G= Plot both G1 vs = B, flfs + 1+√√f/2(g Pr0.57-B) 2/3 1+B, G (1+B)1.07+12.7 √,/2 (Pr2³-1)] E and G2 vs F on the same graph. The solution is where the two lines intersect. The solution should occur at G = 1.23 and 長 √F (1) 8.82 3. With √ √2/f known, calculate St from equation 9.25 of the book using n = 0.57. 4. Solve for A/A, from Eq. 5b (from the appendix of the Chapter 8 problem) evaluated at Q/Qs = 1. Note that ß= ß, for the rib roughness because the heat transfer coefficient is based on the smooth tube area, i.e., A/L = A/L. 5. The associated e/D is calculated from e* = (e/D)Re√√f/2 6. The remaining parameters (N/Ns, L/Ls) are easily calculated.
Design Against Fluctuating Loads
Machine elements are subjected to varieties of loads, some components are subjected to static loads, while some machine components are subjected to fluctuating loads, whose load magnitude tends to fluctuate. The components of a machine, when rotating at a high speed, are subjected to a high degree of load, which fluctuates from a high value to a low value. For the machine elements under the action of static loads, static failure theories are applied to know the safe and hazardous working conditions and regions. However, most of the machine elements are subjected to variable or fluctuating stresses, due to the nature of load that fluctuates from high magnitude to low magnitude. Also, the nature of the loads is repetitive. For instance, shafts, bearings, cams and followers, and so on.
Design Against Fluctuating Load
Stress is defined as force per unit area. When there is localization of huge stresses in mechanical components, due to irregularities present in components and sudden changes in cross-section is known as stress concentration. For example, groves, keyways, screw threads, oil holes, splines etc. are irregularities.
Coding assistance.
fs= .00743
![G" =
Plot both G1 vs
B flfs +.
1+ B. G*
√B
0.57
1+√√f/2(g Prº. B)
(1+B) 1.07 +12.7√√/2 (P-1)
and G2 vs
√√₁
on the same graph. The solution is where the
1√ √=8
two lines intersect. The solution should occur at G* = 1.23 and
(1)
= 8.82
3. With √√2/f known, calculate St from equation 9.25 of the book using n = 0.57.
4.
Solve for A/A, from Eq. 5b (from the appendix of the Chapter 8 problem)
evaluated at Q/Qs = 1. Note that ß= B. for the rib roughness because the heat
transfer coefficient is based on the smooth tube area, i.e., A/L = As/L.
5. The associated e/D is calculated from e+ = (e/D)Re√√f/2
6. The remaining parameters (N/Ns, L/Ls) are easily calculated.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff0e69105-edad-4e40-a986-7d2af8f1b9bd%2F8fbcf618-f8db-41c5-ba47-d86cb2987f45%2F6uttaxx_processed.jpeg&w=3840&q=75)
![Calculate the
material savings for the VG-1 case offered by two-dimensional rib roughness. For this
roughness, the geometrical parameters are e/D, ple, tle, a. Calculate the material
savings for a = 90° with p/e = 10, t/e = 0.5, while using the smooth tube operating
parameters as used in the Chapter 8 problem: Res = 12320, Pr = 62, oil, etc.
Because fand St are functions of e/D and et (or Re), the PEC equation for A/As
implicitly contains two independent variables: e/D and et. Therefore, a range of
solutions exist. Corresponding to each arbitrary value of et is a particular value of Re
(or Gs/G) required to satisfy the constraints (P/Ps = Q/Qs = 1). And, with et and Re
defined, e/D is known and calculated from eq. (9.9) of the book: (e/D=
et/ (Re√√f/2)). Therefore, one should calculate A/As for a range of e* (or Re) to
define the minimum A/As. Webb and Eckert (1972) have calculated A/As vs. Gs/G for
ple = 10, a = 90° roughness over a range of Re (or Gs/G) and Pr and suggest that e* =
20 is a reasonably good design specification for the VG-1 criterion. However, if the
A/As vs. Gs/G curve shows a fairly flat minimum, the required flow frontal area will
increase with increasing e* (or e/D).
For this problem, we will select et = 20 for our design specification, which fixes the
g (et) and B(e) contained in the correlations for fand St. These values are: a = 90°:
at e = 20, B = 4.2, g = 11.
Use the following iterative solution outlined below and given in the notes:
1. Assume G* = G/G. So Re = Res/ G*
2. Solve for G* from eq. (22) of Webb and Eckert (1972):
G
=
Re
2e+
V.₁
exp
Devise a computer code to generate an array for
in increments of 0.01. Use the
√F
-√21f- -3.75
2.5
array Gl. Use the G1 array and the
√
with values from 7 to 11
array to evaluate eq. (22) for G*. This will
give you an array of G* for values associated with the
F
1₁
array. Call this G*
array to evaluate a new G using
eq. (1) of your roughness class notes given below (i.e., eq. (17) of Webb and
Eckert, 1972) with r = Bs, (call this array G2):
¹Please use the Filonenko and Petuknov equations to calculate fs and Sts, respectively, for this problem.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff0e69105-edad-4e40-a986-7d2af8f1b9bd%2F8fbcf618-f8db-41c5-ba47-d86cb2987f45%2Fc0qk8xt_processed.jpeg&w=3840&q=75)
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