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Juli2301 [7.4K]
3 years ago
12

Consider a steady-state experiment in which the observed current due to reduction of Ox to R is 85 mA/cm2. What is the concentra

tion gradient at the electrode surface? Knowing that the concentration of the species being reduced is zero at the electrode surface, and 0.10 mM in the bulk, how thick is the diffusion layer in this experiment?
Engineering
1 answer:
Elanso [62]3 years ago
5 0

Answer: Yes

Explanation:

Surface Concentration: In electrochemistry, there is an important distinction between the concentration of a species at the electrode’s surface and its concentration at some distance from the electrode’s surface (in what we call the bulk solution). Suppose we place an electrode in a solution of Fe3+ and fix the potential at 1.00 V. At this potential Fe3+ is stable—the standard state reduction potential for Fe3+ to Fe2+ is +0.771 V, the concentration of Fe3+ remains the same at all distances from the electrode’s surface.

Bulk Concentraton: If we change the electrode’s potential to +0.500 V, the concentration of Fe3+ at the electrode’s surface decreases to approximately zero. The concentration of Fe3+ increases as we move away from the electrode’s surface until it equals the concentration of Fe3+ in bulk solution. The resulting concentration gradient causes additional Fe3+ from the bulk solution to diffuse to the electrode’s surface.

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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
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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.

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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.

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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 :

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

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

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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.

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