{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

p0404

# p0404 - U o(1 − x/x o − 3 − 1/x o = 2 U 2 o x o(1 −...

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

4.4. CHAPTER 4, PROBLEM 4 487 4.4 Chapter 4, Problem 4 Problem: The axial velocity in a conical nozzle is u = U o (1 x/x o ) 2 i where U o and x o are constant velocity and length scales, respectively. (a) If we treat the flow as being one dimensional, what is the acceleration, a ? (b) If the magnitude of the acceleration at x = x 2 is 1/32 of its value at x = x 1 , what is the ratio of the velocity at x = x 2 to its value at x = x 1 ? Solution: Because the given velocity is one dimensional and independent of time, a = d u dt = u t + u · u = u du dx i (a) So, letting a = a i , the acceleration is a = U o (1 x/x o ) 2 d dx ± U o (1 x/x o ) 2 = = U o (1 x/x o ) 2 ( 2 U
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: U o ) (1 − x/x o ) − 3 ( − 1 /x o ) = 2 U 2 o x o (1 − x/x o ) − 5 Therefore, the acceleration vector for this flow is a = 2 U 2 o x o (1 − x/x o ) − 5 i (b) First, note that (1 − x/x o ) − 1 = ² u U o Thus, the acceleration can be rewritten as a = 2 U 2 o x o w u U o W 5 / 2 So, we can solve for the velocity ratio as follows. w u 2 u 1 W 5 / 2 = a 2 a 1 = ⇒ u 2 u 1 = w a 2 a 1 W 2 / 5 Thus, if a 2 /a 1 = 1 / 32 , the velocity ratio is u 2 u 1 = w 1 32 W 2 / 5 = 1 4...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online