Determine the carburizing time necessary to achieve a carbon concentration of 0.30 wt% at a position 4 mm into an iron–carbon alloy that initially contains 0.10 wt% C. The surface concentration is to be maintained at 0.90 wt% C, and the treatment is to be conducted at 1100°C. Use the diffusion data for γ-Fe.

Answers

Answer 1

Answer:

the carburizing time necessary to achieve a carbon concentration is 31.657 hours

Explanation:

Given the data in the question;

To determine the carburizing time necessary to achieve the given carbon concentration, we will be using the following equation:

(Cs - Cx) / (Cs - C0) = ERF( x / 2√Dt)

where Cs is Concentration of carbon at surface = 0.90

Cx is Concentration of carbon at distance x = 0.30 ; x in this case is 4 mm = ( 0.004 m )

C0 is Initial concentration of carbon = 0.10

ERF() = Error function at the given value

D = Diffusion of Carbon into steel

t = Time necessary to achieve given carbon concentration ,

so

(Cs - Cx) / (Cs - C0) = (0.9 - 0.3) / (0.9 - 0.1)

= 0.6 / 0.8

= 0.75

now, ERF(z) = 0.75; using ERF table, we can say;

Z ~ 0.81; which means ( x / 2√Dt) = 0.81

Now, Using the table of diffusion data

D = 5.35 × 10⁻¹¹ m²/sec at (1100°C) or 1373 K

now we calculate the carbonizing time by using the following equation;

z = (x/2√Dt)

t is carbonizing time

so we we substitute in our values

0.81 = ( 0.004 / 2 × √5.35 × 10⁻¹¹ × √t)

0.81 = 0.004 / 1.4628 × 10⁻⁵ × √t

0.81 × 1.4628 × 10⁻⁵ × √t = 0.004

1.184868 × 10⁻⁵ × √t = 0.004  

√t = 0.004 / 1.184868 × 10⁻⁵

√t = 337.5903

t = ( 337.5903)²  

t = 113967.21 seconds

we convert to hours

t = 113967.21 / 3600

t = 31.657 hours

Therefore, the carburizing time necessary to achieve a carbon concentration is 31.657 hours


Related Questions

A utility generates electricity with a 36% efficient coal-fired power plant emitting the legal limit of 0.6 lb of SO2 per million Btus of heat into the plant. Suppose the utility encourages its customers to replace their 75-W incandescents with 18-W compact fluorescent lamps (CFLs) that produce the same amount of light. Over the 10,000-hr lifetime of a single CFL.

Required:
a. How many kilowatt-hours of electricity would be saved?
b. How many 2,000-lb tons of SO2 would not be emitted?
c. If the utility can sell its rights to emit SO2 at $800 per ton, how much money could the utility earn by selling the SO2 saved by a single CFL?

Answers

Answer:

a) 570 kWh of electricity will be saved

b) the amount of  SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF

c) $1.296 can be earned by selling the SO₂ saved by a single CFL

Explanation:

Given the data in the question;

a) How many kilowatt-hours of electricity would be saved?

first, we determine the total power consumption by the incandescent lamp

[tex]P_{incandescent}[/tex] = 75 w × 10,000-hr = 750000 wh = 750 kWh

next, we also find  the total power consumption by the fluorescent lamp

[tex]P_{fluorescent}[/tex] = 18 × 10000 = 180000 = 180 kWh

So the value of power saved will be;

[tex]P_{saved}[/tex] = [tex]P_{incandescent}[/tex]  - [tex]P_{fluorescent}[/tex]

[tex]P_{saved}[/tex] = 750 - 180

[tex]P_{saved}[/tex]  = 570 kWh

Therefore, 570 kWh of electricity will be saved.

now lets find the heat of electricity saved in Bituminous

heat saved = energy saved per CLF / efficiency of plant

given that; the utility has 36% efficiency

we substitute

heat saved =  570 kWh/CLF / 36%

we know that; 1 kilowatt (kWh) = 3,412 btu per hour (btu/h)

so

heat saved =  570 kWh/CLF / 0.36 × (3412 Btu / kW-hr (

heat saved = 5.4 × 10⁶ Btu/CLF

i.e eat of electricity saved per CLF is 5.4 × 10⁶

b) How many 2,000-lb tons of SO₂ would not be emitted

2000 lb/tons = 5.4 × 10⁶ Btu/CLF

0.6 lb SO₂ / million Btu = x

so

x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ /  million Btu )] / 2000 lb/tons

x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ )] / [ ( 10⁶) × ( 2000 lb/ton) ]

x = 3.24 × 10⁶ / 2 × 10⁹

x = 0.00162 ton/CLF

Therefore, the amount of  SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF

c)  If the utility can sell its rights to emit SO2 at $800 per ton, how much money could the utility earn by selling the SO2 saved by a single CFL?

Amount = ( SO₂ saved per CLF ) × ( rate per CFL )

we substitute

Amount = 0.00162 ton/CLF × $800

= $1.296

Therefore; $1.296 can be earned by selling the SO₂ saved by a single CFL.

An infinite cylindrical rod falls down in the middle of an infinite tube filled with fluidat a constant speed V (terminal velocity). The density of the rod and the fluid are different.Assume that the pressure field is hydrostatic.(a)[5pts] Solve for the velocity profileas a function of rin terms of V and the other variables.(b)[2pts] Calculate upward force per unit length of the rod from the fluid wall shear stress on the rod.(c)[2 pts] Calculate upward force per unit length of the rod from bouyancy.(d)[1pts] Calculate V.VR1

Answers

Answer:

the speed of your poop

Explanation:

A cylindrical specimen of some metal alloy 10 mm in diameter and 150 mm long has a modulus of elasticity of 100 GPa. Does it seem reasonable to expect a tensile stress of 200 MPa to produce a reduction in specimen diameter of 0.08 mm

Answers

Answer:

N0

Explanation:

It does not seem reasonable to expect a tensile stress of 200 MPa to produce a reduction in specimen diameter of 0.08 mm

Given data :

Diameter ( d ) = 10 mm

length ( l ) = 150 mm

elasticity ( ∈ ) = 100 GPa

longitudinal strain ( б ) 200 MPa

Poisson ratio ( μ )  ( assumed ) =0.3

Assumption : deformation totally elastic

attached below is the detailed solution to why it is not reasonable .

The Sd value = 0.08 > the calculated Sd value ( 6*10^-3 ) hence it is not reasonable to expect a tensile stress of 200 MPa to produce a reduction in specimen

You are designing a hydraulic power takeoff for a garden tractor. The hydraulic pump will be directly connected to the motor and supply hydraulic fluid at 250 psi for use by accessories. In order for the tractor to maintain normal operation, the maximum power the hydraulic system can use is limited to 11 hp. For what maximum hydraulic flow rate in gallons per minute (gpm) should you design

Answers

Answer:

required flow rate is 75.44 gal/min  

Explanation:

Given the data in the question;

Power developed = 250 psi = 1.724 × 10⁶ Pa

hydraulic power W = 11 hp = 11 × 746 = 8206 Watt

now, Applying the formula for pump power

W = pgQμ

where p is density of fluid, Q is flow rate, μ is heat and W is power developed;

W = pgQμ

W = pgμ × Q  

W = P × Q -------- let this be equ 1

so we substitute in our values;

8.2027 kW = 1.724 × 10⁶ Pa × Q

Q = 8206 / 1.724 × 10⁶

Q = 4.75986 × 10⁻³ m³/sec

We know that, 1 cubic meter per seconds = 15850.3 US liquid gallon per minute, so

Q = 4.75986 × 10⁻³  × 15850.3 gallon/min

Q = 75.44 gal/min    

Therefore, required flow rate is 75.44 gal/min    

A water jet pump involves a jet cross-sectional area of 0.01 m^2, and a jet velocity of 30 m/s. The jet is surrounded by entrained water. The total cross-sectional area associated with the jet and entrained streams is 0.075 m^2. These two fluid streams leave the pump thoroughly mixed with an average velocity of 6 m/s through a cross-sectional area of 0.075 m^2. Determine the pumping rate (i.e., the entrained fluid flowrate) involved in liters/s.

Answers

Answer:

the entrained fluid flowrate is 150 liters/s

Explanation:

Given the data in the question;

we determine the flow rate of water though the jet by using the following expression;

Q₂ = A₂ × V₂

where Q₂  is the flow rate of water though the jet, A₂ is the cross sectional area of the jet( 0.01 m² ) and V₂ is the jet velocity( 30 m/s )

so we substitute

Q₂ = 0.01 m² × 30 m/s

Q₂ = 0.3 m³/s

Next we determine the flow rate of water through the pump by using the following expression

Q₃ = A₃ × V₃

where Q₃  is the flow rate of water though the pump, A₃ is the cross sectional area of the pump( 0.075 m² ) and V₃ is the average velocity of mixing( 6 m/s )

so we substitute

Q₃ = 0.075 m² × 6 m/s

Q₃ = 0.45 m³/s

so to calculate the flow pumping rate of water into the water jet pump, we use the expression;

Q₁ + Q₂ = Q₃

we substitute

Q₁ + 0.3 m³/s = 0.45 m³/s

Q₁ = 0.45 m³/s - 0.3 m³/s

Q₁ = 0.15 m³/s

we know that 1 m³/s = 1000 Liter/second

so

Q₁ = 0.15 × 1000 Liter/seconds

Q₁ = 150 liters/s

Therefore, the entrained fluid flowrate is 150 liters/s

A cylindrical 1040 steel rod having a minimum tensile strength of 865 MPa (125,000 psi), a ductility of at least 10%EL, and a final diameter of 6.0 mm (0.25 in.) is desired. Some 7.94 mm (0.313 in.) diameter 1040 steel stock, which has been cold worked 20% is available. Describe the procedure you would follow to obtain this material. Assume that 1040 steel experiences cracking at 40%CW

Answers

Answer:

procedure attached below

The material to be used will have a %Cw of 34.5%Cw which is < 40%Cw

Explanation:

Given data:

Minimum tensile strength = 865 MPa

Ductility = 10%EL

Desired Final diameter = 6.0 mm

20% cold worked 7.94 mm diameter 1040 steel stock

Describe the procedure you would follow to obtain this material.

assuming 1040 steel experiences cracking at 40%CW

attached below is a detailed procedure of obtaining the material

The material to be used will have a %Cw of 34.5%Cw which is < 40%Cw

Kim is working on the cost estimate and feasible design options for a building. Which stage of a construction plan is Kim working on now? A. design development B. schematic design C. mechanical D. structural

Answers

Answer:

B. schematic design

Explanation:

This correct for Plato/edmentum

Kim is working on the cost estimate. The stage of a construction plan is Kim working on now is schematic design. The correct option is B.

What is a schematic design?

A schematic design is an outline of a house or a building or another construction thing. The schematic design makes the outline map of the exterior or interior of the building. It is the foremost phase of designing something.

The design expert discusses the project three-dimensionally at this point in the process. To define the character of the finished project and an ideal fulfillment of the project program, a variety of potential design concepts are investigated.

The schematic design consists of a rough sketch with markings and measurements.

Therefore, the correct option is B. schematic design.

To learn more about schematic design, refer to the link:

https://brainly.com/question/14959467

#SPJ5

Determine the convection heat transfer coefficient, thermal resistance for convection, and the convection heat transfer rate that are associated with air at atmospheric pressure in cross flow over a cylinder of diameter D = 100 mm and length L = 2 m. The cylinder temperature is Ts = ° 70 C while the air velocity and temperature are V = 3 m/s and T[infinity] = 20°C, respectively. Plot the convection heat transfer coefficient and the heat transfer rate from the cylinder over the range 0.05 m ≤ D ≤ 0.5 m.

Answers

Answer:

attached below

Explanation:

Attached below is  a detailed solution to the question above

Step 1 : determine the Reynolds number using the characteristics of Air at 45°c

Step 2 : calculate the Nusselt's number

Step 3 : determine heat transfer coefficient

Step 4 : calculate heat transfer ratio and thermal resistance

Repeat steps 1 - 4 for each value of diameter from 0.05 to 0.5 m

attached below is a detailed solution

Because the mechanism of creep deformation is different from the mechanism of slip in most metal deformation processes, one of the fundamental relationships between microstructure and mechanical properties of metals is reversed for creep deformation compared with normal deformation. Is it:________.
A. The Hume-Rothery Rules
B. The Hall-Petch Relation
C. The Schmid Equation

Answers

Answer:

B. The Hall-Petch Relation

Explanation:

The Hall-Petch relation indicates that by reducing the grain size the strength of a material is increased up to the theoretical strength of the material however when the material grain size is reduced below 20 nm the material is more susceptible to creep deformation and displays an "inverse" Hall-Petch Relation as the Hall-Petch relation then has a negative slope (k value)

The Hall-Petch relation can be presented as follows;

[tex]\sigma_y[/tex] = [tex]\sigma_0[/tex] + k·(1/√d)

Where;

[tex]\sigma_y[/tex] = The strength

σ₀ = The friction stress

d = The grain size

k = The strengthening coefficient

The model equation for the reverse Hall-Petch effect is presented here as follows;

[tex]\sigma_y[/tex] = 10.253 - 10.111·(1/√d)

Which option identifies the section of a project charter represented in the following scenario?
Updated POS terminals will be available to the following five departments by July 31, 2015.
O project assumptions
O project deliverables
O project constraints
O project requirements

Answers

it’s b! i know this because i did the test!
Answer B , Just took the test and got 100%

Match the example to the model type it represents.

1. The client complains about the way the keyboard feels
1.mock-up
2. The engineering team tests how the tire treads on a new SUV perform on 2.various road conditions
preproduction model
3. The engineering team performs tests on the efficiency of the manufacturing process used for a recumbent bicycle
3.presentation model

Answers

Answer:

represnt

Explanation:

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