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AysviL [449]
3 years ago
9

Calculate the resolved shear stress on the (111) 01%1 slip system of a unit cell in an FCC nickel single crystal if a stress of

13.7 MPa is applied in the [001] direction of the unit cell.
Engineering
1 answer:
abruzzese [7]3 years ago
4 0

I have attached an image showing the shear stresses and direction.

Answer:

Resolved shear stress = 5.6MPa

Explanation:

From the diagram attached,

Cos λ = [(0,-1,1)•(0,0,1)]/[(√(0^2 + (-1)^2 + 1^2)) + (√(0^2 + 0^2 + 1^2))]

= 1/(√2 x√1) = 1/√2

Also, Cos Φ = [(1,1,1)•(0,0,1)]/[(√(1^2 + 1^2 + 1^2)) + (√(0^2 + 0^2 + 1^2))]

= 1/(√3 x√1) = 1/√3

So resolved shear stress (τ resolved) = 13.7 x (1/√2) x (1/√3) = 13.7 x (1/√6) = 5.593 or approximately 5.6MPa

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The outer surface of an engine is situated in a place where oil leakage can occur. When leaked oil comes in contact with ahot su
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Answer:

TBC thickness of 4 mm is insufficient to prevent fire hazard

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- The thickness of engine cover L_1 = 0.01 m

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- The convection coefficient ambient air h_o = 7 W/m^2K

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Would a TBC layer of 4 mm thickness be sufficient to keep the engine cover surface below autoignition temperature of 200°C to prevent fire hazard?

Solution:

- We will use thermal circuit analogy for the 1-D problem and steady state conduction with no heat generation in the cover or TBC layer.

 The temperature at each medium interface and the Thermal resistance for each medium is given in the attachment schematic and circuit analogy.

 - We will calculate the total heat flux for the entire system q:

                       q = ( T_i - T_o ) / R_total

- R_total is the equivalent thermal resistance of the entire circuit. Since all resistances are in series we have:

                       R_total = 1 / h_i + L_1 / k_1 + L_2 / k_2 + 1 / h_o

- Plug in the values and compute:

                       R_total = 1 / 7 + 0.01 / 14 + 0.004 / 1.1 + 1 / 7

                       R_total = 0.2900649351 T-m^2 / W

- Calculate the Total heat flux q:

                       q = ( 333 - 69 ) / 0.2900649351

                       q = 910.141 W / m^2

- Just like the total current in a circuit remains same, the total heat flux remains same. We will use the total heat flux q to calculate the temperature of outer engine surface T_2 as follows:

                      q = ( T_i - T_2 ) / R_i2

Where,

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                      R_i2 = 1 / 7 + 0.01 / 14 = 0.14357 T-m^2 / W

Hence,

                      ( T_i - T_2 ) = q*R_i2

                        T_2 = T_i - q*R_i2

Plug the values in:

                        T_2 = 333 - 910.141*0.14357

                        T_2 = 202.33°C

- The outer surface of the engine cover has a temperature above T_ignition = 200°C. Hence, the TBC thickness of 4 mm is insufficient to prevent fire hazard

3 0
3 years ago
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