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Congratulations…Welcome to UTHM PARIT RAJA….A Place To Be..

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Presentation on theme: "Congratulations…Welcome to UTHM PARIT RAJA….A Place To Be.."— Presentation transcript:

1 Congratulations…Welcome to UTHM PARIT RAJA….A Place To Be..
DR. ZAMRI BIN OMAR Department of Aeronautical Engineering Faculty of Mechanical & Manufacturing Eng. Phone : /7711 Room: C & A4, 2nd Floor

2 ~(6) Performance Analysis :
Objective : To check whether our airplane sketched in step 4 meet or exceed the requirement stated in step 1. Now we’ll use better weight estimates obtained in Step 5. Here the updated performance parameters; But we choose not to re-estimate the aerodynamics coefficients even we alreafy have the layout. In more ‘high-level’ design process, these coefficents would be re-estimated, with the of configuration layout depicted in Step 4 (use datcom, Roskam methods).

3 ~(6) Performance Analysis : Power required (PR) & Power available (PA) curve
PR & PA should be calculated at cruise altitude of 20,000 ft. Drag breakdown,

4 ~(6) Performance Analysis : Power required (PR) & Power available (PA) curve
PR & PA should be calculated at cruise altitude of 20,000 ft. Drag vs velocity plot,

5 ~(6) Performance Analysis : Power required (PR) & Power available (PA) curve Required Vmax = 200 mi/h (midcruise) PR & PA should be calculated at cruise altitude of 20,000 ft. PR vs PA plot, From the plot, Vmax = 437 ft/s = 291 mi/h (r’qment says 250 mi/h) Interestingly, this Vmax assumes total Wo = 4,100 lb (not weight at midcruise) So Vmax at midcruise would be even higher ! CONCLUSION : Our airplane design exceeds Vmax specification. Vmax = 437 ft/s

6 ~(6) Performance Analysis : Rate of Climb (RoC) Required RoC = 1000 ft/min
Variation of max RoC with altitude. Wo is assumed constant throughout. At SL, RoCmax=1,572 ft/min. This RoC far exceeds the requirement. Not at 18K ft, there’s a kink in RoC. Reason; engine being supercharged ! Conclusion : Our airplane design far exceeds the RoC specification.

7 ~(6) Performance Analysis : Range Required Range = 1,200 miles.
In step 2, the calculated Wf/Wo = is based on the range of 1,200 miles. Now we found our gross weight is lighter, so Wf is smaller. Old Wf = 820 lb. New Wf = 652 lb. So we know that the range of 1,200 miles will be met by lesser fuel. Conclusion : Our airplane meets the range specification.

8 ~(6) Performance Analysis : Stall Speed Required Vs = 70 mi/h
The (CL)max = 2.34 remains unchanged until the end because we assume the same aerodynamic characteristic as adopted in Step 3. BUT…because of we have new W/S (23.3)  the stall speed would be different. The new Vs is much better. Conclusion : Our airplane design exceeds Vs specification.

9 ~(6) Performance Analysis : Landing Distance. Required SL = 2,200 ft
We’ll again take approach angle of 3o. The new Vf is Here’s the updated calculations; CONCLUSION: Our airplane design exceeds the SL specification.

10 ~(6) Performance Analysis : Take-off distance. Required STO = 2,500 ft

11 ~(6) Performance Analysis : Take-off distance. Required STO = 2,500 ft
CONCLUSION : Our airplane design far exceeds the STO specification.

12 ~Conceptual Design : A summary
We’ve just completed 6 steps (out of 7) in the conceptual design process. In Step 6 : Does our airplane design meet or exceed the requirement ? YES (superb !) Reason : Gross weight reduced: 5,158 lb  4,100 lb. The engine was originally selected for 5,158 lb weight. Since we have a lighter weight, so we have a smaller power loading (11.39 lb/hp) To conclude…our airplane is ‘too good’. THUS..no need iteration to find better performance. Now to do last step. STEP 7: To answer this  Is this the BEST design ? Answer; We’ve to do optimization ! But it’s virtually certain that we DO NOT have the best design for GIVEN specifications in Step 1. Our airplane is OVERDESIGNED…money issue.. Eg: For less weight of 4,100 lb : we’re able to find smaller engine and STILL can meet the specs. Optim parameters : W/S, W/P, etc.. Optimization is IMPORTANCE but beyond our scope !


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