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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
(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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

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.

Ask a homework question - tutors are online