chpt 3 suggested qns

chpt 3 suggested qns - Chapter 3 Integral Relations for a...

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Chapter 3 Integral Relations for a Control Volume P. 3.6 Solution: Let the slot width be b into the paper. Then the volume flow from Eq. (3.7) is +L 1/2 L Q udA [2g(h z)] bdz . Ans   3/2 3/2 2b (2g)[(h L) (h L) ] 3  In the limit of this formula reduces to L<<h, . Ans Q (2Lb) (2gh) P3.7 Solution: If the control volume surrounds the tank and cuts through the exit flow, ) .( ) ( ) ( fo Solve ) ( ) ( ) ( 0 | tank tank tank tank tank tank a Ans AV dt d r AV dt d m dt d dt dm out out out system   (b) For the given data, we calculate ) .( ) 35 . 0 )( 6 / ( ) / 360 ]( ) 005 . 0 )( 4 / )[( / 5 . 2 ( 3 2 3 tank b Ans m s m m m kg dt d s kg/m 0.79 3 P. 3.8 Solution: (a) For steady flow we have Q1 + Q2 + Q3 Q4, or 11 2 2 33 4 4 VA VA VA  (1) Since 0.2Q3 0.1Q4, and Q4 (120 m 3 /h)(1 h/3600 s) 0.0333 m 3 /s, 3 n s . 4 3 2 3 (0.0333 m /s) (b) 2 (0.06 ) 2 Q A 5.89 m/s
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2 Solutions Manual Fluid Mechanics, Seventh Edition Substituting into (1), 22 2 1 (0.04 ) (5) (0.05 ) (5.89) (0.06 ) 0.0333 (a) 44 4 VA      1 V5 . 4 5 m / s n s . From mass conservation, Q4 V4A4 32 4 (0.0333 m /s) V ( )(0.06 )/4 (c) Ans. 4 . 2 4 m / s 3.10 Solution: (a) For a suction velocity of vw 0.15 m/s, and a cylindrical suction surface area, 2 w A 2 (0.04)(1.2) 0.3016 m  1w QQ Q 2 2 (12)( )(0.08 )/4 (0.15)(0.3016) V ( )(0.08 )/4 (a) Ans.  2 V3 m / s (b) For an injection velocity into the pipe, vw 0.10 m/s, Q 1 + Q w = Q 2 , or: 2 (12)( )(0.08 )/4 (0.10)(0.3016) V ( )(0.08 )/4 (b) Ans.  2 V=1 8 m / s (c) Setting the outflow V2 to 9 m/s, the wall suction velocity is, w (12)( )(0.08 )/4 (v )(0.3016) (9)( )(0.08 )/4 w v0 . 0 5 m / s 5 c m / s out P3.11
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3 Chapter 3 Integral Relations for a Control Volume Solution: ( a ) Convert 25 ft 3 /min to 25/60 = 0.417 ft 3 /s. Then the inlet velocity is 3 0.417 / 12.0 0.3048 .( ) (1/12 )(5/12 ) inlet inlet ft s Qf t m VA Af t f t s f t s  3.66 m n s a ( b ) At sea level, air = 1.2255 kg/m 3 . Convert 25 ft 3 /min to 0.0118 m 3 /s. Then 3 3 (1.2255 )(0.0118 ) .( ) air air kg m kg mQ A n ss m 0.0145 s b _______________________________________________________________________ 3.12 Solution: For a control volume enclosing the tank and the portion of the pipe below the tank, 0 out in d dv m m dt     2 ()() out in dh RA V A V dt  0 2 4 998 (0.12 )(2.5 1.9) 0.0153 / , dh ms   2 4 998( )(0.75 ) 0.7/0.0153 dt Ans. ts     46
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4 Solutions Manual Fluid Mechanics, Seventh Edition 3.13 Solution: We could simply note that dh/dt is the same as the water velocity at the surface and use Q 1 = Q 2 , or, more instructive, approach it as a control volume problem. Let the control volume encompass the entire container. Then the mass relation is . )] 072 . 0 ( [ ) 3 20 ( : data the Introduce ) ( ) ( : 4 Cancel 0 4 4 : or , 4 | ) 4 ( ) ( 0 | 2 2 2 2 2 2 Ans s m cm cm V dt dh D D V V D dt dh D V D h D dt d m d dt d dt dm exit exit exit cone CV out system s m 3.2   P3.14 Solution: For a control volume enclosing the tank, 2 ) ,  213 () ( 4 CV dd d h dQ Q Q Q Q Q dt dt    . (a) solve Ans 132 2 QQQ dh dt (d/ 4 ) If h is constant, then 22 2 0.01 (0.05) (3.0) 0.0159 (0.07) , 44 QQ Q V  2 . (b) solve Ans V4 . 1 3 m / s
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P3.16 Solution: For the given control volume and incompressible flow, we obtain 3  
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This note was uploaded on 02/20/2011 for the course MEC 516 taught by Professor Ziad during the Fall '10 term at Ryerson.

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chpt 3 suggested qns - Chapter 3 Integral Relations for a...

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