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CALCULATION OF MAXIMUM RANGE Range is the distance traversed by an airplane on one load of fuel. The maximum range is given by; (Anderson 5.156) 2 C L/ 2 W 1 2 R W1 S CD 0 ct Here, the specific fuel consumption, c t , of our engine is 0.4571/hr. S is the wing planform area as calculated in study 3; S 19.91m 2 . The value of C L/ 2 / C D is; 1 (Anderson 5.44) 1/ 4 C L/ 2 1 3 1 C 4 3KC D ,0 3 D m ax Assuming a cruise altitude of 3000m, 0.91kg / m 3 . The parasite drag coefficient is calculated as C D ,0 0.02242 and drag due-to-lift coefficient as K 0.0746 . Hence, C L/ 2 1/ 4 1 3 1 C 3 18.82 D max 4 3 0.0746 0.02242 Without external fuel tanks, W fuel 1018 .3kg and recalling that W0 5337 kg , maximum range is; 2 C L/ 2 W 1 2 R W1 ct S CD 0 R 2 2 0.457 / 3600 0.91 19.91 18.82 5337 9.81 5337 1018.3 9.81 R 2263km With external fuel tanks with a total fuel capacity of 465.6kg (150 gallon-JP-5), W fuel 1483 .9kg , and the improved range is; R 2 0.457 / 3600 2 0.91 19 .91 18 .82 5337 9.81 5337 1483 .9 9.81 R 3388km CALCULATION OF MAXIMUM ENDURANCE Maximum endurance comes into the scene when the main concern is staying in the air for the longest possible time. Hence, as the word itself entails, endurance is defined as the amount of time that an airplane can stay in the air on one load of fuel. Endurance for jet aircraft is expressed as; (Anderson 5.167) 1 L W0 E ln ct D W1 To maximize this expression, L/D should be taken as L / D m ax which is previously found to be 12.23. Without external fuel tanks, W fuel 1018 .3kg and recalling that W0 5337 kg , maximum endurance is; 1 L W0 1 5337 E ln 12 .23 ln ct D W1 0.457 5337 1018 .3 E 5.67hr With external fuel tanks, W fuel 1483 .9kg and the improved endurance is; 1 L W0 1 5337 E ln 12 .23 ln ct D W1 0.457 5337 1483 .9 E 8.72hr CALCULATION OF THE STALLING SPEED The stalling speed of an aircraft is given by; (Anderson 5.67) 2 W 1 Vstall S (C L ) max From study 3, C L m ax 2.404 with full (60) flap deflection and the wing loading was approximated as; W 2600 N / m2 S Assuming sea level conditions, 1.225 kg / m 3 . Hence the stalling speed is; 2 W 1 2 2600 Vstall 42.02m / s S (CL ) max 1.225 2.404 CALCULATION OF THE GROUND ROLL AND THE TOTAL DISTANCE FOR TAKEOFF The ground roll for takeoff is approximated as; (Anderson 6.95) 1.21(W / S ) sg g cL max (T / W ) From study 3, thrust to weight ratio is 0.25 and CL m ax 2.173 when the flaps are partially deflected (30 deflection). Assuming sea level conditions; 1.21(W / S ) 1.21(2600) sg 482m g cL max (T / W ) 9.811.225 2.173 (0.25) The distance required while airborne to clear an obstacle is expressed as; (Anderson 6.98) sa R sin OB where R is the turn radius; (Anderson 6.98) R 2 6.96 Vstall g recalculating Vstall for sea level conditions and with flaps partially deflected; 2 W 1 2 2600 Vstall 44.2m / s S (C L ) max 1.225 2.173 thus R turns out to be; R 2 6.96 Vstall 6.96 (44 .2) 31 .36 m g 9.81 and assuming an obstacle height of 10 m; hOB 1 10 OB cos1 1 cos 1 47 .07 R 31 .36 Hence, sa R sin OB 31 .36 sin 47 .07 23 m The total takeoff distance is the sum of the ground roll and the distance required while airborne to clear an obstacle, s g s a . Total takeoff distance s g sa 482 23 505 m CALCULATION OF THE GROUND ROLL AND THE TOTAL DISTANCE FOR LANDING The landing ground roll is given by; (Anderson 8.28) 2 W 1 j 2 (W / S ) S g jN S Cl max g Cl max r Assuming sea level conditions, =1.225 kg/m3. j n be taken as 1.1 for military jets. For flaps in fully deflected position, CL max 2.404 . Taking N=2 s and r =0.4 then; 2 1 1.12 (2600) S g 1.1 2 2600 364.7 m 1.225 2.404 9.81 1.225 2.404 0.4 Assuming an approach angle of a 3 , and an average velocity during flare as; V f 1.23Vstall 1.23 42 .02 51 .68 m / s . The resulting flight path radius during flare is; (Anderson 6.107) V f2 51.82 R 1367.6m 0.2 g 0.2 9.81 Using Eq. 6.106 from Anderson, the flare height can be calculated as; h f R 1 cos a 1367 .61 cos 3 1.9m Then, the approach distance to clear a 15m can be found from Eq. 6.108; 15 h f 15 1.9 sa 250m Tan a Tan3 And the flare distance can be obtained by multiplying the flight path radius, R, by sin a ; s f R sin a 1367 .6 sin 3 71 .6m Hence, the total landing distance is found by summing the ground roll, and the approach and flare distances; Total Landing Distance s g s f sa 364 .7 250 71 .6 686 .3m VELOCITY MODIFICATION The velocity used for maximum range calculation is given by; (Anderson 5.45) 1/ 2 2 3K W V(C1 / 2 / C ) D max C D,0 S L Substituting 0.91kg / m 3 , C D ,0 0.02242 , K 0.0746 , and W / S 2600 N / m2 ; 1/ 2 2 3 0.0746 V( C 1 / 2 / C 0.91 0.02242 2600 134 .4m / s D ) m ax L Recall that Vstall 42 .02 m / s . Thus, V(C1 / 2 / C Vstall . L D ) max The velocity used for the maximum endurance calculation corresponds to the velocity at L / D m ax and can be calculated as; (Anderson 5.34) 1/ 2 2 K W V( L / D ) max C D,0 S Sea level conditions apply for maximum endurance calculation, hence should be taken as 1.225 kg / m 3 ; 1/ 2 2 0.0746 V( L / D ) m ax 1.225 2600 88 m / s 0.02242 Thus V( L / D) max Vstall . Coming to the velocity used at rate of climb calculations, for jet propelled airplane, Anderson gives V R / C m ax as (5.114); 1/ 2 (T / W )(W / S ) Z V( R / C ) m ax 3 C D , 0 3 where Z 1 1 . Recall that T / W 0.25 and L / D T / W 2 2 max L / D m ax 12 .23 . Hence, 3 Z 1 1 2.15 12.23 0.25 2 2 1/ 2 (0.25)(2600)2.15 V( R / C ) max 130.2m / s 3 1.225 0.02242 Thus, V( R / C ) max Vstall as expected. As illustrated, the results favored the design and there is no need to make any modifications.