Answer:
If the turbulent velocity profile in a pipe of diameter 0.6 m may be approximated by u/U=(y/R)^(1/7), where u is in m/s and y is in m and 0.15 m from the pipe.
Explanation:
hope it helps
Answer:
you know what sounds rlly nice? first have you seent he new corvettes? next u should here a nisan gtr. (like tanner foxes)
Explanation:
Cut that photo by
1. Left click your mouse on the photo
2. Click cut
Then enter the file where you want to transfer and press
1. ctrl+v
Answer:
vec(a) = 16 i + 16 j
mag(a) = 22.63 ft/s^2
Explanation:
Given,
- The two components of velocity are given for fluid flow:
u = 4*y ft/s
v = 4*x ft/s
Find:
What is the time rate of change of the velocity vector V (i.e., the acceleration vector) for a fluid particle at x = 1 ft. and y = 1 ft. at time t = 1 second?
Solution:
- The rate of change of velocity is given to be acceleration. We will take derivative of each components of velocity with respect to time t:
a_x = du / dt
a_x = 4*dy/dt
a_y = dv/dt
a_y = 4*dx/dt
- The expressions dx/dt is the velocity component u and dy/dt is the velocity component v:
a_x = 4*(4*y) = 16y
a_y = 4*(4*x) = 16x
- The acceleration vector can be expressed by:
vec(a) = 16y i + 16x j
- Evaluate vector (a) at x = 1 and y = 1:
vec(a) = 16*1 i + 16*1 j = 16 i + 16 j
- The magnitude of acceleration is given by:
mag(a) = sqrt ( a^2_x + a^2_y )
mag(a) = sqrt ( 16^2 + 16^2 )
mag(a) = 22.63 ft/s^2
