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adell [148]
4 years ago
10

"It is necessary to select a metal alloy for an application that requires a yield strength of at least 345 MPa (50,000 psi) whil

e maintaining a minimum ductility (%EL) of 20%. If the metal may be cold worked, decide which of the following are candidates: copper, brass, and a 1040 steel."

Engineering
1 answer:
nataly862011 [7]4 years ago
4 0

Answer:

Both Brass and 1040 Steel maintain the required ductility of 20%EL.

Explanation:

Solution:-

- This questions implies the use of empirical results for each metal alloy plotted as function of CW% and Yield Strength.

- So for each metal alloy use the attached figures as reference and determine the amount of CW% required for a metal alloy to maintain a Yield Strength Y = 345 MPa.

- Left Figure (first) at Y = 345 MPa ( y -axis ) and read on (x-axis):

                        1040 Steel --------> 0% CW

                        Brass ---------------> 22% CW

                        Copper ------------> 66% CW

The corresponding ductility (%EL) for cold Worked metal alloys can be determined from the right figure. Using the %CW for each metal alloy determined in first step and right figure to determine the resulting ductility.

- Right Figure (second) at respective %CW (x-axis) read on (y-axis)

                       1040 Steel (0% CW) --------> 25% EL

                        Brass (22% CW) -------------> 21% EL

                        Copper (66% CW) ----------> 4% EL

We see that both 1040 Steel and Brass maintain ductilities greater than 20% EL at their required CW% for Yield Strength = 345 MPa.

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Answer:

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Explanation:

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Dynamic viscosity (μ) = 22.87 *10^-6

Prandtl number (pr) = 0.688

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Specific heat (Cp) = 1.013kj/kg.K

Step 1: Calculating Reynolds number using the formula;

Re = 4m/πDμ

     = (4*140*10^-6)/(π* 0.005*22.87 *10^-6)

     = 5.6*10^-4/3.59*10^-7

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Step 2: Calculating the thermal entry length using the formula

Le = 0.05*Re*Pr*D

Substituting, we have

Le = 0.05 * 1559 * 0.688 *0.005

Le = 0.268

Step 3: Calculate the heat transfer coefficient  using the formula;

Nu = hD/k

h = Nu*k/D

Since Le is less than given length, Nusselt number (Nu) for fully developed flow and uniform surface heat flux is 4.36.

h = 4.36 * 33.65 *10^-3/0.005

h = 0.1467/0.005

h = 29.34 W/m²-k

Step 4: Calculating the surface area using the formula;

A = πDl

   =π * 0.005 * 2

    =0.0314 m²

Step 5: Calculating the temperature Tm

For energy balance,

Qc = Qh

Therefore,

H*A(Te-Tm) = MCp(Tm - Ti)

29.34* 0.0314(160-Tm) =  140 × 10-6* 1.013*10^3 (Tm-100)

0.921(160-Tm) = 0.14182(Tm-100)

     147.36 -0.921Tm = 0.14182Tm - 14.182

1.06282Tm = 161.542

Tm = 161.542/1.06282

Tm = 151.99 K

Step 6: Calculate the rate of heat transferred using the formula

Q = H*A(Te-Tm)

   = 29.34* 0.0314(160-151.99)

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We know that for a simple undamped system spring mass system

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now by putting the values

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Answer:

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• State 2 is in saturation state

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Once all enthalpies are found, the following equations are used using the first law of thermodynamics

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Qin = m (h1-h4)

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The efficiency is calculated as the power obtained on the heat that enters

Efficiency = Wout / Qin

Efficiency = (h1-h2) / (h1-h4)

For this problem, we will first find the enthalpies in all states

h1=3231kJ/Kg

h2=2310kJ/Kg

h3=h4=272kJ/Kg

A) using the eficiency ecuation

Efficiency = (h1-h2) / (h1-h4)

Efficiency =(3231-2310)/(3231-272)=0.31=<u>31%</u>

b)using ecuation for Wout

Wout = m (h1-h2)

Wout=37(3231-2310)=34077KW=<u>34.077MW</u>

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