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adell [148]
3 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]3 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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A proposed piping and pumping system has 20-psig static pressure, and the piping discharges to atmosphere 160 ft above the pump.
larisa86 [58]

Answer: (B) 100

Explanation:

Given that;

Pstatic = 20 psig , hz = 160ft, hf = 20ft

Now total head will be;

T.h = hz + hf

T.h= 160 + 20

T.h = 180ft

Minimum pressure = Psatic + egh

we know that specific weight of water is 62.4 (lb/ft3)

so

P.min = (20 bf/in² ) + (62.4 b/ft³ × 180 fr

P.min = (20 bf/in² ) + ( 62.4 × 180 × 1 ft²/144 in²)

P.min = 20 + 78

P.min = 98 lbf/in²

Therefore the minimum pressure rating (psi) of the piping system is most nearly B) 100

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3 years ago
Which one of these is not a successful budgeting strategy
Amiraneli [1.4K]

Answer:

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6 0
3 years ago
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A well-insulated tank in a vapor power plant operates at steady state. Saturated liquid water enters at inlet 1 at a rate of 125
Gekata [30.6K]

Answer:

a) \dot m_{3} = 135\,\frac{lbm}{s}, b) h_{3}=168.965\,\frac{BTU}{lbm}, c) T = 200.829\,^{\textdegree}F

Explanation:

a) The tank can be modelled by the Principle of Mass Conservation:

\dot m_{1} + \dot m_{2} - \dot m_{3} = 0

The mass flow rate exiting the tank is:

\dot m_{3} = \dot m_{1} + \dot m_{2}

\dot m_{3} = 125\,\frac{lbm}{s} + 10\,\frac{lbm}{s}

\dot m_{3} = 135\,\frac{lbm}{s}

b) An expression for the specific enthalpy at outlet is derived from the First Law of Thermodynamics:

\dot m_{1}\cdot h_{1} + \dot m_{2} \cdot h_{2} - \dot m_{3}\cdot h_{3} = 0

h_{3} = \frac{\dot m_{1}\cdot h_{1}+\dot m_{2}\cdot h_{2}}{\dot m_{3}}

Properties of water are obtained from tables:

h_{1}=180.16\,\frac{BTU}{lbm}

h_{2}=28.08\,\frac{BTU}{lbm} + \left(0.01604\,\frac{ft^{3}}{lbm}\right)\cdot (14.7\,psia-0.25638\,psia)

h_{2}=29.032\,\frac{BTU}{lbm}

The specific enthalpy at outlet is:

h_{3}=\frac{(125\,\frac{lbm}{s} )\cdot (180.16\,\frac{BTU}{lbm} )+(10\,\frac{lbm}{s} )\cdot (29.032\,\frac{BTU}{lbm} )}{135\,\frac{lbm}{s} }

h_{3}=168.965\,\frac{BTU}{lbm}

c) After a quick interpolation from data availables on water tables, the final temperature is:

T = 200.829\,^{\textdegree}F

8 0
3 years ago
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if you had 100 B size sheets and you cut them into A size sheets, how many sheets of A size paper would you have
castortr0y [4]

Answer:

200

Explanation:

A size sheets (also known as letter size) are 8.5 inches by 11 inches.

B size sheets (also known as ledger size) are 11 inches by 17 inches.

One B size sheet is twice as large as a A size sheet.  So if you have 100 B size sheets and cut each one in half, you'll get 200 A size sheets.

8 0
3 years ago
A fatigue test was conducted in which the mean stress was 46.2 MPa and the stress amplitude was 219 MPa.
sleet_krkn [62]

Answer:

a)σ₁ = 265.2 MPa

b)σ₂ = -172.8 MPa

c)Stress\ ratio =-0.65

d)Range = 438 MPa

Explanation:

Given that

Mean stress ,σm= 46.2 MPa

Stress amplitude ,σa= 219 MPa

Lets take

Maximum stress level = σ₁

Minimum stress level =σ₂

The mean stress given as

\sigma_m=\dfrac{\sigma_1+\sigma_2}{2}

2\sigma_m={\sigma_1+\sigma_2}

2 x 46.2 =  σ₁ +  σ₂

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The amplitude stress given as

\sigma_a=\dfrac{\sigma_1-\sigma_2}{2}

2\sigma_a={\sigma_1-\sigma_2}

2 x 219 =  σ₁ -  σ₂

 σ₁ -  σ₂ = 438 MPa    --------2

By adding the above equation

2  σ₁ = 530.4

σ₁ = 265.2 MPa

-σ₂ = 438 -265.2 MPa

σ₂ = -172.8 MPa

Stress ratio

Stress\ ratio =\dfrac{\sigma_{min}}{\sigma_{max}}

Stress\ ratio =\dfrac{-172.8}{265.2}

Stress\ ratio =-0.65

Range = 265.2 MPa - ( -172.8 MPa)

Range = 438 MPa

8 0
3 years ago
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