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Liula [17]
3 years ago
5

Quantitative meaning

Engineering
2 answers:
emmasim [6.3K]3 years ago
8 0

Answer: quantitative is measering by quantity. (・_・;)

pychu [463]3 years ago
6 0

Answer:

Explanation:

relating to, measuring, or measured by the quantity of something rather than its quality.Often contrasted with qualitative.

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Answer true


Explanation
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3 years ago
For some transformation having kinetics that obey the Avrami equation, the parameter n is known to have a value of 2. If, after
kotegsom [21]

This question is incomplete, the complete question is;

For some transformation having kinetics that obey the Avrami equation, the parameter n is known to have a value of 2. If, after 100 s, the reaction is 40% complete, how long (total time in seconds) will it take the transformation to go to 95% completion

y = 1 - exp( -ktⁿ )

Answer: the time required for 95% transformation is 242.17 s

Explanation:

First, we calculate the value of k which is the dependent variable in Avrami equation

y = 1 - exp( -ktⁿ )

exp( -ktⁿ ) = 1 - y

-ktⁿ = In( 1 - y )

k = - In( 1 - y ) / tⁿ

now given that; n = 2, y = 40% = 0.40, and t = 100 s

we substitute

k = - In( 1 - 0.40 ) / 100²

k = - In(0.60) / 10000

k = 0.5108 / 10000

k = 0.00005108 ≈ 5.108 × 10⁻⁵

Now calculate the time required for 95% transformation

tⁿ = - In( 1 - y ) / k

t = [- In( 1 - y ) / k ]^1/n

n = 2, y = 95% = 0.95 and k = 5.108 × 10⁻⁵

we substitute our values

t = [- In( 1 - 0.95 ) / 5.108 × 10⁻⁵ ]^1/2

t = [2.9957 / 5.108 × 10⁻⁵]^1/2

t = [ 58647.22 ]^1/2

t = 242.17 s

Therefore the time required for 95% transformation is 242.17 s

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3 years ago
As the junior engineer at the Mesabi Range Hydraulic Engineering Company located in Ely, Minnesota, you have been tasked with de
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yes it will

Explanation:

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3 years ago
For a cylindrical annulus whose inner and outer surfaces are maintained at 30 ºC and 40 ºC, respectively, a heat flux sensor mea
miskamm [114]

Answer:

k=0.12\ln(r_2/r_1)\frac {W}{ m^{\circ} C}

where r_1 and r_2 be the inner radius, outer radius of the annalus.

Explanation:

Let r_1, r_2 and L be the inner radius, outer radius and length of the given annulus.

Temperatures at the inner surface, T_1=30^{\circ}C\\ and at the outer surface, T_2=40^{\circ}C.

Let q be the rate of heat transfer at the steady-state.

Given that, the heat flux at r=3cm=0.03m is

40 W/m^2.

\Rightarrow \frac{q}{(2\pi\times0.03\times L)}=40

\Rightarrow q=2.4\pi L \;W

This heat transfer is same for any radial position in the annalus.

Here, heat transfer is taking placfenly in radial direction, so this is case of one dimentional conduction, hence Fourier's law of conduction is applicable.

Now, according to Fourier's law:

q=-kA\frac{dT}{dr}\;\cdots(i)

where,

K=Thermal conductivity of the material.

T= temperature at any radial distance r.

A=Area through which heat transfer is taking place.

Here, A=2\pi rL\;\cdots(ii)

Variation of temperature w.r.t the radius of the annalus is

\frac {T-T_1}{T_2-T_1}=\frac{\ln(r/r_1)}{\ln(r_2/r_1)}

\Rightarrow \frac{dT}{dr}=\frac{T_2-T_1}{\ln(r_2/r_1)}\times \frac{1}{r}\;\cdots(iii)

Putting the values from the equations (ii) and (iii) in the equation (i), we have

q=\frac{2\pi kL(T_1-T_2)}{\LN(R_2/2_1)}

\Rightarrow k= \frac{q\ln(r_2/r_1)}{2\pi L(T_2-T_1)}

\Rightarrow k=\frac{(2.4\pi L)\ln(r_2/r_1)}{2\pi L(10)} [as q=2.4\pi L, and T_2-T_1=10 ^{\circ}C]

\Rightarrow k=0.12\ln(r_2/r_1)\frac {W}{ m^{\circ} C}

This is the required expression of k. By putting the value of inner and outer radii, the thermal conductivity of the material can be determined.

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