Ask question

Ask question

All categories

3 years ago

8

A hypothetical metal alloy has a grain diameter of 1.7 102 mm. After a heat treatment at 450C for 250 min, the grain diameter ha

s increased to 4.5 102 mm. Compute the time required for a specimen of this same material (i.e., d 0 1.7 102 mm) to achieve a grain diameter of 8.7 102 mm while being heated at 450C. Assume the n grain diameter exponent has a value of 2.1.

1 answer:

Romashka-Z-Leto [24]3 years ago

4 0

Answer:

the required time for the specimen is 1109.4 min

Explanation:

Given that;

diameter of metal alloy d₀ = 1.7 × 10² mm

Temperature of heat treatment T = 450°C = 450 + 273 = 723 K

Time period of heat treatment t = 250 min

Increased grain diameter 4.5 × 10² mm

grain diameter exponent n = 2.1

First we calculate the time independent constant K

dⁿ - d₀ⁿ = Kt

K = (dⁿ - d₀ⁿ) / t

we substitute

K = (( 4.5 × 10² )²'¹ - ( 1.7 × 10² )²'¹) / 250

K = (373032.163378 - 48299.511117) / 250

K = 1298.9306 mm²/min

Now, we calculate the time required for the specimen to achieve the given grain diameter ( 8.7 × 10² mm )

dⁿ - d₀ⁿ = Kt

t = (dⁿ - d₀ⁿ) / K

t = (( 8.7 × 10² )²'¹ - ( 1.7 × 10² )²'¹) / 1298.9306

t = ( 1489328.26061158 - 48299.511117) / 1298.9306

t = 1441028.74949458 / 1298.9306

t = 1109.4 min

Therefore, the required time for the specimen is 1109.4 min

You might be interested in

What is the connection between the air fuel ratio and an engine running rich/poor? please give clear examples and full sentances

gavmur [86]

Explanation:

Air fuel ratio:

Air fuel ratio is the ratio of mass of air to the mass of fuel.So we can say that

$Air\ fuel\ ratio=\dfrac{mass\ of\ air}{mass\ of\ fuel}$

As we know that fuel burn in the presence of air that is why we have to maintain a proper amount of air fuel ratio.

When we need more power then we have supply more fuel and to burn this fuel ,require a specified amount of air.So for different loading condition of engine different air fuel ratio is required.

When air is less and fuel is more then it is called rich air fuel ratio .when air is more and fuel is less then it is called poor air fuel ratio.

5 0

3 years ago

(a) Calculate the heat flux through a sheet of steel that is 10 mm thick when the temperatures oneither side of the sheet are he

aleksandr82 [10.1K]

Answer:

do the wam wam

Explanation:

6 0

3 years ago

While playing a game of catch on the quadrangle, you throw a ball at an initial velocity of 17.6 m/s (approximately 39.4 mi/hr),

MAXImum [283]

Answer:

a) The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) The ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

Explanation:

a) The ball experiments a parabolic motion, which is a combination of horizontal motion at constant velocity and vertical motion at constant acceleration. First, we calculate the time taken by the ball to hit the ground:

$y = y_{o} + (v_{o}\cdot \sin \theta) \cdot t+\frac{1}{2}\cdot g\cdot t^{2}$ (1)

Where:

$y_{o}$ , $y$ - Initial and final vertical position, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

$\theta$ - Launch angle, measured in sexagesimal degrees.

$g$ - Gravitational acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we know that $y_{o} = 2\,m$ , $y = 0\,m$ , $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $g = -9.807\,\frac{m}{s^{2}}$ , then the time taken by the ball is:

$-4.904\cdot t^{2}+13.482\cdot t +2 = 0$ (2)

This second order polynomial can be solved by Quadratic Formula:

$t_{1} \approx 2.890\,s$ and $t_{2} \approx -0.141\,s$

Only the first root offers a solution that is physically reasonable. That is, $t \approx 2.890\,s$ .

The vertical velocity of the ball is calculated by this expression:

$v_{y} = v_{o}\cdot \sin \theta +g\cdot t$ (3)

Where:

$v_{o,y}$ , $v_{y}$ - Initial and final vertical velocity, measured in meters per second.

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ , $g = -9.807\,\frac{m}{s^{2}}$ and $t \approx 2.890\,s$ , then the final vertical velocity is:

$v_{y} = -14.860\,\frac{m}{s}$

The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) From a) we understand that ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball ( $x$ ) is determined by the following expression:

$x = (v_{o}\cdot \cos \theta)\cdot t$ (4)

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $t \approx 2.890\,s$ , then the distance covered by the ball is:

$x = 32.695\,m$

The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before hitting the ground ( $v$ ), measured in meters per second, is determined by the following Pythagorean identity:

$v = \sqrt{(v_{o}\cdot \cos \theta )^{2}+v_{y}^{2}}$ (5)

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $v_{y} = -14.860\,\frac{m}{s}$ , then the magnitude of the velocity of the ball is:

$v \approx 18.676\,\frac{m}{s}$ .

The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is defined by the following trigonometric relationship:

$\tan \theta = \frac{v_{y}}{v_{o}\cdot \cos \theta_{o}}$

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta_{o} = 50^{\circ}$ and $v_{y} = -14.860\,\frac{m}{s}$ , the angle of the total velocity of the ball just before hitting the ground is:

$\theta \approx -52.717^{\circ}$

The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

3 0

3 years ago

Read 2 more answers

Is an isothermal process necessarily internally reversible? Explain your answer with an example

torisob [31]

Answer:

please give me brainlist and follow

Explanation:

Example of an irreverseble isothermal process is mixing of two fluids on the same temperature - it requires a lot of energy to unmix Jack and coke. ... Example of an reversible process with changing temperature is isentropic expansion.

5 0

2 years ago

A 4-kW electric heater runs for 2 hours to raise the room temperature to the desired level. Determine the amount of electric ene

Anna35 [415]

Answer:

Q' = 8 KW.h

Q'=28800 KJ

Explanation:

Given that

Heat Q= 4 KW

time ,t = 2 hours

The amount of energy used in KWh given as

Q ' = Q x t

Q' = 4 x 2 KW.h

Q' = 8 KW.h

We know that

1 h = 60 min = 60 x 60 s = 3600 s

We know that W = 1 J/s

The amount of energy used in KJ given as

Q' = 8 x 3600 = 28800 KJ

Therefore

Q' = 8 KW.h

Q'=28800 KJ

6 0

3 years ago