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Work Done Problem (20 分)

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A conical tank has a diameter of 8 feet and is 10 feet deep. If the tank is filled to a depth of 6 feet with water of density 62.4 lbs/ft^3, how much work is required to pump the water over the top? 有冇高手可以幫幫手解答？thx 更新: umm~我諗呢條數係要用integration 的~

最佳解答:

With reference to the following diagram: 圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Feb08/Crazyint1-1.jpg From the diagram, by similar triangle, we have: r = 0.4h Then, the volume of the disc-shaped water with thickness dh is: dV = πr2dh (in ft3) with weight equal to: dW = 62.4πr2dh (in lbs) = 62.4π(0.4h)2dh = 9.984πh2dh = 4.538πh2dh (in kg) And when bringing this disc-shaped water from height h to the top of the tank, work done is equal to: 4.538πh2dh × g × (10 - h) = 45.38π(10 - h)h2dh (Taking g = 10) Hence, the total work done can be found by integrating with respect to h from h = 0 to h = 6: Work done = ∫(h = 0 → 6)45.38π(10 - h)h2dh = 45.38π∫(h = 0 → 6)(10h2 - h3)dh = 45.38π [10h3/3 - h4/4](h = 0 → 6) = 45.38π (720 - 324) = 17970π (in kg m s-2 ft) To convert back the unit to J, we use the relation 1 ft = 0.3048 m which gives the work done equal to: 17970π × 0.3048 = 5477π J

其他解答:7638E7481407D16B

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