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Unformatted text preview: DATE: November 25, 2011 DUE: November 30 (at or before 2:00pm), 2011 ENG102, Solution to homework set #9. 1. Look at Problem 4/71 on page 315 in the book. a. Solve the problem. The momentum balance in a short slice of time is: dp = mdv v dm u dm = P dt (1) where u is the speed of water flow. The mass of the tank with water is m . P = dp dt = ma m ′ ( u + v ) (2) where both v and u are positive quantities and m ′ = ρA ( v + u ) . P = dp dt = ma ρA ( u + v ) 2 = 76 . 5N (3) with a = . 4 m/s 2 and m = 130 kg. We now assume that a = 0 and that the tank is empty at t = 0 . b. Determine the necessary external force. No acceleration gives: P = ρA ( v + u ) 2 = 24 . 5N (4) c. How long time does it take to fill the tank with 80kg water? What distance has the tank traveled? dm dt = ρA ( v + u ) (5) ⇒ Δ t = Δ m ρA ( v + u ) = 11 . 43s (6) ⇒ Δ x = v Δ t = 22 . 86m (7) d. What is the power supplied to the tank? W = Pv = ρA ( v + u ) 2 v = 49J / s (8) e. What is the energy loss in scooping the 80kg water? Initially, the Δ m = 80 kg has the kinetic energy E k = 1 2 Δ mu 2 . We have supplied the work U = P Δ x = Δ mv ( u + v ) . The energy loss is therefore: Δ E = 1 2 Δ mu 2 bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright kinetic energy before + work done bracehtipdownleft bracehtipuprightbracehtipupleft bracehtipdownright Δ mv ( v + u ) 1 2 Δ mv 2 bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright kinetic energy after (9) = 1 2 Δ m ( u + v ) 2 = 490J DEPARTMENT OF MECHANICAL AND AERONAUTICAL ENGINEERING  UNIVERSITY OF CALIFORNIA  DAVIS, CALIFORNIA 95616 TEL: 530.752.5335 ENG102, Solution to Homework set #9 Page 2 m M m u u 1 2 u M u m m 1 2 Notice that m 2 ’s velocity after the collision should have been labeled u ′ ....
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This note was uploaded on 01/12/2012 for the course ENG 102 taught by Professor Eke during the Winter '08 term at UC Davis.
 Winter '08
 Eke

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