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Nata [24]
3 years ago
15

The yield strength of mild steel is 150 MPa for an average grain diameter of 0.038 mm ; yield strength is 250 MPa for average gr

ain diameter 0.009 mm.
What is the yield strength for the same steel having an average grain diameter 0.004mm?

Hint: Assume Hall-Petch equation is valid.
Engineering
1 answer:
djyliett [7]3 years ago
6 0

Answer:

Explanation:

Hall-Petch equation provides direct relations between the strength of the material and the grain size:

σ=σ0+k/√d , where d- grain size, σ- strength for the given gran size, σ0 and k are the equation constants.

As in this problem, we don't know the constants of the equation, but we know two properties of the material, we are able to find them from the system of equations:

σ1=σ0+k/√d1

σ2=σ0+k/√d2 , where 1 and 2 represent 150MPa and 250MPa strength of the steel.

Note, that for the given problem, there is no need to convert units to SI, as constants can have any units, which are convenient for us.

From the system of equations calculations, we can find constant: σ0=55.196 MPa, k=18.48 MPa*mm^(0.5)

Now we are able to calculate strength for the grain diameter of 0.004 mm:

σ=55.196+18.48/(√0.004)=347.39 MPa

The strength of the steel with the grais size of 0.004 mm is 347.39 MPa.

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Answer:

16.2 cents

Explanation:

Given that a homeowner consumes 260 kWh of energy in July when the family is on vacation most of the time.

Where Base monthly charge of $10.00. First 100 kWh per month at 16 cents/kWh. Next 200 kWh per month at 10 cents/kWh. Over 300 kWh per month at 6 cents/kWh.

For the first 100 kWh:

16 cent × 100 = 1600 cents = 16 dollars

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The total cost = 10 + 16 + 16 = 42 dollars

Note that the base monthly of 10 dollars is added.

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That is, 42 / 260 = 0.1615 dollars

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7 0
3 years ago
Close to 16 billion pounds of ethylene glycol (EG) were produced in 2013. It previously ranked as the twenty-sixth most produced
bekas [8.4K]

Answer:

a) 0.684

b) 0.90

Explanation:

Catalyst

EO + W → EG

<u>a) calculate the conversion exiting the first reactor </u>

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Given that there are two stream one  contains 16.1 mol/dm^3 while the other contains   0.9 wt% catalyst

Vo = 7.24 dm^3/s

Vm = 800 gal = 3028 dm^3

hence Im = Vin/ Vo = (3028 dm^3) / (7.24dm^3/s) = 418.232 secs = 6.97 mins

next determine the value of conversion exiting the reactor ( Xai ) using the relation below

KIm = \frac{Xai}{1-Xai}  ------ ( 1 )

make Xai subject of the relation

Xai = KIm / 1 + KIm  ---  ( 2 )

<em>where : K = 0.311 ,  Im = 6.97   ( input values into equation 2 )</em>

Xai = 0.684

<u>B) calculate the conversion exiting the second reactor</u>

CA1 = CA0 ( 1 - Xai )

therefore CA1 = 2.5438 mol/dm^3

Vo = 7.24 dm^3/s

To determine the value of the conversion exiting the second reactor  ( Xa2 ) we will use the relation below

XA2 = ( Xai + Im K ) / ( Im K + 1 ) ----- ( 3 )

<em> where : Xai = 0.684 , Im = 6.97,  and K = 0.311  ( input values into equation 3 )</em>

XA2 = 0.90

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