Which of the following is an example of an iterative process?

You are evaluating the lifetime of a turbine blade. The blade is 4 cm long and there is a gap of 0.16 cm between the tip of the

Tcecarenko [31]

Answer:

Explanation:

Given conditions

1)The stress on the blade is 100 MPa

2)The yield strength of the blade is 175 MPa

3)The Young’s modulus for the blade is 50 GPa

4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.

5)The temperature of the blade is 800°C.

6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)

where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K

Young Modulus, E = Stress, $\sigma$ /Strain, ∈

initial Strain, $\epsilon_i = \frac{\sigma}{E}$

$\epsilon_i = \frac{100\times 10^{6} Pa}{50\times 10^{9} Pa}$

$\epsilon_i = 0.002$

creep rate in the steady state

$\frac{\delta \epsilon}{\delta t} = (1 \times {10}^{-5})\sigma^4 exp^(\frac{-2eV}{kT} )$

$\frac{\epsilon_{initial} - \epsilon _{primary}}{t_{initial}-t_{final}} = 1 \times 10^{-5}(100)^{4}exp(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )(800+273)K} )$

but Tinitial = 0

$\epsilon_{initial} - \epsilon _{primary}} = 0.002 - 0.003 = -0.001$

$\frac{-0.001}{-t_{final}} = 1 \times 10^{-5}(100)^{4}\times 10^{(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )1073K} )}$

solving the above equation,

we get

Tfinal = 2459.82 hr

3 0

2 years ago

Need help solving math problem using integration

notka56 [123]

Ummm did you try to add or subtract and multiply or divide that can get your answer

8 0

1 year ago

How do you build a engine best anwser will get marked as brainlist

marysya [2.9K]

Answer:

It is best to soak the new lifters in engine oil overnight in order to build a car engine from scratch, regardless of its length. Ensure all rings of the pistons are installed correctly. You should flip the block over and carefully install the top half of the main rod and bearing parts. Get the block out of your way and flip it over again. By flipping over the block, you’ll see the full effect. Ensure that the lifters are installed correctly.

Explanation:

5 0

1 year ago

Determine the percent increase in the nominal moment capacity of the section in Problem 2 when including compression steel at to

sergejj [24]

Explanation:

Please kindly share your problem two with us as to know the actual problem we are dealing with, the question looks incomplete

5 0

2 years ago

Suppose we have a database for an investment firm, consisting of the following attributes: B (broker), O (office of a broker), I

Snowcat [4.5K]

Answer:

Given, FDs are:

S -> D

I -> B

IS -> Q

B -> O

a)

"I" and "S" must be there in any candidate key because they do not appear on the right side of any functional dependency.

The only candidate key is: IS

IS -> ISBDQO

b)

Decomposition of R into 3NF: (I, B), (S, D), (B, O), (I, S, Q)

c)

Decomposition of R into BCNF:

Decompose R by I → B into R1 = (I, B) and R2 = (I, O, S, Q, D).

R1 is in BCNF

Decompose R2 by S → D into R21 = (S, D) and R22 = (O, I, S, Q).

R21is in BCNF

Decompose R22 by I → O into R221 = (I, O) and R222 = (I, S, Q).

R221 is in BCNF.

R222 is in BCNF.

The decomposition is: (I, B), (S, D), (I, O), (I, S, Q)

We can also write it as: (I, B), (S, D), (B, O), (I, S, Q)

Explanation:

The answer above is rendered in a very explanatory way.

8 0

2 years ago