Propeller Performance: Blade Numbers 3 < N > 4

Using the graphs, a linear extrapolation can be made for two – and five-bladed pro­pellers with a similar AF. Reference [18] discusses the subject in detail with pro­peller charts for other AF.

10.10.1 Propeller Performance at STD Day: Worked-Out Example

In a stepwise manner, thrust from a propeller is worked out as a coursework exer­cise. The method uses the Hamilton Standard charts intended for constant-speed propellers. These charts also can be used for fixed-pitch propellers when the pitch of the propeller should match the best performance at a specific speed: either cruise speed or climb speed. Two forms are shown: (1) from the given thrust, compute the HP (in turboprop case SHP) required; and (2) from the given HP (or SHP), compute the thrust. The starting point is the Cp. Typically, at sea-level takeoff rating at static condition, one SHP produces about 4 pounds on STD day. At the first guesstimate, a factor of 4 is used to obtain SHP to compute the Cp. One iteration may prove sufficient to refine the SHP.

Problem description. Consider a single, 4-bladed, turboprop military trainer air­craft of the class RAF Tucano operating with a constant speed propeller at N = 2,400 rpm giving installed TSLS = 4,000 lbs. The specified aircraft speed is 320 mph at a 20,000-ft altitude (i. e., Mach 0.421). For the aircraft speed, the blade AF is taken as 180. Establish its rated SHP at sea-level static condition and thrusts at various speeds and altitudes. All computations are in STD day.

Case I: Takeoff performance (HP from thrust). This is used to compute the SHP at sea-level takeoff. Guesstimate installed SHP = 4,000/4 = 1,000 SHP.

From Figure 10.39, for a four-bladed propeller, the diameter is taken to be 96 inches, or 8 ft. Figure 10.38 establishes the integrated design Cu; the ratio of the speed of sound at STD day sea level to the altitude, fc = 1.0.

The factor ND x fc = 2,400 x 8 x 1.0 = 19,200. Corresponding to aircraft speed of Mach 0 and the factor NDx fc = 19,200; Figure 10.38 gives the integrated design Cu ^ 0.5.

Equation 10.44 gives:

Cp = (550 x SHP)/(pn3 D5)

or Cp = (550 x 1,000)/(0.00238 x 403 x 85) = 5,50,000/49,91,222 = 0.11

Figure 10.36 (4-blade, AF = 180, CL = 0.5) gives CT/CP = 2.4 corresponding to integrated design Cui = 0.5 and CP = 0.11.

Therefore, installed static thrust, TSLS = (CT/CP)(33,000 x SHP)/ND = (2.4 x 550 x 1,000)/(40 x 8) = 1,320,000/320 = 4,125 lb.

The installed SHP is revised to 970 giving installed thrust TSLS = 4,000 lb. It is close enough to avoid any further iteration.

Taking into account a 7% installation loss at takeoff, the uninstalled TSLS = 4,000/0.93 = 4,300 lb, giving the uninstalled SHP = 1,043. It may now be summa­rized that to obtain 4,000 lb installed thrust, the uninstalled rated power is 1,043 SHP.

Aircraft configuration must ensure ground clearance at a collapsed nose-wheel tire and oleo. A higher number of blades (i. e., higher solidity) could reduce the diameter, at the expense of higher cost. For this aircraft class, it is best to retain the largest propeller diameter permissible, keeping the number of blades to four or five.

If ground clearance is required, then a 1.5-inch radius can be cut off from the tip (i. e., a 3% reduction to a 93-inch diameter), which requires slightly higher

Table 10.5. Propeller installed thrust results 20 mph* 50 mph 80 mph 120 mph 160 mph J = 0.00463 x mph 0.092 0.23 0.37 0.55 0.74 cp 0.11 0.11 0.11 0.11 0.11 Installed SHP 1000 1000 1000 1000 1000 From Figure 10.37, nprop 0.19 0.4 0.56 0.7 0.77 Installed thrust, T lb 3820 3225 2820 2350 1940 * Too low velocity for Figure 10.37. It is close to static condition and agrees.

uninstalled power to « 1,043/0.97 = 1,075 SHP to obtain 4,000 lb installed thrust corresponding to installed 1,000 SHP.

Check the diameter with the empirical relation, D = K( P)025 = 18 x (1,050)°-25 = 102.5 inch. The empirical relation is close to the cropped 93-inch diameter computed previously. The smaller diameter is retained for better ground clearance.

Case II: Thrust from HP (worked out with 1,000 SHP installed as maximum takeoff rating at sea-level static condition on STD day). Once the installed SHPsls is known, the thrust for the takeoff rating can be computed. The turboprop fuel control maintains a constant SHP at takeoff rating, keeping it almost invariant. This section computes the thrust available at speeds up to 160 mph to cover liftoff and enter the enroute climb phase. Available thrust is computed at 20, 50, 80,120, and 160 mph, as shown in Table 10.5.

Compute:

J = V/nD = (1.467 x V)/(40 x 8) = 0.00458 x V, where V is in mph

Power coefficient:

Cp = (550 x SHP)/(pn3D5) = (550 x 1,000)/(0.00238 x 403 x 85) = 0.11

For an integrated design, Си = 0.5 and CP = 0.11. The propeller nprop corre­sponding to J and CP is obtained from Figure 10.37 (4-blade, AF = 180, Си = 0.5), as shown in Table 10.5.

Compute thrust: T = (550 nprop x 1,000)/V = 550,000 x nprop/ V, where V is in mph (see Table 10.5). Use Equation 10.46 for the FPS system.

Figure 10.40 plots thrust versus speed at the takeoff rating (see Section 10.11). In a similar manner, thrust at any speed, altitude, and engine rating can be determined from the relevant graphs (Figures 10.41 through 10.44). Section 13.3.4 works out the installed turboprop thrust.

Refer to Section 10.11.2 to obtain the SHP at various engine ratings. For exam­ple, an engine throttles back from the takeoff rating to the maximum climb rating for an enroute climb. Up to about 4,000-ft altitude, it is kept at around 85% of the maximum power and goes down with altitude.