chap10 - PROBLEM 10.1 KNOWN Flow of air through a pipe U(r...

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PROBLEM 10.1 KNOWN: Flow of air through a pipe U(r) = 25[1 - (r/r 1 ) 2 ] cm/s p = 1 bar abs = 100,000 N/m 2 abs T = 5 o C = 278 K d 1 = 2r 1 = 5 cm FIND: mass flow rate ASSUMPTIONS: Steady, incompressible, axisymmetric flow of a perfect gas. SOLUTION Conservation of mass gives which for steady, incompressible, axisymmetric flow becomes The velocity can be written in vector form as resulting in For a perfect gas, the density can be estimated by ρ = p/RT = 1.16 kg/m 3 so that (with 1 m = 100 cm): 0 dA n ˆ U ρ ρ d t CV CS = + ∫∫∫ ∫∫ ∫ ∫ = 2 π 0 r1 0 ρ U(r)rdrd θ m z e 2 1 ) r r 25(1 U = /s cm 39.3 ρ ρπ r 2 25 θ d dr r ) ) r r ( 25(1 ρ m 3 2 1 2 π 2 0 r1 0 1 = = = ∫ ∫ kg/hr 0.16 kg/s 10 4.5 m 5 = × =
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PROBLEM 10.2 KNOWN: Air flow through a pipe diameter = 2r 1 = 20 cm N = 5 velocity measurements per cross-sectional traverse M = 3 cross-sectional traverses B Q /Q = ±2% FIND: Q' ASSUMPTIONS: Steady, incompressible flow of a perfect gas. Flow rate remains perfectly controlled (constant) during all measurements (equivalent to the assumption that all measurements are taken simultaneously). SOLUTION The flow rate along each traverse line (j = 1, 2, 3) can be approximated by 1 5 1 0 2 2 r j ij i Q Urdr U r r π π = = i = 1,2,3,4,5 where r = 2 cm and j = 1, 2, 3. Then, [ ] 1 2 (25.31)(1)(2) (22.48)(3)(2) (21.66)(5)(2) (15.24)(7)(2) (5.12)(9)(2) Q π = + + + + = 4446 cm 3 /s Similarly, Q 2 = 4421 cm 3 /s, Q 3 = 4400 cm 3 /s. The mean flow rate is Q = (1/3)[4446 + 4421 + 4400]cm 3 /s = 4423 cm 3 /s with standard deviation, 1/ 2 3 2 1 1 ( ) 2 Q j j S Q Q = = = 23 cm 3 /s and 1/ 2 /3 Q Q S S = = 13.3 cm 3 /s with ν = 2
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