ch14-p064

# ch14-p064 - 2 600 m/s = 6.12 m 3/s If the more accurate...

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(b) Using Bernoulli’s equation (in the form Eq. 14-30) we find the pressure difference may be written in the form a straight line: mx + b where x is inverse-area-squared (the horizontal axis in the graph), m is the slope, and b is the intercept (seen to be –300 kN/m 2 ). Specifically, Eq. 14-30 predicts that b should be 1 2 ρ v 2 2 . Thus, with ρ = 1000 kg/m 3 we obtain v 2 = 600 m/s. Then the volume flow rate (see Eq. 14-24) is R = A 2 v 2 = (0.25 m
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Unformatted text preview: 2 )( 600 m/s) = 6.12 m 3 /s. If the more accurate value (see Table 14-1) ρ = 998 kg/m 3 is used, then the answer is 6.13 m 3 /s. 64. (a) We note (from the graph) that the pressures are equal when the value of inverse-area-squared is 16 (in SI units). This is the point at which the areas of the two pipe sections are equal. Thus, if A 1 = 1/ 16 when the pressure difference is zero, then A 2 is 0.25 m 2 ....
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## This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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