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dsp73
4 years ago
14

For a 68 wt% Zn-32 wt% Cu alloy, make schematic sketches of the microstructure that would be observed for conditions of very slo

w cooling at the following temperatures: 1000C (1830F), 760C (1400F), 600C (1110F), and 400C (750F). Label all phases and indicate their approximate compositions.

Engineering
1 answer:
Rina8888 [55]4 years ago
6 0

Answer:

-At 1000°C, liquid phase at 68 wt% Zn

-At 760°C, liquid+γ phase, liquid at 74 wt% and 66 wt% γ phase

-At 600°C, γ phase

-At 400°C, γ phase

Explanation:

The phase diagram will be used for the combination of the copper-zinc alloy, first 68% by weight of Zn will be located on the axis corresponding to the composition. Draw a vertical line upwards and go through the temperatures given by the exercise 1000, 760, 600 and 400 ° C. Later you have to know the phases for each of the temperatures. Draw a horizontal line within each phase and look at the adjacent phases and the composition of the alloy. For each of the temperatures, the phases are in response and the drawings are attached to the image.

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Answer:

1st one.

Explanation:

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7 0
3 years ago
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A 50 mm diameter shaft is subjected to a static axial load of 160 kN. If the yield stress of the material is 350 MPa, the ultima
zvonat [6]

In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.

With the given data we can proceed to calculate the compression stress:

\sigma_c = \frac{P}{A}

\sigma_c = \frac{160*10^3}{\pi/4*0.05^2}

\sigma_c = 81.5MPa

Through Goodman's equations the combined effort by fatigue and compression is expressed as:

\frac{\sigma_a}{S_e}+\frac{\sigma_c}{\sigma_u}=\frac{1}{Fs}

Where,

\sigma_a=Fatigue limit for comined alternating and mean stress

S_e =Fatigue Limit

\sigma_c=Mean stress (due to static load)

\sigma_u = Ultimate tensile stress

Fs =Security Factor

We can replace the values and assume a security factor of 1, then

\frac{\sigma_a}{320}+\frac{81.5}{400}=\frac{1}{1}

Re-arrenge for \sigma_a

\sigma_a = 254.8Mpa

We know that the stress is representing as,

\sigma_a = \frac{M_c}{I}

Then,

Where M_c=Max Moment

I= Intertia

The inertia for this object is

I=\frac{\pi d^4}{64}

Then replacing and re-arrenge for M_c

M_c = \frac{\sigma_a*\pi*d^3}{32}

M_c = \frac{260.9*10^6*\pi*0.05^3}{32}

M_c = 3201.7N.m

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4 years ago
Two vertical, parallel clean glass plates are spaced a distance of 2mm apart. if the plates are placed in water, how high will t
Ulleksa [173]

Answer with Explanation:

The capillary rise in 2 parallel plates immersed in a liquid is given by the formula

h=\frac{2\sigma cos(\alpha )}{\rho gd}

where

\sigma is the surface tension of the liquid

\alpha is the contact angle of the liquid

\rho is density of liquid

'g' is acceleratioj due to gravity

'd' is seperation between thje plates

Part a) When the liquid is water:

For water and glass we have

\sigma =7.28\times 10^{-2}N/m

\alpha =0

\rho _{w}=1000kg/m^3

Applying the values we get

h=\frac{2\times 7.28\times 10^{-2}cos(0)}{1000\times 9.81\times 2\times 10^{-3}}=7.39mm

Part b) When the liquid is mercury:

For mercury and glass we have

\sigma =485.5\times 10^{-3}N/m

\alpha =138^o

\rho _{w}=13.6\times 10^{3}kg/m^3

Applying the values we get

h=\frac{2\times 485.5\times 10^{-3}cos(138)}{13.6\times 1000\times 9.81\times 2\times 10^{-3}}=-2.704mm

The negative sign indicates that there is depression in mercury in the tube.

4 0
3 years ago
A work element in a manual assembly task consists of the following MTM-1 elements: (1) R16C, (2) G4A, (3) M10B5, (4) RL1, (5) R1
ella [17]

Answer:

a)

1) R16C ; Tn = 17 TMU

2) G4A ; Tn = 7.3 TMU

3) M10B5 ; Tn = 15.1 TMU

4) RL1 ; Tn = 2 TMU

5) R14B ; Tn = 14.4 TMU

6) G1B ; Tn = 3.5 TMU

7) M8C3 ; Tn = 14.7 TMU

8) P1NSE ; Tn = 10.4 TMU

9) RL1 ; Tn = 2 TMU

b) 3.1 secs

Explanation:

a) Determine the normal times in TMUs for these motion elements

1) R16C ; Tn = 17 TMU

2) G4A ; Tn = 7.3 TMU

3) M10B5 ; Tn = 15.1 TMU

4) RL1 ; Tn = 2 TMU

5) R14B ; Tn = 14.4 TMU

6) G1B ; Tn = 3.5 TMU

7) M8C3 ; Tn = 14.7 TMU

8) P1NSE ; Tn = 10.4 TMU

9) RL1 ; Tn = 2 TMU

b ) Determine the total time for this work element in seconds

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note ; 1 TMU = 0.036 seconds

hence the total time for the work in seconds = 86.4 * 0.036 = 3.1 seconds

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The purpose of pasteurizing milk is to A. Kill pathogens B. Break down milk fat C. Add vitamins and minerals D. Prevent spoilage
motikmotik
Answer: A Kill pathogens
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3 years ago
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