p0192 - μL = u m R 2 = ⇒ Q = π 2 u m R 2 The given...

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100 CHAPTER 1. INTRODUCTION 1.92 Chapter 1, Problem 92 1.92(a): As stated, the volume flow rate, Q ,isg ivenby Q =2 π 8 R 0 u ( r ) rdr =2 π p 1 p 2 4 μL 8 R 0 p R 2 r 2 Q rdr =2 π ( p 1 p 2 ) R 4 4 μL 8 R 0 X 1 r 2 R 2 ~ w r R W d w r R W Now, let η = r/R .Th en , Q = π ( p 1 p 2 ) R 4 2 μL 8 1 0 p 1 η 2 Q η d η = π ( p 1 p 2 ) R 4 2 μL } 1 2 η 2 1 4 η 4 ] η =1 η =0 ± 1 =1 / 4 Therefore, for a given pressure difference, the volume flux is Q = π ( p 1 p 2 ) 8 μL R 4 which is clearly proportional to R 4 . 1.92(b): We know from the pipe-flow solution given in the text that ( p 1 p
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Unformatted text preview: μL = u m R 2 = ⇒ Q = π 2 u m R 2 The given conditions are u m = 0.3 m/sec = 30 cm/sec and R = 1.5 cm. Thus, the volume flow rate is Q = π 2 (30 cm / sec)(1 . 5 cm) 2 = 106 cm 3 sec Therefore, the time required for 100 cm 3 of blood to pass through the aorta is t = 100 cm 3 Q = 100 cm 3 106 cm 3 / sec = 0 . 94 sec...
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This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

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