 2nnQ Therefore 750 000 x 60 - 5730 - torque Q = 2n x 1250 n=- 1250 = 20.83 rps 60 Therefore = 0.0368 Q 5730 kQ=T= Pn D 0.639 x 435 x (3.4)5 x 1.226 Plotting the given values of ke and q against J shows that, for ke = 0.0368, J = 1.39 and q = 0.848. Now J = V/nD, and therefore V = JnD = 1.39 x 20.83 x 3.4 = 98.4ms-’ Since the efficiency is 0.848 (or 84.8%), the thrust power is 750 x 0.848 = 635 kW Therefore the thrust is Power 635 000 - 6460 T=- - Speed 98.4 For the ideal actuator disc = 0.0434 T 6460 a(l +a) = - - 2Psv’ - 2 x 0.693 x “3.4)’ x (98.4)’ x 1.226 4 whence a = 0.0417 Thus the ideal efficiency is 1 1.0417 71 =- = 0.958 or 95.8% Thus the efficiency of the practical airscrew is (0.848/0.958) of that of the ideal actuator disc. Therefore the relative efficiency of the practical airscrew is 0.885, or 88.5%. Example 9.4 An aeroplane is powered by a single engine with speed-power characteristic: Speed (rpm) 1800 1900 2000 2100 Power (kW) 1072 1113 1156 1189 The fixed-pitch airscrew of 3.05 m diameter has the following characteristics: J 0.40 0.42 0.44 0.46 0.48 0.50 kT 0.118 0.115 0.112 0.109 0.106 0.103 kQ 0.0157 0.0154 0.0150 0.0145 0.0139 0.0132 538 Aerodynamics for Engineering Students and is directly coupled to the engine crankshaft. What will be the airscrew thrust and efficiency during the initial climb at sea level, when the aircraft speed is 45 m s-l? Preliminary calculations required are: Q = kQpn2DS = 324.2 kQn2 after using the appropriate values for p and D. J = V/nD = 14.751n The power required to drive the airscrew, P,, is P, = 27rnQ With these expressions, the following table may be calculated: rPm 1800 Pa(kW) 1072 n (rps) 30.00 n2(rps)* 900 J 0.492 0.013 52 3950 kQ Q(Nm) Pr(kW) 745 1900 2000 1113 1156 31.67 33.33 1003 1115 0.465 0.442 0.01436 0.01494 4675 5405 930 1132 2100 1189 35.00 1225 0.421 0.015 38 6100 1340 In this table, Pa is the brake power available from the engine, as given in the data, whereas the values of kQ for the calculated values of J are read from a graph. A graph is now plotted of Pa and P, against rpm, the intersection of the two curves giving the equilibrium condition. This is found to be at a rotational speed of 2010rpm, i.e. n = 33.5 rps. For this value of n, J = 0.440 giving kT = 0.1 12 and kQ = 0.0150. Then T = 0.112 x 1.226 x (33.5)2 x (3.05)4 = 13330N and 1 0.112 x 0,440 = 0.523 or 52.3% lk 77 =-L J = 2Tk~ 2~0.0150 As a check on the correctness and accuracy of this result, note that thrust power = TV = 13 300 x 45 = 599 kW At 2010rpm the engine produces 1158 kW (from engine data), and therefore the efficiency is 599 x 100/1158 = 51.6%, which is in satisfactory agreement with the earlier result. 9.3 Airscrew pitch By analogy with screw threads, the pitch of an airscrew is the advance per revolution. This definition, as it stands, is of little use for airscrews. Consider two extreme cases. If the airscrew is turning at, say, 2000 rpm while the aircraft is stationary, the advance per revolution is zero. If, on the other hand, the aircraft is gliding with the engine stopped the advance per revolution is infinite. Thus the pitch of an airscrew can take any value and is therefore useless as a term describing the airscrew. To overcome this difficulty two more definite measures of airscrew pitch are accepted. Propellers and propulsion 539 Fig. 9.4 9.3.1 Geometric pitch Consider the blade section shown in Fig. 9.4, at radius r from the airscrew axis. The broken line is the zero-lift line of the section, i.e. the direction relative to the section of the undisturbed stream when the section gives no lift. Then the geometric pitch of the element is 27rr tan 8. This is the pitch of a screw of radius r and helix angle (90 - 8) degrees. This geometric pitch is frequently constant for all sections of a given air- screw. In some cases, however, the geometric pitch varies from section to section of the blade. In such cases, the geometric pitch of that section at 70% of the airscrew radius is taken, and called the geometric mean pitch. The geometric pitch is seen to depend solely on the geometry of the blades. It is thus a definite length for a given airscrew, and does not depend on the precise conditions of operation at any instant, although many airscrews are mechanically variable in pitch (see Section 9.3.3). 9.3.2 Experimental mean pitch The experimental mean pitch is defined as the advance per revolution when the airscrew is producing zero net thrust. It is thus a suitable parameter for experimental measurement on an existing airscrew. Like the geometric pitch, it has a definite value for any given airscrew, provided the conditions of test approximate reasonably well to practical flight conditions. 9.3.3 Effect of geometric pitch on airscrew performance Consider two airscrews differing only in the helix angles of the blades and let the blade sections at, say, 70% radius be as drawn in Fig. 9.5. That of Fig. 9.5a has a fine pitch, whereas that of Fig. 9.5b has a coarse pitch. When the aircraft is at rest, e.g. at the start of the take-off run, the air velocity relative to the blade section is the resultant VR of the velocity due to rotation, 27rnr, and the inflow velocity, Vi,. The blade section of the fine-pitch airscrew is seen to be working at a reasonable incidence, the lift SL will be large, and the drag SD will be small. Thus the thrust ST will be large and the torque SQ small and the airscrew is working efficiently. The section of the coarse-pitch airscrew, on the other hand, is stalled and therefore gives little lift and much drag. Thus the thrust is small and the torque large, and the airscrew is inefficient. At high flight speeds the situation is much changed, as shown in Fig. 9.5c,d. Here the section of the coarse- pitch airscrew is working efficiently, whereas the fine-pitch airscrew is now giving a negative thrust, a situation that might arise in a steep dive. Thus an airscrew that has 540 Aerodynamics for Engineering Students 2mr 2unr Fig. 9.5 Effect of geometric pitch on airscrew performance a pitch suitable for low-speed fight and take-off is liable to have a poor performance at high forward speeds, and vice versa. This was the one factor that limited aircraft performance in the early days of powered fight. A great advance was achieved consequent on the development of the two-pitch airscrew. This is an airscrew in which each blade may be rotated bodily, and set in either of two positions at will. One position gives a fine pitch for take-off and climb, whereas the other gives a coarse pitch for cruising and high-speed flight. Consider Fig. 9.6 which shows typical variations of efficiency 7 with J for (a) a fine-pitch and (b) a coarse-pitch airscrew. For low advance ratios, corresponding to take-off and low-speed flight, the fine pitch is obviously better whereas for higher speeds the coarse pitch is preferable. If the pitch may be varied at will between these two values the overall performance J Fig. 9.6 Efficiency for a two-pitch airscrew Propellers and propulsion 541 J Fig. 9.7 Efficiency for a constant-speed airscrew attainable is as given by the hatched line, which is clearly better than that attainable from either pitch separately. Subsequent research led to the development of the constant-speed airscrew in which the blade pitch is infinitely variable between predetermined limits. A mech- anism in the airscrew hub varies the pitch to keep the engine speed constant, per- mitting the engine to work at its most efficient speed. The pitch variations also result in the airscrew working close to its maximum efficiency at all times. Figure 9.7 shows the variation of efficiency with J for a number of the possible settings. Since the blade pitch may take any value between the curves drawn, the airscrew efficiency varies with J as shown by the dashed curve, which is the envelope of all the separate q, J curves. The requirement that the airscrew shall be always working at its optimum efficiency while absorbing the power produced by the engine at the predetermined constant speed calls for very skilful design in matching the airscrew with the engine. The constant-speed airscrew, in turn, led to the provision of feathering and reverse- thrust facilities. In feathering, the geometric pitch is made so large that the blade sections are almost parallel to the direction of flight. This is used to reduce drag and to prevent the airscrew turning the engine (windmilling) in the event of engine failure. For reverse thrust, the geometric pitch is made negative, enabling the airscrew to give a negative thrust to supplement the brakes during the landing ground run, and also to assist in manoeuvring the aircraft on the ground. 9.4 Blade element theory This theory permits direct calculation of the performance of an airscrew and the design of an airscrew to achieve a given performance. 9.4.1 The vortex system of an airscrew An airscrew blade is a form of lifting aerofoil, and as such may be replaced by a hypothetical bound vortex. In addition, a trailing vortex is shed from the tip of each blade. Since the tip traces out a helix as the airscrew advances and rotates, the trailing vortex will itself be of helical form. A two-bladed airscrew may therefore be con- sidered to be replaced by the vortex system of Fig. 9.8. Photographs have been taken of aircraft taking off in humid air that show very clearly the helical trailing vortices behind the airscrew. 542 Aerodynamics for Engineering Students Rei11 trailing helical vortices Hypothetical bound vortex I\ A Fig. 9.8 Simplified vortex system for a two-bladed airscrew Rotational interference The slipstream behind an airscrew is found to be rotating, in the same sense as the blades, about the airscrew axis. This rotation is due in part to the circulation round the blades (the hypothetical bound vortex) and the remainder is induced by the helical trailing vortices. Consider three planes: plane (i) immediately ahead of the airscrew blades; plane (ii), the plane of the airscrew blades; and plane (iii) immediately behind the blades. Ahead of the airscrew, in plane (i) the angular velocity of the flow is zero. Thus in this plane the effects of the bound and trailing vortices exactly cancel each other. In plane (ii) the angular velocity of the flow is due entirely to the trailing vortices, since the bound vortices cannot produce an angular velocity in their own plane. In plane (iii) the angular velocity due to the bound vortices is equal in magnitude and opposite in sense to that in plane (i), and the effects of the trailing and bound vortices are now additive. Let the angular velocity of the airscrew blades be a, the angular velocity of the flow in the plane of the blades be bay and the angular velocity induced by the bound vortices in planes ahead of and behind the disc be &pa2. This assumes that these planes are equidistant from the airscrew disc. It is also assumed that the distance between these planes is small so that the effect of the trailing vortices at the three planes is practically constant. Then, ahead of the airscrew (plane (i)): (b - ,O)O = 0 i.e. b=P Behind the airscrew (plane (iii)), if w is the angular velocity of the flow w = (b + ,B)O = 2b0 Thus the angular velocity of the flow behind the airscrew is twice the angular velocity in the plane of the airscrew. The similarity between this result and that for the axial velocity in the simple momentum theory should be noted. 9.4.2 The performance of a blade element Consider an element, of length Sr and chord cy at radius r of an airscrew blade. This element has a speed in the plane of rotation of ar. The flow is itself rotating in the same plane and sense at bay and thus the speed of the element relative to the air in Propellers and propulsion 543 this plane is Rr(1 - b). If the airscrew is advancing at a speed of V the velocity through the disc is V(l +a), a being the inflow at the radius r. Note that in this theory it is not necessary for u and b to be constant over the disc. Then the total velocity of the flow relative to the blade is VR as shown in Fig. 9.9. If the line CC’ represents the zero-lift line of the blade section then t9 is, by definition, the geometric helix angle of the element, related to the geometric pitch, and a is the absolute angle of incidence of the section. The element will therefore experience lift and drag forces, respectively perpendicular and parallel to the relative velocity VR, appropriate to the absolute incidence a. The values of CL and CD will be those for a two-dimensional aerofoil of the appropriate section at absolute incidence a, since three-dimensional effects have been allowed for in the rotational interference term, bR. This lift and drag may be resolved into components of thrust and ‘torque-force’ as in Fig. 9.9. Here SL is the lift and SD is the drag on the element. SR is the resultant aerodynamic force, making the angle y with the lift vector. SR is resolved into components of thrust 6T and torque force SQ/r, where SQ is the torque required to rotate the element about the airscrew axis. Then (9.24) (9.25) (9.26) tanr = SD/SL = CD/CL VR = V( 1 + a)cosec $ = Rr( 1 - b) sec $ ST = SRCOS($ + 7) SQ - = SR sin($ + y) r V( 1 + a) tan$ = Rr( 1 - b) (9.27) (9.28) The efficiency of the element, 71, is the ratio, useful power out/power input, i.e. V ST Vcos($ + y) R SQ Rr sin($ + y) VI= = Now from the triangle of velocities, and Eqn (9.28): V 1-b -=-tan$ Rr l+a C’ / (9.29) V(I+d Fig. 9.9 The general blade element 544 Aerodynamics for Engineering Students whence, by Eqn (9.29): 1 - b tan4 rll =- l+atan(++y) (9.30) Let the solidity of the annulus, u, be defined as the ratio of the total area of blade in annulus to the total area of annulus. Then where B is the number of blades. Now From Fig. 9.9 ST = SLcosqb - SDsin+ 1 2 = BcSr - p Vi ( CL cos 4 - CD sin 4) Therefore dT 1 - = BC- pvi( CL cos 4 - CD sin 4) dr 2 1 =2.rrra-p~i(~r.cosqb- 2 ~Dsinqb) Bearing in mind Eqn (9.24), Eqn (9.33) may be written as (9.31) (9.32a) (9.32b) (9.33) dT - = mupV;tCL(cos4 - tanysinqb) dr = 7rrupViCL sec y (cos 4 cos y - sin 4 sin 7) Now for moderate incidences of the blade section, tany is small, about 0.02 or so, i.e. LID -h 50, and therefore sec y 5 1 , when the above equation may be written as dT - = mup vi CL cos (4 + y) dr Writing t = CL cos($ + 7) Then dT - = nrutpVi for the airscrew dr 1 2 = Bc-pVit for the airscrew (9.34) (9.3 5a) (9.35b) (9.35c) = c - 1 p Vi t per blade 2 Propellers and propulsion 545 Similarly - 'Q = SL sin4 + SD cos4 r whence, using Eqn (9.32a and b) dQ 1 - = 2nr2a-p~i(~Lsin4+ CDCOS~) dr 2 Writing now leads to q = CL sin(+ + 7) (9.36) dQ - = 7rr2aqpVi totd dr (9.37a) I = Bcr-pViq total = cr-pViq per blade 2 1 2 (9.37b) (9.37c) The quantities dT/dr and dQ/dr are known as the thrust grading and the torque grading respectively. Consider now the axial momentum of the flow through the annulus. The thrust ST is equal to the product of the rate of mass flow through the element with the change in the axial velocity, i.e. ST = mSV. Now = area of annulus x velocity through annulus x density = (27rrSr) [ V( 1 + a)]p = 2mpSr V( 1 + a) A V = V, - V = V( 1 + 2~) - V = 2~ V whence ST = 2nrp~r~~2a(l+ a) giving dT - = 47rpr V2a( 1 + a) dr Equating Eqn (9.38) and (9.35a) and using also Eqn (9.25), leads to: 4.rrprV2a(l + a) = 7rrarpV2(1 + a)2cosec2q5 (9.38) whence a1 - = -at cosec2+ l+a 4 (9.39) 546 Aerodynamics for Engineering Students In the same way, by considering the angular momentum SQ = maw? where Aw is the change in angular velocity of the air on passing through the airscrew. - Then SQ = (27rr~r)[Pv(1+ u)](2bfl)r2 = 4xr3pVb(l + u)RS~ whence 9 = 47rr3pVb( 1 + u)QS dr Now, as derived previously, dQ - = nr2uqpVi (Eqn (9.37a)) dr Substituting for VR both expressions of Eqn (9.25), this becomes dQ -= dup[V(l +a)cosecq5][Rr(l - b)sec4]q dr (9.40) (9.41) Equating this expression for dQ/dr to that of Eqn (9.41) gives after manipulation b1 1-b 4 1 2 - = -uq cosec q5secq5 (9.42) = -uq cosec 24 The local efficiency of the blade at the element, ql, is found as follows. dT Useful power output = VST = V-Sr dr dQ dr Power input = 27m SQ = 2.nn - Sr Therefore V dT/dr VI= 27rn dQ/dr (9.43) which is an alternative expression to Eqn (9.30). With the expressions given above, dT/dr and dQ/dr may be evaluated at several radii of an airscrew blade given the blade geometry and section characteristics, the forward and rotational speeds, and the air density. Then, by plotting dT/dr and [...]... mt) Therefore which, on integrating by parts, gives 1 1 G =-, [eY(y-l)]~~=T[eY(l-y)]~~ rn m Substituting back for y in terms of Mo, my and t gives 1 G = T[(Mo- &)(l - ~ { M - lizt))]; o m 1 =-[ M(l -1 nM) -Mo(l -1 nMo)l m where M = MO- mt Thus finally G= 1 [ ( M - Mo) - M l n M rn + MolnMo] (9.67) Propellers and propulsion 557 Substituting this value of the integral back into Eqn (9.67) gives, for the distance... vlnMo - vln(M0 - rizizt) 556 Aerodynamics for Engineering Students Now if the distance travelled from the instant of firing is x in time t: x= Lt Y dt 1 t =v [InMo - ln(M0 - ht)]dt -v = vtlnMo 1’ ln(M0 - rizt)dt To solve the integral (G, say) in Eqn (9.67), let y = ln(M0 - rizt) Then exp(y)=Mo-rizt 1 t=:(Mo-eY) m and whence Then 1 t G= = ln(M0 - rizt)dt l (y y kgdy) where yo = lnM0 and y1 = ln(M0 - mt)... v2+ 2vvh + vi + v: Substituting for v, and fi, and multiplying by U2 gives Introducing the effective disc loading, Ide, from Eqn (9.48) leads to 1 2 v4 - u2v2 -c#u - 1 16 = - c p 4 -k (9.57) 552 Aerodynamics for Engineering Students a quartic equation for U in terms of given quantities Since, from Eqn (9.56), u; = v2+ 4vv, + 4 4 + 4 4 Then 1 1 P = - p A U ( U ? - V2) =-pAU[4vv,+4v;+4v,] 2 2 (9.58) which,... the time from firing in seconds, or M = 1100 0-1 OOONkg where N is the number of half-minute periods elapsed since firing _ - l100/3 - 330 seconds M o 1 Oo0 + I (9.65a) 558 Aerodynamics for Engineering Students Table 9.1 t (min) 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 M (100 0kg) Acceleration (m sP2) V (m s-') x (km) 11 10 9 8 7 6 5 4 3 2 1 4.55 5.00 5.55 6.25 7.15 8.33 10. 0 12.5 16.7 25.0 50.0 0 0 143 300 478... n V m ( ( M - Mo) - Mln M + Mo In Mo} (9.68) Now, if m is constant: 1 t = T(M0 - M ) m which, substituted into Eqn (9.68), gives V x = %{ (Mo - M ) lnMo - ( M - Mo) (9.69) + M l n M - Mo In&} (9.70) For the distance at all-burnt, when x = X and M = M I = Mo/R: (9.71) m Alternatively, this may be written as V M O X= [(RmR 1)-lnR] i.e X =v M [(R - 1) - In R] 1 m (9.72) Example 9.7 A rocket-propelled missile... the axes Then, at time t, the total forward momentum is Hl=MV-I (9.64) + At time ( t St) the mass of the body plus unburnt fuel is ( M - rizSt) and its speed is (V + SV), while a mass of fuel hSt has been ejected rearwards with a mean speed, relative to the axes, of (v - V - ;ST/? The total forward momentum is then H2 = ( M - rizbt)(V 1 + SV) - rizSt(v - V - -SV) - I 2 Now, by the conservation of momentum... 1100 0 kg Of this mass, 10 000 kg is fuel which is completely consumed in 5 minutes burning time The exhaust is 1500m s-l relative to the rocket Plot curves showing the variation of acceleration, speed and distance with time during the burning period, calculating these quantities at each half-minute For the acceleration dV m _M V dt Now m =-= ' loooo 5 M = Mo 2000kg min-' = 33.3 kgs-' - ht = 1100 0 -. .. rearranged as dV -v - m dt M i.e 1 m -dV = -dt V A 4 Now riz = -dM/dt, since ni is the rate of which fuel is burnt, and therefore 1 1d - d V = -dt= M V M dt dM M Therefore V/v = -M + constant assuming v, but not necessarily myto be constant If the rate of fuel injection into the combustion chamber is constant, and if the pressure into which the nozzle exhausts is also constant, e.g the near-vacuum implicit... Allowing for both axial and rotational interference find the local efficiency of the (Answer: 0.885) (CU) element 7 The thrust and torque gradings at 1.22m radius on each blade of a 2-bladed airscrew are 2120Nm-' and 778Nmm-' respectively Find the speed of rotation (in rads-') of the airstream immediately behind the disc at 1.22m radius (Answer: 735 rads-') 562 Aerodynamics for Engineering Students 8 A 4-bladed... conservation of momentum of a closed system: Hi = H2 i.e M Y - I =MV + MSV - %VSt - rizStSV - rizvSt + hVSt 1 + -rizStSV - I 2 which reduces to 1 MSV - -rizStSV - mvSt = 0 2 Dividing by St and taking the limit as St + 0, this becomes dV M ~v=O dt (9.65) Propellers and propulsion 555 Note that this equation can be derived directly from Newton’s second law, force = mass x acceleration, but it is not always immediately . (9.48) leads to 1 1 2 16 v4 - u2v2 - -c#u = -cp4 -k (9.52a) (9.54a) (9.57) 552 Aerodynamics for Engineering Students a quartic equation for U in terms of given quantities 2 - 2v V - - v2 - 2vV - (v/V) - 2 (9.61) Now suppose v/V = 4. Then *=4_ 2- - 1 or 100 % If v/V < 4, i.e. V > v/4, the propulsive efficiency exceeds 100 % back for y in terms of Mo, my and t gives 1 m 1 m G = T [(Mo - &)(l - ~{Mo - lizt))]; =-[ M(l -1 nM) -Mo(l -1 nMo)l where M = MO - mt. Thus finally 1 G = [(M -