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4) Depreciation per annum using straight line = (Cost - Residual Value)/Useful Life

= ($40,000,000 - $40,000)/6 years = $6,660,000 per year

Depreciation Schedule using the straight line method (Amounts in $)

DateAsset Cost (A)Depreciation for the year (B)Accumulated Depreciation (C)Book Value (A-C)
01/01/1940,000,00040,000,000
12/31/1940,000,0006,660,0006,660,00033,340,000
12/31/2040,000,0006,660,00013,320,00026,680,000
12/31/2140,000,0006,660,00019,980,00020,020,000
12/31/2240,000,0006,660,00026,640,00013,360,000
12/31/2340,000,0006,660,00033,300,0006,700,000
12/31/2440,000,0006,660,00039,960,00040,000
Total39,960,000

5) Depreciation per hour = (Cost - Residual Value)/Useful Life in hours

= ($40,000,000 - $40,000)/24,000 hours = $1,665 per hour

Depreciation Schedule using the straight line method (Amounts in $)

DateAsset Cost (A)Depreciation for the year (B)Accumulated Depreciation (C)Book Value (A-C)
01/01/1940,000,00040,000,000
12/31/1940,000,000($1,665*4,500 hrs) = 7,492,5007,492,50032,507,500
12/31/2040,000,000($1,665*6,000 hrs) = 9,990,00017,482,50022,517,500
12/31/2140,000,000($1,665*5,200 hrs) = 8,658,00026,140,50013,859,500
12/31/2240,000,000($1,665*4,300 hrs) = 7,159,50033,300,0006,700,000
12/31/2340,000,000($1,665*2,000 hrs) = 3,330,00036,630,0003,370,000
12/31/2440,000,000($1,665*2,000 hrs) = 3,330,00039,960,00040,000
Total39,960,000

6) Depreciation rate using double declining balance = (1/Useful Life in years)*2

= (1/6 yrs)*2 = 0.333 i.e. 1/3

Depreciation Schedule using the Double Declining method (Amounts in $)

DateOpening Book value (A)Depreciation for the year (B = A*1/3)Accumulated Depreciation (C)Closing Book Value (A-C)
12/31/1940,000,00013,333,33313,333,33326,666,667
12/31/2026,666,6678,888,88922,222,22217,777,778
12/31/2117,777,7785,925,92628,148,14811,851,852
12/31/2211,851,8523,950,61732,098,7657,901,235
12/31/237,901,2352,633,74534,732,5105,267,490
12/31/245,267,4901,755,83036,488,3403,511,660
Total36,488,340

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QUESTION 6 (a) The bar shown in Figure Q2(a) is subjected to tensile load of 150 Kn. If the stress in the middle portions is limited to 160 N/mm², determine the diameter of the middle portion. Find also the length of the middle portion if the total elongation of the bar is to be 0.25 mm. E E = 2.0 + 105N/mm². (12 marks) 150 KN 10 cm DIA 10 cm DIA 150 KN 45 cm Figure Q6(a) (b) A brass bar, having cross-section area of 900 mm², is subjected to axial forces as shown in Figure Q2(b), in which AB = 0.6 m, BC = 0.8 m, and CD = 1.0 m. Find the total elongation of the bar. E = 1.0 + 105N/mm2 . (8 marks) Page 4 of 5 B D 40 KN 70 KN 20 KN 10 KN Figure Q6(b) (TOTAL = 20 MARKS)

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Question: A 250-V, 4-pole, wave-wound d.c. series motor has 782 conductors on itsarmature. It has armature and series field resistance of 0.75 ohm. The motor takesa current of 40 A. Estimate its speed and gross torque developed if it has a flux per pole of 25 mWb Answer: Step 1 Mechanical Engineering homework question answer, step 1, image 1 Mechanical Engineering homework question answer, step 1, image 2 Step 2 Mechanical Engineering homework question answer, step 2, image 1 Step 3 Mechanical Engineering homework question answer, step 3, image 1

Question: A 250-V, 4-pole, wave-wound d.c. series motor has 782 conductors on itsarmature. It has armature and series field resistance of 0.75 ohm. The motor takesa current of 40 A. Estimate its speed and gross torque developed if it has a flux per pole of 25 mWb Answer: Step 1 Step 2 Step 3

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