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grin007 [14]
3 years ago
7

Fluid Dynamics: How do I find gauge pressure of an air current at various points around a cylinder?A freestream air current of v

elocity 15m/s flows through a wind tunnel. The current hits a cylinder of diameter 19mm, and has a laminar boundary layer. The stagnation pressure in the test section of the wind tunnel is 1 atm. Determine the minimum and maximum gauge pressures of the flow around the cylinder.I'm given the equation Cp=2(P-Po)/(?V^2), where Cp is the pressure coefficient, Po is the upstream static pressure. My problem is largely confusion over what to do with the diameter of the cylinder and where it comes into play here.
Engineering
1 answer:
DaniilM [7]3 years ago
7 0

Answer:

The answer is as given in the explanation.

Explanation:

The 1st thing to notice is the assumptions required. Thus as the diameter of the cylinder and the wind tunnel are given such that the difference is of the orders of the magnitude thus the assumptions as given below are validated.

  1. Flow is entirely laminar, there's no boundary layer release.
  2. Flow is streamlined, ie, it follows the geometrical path imposed by the curvature.

By D'alembert's paradox, "The net pressure drag exerted on a circular cylinder that moves in an inviscid fluid of large extent is identically zero".Just in the surface of the cylinder, the velocity profile can be given in the next equation:

V=2Usin\theta

And the pressure P on the surface of cylinder is given by Bernoulli's equation along the streamline through that point:

P=P_{_{\infty }}+\frac{1}{2}\rho U^{2}(1-4sin^{2}\Theta ))

where P_∞ is  Pressure at stagnation point, U is the velocity given, ρ is the density of the fluid (in this case air) and θ is the angle measured from the center of cylinder to the adjacent point where your pressure point will be determine.

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(25) Consider the mechanical system below. Obtain the steady-state outputs x_1 (t) and x_2 (t) when the input p(t) is the sinuso
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3 years ago
(a) Determine the dose (in mg/kg-day) for a bioaccumulative chemical with BCF = 103 that is found in water at a concentration of
solmaris [256]

Answer:

0.064 mg/kg/day

6.25% from water, 93.75% from fish

Explanation:

Density of water is 1 kg/L, so the concentration of the chemical in the water is 0.1 mg/kg.

The BCF = 10³, so the concentration of the chemical in the fish is:

10³ = x / (0.1 mg/kg)

x = 100 mg/kg

For 2 L of water and 30 g of fish:

2 kg × 0.1 mg/kg = 0.2 mg

0.030 kg × 100 mg/kg = 3 mg

The total daily intake is 3.2 mg.  Divided by the woman's mass of 50 kg, the dosage is:

(3.2 mg/day) / (50 kg) = 0.064 mg/kg/day

b) The percent from the water is:

0.2 mg / 3.2 mg = 6.25%

And the percent from the fish is:

3 mg / 3.2 mg = 93.75%

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A steel bar is 150 mm square and has a hot-rolled finish. It will be used in a fully reversed bending application. Sut for the s
Xelga [282]

Answer:

See explanation

Explanation:

Given The bar is square and has a hot-rolled finish. The loading is fully reversed bending.

Tensile Strength

Sut: 600 MPa

Maximum temperature

Tmax: 500 °C

Bar side dimension

b: 150 mm

Alternating stress

σa: 100 MPa

Reliability

R: 0.999 Note 1.

Assumptions Infinite life is required and is obtainable since this ductile steel will have an endurance limit. A reliability factor of 99.9% will be used.

Solution See Excel file Ex06-01.xls.

1 Since no endurance-limit or fatigue strength information is given, we will estimate S'e based on the ultimate tensile strength using equation 6.5a.

S'e: 300 MPa = 0.5 * Sut

2 The loading is bending so the load factor from equation 6.7a is

Cload: 1

3 The part size is greater than the test specimen and the part is not round, so an equivalent diameter based on its 95% stressed area must be determined and used to find the size factor. For a rectangular section in nonrotating bending, the A95 area is defined in Figure 6-25c and the equivalent diameter is found from equation 6.7d

A95: 1125 mm2 = 0.05 * b * b Note 2.

dequiv: 121.2 mm = SQRT(A95val / 0.0766)

and the size factor is found for this equivalent diameter from equation 6.7b, to be

Csize: 0.747 = 1.189 * dequiv^-0.097

4 The surface factor is found from equation 6.7e and the data in Table 6-3 for the specified hot-rolled finish.

Table 6-3 constants

A: 57.7

b: -0.718 Note 3.

Csurf: 0.584 = Acoeff * Sut^bCoeff

5 The temperature factor is found from equation 6.7f :

Ctemp: 0.710 = 1 - 0.0058 * (Tmax - 450)

6 The reliability factor is taken from Table 6-4 for R = 0.999 and is

Creliab: 0.753

7 The corrected endurance limit Se can now be calculated from equation 6.6:

Se: 69.94 MPa = Cload * Csize * Csurf * Ctemp *

Creliab * Sprme

Let

Se: 70 MPa

8 To create the S-N diagram, we also need a value for the estimated strength Sm at 103 cycles based on equation 6.9 for bending loading.

Sm: 540 MPa = 0.9 * Sut

9 The estimated S-N diagram is shown in Figure 6-34 with the above values of Sm and Se. The expressions of the two lines are found from equations 6.10a through 6.10c assuming that Se begins at 106 cycles.

b: -0.2958 Note 4.

a: 4165.7

Plotting Sn as a function of N from equation 6.10a

N Sn (MPa)

1000 540 =aa*B73^bb

2000 440

4000 358

8000 292

16000 238

32000 194

64000 158

128000 129

256000 105

512000 85

1000000 70

FIGURE 6-34. S-N Diagram and Alternating Stress Line Showing Failure Point

10 The number of cycles of life for any alternating stress level can now be found from equation 6.10a by replacing σa for Sn.

At N = 103 cycles,

Sn3: 540 MPa = aa * 1000^bb

At N = 106 cycles,

Sn6: 70 MPa = aa * 1000000^bb

The figure above shows the intersection of the alternating stress line (σa = 100 MPa) with the failure line at N = 3.0 x 105 cycles.

8 0
3 years ago
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