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4 years ago

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For a steel alloy it has been determined that a carburizing heat treatment of 14 h duration at 809°C will raise the carbon conce

ntration to 0.54 wt% at a point 3.6 mm from the surface. Estimate the time necessary to achieve the same concentration at a 6.0 mm position for an identical steel and at a carburizing temperature of 1100°C. Assume that D0 is 2.5 × 10-5 m2/s and Qd is 120 kJ/mol.

1 answer:

Yakvenalex [24]4 years ago

3 0

Answer:

$t_2 = 27.7 hr$

Explanation:

Given data:

carbon concentration = 0.54%

from the relation given below calculate the time required to achieve concentration at 6.00 mm from surface

$\frac{x^2}{Dt} = constant$

D considered constant

$\frac{x^2}{t} = constant$

here, x POSITION FROM SURFACE, t is time required to achieve concentration

$\frac{x_1^2}{t_1} = \frac{x_2^2}{t_2}$

$\frac{3.6^2}{14} = \frac{6^2}{t_2}$

$t_2 = 27.7 hr$

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Explain why advances in technology can actually create more problems for engineers.

andrew11 [14]

Explanation:

Engineering is science in practical terms. It is the application of scientific findings in problem solving and creating a better world.

How does technological advancements create more problems for engineers?

Loss of job to automation: the world is driving at automating work processes through the use of specially designed and crafted machinery. Work is now properly being done using machines with little to no human input in the whole process. This is a huge let off for engineers. Engineers have to compete with machines which are their own inventions for jobs now.
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3 years ago

A soil is at a void ratio e = 0.90 with a specific gravity of the solid particles Gs = 2.70.

Alexus [3.1K]

Answer:

The correct answers are:

a. % w = 33.3%

b. mass of water = 45g

Explanation:

First, let us define the parameters in the question:

void ratio e = $\frac{V_v}{V_s}$ = $\frac{\left\begin{array}{ccc}volume&of&void\end{array}\right}{\left\begin{array}{ccc}volume&of&solid\end{array}\right}------ (1)$

Specific gravity $G_{s}$ = $\frac{P_s}{P_w} =$ $\frac{\left\begin{array}{ccc}density&of&soil\end{array}\right}{\left\begin{array}{ccc}density&of&water\end{array}\right}------ (2)$

% Saturation S = $\frac{V_w}{Vv}$ × $\frac{100}{1}$ = $\frac{\left\begin{array}{ccc}volume&of&water\end{array}\right}{\left\begin{array}{ccc}volume&of&void\end{array}\right}$ × $\frac{100}{1}--------(3)$

water content w = $\frac{M_w}{M_s}$ = $\frac{\left\begin{array}{ccc}mass&of&water\end{array}\right}{\left\begin{array}{ccc}mass&of&solid\end{array}\right} ------(4)$

a) To calculate the lower and upper limits of water content:

when S = 100%, it means that the soil is fully saturated and this will give the upper limit of water content.

when S < 100%, the soil is partially saturated, and this will give the lower limit of water content.

Note; S = 0% means that the soil is perfectly dry. Hence, when s = 1 will give the lowest limit of water content.

To get the relationship between water content and saturation, we will manipulate the equations above;

w = $\frac{M_w}{Ms}$

Recall; mass = Density × volume

w = $\frac{V_wP_w}{V_sP_s} ------(5)$

From eqn. (2) $G_{s}$ = $\frac{P_s}{P_w}$

∴ $\frac{1}{G_s} = \frac{P_w}{P_s} ------(6)$

putting eqn. (6) into (5)

w = $\frac{V_w}{V_sG_s} -----(7)$

Again, from eqn (1)

$V_s = \frac{V_v}{e}$

substituting into eqn. (7)

$w = \frac{V_w}{\frac{V_v}{e}{G_s} } = \frac{V_w e}{V_vG_s} \\ but \frac{V_w}{V_v} = S$

∴ $w = \frac{Se}{G_s} -----(8)$

With eqn. (7), we can calculate

upper limit of water content

when S = 100% = 1

Given, $G_{s} = 2.7, e= 0.9$

∴ $w= \frac{0.9*1}{2.7} = 0.333$

∴ %w = 33.3%

Lower limit of water content

when S = 1% = 0.01

$w= \frac{0.01*0.9 }{2.7} = 0.0033$

∴ % w = 0.33%

b) Calculating mass of water in 100 cm³ sample of soil ( $P_w=\frac{1_g}{cm^{3} }$ )

Given, $V_{s} = 100 cm^{3 }$ , S = 50% = 0.5

%S = $\frac{V_w}{V_v}$ × $\frac{100}{1}$ = $\frac{V_w}{eV_s}$ × $\frac{100}{1}$

0.50 = $\frac{V_w}{0.9* 100} = 45cm^{3}$

mass of water = $P_wV_w= 1 * 45 = 45_{g}$

7 0

4 years ago

Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe

NemiM [27]

Answer:

(a) attached below

(b) $pf_{C}=0.85$ $lagging$

(c) $I_{C} =32.37 A$

(d) $X_{C} =49.37$ Ω

(e) $I_{cap} =9.72 A$ and $I_{line} =27.66 A$

Explanation:

Given data:

$P_{1}=15 kW$

$S_{2} =10 kVA$

$pf_{1} =0.6$ $lagging$

$pf_{2}=0.8$ $leading$

$V=480$ $Volts$

(a) Draw the power triangle for each load and for the combined load.

$\alpha_{1}=cos^{-1} (0.6)=53.13$ °

$\alpha_{2}=cos^{-1} (0.8)=36.86$ °

$S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA$

$Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99$ ≅ $20kVAR$

$P_{2} =S_{2}*pf_{2} =10*0.8=8 kW$

$Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99$ ≅ $-6 kVAR$

The negative sign means that the load 2 is providing reactive power rather than consuming

Then the combined load will be

$P_{c} =P_{1} +P_{2} =15+8=23 kW$

$Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR$

(b) Determine the power factor of the combined load and state whether lagging or leading.

$S_{c} =P_{c} +jQ_{c} =23+14j$

or in the polar form

$S_{c} =26.92$ °

$pf_{C}=cos(31.32) =0.85$ $lagging$

The relationship between Apparent power S and Current I is

$S=VI^{*}$

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

(c) Determine the magnitude of the line current from the source.

Current of the combined load can be found by

$I_{C} =S_{C}/\sqrt{3}*V$

$I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A$

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

$Q_{C} =3*V^2/X_{C}$

$X_{C} =3*V^2/Q_{C}$

$X_{C} =3*(480)^2/14*10^3$ Ω

(e) Compute the magnitude of the current in each capacitor and the line current from the source.

Current flowing in the capacitor is

$I_{cap} =V/X_{C} =480/49.37=9.72 A$

Line current flowing from the source is

$I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A$

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Thepotemich [5.8K]

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Aliun [14]

Answer:

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Explanation:

That's it

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