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pychu [463]
3 years ago
8

Estimate the theoretical fracture strength (in MPa) of a brittle material if it is known that fracture occurs by the propagation

of an elliptically shaped surface crack of length 0.25 mm that has a tip radius of curvature of 0.004 mm when a stress of 1060 MPa is applied.

Engineering
1 answer:
Annette [7]3 years ago
6 0

Answer:

σ =20700 MPa

Explanation:

Answer is in the following attachment

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Velocity and temperature profiles for laminar flow in a tube of radius ro = 10 mm have the form: u(r) = 0.15[1 − (r/ro ) 2 ] T(r
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Convert 0.025 in into mm.
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Answer:  0.025 in = 0.065 mm

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The emissivity of galvanized steel sheet, a common roofing material, is ε = 0.13 at temperatures around 300 K, while its absorpt
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Answer:

759.99W/m²

Explanation:

Question: If the temperature of the sheet is 77C,what is the incident solar radiation on aday with Tinf= Tsurr= 16°C?

Given

Energy Equation of the Gas

αs * Gs * A + h * A * (T inf - Tg) + εσA (Tsurr⁴- Tg⁴) = 0

Where σ= 5.67 *10^-8 W/m²K⁴ (Stefan-Boltzmann constant)

ε = 0.13 (Emisivity)

αs = 0.65 (Absorptivity for solar radiation)

h = 7W/m²K⁴

Tg = 77 + 273.15K = 350.15K

T inf = 16 + 273.15 = 288.15K

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Substitute the above values in the Gas Equation, we have

0.65 * Gs * A + 7 * A * (288.15 - 350.15) + 0.13 * 5.67 * 10^-8 * A * (288.15⁴ - 350.15⁴) = 0

0.65 * Gs * A = - 7 * A * (288.15 - 350.15) - 0.13 * 5.67 * 10^-8 * A * (288.15⁴ - 350.15⁴)

A cancels out, so we are left with

0.65 * Gs = - 7 * (288.15 - 350.15) - 0.13 * 5.67 * 10^-8 * (288.15⁴ - 350.15⁴)

0.65Gs = 434 - 0.7372 * 10^-8(−8,137,940,481.697)

0.65Gs = 434 + 0.7372 * 81.37940481697

0.65Gs = 493.992897231070284

Gs = 493.992897231070284/0.65

Gs = 759.9890726631850

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