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PROBLEM 8.1 KNOWN: Flowrate and temperature of water in fully developed flow through a tube of prescribed diameter. FIND: Maximum velocity and pressure gradient. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Isothermal flow. PROPERTIES: Table A-6 , Water (300K): = 998 kg/m 3 , = 855 10-6 N s/m 2 . ANALYSIS: From Eq. 8.6, ( 29 D 6 4 m 4 0.01 kg/s R e 596. D 0.025m 855 1 kg m/s p m p- = = = & Hence the flow is laminar and the velocity profile is given by Eq. 8.15, ( 29 ( 29 2 o m u r 2 1 r/ r . u =- The maximum velocity is therefore at r = 0, the centerline, where ( 29 m u 0 2 u . = From Eq. 8.5 ( 29 m 2 2 3 m 4 0.01 kg/s u 0.020 m/s, D / 4 998 kg/ m 0.025m rp p = = = & hence ( 29 u 0 0.041 m/s. = Combining Eqs. 8.16 and 8.19, the pressure gradient is 2 m D d p 6 4 u d x R e 2D r = - ( 29 2 3 2 2 998 kg/m 0.020 m/s d p 64 0.86 kg/ m s d x 59 6 2 0.025 m =- =- 2-5 dp 0.86N/ m m 0.86 10 bar/m. dx =- =- < PROBLEM 8.2 KNOWN: Temperature and mean velocity of water flow through a cast iron pipe of prescribed length and diameter. FIND: Pressure drop. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Fully developed flow, (3) Constant properties. PROPERTIES: Table A-6 , Water (300K): = 997 kg/m 3 , = 855 10-6 N s/m 2 . ANALYSIS: From Eq. 8.22, the pressure drop is 2 m u p f L. 2D r = With 3 4 m D- 6 2 u D 997 kg/ m 0.2 m/s 0.15 m R e 3.50 10 855 10 N s/m r m = = = the flow is turbulent and with e = 2.6 10-4 m for cast iron (see Fig. 8.3), it follows that e/D = 1.73 10-3 and f 0.027. Hence, ( 29 ( 29 2 3 997 kg/ m 0.2 m/s p 0.02 7 600m 2 0.15 m = 2 2 p 2154 kg/ s m 2154 N/m = = p 0.0215 bar. = < COMMENTS: For the prescribed geometry, L/D = (600/0.15) = 4000 >> (x fd,h /D) turb 10, and the assumption of fully developed flow throughout the pipe is justified. PROBLEM 8.3 KNOWN: Temperature and velocity of water flow in a pipe of prescribed dimensions.... View Full Document

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