Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-mm-thick steel plate which is initially at a uniform temperature of Ti = 200 degree C and is to be heated to a minimum temperature of 550 degree C. Heating is effected in a gas-fired furnace, where products of combustion at T infinity = 800 degree C maintain a convection coefficient of h = 250 W/m2 K on both surfaces of the plate How long should the plate be left in the furnace?

Answers

Answer 1

Complete question is;

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100 mm thick plate (ρ = 7830 kg/m3, Cp = 550 J/kg K, k = 48 W/m K). The plate initially is at 200 °C and is to be heated to a minimum temperature of 550 °C. Heating is effected in a gas-fired furnace where the products of combustion at T∞ = 800 °C maintain a convection heat transfer coefficient of h = 250 W/m.K on both surfaces of the plate. How long should the plate be left in the furnace?

Answer:

860 seconds

Explanation:

We are given;

Initial Temperature; Ti = 200 °C

Minimum Temperature; T_i = 550 °C

T∞ = 800 °C

convection coefficient; h = 250 W/m².K

ρ = 7830 kg/m³

Cp = 550 J/kg K

k = 48 W/m K

Plate thickness = 100mm

Thus,L = 100/2 = 50mm = 0.05 m

Let's find the biot number from the formula;

Bi = hL/K

Bi = (250 × 0.05)/48

Bi = 0.2604

Now, lowest temperature in the slab is given as;

θ_o = (T_o - T∞)/(T_i - T∞)

θ_o = (550 - 800)/(200 - 800)

θ_o = 0.4167

Now, from online tables calculation, we can find the root of the biot number.

Thus, root of the biot number Bi = 0.2604 is;

ζ1 = 0.488 rad

Also, C1 is gotten as 1.0396

Now,formula for thermal diffusivity is;

α = k/ρc

α = 48/(7830 × 550)

α = 1.115 × 10^(-5) m²/s

Also, from online tables, f(ζ1) = 0.401

Thus, we can find the time the plate should the plate be left in the furnace from;

-(ζ1)²(αt/L²) = In 0.401

-(ζ1)²(αt/L²) = -0.9138

t = (-0.9138 × 0.05²)/-(0.488² × 1.115 × 10^(-5))

t ≈ 860 s

Answer 2

In this exercise we want to calculate the time, in seconds, of the time left on the plate in the furnace, like this:

860 seconds

Organizing the information given in the statement we find that:

Initial Temperature; Ti = 200 °CMinimum Temperature; T_i = 550 °CInfinity Temperature: T=800°Cconvection coefficient; h = 250 W/m².Kρ = 7830 kg/m³Cp = 550 J/kg Kk = 48 W/m KPlate thickness = L = 0.05 m

Using the formula given below, we will have how to calculate the number of biot, like this:

[tex]B = hL/K\\B = (250 * 0.05)/48\\B = 0.2604[/tex]

calculating the angle that corresponds to the temperature difference as:

[tex]\theta_o = (T_o - T)/(T_i - T)\\\theta_o = (550 - 800)/(200 - 800)\\\theta_o = 0.4167[/tex]

Using the formula below, we will have:

[tex]\alpha = k/\rho c\\\alpha = 48/(7830 * 550)\\\alpha = 1.115 * 10^{-5}[/tex]

Combining all the information from the previous calculations, we have that the time will be calculated as:

[tex]-(\phi)^2(\alpha t/L^2) = In 0.401\\-(\phi )^2(\apha t/^2) = -0.9138\\t = (-0.9138 * 0.05^2)/-(0.488^2 * 1.115 * 10^{-5})\\t = 860 s[/tex]

See more about time at brainly.com/question/2570752


Related Questions

Consider a 1.5-m-high and 2.4-m-wide glass window whose thickness is 6 mm and thermal conductivity is k = 0.78 W/m⋅K. Determine the steady rate of heat transfer through this glass window and the temperature of its inner surface for a day during which the room is maintained at 24°C while the temperature of the outdoors is −5°C. Take the convection heat transfer coefficients on the inner and outer surfaces of the window to be h1 = 10 W/m2⋅K and h2 = 25 W/m2⋅K, respectively, and disregard any heat transfer by radiation.

Answers

Answer:

The steady rate of heat transfer through the glass window is 707.317 watts.

Explanation:

A figure describing the problem is included below as attachment. From First Law of Thermodynamics we get that steady rate of heat transfer through the glass window is the sum of thermal conductive and convective heat rates, all measured in watts:

[tex]\dot Q_{total} = \dot Q_{cond} + \dot Q_{conv, in} + \dot Q_{conv, out}[/tex] (Eq. 1)

Given that window is represented as a flat element, we can expand (Eq. 1) as follows:

[tex]\dot Q_{total} = \frac{T_{i}-T_{o}}{R}[/tex] (Eq. 2)

Where:

[tex]T_{i}[/tex], [tex]T_{o}[/tex] - Indoor and outdoor temperatures, measured in Celsius.

[tex]R[/tex] - Overall thermal resistance, measured in Celsius per watt.

Now, we know that glass window is configurated in series and overall thermal resistance is:

[tex]R = R_{cond} + R_{conv, in}+R_{conv, out}[/tex] (Eq. 3)

Where:

[tex]R_{cond}[/tex] - Conductive thermal resistance, measured in Celsius per watt.

[tex]R_{conv, in}[/tex], [tex]R_{conv, out}[/tex] - Indoor and outdoor convective thermal resistances, measured in Celsius per watt.

And we expand the expression as follows:

[tex]R = \frac{l}{k\cdot w\cdot d} + \frac{1}{h_{i}\cdot w\cdot d} + \frac{1}{h_{i}\cdot w\cdot d}[/tex]

[tex]R = \frac{1}{w\cdot d}\cdot \left(\frac{l}{k}+\frac{1}{h_{i}}+\frac{1}{h_{o}} \right)[/tex] (Eq. 4)

Where:

[tex]w[/tex] - Width of the glass window, measured in meters.

[tex]d[/tex] - Length of the glass window, measured in meters.

[tex]l[/tex] - Thickness of the glass window, measured in meters.

[tex]k[/tex] - Thermal conductivity, measured in watts per meter-Celsius.

[tex]h_{i}[/tex], [tex]h_{o}[/tex] - Indoor and outdoor convection coefficients, measured in watts per square meter-Celsius.

If we know that [tex]w = 2.4\,m[/tex], [tex]d = 1.5\,m[/tex], [tex]l = 0.006\,m[/tex], [tex]k = 0.78\,\frac{W}{m\cdot ^{\circ}C}[/tex], [tex]h_{i} = 10\,\frac{W}{m^{2}\cdot ^{\circ}C}[/tex] and [tex]h_{o} = 25\,\frac{W}{m^{2}\cdot ^{\circ}C}[/tex], the overall thermal resistance is:

[tex]R = \left[\frac{1}{(2.4\,m)\cdot (1.5\,m)}\right] \cdot \left(\frac{0.006\,m}{0.78\,\frac{W}{m\cdot ^{\circ}C} }+\frac{1}{10\,\frac{W}{m^{2}\cdot ^{\circ}C} }+\frac{1}{25\,\frac{W}{m^{2}\cdot ^{\circ}C} } \right)[/tex]

[tex]R = 0.041\,\frac{^{\circ}C}{W}[/tex]

Now, we obtain the steady rate of heat transfer from (Eq. 2): ([tex]R = 0.041\,\frac{^{\circ}C}{W}[/tex], [tex]T_{i} = -5\,^{\circ}C[/tex], [tex]T_{o} = 24\,^{\circ}C[/tex])

[tex]\dot Q_{total} = \frac{24\,^{\circ}C-(-5\,^{\circ}C)}{0.041\,\frac{^{\circ}C}{W} }[/tex]

[tex]\dot Q_{total} = 707.317\,W[/tex]

The steady rate of heat transfer through the glass window is 707.317 watts.

3.94 x 105) + (2.04 x 105)

Answers

627.9 is the answer

2
A spring balance pulls with 5 N on a beam of 0.5 m.
What is the torque at the end of the beam?

Answers

Answer:

The torque at the end of the beam is 2.5 Nm

Explanation:

Given;

length of beam, r = 0.5 m

applied force, F = 5 N

The torque at the end of the beam is given by;

τ = F x r

where;

τ  is the torque

F is applied force

r is length of the beam

τ = 5 x 0.5

τ = 2.5 Nm

Therefore, the torque at the end of the beam is 2.5 Nm

Think about a good game story that made you feel a mix of positive and negative emotions. What was the story, what emotions did you feel, and how did it make you feel them? Why did those emotions draw you into the story?

Answers

when my brother got into a far away college. it made my happy that he got accepted but it made me sad thinking about how he wouldn’t be here w me

Which of the following best describes empathy?

the understanding of the feelings and beliefs of others
the lack of pride or boastfulness
the courage to speak up with one’s ideas
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Answers

Answer:

the first one is the correct answer

Answer:

the first one would be correct

Explanation:

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