Answer:
Check the attached image below
Explanation:
Kindly check the attached image below to get the step by step explanation to the question above.
Consider a very long, slender rod. One end of the rod is attached to a base surface maintained at Tb, while the surface of the r
od is exposed to an air temperature of 400°C. Thermocouples imbedded in the rod at locations 25 mm and 120 mm from the base surface register temperatures of 335°C and 375°C, respectively.
(a) Calculate the rod base temperature (°C).
(b) Determine the rod length (mm) for the case where the ratio of the heat transfer from a finite length fin to the heat transfer from a very long fin under the same conditions is 99 percent.
(a) Calculate the rod base temperature (°C).
(b) Determine the rod length (mm) for the case where the ratio of the heat transfer from a finite length fin to the heat transfer from a very long fin under the same conditions is 99 percent.
1 answer:
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Answer:
the ratio of the etched to the original crack tip radius is 30.24
Explanation:
Given the data in the question;
we determine the initial fracture stress using the following expression;
(σf)₁ = 2(σ₀)₁ α₁/(
)₁
----- let this be equation 1
where; (σ₀)₁ is the initial fracture strength
()₁ is the original crack tip radius
α₁ is the original crack length.
first, we determine the final crack length;
α₂ = α₁ - 16% of α₁
α₂ = α₁ - ( 0.16 × α₁)
α₂ = α₁ - 0.16α₁
α₂ = 0.84α₁
next, we calculate the final fracture stress;
the fracture strength is increased by a factor of 6;
(σ₀)₂ = 6( σ₀ )₁
Now, expression for the final fracture stress
(σf)₂ = 2(σ₀)₂ α₂/(
)₂
------- let this be equation 2
where ()₂ is the etched crack tip radius
value of fracture stress of glass is constant
Now, we substitute 2(σ₀)₁ α₁/(
)₁
from equation for (σf)₂ in equation 2.
0.84α₁ for α₂.
6( σ₀ )₁ for (σ₀)₂.
∴
2(σ₀)₁ α₁/(
)₁
= 2(6( σ₀ )₁)
0.84α₁/(
)₂
divide both sides by 2(σ₀)₁
α₁/(
)₁
= 6
0.84α₁/(
)₂
1/(
)₁
= 6
0.84/(
)₂
1/(
)₁
= 36
0.84/(
)₂
1 / ()₁ = 30.24 / (
)₂
()₂ = 30.24(
)₁
()₂/(
)₁ = 30.24
Therefore, the ratio of the etched to the original crack tip radius is 30.24
Answer:
the only one that meets the requirements is option C .
Explanation:
The tolerance of a quantity is the maximum limit of variation allowed for that quantity.
To find it we must have the value of the magnitude, its closest value is the average value, this value can be given or if it is not known it is calculated with the formula
x_average = ∑ / n
The tolerance or error is the current value over the mean value per 100
Δx₁ = x₁ / x_average
tolerance = | 100 -Δx₁ 100 |
bars indicate absolute value
let's look for these values for each case
a)
x_average = (2.1700000+ 2.258571429) / 2
x_average = 2.2142857145
fluctuation for x₁
Δx₁ = 2.17000 / 2.2142857145
Tolerance = 100 - 97.999999991
Tolerance = 2.000000001%
fluctuation x₂
Δx₂ = 2.258571429 / 2.2142857145
Δx2 = 1.02
tolerance = 100 - 102.000000009
tolerance 2.000000001%
b)
x_average = (2.2 + 2.29) / 2
x_average = 2,245
fluctuation x₁
Δx₁ = 2.2 / 2.245
Δx₁ = 0.9799554
tolerance = 100 - 97,999
Tolerance = 2.00446%
fluctuation x₂
Δx₂ = 2.29 / 2.245
Δx₂ = 1.0200445
Tolerance = 2.00445%
c)
x_average = (2.211445 +2.3) / 2
x_average = 2.2557225
Δx₁ = 2.211445 / 2.2557225 = 0.9803710
tolerance = 100 - 98.0371
tolerance = 1.96%
Δx₂ = 2.3 / 2.2557225 = 1.024624
tolerance = 100 -101.962896
tolerance = 1.96%
d)
x_average = (2.20144927 + 2.29130435) / 2
x_average = 2.24637681
Δx₁ = 2.20144927 / 2.24637681 = 0.98000043
tolerance = 100 - 98.000043
tolerance = 2.000002%
Δx₂ = 2.29130435 / 2.24637681 = 1.0200000017
tolerance = 2.0000002%
e)
x_average = (2 +2,3) / 2
x_average = 2.15
Δx₁ = 2 / 2.15 = 0.93023
tolerance = 100 -93.023
tolerance = 6.98%
Δx₂ = 2.3 / 2.15 = 1.0698
tolerance = 6.97%
Let's analyze these results, the result E is clearly not in the requested tolerance range, the other values may be within the desired tolerance range depending on the required precision, for the high precision of this exercise the only one that meets the requirements is option C .