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Anna [14]
3 years ago
14

Estimate the endurance strength, Se, of a 37.5-mm- diameter rod of AISI 1040 steel having a machined finish and heat-treated to

a tensile strength of 760 MPa.
Engineering
1 answer:
7nadin3 [17]3 years ago
4 0

Answer:

endurance length is 236.64 MPa

Explanation:

data given:

d = 37.5 mm

Sut = 760MPa

endurance limit is

Se = 0.5 Sut

   = 0.5*760 = 380 MPa

surface factor is

Ka = a*Sut^b

where

Sut is ultimate strength

for AISI 1040 STEEL

a = 4.51, b = -0.265

Ka = 4.51*380^{-0.265}

Ka = 0.93

size factor is given as

Kb =1.29 d^{-0.17}

Kb = 0.669

Se = Sut *Ka*Kb

    = 380*0.669*0.93

Se = 236.64 MPa

therefore endurance length is 236.64 MPa

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Answer:

Steps:

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2.  load the data file for example, in matlab, use ('fileame.txt') to load the file

3. create variables from each column of your data

  for example, in matlab,  

     x=t{1}

     y=t{2}

4. plot the wind velocity and electricity generated.

   plot(x, y)

5. Label the individual axis and name the graph title.

    title('Graph of wind velocity vs approximate electricity generated for the year')

     xlabel('wind velocity')

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3 years ago
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Determine the reactor volume (assume a CSTR activated sludge aerobic reactor at steady state) required to treat 5 MGD of domesti
12345 [234]

Answer:

1.0MG

Explanation:

to solve this problem we use this formula

S₀-S/t = ksx --- (1)

the values have been given as

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effluent concentration = S= 10mg

value of K = 0.04L/day

x = 3000 mg

when we put these values into this equation,

250-10/t = 0.04x10x3000

240/t = 1200

we cross multiply from this stage

240 = 1200t

t = 240/1200

t = 0.2

remember the question says that 5MGD is required to be treated

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4 0
2 years ago
In remote areas, your gps devices may lose reception. It’s a good idea to have a
Katarina [22]

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4 0
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Select the right answer<br>​
Kruka [31]

Answer:

for 1st question the answer is 5th option.

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5 0
3 years ago
There are three options for heating a particular house: a. Gas: $1.33/therm where 1 therm=105,500 kJ b. Electric Resistance: $0.
sergejj [24]

Answer:

Option ‘a’ is the cheapest for this house.

Explanation:

Cheapest method of heating must have least cost per kj of energy. So, convert all the energy in the same unit (say kj) and take select the cheapest method to heat the house.

Given:

Three methods are given to heat a particular house are as follows:

Method (a)

Through Gas, this gives energy of amount $1.33/therm.

Method (b)

Through electric resistance, this gives energy of amount $0.12/KWh.

Method (c)

Through oil, this gives energy of amount $2.30/gallon.

Calculation:

Step1

Change therm to kj in method ‘a’ as follows:

C_{1}=\frac{\$ 1.33}{therm}\times(\frac{1therm}{105500kj})

C_{1}=1.2606\times10^{-5} $/kj.

Step2

Change kWh to kj in method ‘b’ as follows:

C_{2}=\frac{\$ 0.12}{kWh}\times(\frac{1 kWh }{3600kj})

C_{2}=3.334\times10^{-5} $/kj.

Step3

Change kWh to kj in method ‘c’ as follows:

C_{3}=\frac{\$ 2.30}{gallon}\times(\frac{1 gallon }{138500kj})

C_{3}=1.66\times10^{-5} $/kj.

Thus, the method ‘a’ has least cost as compare to method b and c.

So, option ‘a’ is the cheapest for this house.

 

5 0
3 years ago
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