chap10 - PROBLEM 10.1 KNOWN: Flow of air through a pipe...

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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 22 r ji j i QU r d rU r r ππ = =≈∆ 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 Qj j SQ Q = ⎡⎤ =− ⎢⎥ ⎣⎦ = 23 cm 3 /s and /3 Q Q SS = = 13.3 cm 3 /s with ν = 2
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Then, t 2,95 = 4.303. With P = Q S , tP=57.2 cm 3 /s and with B = 0.02 Q = 88.5 cm 3 /s, then u Q = [88.5 2 + 57.2 2 ] 1/2 = 105.3 cm 3 /s. So, Q' = 4423 ± 105.3 cm 3 /s (95%) COMMENT Sources of systematic error include: instrument errors, errors in the control of operating conditions for all three traverses, and the approximation for the integral expression at the start of the solution.
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PROBLEM 10.3 KNOWN: Manometer measuring the pressure drop of flowing water H = 10.16 cm Hg d 1 = 5.1 cm γ = 9800 N/m 3 (water) S m = 13.57 (mercury) so that mm S = FIND: p 1 – p 2 ASSUMPTIONS: p 1 and p 2 taps are located along the same horizontal datum line. SOLUTION Applying the hydrostatic equation between points 1 and 2 yields p 1 + L + H m - (L+H) = p 2 p 1 – p 2 = H( m - ) = H(13.57 - ) = H 12.57 = (0.1016 m)(12.57)(9800 N/m 2 )(1 Pa/N/m 2 ) = 12.516 kPa
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PROBLEM 10.4 KNOWN: Air flow through orifice meter p 1 – p 2 = 69 kPa d 1 = 25.4 cm T = 32 o C FIND: H SOLUTION p 1 – p 2 = γ H H = (p 1 – p 2 )/ with = 9750 N/m 3 (Appendix) H = (69,000 N/m 2 )/9750 N/m 3 = 7.041 m H 2 O = 704.1 cm H 2 O
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PROBLEM 10.5 KNOWN: Water flow through orifice meter using flange taps. d 1 = 3 in. T = 60 o F p 1 = 100 psi p 2 = 76 psi d o = 1.5 in R O2 = 48.3 ft-lb/lb m - o R FIND: Is Y < 1? ASSUMPTIONS:& Steady flow SOLUTION Y = f(k, 1 ,/ p p β ) For a diatomic gas, such as O 2 , k = 1.4. 1 / o dd = = 0.5 1 / p p = 0.24 Using Figure 10.7, Y = 0.92 There is an 8% reduction in flow rate due to compressibility effects. COMMENT Aside from the usual sources of error in an measurement, data reduction errors enter into the obstruction meter relations from the assumed values of the various coefficients and from the ability to read these values from the tables and charts. Normally, these are treated as systematic errors unless additional information about how they were estimated is known.
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PROBLEM 10.6
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chap10 - PROBLEM 10.1 KNOWN: Flow of air through a pipe...

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