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allsm [11]
3 years ago
5

For a steel alloy it has been determined that a carburizing heat treatment of 15 h duration will raise the carbon concentration

to 0.35 wt% at a point 2.0 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 the same carburizing temperature.
Engineering
1 answer:
Free_Kalibri [48]3 years ago
5 0

Answer:

135 hour

Explanation:

It is given that a carburizing heat treatment of 15 hour will raise the carbon concentration by 0.35 wt% at a point of 2 mm from the surface.

We have to find the time necessary to achieve the same concentration at a 6 mm position.

we know that \frac{x_1^2}{Dt}=constant where x is distance and t is time .As the temperature is constant so D will be also constant

So \frac{x_1^2}{t}=constant

then \frac{x_1^2}{t_1}=\frac{x_2^2}{t_2} we have given x_1=2 mm\ ,t_1=15 hour\ ,x_2=6\ mm and we have to find t_2 putting all these value in equation

\frac{2^2}{15}=\frac{6^2}{t_2}

so t_2=135\ hour

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a buyer can purchase 70 screwdrivers ten 4-inch length twelve 6 inch length twenty 8-inch length are needed. how many heavy 24-i
jasenka [17]

Answer:

28 , 24-inch screwdrivers

Explanation:

The total number of screwdrivers that can be purchased is = 70

4 - inch length screwdrivers = 10

6- inch length screwdrivers = 12

8- inch length screwdrivers = 20

Total = 20 +12 +10 = 42

Remaining = 70-42 = 28

So, heavy 24-inch screwdrivers = 28

3 0
3 years ago
Steam enters a turbine steadily at 7 MPa and 600°C with a velocity of 60 m/s and leaves at 25 kPa with a quality of 95 percent.
Rufina [12.5K]

Answer:

a) \dot m = 16.168\,\frac{kg}{s}, b) v_{out} = 680.590\,\frac{m}{s}, c) \dot W_{out} = 18276.307\,kW

Explanation:

A turbine is a steady-state devices which transforms fluid energy into mechanical energy and is modelled after the Principle of Mass Conservation and First Law of Thermodynamics, whose expressions are described hereafter:

Mass Balance

\frac{v_{in}\cdot A_{in}}{\nu_{in}} - \frac{v_{out}\cdot A_{out}}{\nu_{out}} = 0

Energy Balance

-q_{loss} - w_{out} + h_{in} - h_{out} = 0

Specific volumes and enthalpies are obtained from property tables for steam:

Inlet (Superheated Steam)

\nu_{in} = 0.055665\,\frac{m^{3}}{kg}

h_{in} = 3650.6\,\frac{kJ}{kg}

Outlet (Liquid-Vapor Mix)

\nu_{out} = 5.89328\,\frac{m^{3}}{kg}

h_{out} = 2500.2\,\frac{kJ}{kg}

a) The mass flow rate of the steam is:

\dot m = \frac{v_{in}\cdot A_{in}}{\nu_{in}}

\dot m = \frac{\left(60\,\frac{m}{s} \right)\cdot (0.015\,m^{2})}{0.055665\,\frac{m^{3}}{kg} }

\dot m = 16.168\,\frac{kg}{s}

b) The exit velocity of steam is:

\dot m = \frac{v_{out}\cdot A_{out}}{\nu_{out}}

v_{out} = \frac{\dot m \cdot \nu_{out}}{A_{out}}

v_{out} = \frac{\left(16.168\,\frac{kg}{s} \right)\cdot \left(5.89328\,\frac{m^{3}}{kg} \right)}{0.14\,m^{2}}

v_{out} = 680.590\,\frac{m}{s}

c) The power output of the steam turbine is:

\dot W_{out} = \dot m \cdot (-q_{loss} + h_{in}-h_{out})

\dot W_{out} = \left(16.168\,\frac{kg}{s} \right)\cdot \left(-20\,\frac{kJ}{kg} + 3650.6\,\frac{kJ}{kg} - 2500.2\,\frac{kJ}{kg}\right)

\dot W_{out} = 18276.307\,kW

6 0
3 years ago
An inductor and resistor are connected in parallel to a 120-V, 60-Hz line. The resistor has a resistance of 50 ohms, and the ind
BARSIC [14]

Answer:

hi

Explanation:

the answer would be I dont know

6 0
3 years ago
Whats 47,000 resistance converted to kilo or mega ohms.
Julli [10]
  1. The conversion of 47,000 Ohms to kilo-ohms is equal to 47 kilo-ohms.
  2. The conversion of 47,000 Ohms to mega-ohms is equal to 0.047 kilo-ohms.

<h3>What is resistance?</h3>

Resistance can be defined as an opposition to the flow of current in an electric circuit. Also, the standard unit of measurement of the resistance of an electric component is Ohms, which can be converted to kilo-ohms or mega-ohms.

For Ohms to kilo-ohms, we have:

1 Ohms = 0.001 kilo-ohms

47,000 Ohms = X kilo-ohms

Cross-multiplying, we have:

X = 0.001 × 47000

X = 47 kilo-ohms.

For Ohms to mega-ohms, we have:

1,000,000 ohms = 1 mega-ohms

47,000 Ohms = X mega-ohms

Cross-multiplying, we have:

X1,000,000 = 47,000

X = 47,000/1,000,000

X = 0.047 kilo-ohms.

Read more resistance here: brainly.com/question/19582164

#SPJ1

7 0
1 year ago
An asbestos pad is square in cross section, measuring 5 cm on a side at its small end increasing linearly to 10 cm on a side at
polet [3.4K]

Answer:

q = 1.73 W

Explanation:

given data

small end  = 5 cm

large end = 10 cm

high = 15 cm

small end is held = 600 K

large end at = 300 K

thermal conductivity of asbestos  = 0.173 W/mK

solution

first we will get here side of cross section that is express as

S = S1 + \frac{S2-S1}{L} x     ...............1

here x is distance from small end and S1 is side of square at small end

and S2 is side of square of large end and L is length

put here value and we get

S = 5 + \frac{10-5}{15} x

S = \frac{0.15 + x}{3}    m

and  

now we get here Area of section at distance x is

area A = S²    ...............2

area A = (\frac{0.15 + x}{3})^2    m²

and

now we take here small length dx and temperature difference is dt

so as per fourier law

heat conduction is express as

heat conduction q = \frac{-k\times A\  dt}{dx}      ...............3

put here value and we get

heat conduction q = -k\times (\frac{0.15 + x}{3})^2 \   \frac{dt}{dx}  

it will be express as

q \times \frac{dx}{(\frac{0.15 + x}{3})^2} = -k (dt)  

now we intergrate it with limit 0 to 0.15 and take temp 600 to 300 K

q \int\limits^{0.15}_0 {\frac{dx}{(\frac{0.15 + x}{3})^2 } = -0.173 \int\limits^{300}_{600} {dt}          

solve it and we get

q (30)  = (0.173) × (600 - 300)

q = 1.73 W

5 0
3 years ago
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