# hw 38 - homework 38 – HETTIGER CHRISTOF – Due May 3...

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Unformatted text preview: homework 38 – HETTIGER, CHRISTOF – Due: May 3 2008, 4:00 am 1 Question 1, chap 18, sect 6. part 1 of 2 10 points Given: The viscosity is negligible. Atmo- spheric pressure is 101300 Pa. Water flows at speed of 4 . 9 m / s through a horizontal pipe of diameter 3 . 3 cm. The gauge pressure P 1 of the water in the pipe is 1 . 4 atm. A short segment of the pipe is constricted to a smaller diameter of 2 . 4 cm. P 1 4 . 9 m / s 3 . 3 cm 1 . 4 atm P 2 v 2 2 . 4 cm What is the speed v 2 of the water flowing through the constricted segment? Correct answer: 9 . 26406 m / s (tolerance ± 1 %). Explanation: Let : P atm = 101300 Pa , ρ = 1000 kg / m 3 , η = 0 , v 1 = 4 . 9 m / s , d 1 = 3 . 3 cm , d 2 = 2 . 4 cm , and P gauge 1 = 1 . 4 atm . To a good approximation, water is incom- pressible (at ordinary pressures) and thus must obey the continuity equation F = A v = π parenleftbigg d 2 parenrightbigg 2 v = constant . In other words, d 2 2 v 2 = d 2 1 v 1 , and hence the speed of the water flow through the constricted segment of the pipe is v 2 = v 1 d 2 1 d 2 2 = (4 . 9 m / s) (3 . 3 cm) 2 (2 . 4 cm) 2 = 9 . 26406 m / s . Question 2, chap 18, sect 6. part 2 of 2 10 points What is the gauge pressure of the water flowing through the constricted segment? Correct answer: 1 . 0949 atm (tolerance ± 1 %). Explanation: In the absence of viscosity, the water pres- sure and speed obey the Bernoulli equation P + ρ g h + 1 2 ρ v 2 = constant , which for a horizontal pipe ( h 1 = h 2 ) gives us P 2 + 1 2 ρ v 2 2 = P 1 + 1 2 ρ v 2 1 , or P 2 = P 1 − 1 2 ρ ( v 2 2 − v 2 1 ) . Strictly speaking, the pressures P 1 and P 2 in the Bernoulli equation are the absolute pres- sures rather than gauge pressures. However, since the air pressure outside the pipe is con- stant, the difference P gauge 2 − P gauge 1 is the same as the difference between the absolute pressures P 2 − P 1 . Therefore, P gauge 2 = P gauge 1 − 1 2 ρ ( v 2 2 − v 2 1 ) = (141820 Pa) − parenleftbigg 1000 kg / m 3 2 parenrightbigg × bracketleftbig (9 . 26406 m / s) 2 − (4 . 9 m / s) 2 bracketrightbig = [(141820 Pa) − (30906 . 4 Pa)] × (9 . 87167 × 10- 6 atm / Pa) = 1 . 0949 atm . Dimensional analysis of ρ v 2 kg m 3 · parenleftBig m s parenrightBig 2 ≡ Pa · 1 atm 1 . 013 × 10 5 Pa ≡ atm Question 3, chap 18, sect 6....
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hw 38 - homework 38 – HETTIGER CHRISTOF – Due May 3...

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