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• The table lists the values of the torque curve. The nominal speed of the engine is to be 250 rad/s. 1) e degerees T, Nm 0 15 30 39 45 = 60 = 75 239 90 7 105 200 120 135 150 165 180 0 degerees T, Nm 0 2800 2090 2090 2430 2100 2160 1840 2010 1590 1590 1210 1210 1066 803 532 184 0 Relation between torque and crank angle for a one-cylinder, four-stroke-cycle internal combustion engine. 195 210 225 co 240 255 2 270 y 285 500 300 315 330 345 360 Crank torque T -107 -206 -260 -323 -323 -310 -242 Deve -126 3120 -8 89 125 85 0 180 e degerees T, Nm 375 390 405 Boe 420 Byko 453 435 450 Bygg 465 Boe 480 1400 495 510 525 540 360° Crank angle -85 -125 -89 103 8 120 126 242 310 323 32 280 280 206 107 0 0 degerees T, Nm 555 570 585 393 1000 600 199 615 630 b 645 660 675 690 705 720 720° -107 -206 -292 -355 Bee -371 -371 -362 1992 -312 -272 -274 -548 -760 0 Integrate the torque displacement function and find the energy that can be delivered to a load during the cycle.(3368J) Determine the mean torque. (268 Nm) The greatest energy fluctuation is approximately between 0 = 0° and 0 = 180° (see torque diagram and note that To-Tm). Using a largest coefficient of speed fluctuation C = 0,1 find a suitable value for the flywheel inertia. (0,6565) 4) Find w₂ and w₁. (262,5 and 237,5) 2) 3) Theta T ol 15 30 45 60 75 90 105 120 135 150 165 180 0 2800 2090 2430 2160 1840 1590 1210 1066 803 532 184 0 Tavg rad E 1400 0.261799 366.5191 2445 0.261799 640.0995 2260 0.261799 591.6666 2295 0.261799 600.8296 2000 0.261799 523.5988 1715 0.261799 448.986 1400 0.261799 366.5191 1138 0.261799 297.9277 934.5 0.261799 244.6515 667.5 0.261799 174.7511 358 0.261799 93.72418 92 0.261799 24.08554 0 0.261799

• The table lists the values of the torque curve. The nominal speed of the engine is to be 250 rad/s. 1) e degerees T, Nm 0 15 30 39 45 = 60 = 75 239 90 7 105 200 120 135 150 165 180 0 degerees T, Nm 0 2800 2090 2090 2430 2100 2160 1840 2010 1590 1590 1210 1210 1066 803 532 184 0 Relation between torque and crank angle for a one-cylinder, four-stroke-cycle internal combustion engine. 195 210 225 co 240 255 2 270 y 285 500 300 315 330 345 360 Crank torque T -107 -206 -260 -323 -323 -310 -242 Deve -126 3120 -8 89 125 85 0 180 e degerees T, Nm 375 390 405 Boe 420 Byko 453 435 450 Bygg 465 Boe 480 1400 495 510 525 540 360° Crank angle -85 -125 -89 103 8 120 126 242 310 323 32 280 280 206 107 0 0 degerees T, Nm 555 570 585 393 1000 600 199 615 630 b 645 660 675 690 705 720 720° -107 -206 -292 -355 Bee -371 -371 -362 1992 -312 -272 -274 -548 -760 0 Integrate the torque displacement function and find the energy that can be delivered to a load during the cycle.(3368J) Determine the mean torque. (268 Nm) The greatest energy fluctuation is approximately between 0 = 0° and 0 = 180° (see torque diagram and note that To-Tm). Using a largest coefficient of speed fluctuation C = 0,1 find a suitable value for the flywheel inertia. (0,6565) 4) Find w₂ and w₁. (262,5 and 237,5) 2) 3) Theta T ol 15 30 45 60 75 90 105 120 135 150 165 180 0 2800 2090 2430 2160 1840 1590 1210 1066 803 532 184 0 Tavg rad E 1400 0.261799 366.5191 2445 0.261799 640.0995 2260 0.261799 591.6666 2295 0.261799 600.8296 2000 0.261799 523.5988 1715 0.261799 448.986 1400 0.261799 366.5191 1138 0.261799 297.9277 934.5 0.261799 244.6515 667.5 0.261799 174.7511 358 0.261799 93.72418 92 0.261799 24.08554 0 0.261799

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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The table lists the values of the torque curve. The nominal speed of the engine is to be 250 rad/s. 1) 0 degerees T, Nm 0 15 30 45 60 75 90 105 120 135 150 165 180 0 2800 2090 2430 2160 Dave 1840 come 1590 www 1210 1066 803 532 184 0 0 degerees T, Nm Relation between torque and crank angle for a one-cylinder, four-stroke-cycle internal combustion engine. 195 210 225 240 255 270 285 7 300 315 330 345 360 Crank torque T -107 -206 -260 -323 -310 -242 -126 -8 89 125 85 0 180 0 degerees T, Nm 375 390 405 420 435 450 465 480 495 510 525 540 360° Crank angle -85 -125 -89 8 126 242 310 323 280 206 107 0 0 degerees T, Nm 555 570 585 600 615 630 645 660 675 690 705 720 720° -107 -206 -292 -355 -371 -362 -312 -272 -274 -548 -760 0 2) 3) 4) Integrate the torque displacement function and find the energy that can be delivered to a load during the cycle. (3368J) Determine the mean torque. (268 Nm) The greatest energy fluctuation is approximately between 0 = 0° and 0 = 180° (see torque diagram and note that To-Tm). Using a largest coefficient of speed fluctuation C = 0,1 find a suitable value for the flywheel inertia. (0,6565) Find W₂ and w₁. (262,5 and 237,5) Theta T 0 15 30 45 60 75 90 105 120 135 150 165 180 0 2800 2090 2430 2160 1840 1590 1210 1066 803 532 184 0 Tavg rad E 1400 0.261799 366.5191 2445 0.261799 640.0995 2260 0.261799 591.6666 2295 0.261799 600.8296 2000 0.261799 523.5988 1715 0.261799 448.986 1400 0.261799 366.5191 1138 0.261799 297.9277 934.5 0.261799 244.6515 667.5 0.261799 174.7511 358 0.261799 93.72418 92 0.261799 24.08554 0 0.261799
Transcribed Image Text:The table lists the values of the torque curve. The nominal speed of the engine is to be 250 rad/s. 1) 0 degerees T, Nm 0 15 30 45 60 75 90 105 120 135 150 165 180 0 2800 2090 2430 2160 Dave 1840 come 1590 www 1210 1066 803 532 184 0 0 degerees T, Nm Relation between torque and crank angle for a one-cylinder, four-stroke-cycle internal combustion engine. 195 210 225 240 255 270 285 7 300 315 330 345 360 Crank torque T -107 -206 -260 -323 -310 -242 -126 -8 89 125 85 0 180 0 degerees T, Nm 375 390 405 420 435 450 465 480 495 510 525 540 360° Crank angle -85 -125 -89 8 126 242 310 323 280 206 107 0 0 degerees T, Nm 555 570 585 600 615 630 645 660 675 690 705 720 720° -107 -206 -292 -355 -371 -362 -312 -272 -274 -548 -760 0 2) 3) 4) Integrate the torque displacement function and find the energy that can be delivered to a load during the cycle. (3368J) Determine the mean torque. (268 Nm) The greatest energy fluctuation is approximately between 0 = 0° and 0 = 180° (see torque diagram and note that To-Tm). Using a largest coefficient of speed fluctuation C = 0,1 find a suitable value for the flywheel inertia. (0,6565) Find W₂ and w₁. (262,5 and 237,5) Theta T 0 15 30 45 60 75 90 105 120 135 150 165 180 0 2800 2090 2430 2160 1840 1590 1210 1066 803 532 184 0 Tavg rad E 1400 0.261799 366.5191 2445 0.261799 640.0995 2260 0.261799 591.6666 2295 0.261799 600.8296 2000 0.261799 523.5988 1715 0.261799 448.986 1400 0.261799 366.5191 1138 0.261799 297.9277 934.5 0.261799 244.6515 667.5 0.261799 174.7511 358 0.261799 93.72418 92 0.261799 24.08554 0 0.261799
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