p0746 - 7.46. CHAPTER 7, PROBLEM 46 1009 7.46 Chapter 7,...

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Unformatted text preview: 7.46. CHAPTER 7, PROBLEM 46 1009 7.46 Chapter 7, Problem 46 Problem: What head, hp , must be supplied by the pump in order to pump water of density ρ at a rate m from the lower to the upper reservoir? Assume α1 = α2 = 1 and ˙ 2L that the head loss in a pipe of length L and diameter D is given by hL = 0.028 Ug D . 2 For the two pipes, we have D1 = D, D2 = D/2, L1 /D1 = 20 and L2 /D2 = 300. ˙ Express your solution for hp in terms of ρ, ∆z , m, D and gravitational acceleration, g. z = zo + ∆z z = zo j 0 P pe 1 j 2 P pe 2 ................................................................. .......................................................................................................................................................... ....... ........................................................................................................................................................ .................................................................................................................... ...................................................................................................... ......... .................................................................................................. .. ......................................................................................................... ... ............................................................... .... .... .................................................................................................................... . .................................................................................................................................................. ............................................................... .. ....... .................................................................................................................................................. .............................................................. ....... ................................................................................................................ ......... ......... ............................................................. ... ..................................................................................................... ...... ..... . . ........................................................................................................................................................... .... . .................................................................................................1............................................... ........... ....................................................................................................................... . . ..... . . ................ .................................................................................................................. .................................................................................................... .... ................................................................................................................ ................................... .. . ................................................................................................................ ................................... ........................................................................................................... .................................. ρ z=0 . . .. ............................................................................................. ........... ............. .... ............. .... ...... .... ................. .... ............. .... .......... .... .................................................................................................. .... .. . ... ..... ..... .... . ...... ........ .. ........ ..... ....... ........ ...... ........ .... ........ ........ ........ ........ . ........ ........ . .... ....... ........ .... . ........ ........ ........ ........ .. ....... ...... .... . .... ..... | j Pump Solution: Mass Principle: First, note that the mass principle tells us that π π m = ρ U1 D2 = ρ U2 ˙ 4 4 D 2 2 Therefore, the velocities are U1 = 4m ˙ πρD2 and U2 = 4U1 = 16m ˙ πρD2 So we have 2 2 8m2 ˙ 128m2 ˙ U2 U1 = 2 2 4 and = 22 4 2g π ρ gD 2g π ρ gD Energy Principle: Applying the approximate energy equation between the left reservoir free surface (Point 0) and the pipe outlet (Point 2), we have u2 u2 p0 p2 + α0 0 + z0 + hp = + α2 2 + z2 + hL ρg 2g ρg 2g Now, the pressure is atmospheric at both Points 0 and 2 so that p0 = p2 . Also, u0 ≈ 0 and u2 = U2 , while z0 = zo and z2 = zo + ∆z . Finally, we assume α ≈ 1.0. Hence, zo + hp = 2 U2 + (zo + ∆z ) + hL 2g so that the head supplied by the pump is hp = 2 U2 + ∆z + hL 2g 1010 CHAPTER 7. ENERGY PRINCIPLE 2 Now, we are given that the head loss in the pipe is hL = 0.028 Ug 2 head loss, we have contributions from both pipes. Hence, hL = 0.028 L . D To determine the 2 2 16U1 . U1 L1 U 2 L2 U2 U2 + 0.028 2 = 0.028 20 1 + 300 = 135 1 2g D1 2g D2 2g 2g 2g So, the head supplied by the pump is hp = ∆z + 2 U2 U2 U2 U2 + hL = ∆z + 16 1 + 135 1 = ∆z + 151 1 2g 2g 2g 2g Finally, substituting for U1 from above, 8m2 ˙ 1208m2 ˙ hp = ∆z + 151 2 2 4 = ∆z + 2 2 4 π ρ gD π ρ gD ...
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