In Chap. 9(Example 9-15), we generated an “exact” solution of the Navier-Stokes equation for fully developed Couette flow between two horizonatal flat plates (Fig. 10-15), with gravity acting in the negative z- direction (into the page of Fig. 10-15). We used the actual pressure in that example. Repeat the soluation for the x- component of velocity u and pressure P, but use the modified pressure in your equation. The pressure is P 0 at z =0. Show that you get the same result as previoudly. Discuss. Answer: u = V y / h , P = P 0 − ρ g z FIGURE P10-15
In Chap. 9(Example 9-15), we generated an “exact” solution of the Navier-Stokes equation for fully developed Couette flow between two horizonatal flat plates (Fig. 10-15), with gravity acting in the negative z- direction (into the page of Fig. 10-15). We used the actual pressure in that example. Repeat the soluation for the x- component of velocity u and pressure P, but use the modified pressure in your equation. The pressure is P 0 at z =0. Show that you get the same result as previoudly. Discuss. Answer: u = V y / h , P = P 0 − ρ g z FIGURE P10-15
In Chap. 9(Example 9-15), we generated an “exact” solution of the Navier-Stokes equation for fully developed Couette flow between two horizonatal flat plates (Fig. 10-15), with gravity acting in the negative z-direction (into the page of Fig. 10-15). We used the actual pressure in that example. Repeat the soluation for the x-component of velocity u and pressure P, but use the modified pressure in your equation. The pressure is
P
0
at z=0. Show that you get the same result as previoudly. Discuss. Answer:
u
=
V
y
/
h
,
P
=
P
0
−
ρ
g
z
I need handwritten solution with sketches for each
Given answers to be: i) 14.65 kN; 6.16 kN; 8.46 kN ii) 8.63 kN; 9.88 kN iii) Bearing 6315 for B1 & B2, or Bearing 6215 for B1
(b)
A steel 'hot rolled structural hollow section' column of length 5.75 m, has
the cross-section shown in Figure Q.5(b) and supports a load of 750 kN.
During service, it is subjected to axial compression loading where one end
of the column is effectively restrained in position and direction (fixed) and
the other is effectively held in position but not in direction (pinned).
i)
Given that the steel has a design strength of 275 MN/m², determine
the load factor for the structural member based upon the BS5950
design approach using Datasheet Q.5(b).
[11]
ii)
Determine the axial load that can be supported by the column
using the Rankine-Gordon formula, given that the yield strength of
the material is 280 MN/m² and the constant *a* is 1/30000.
[6]
300
600
2-300 mm
wide x 5 mm
thick plates.
Figure Q.5(b)
L=5.75m
Pinned
Fixed
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