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lara [203]
3 years ago
8

A cylindrical aluminum core is surrounded by a titanium sleeve, and both are attached at each end to a rigid end-plate. Set up t

he fundamental equations and combine them in displacement-method order to solve for the following:
a) the elongation of the composite bar if the entire structure is heated by 1000F. Assume that the core and the outer sleeve are both stress free at the reference temperature.
b) the axial stress induced in each material.

Engineering
1 answer:
NeX [460]3 years ago
3 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) The elongation of the composite bar is given as δ = 0.072 in

b) The axial stress induced in each material is = 5485.7 psi

Explanation:

The explanation to the answer above is shown on the second uploaded image

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Answer:

a. 121 Btu/lb

b. 211.8lb

c. 2.69/pc

Explanation:

See the attachments please

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Imagine you are designing a new backpack. Which of the following statement(s) about
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The answer is D.) both A and C
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If a material is found to be in the tertiary phase of creep, the following procedure should be implemented:
kirill [66]

When a material is found to be in the tertiary phase of creep, the following procedure should be implemented that is the component should be replaced immediately. Therefore, Option C is correct.

<h3>What do you mean by a tertiary degree of creep?</h3>

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8 0
2 years ago
A three-phase line has a impedance of 0.4+j2.7 per phase. The line feeds 2 balanced three-phase loads that are connected in para
mamaluj [8]

Answer:

a) 4160 V

b) 12 kW and 81 kVAR

c)  54 kW and 477 kVAR

Explanation:

1) The phase voltage is given as:

V_p=\frac{3810.5}{\sqrt{3} }=2200 V

The complex power S is given as:

S=560.1(0.707 +j0.707)+132=660\angle 36.87^o \ KVA

where\ S^*\ is \ the \ conjugate\ of \ S\\Therefore\ S^*=660\angle -36.87^oKVA

The line current I is given as:

I=\frac{S^*}{3V}=\frac{660000\angle -36.87}{3(2200)}  =100\angle -36.87^o\ A

The phase voltage at the sending end is:

V_s=2200\angle 0+100\angle -36.87(0.4+j2.7)=2401.7\angle 4.58^oV

The magnitude of the line voltage at the source end of the line (V_{sL}=\sqrt{3} |V_s|=\sqrt{3} *2401.7=4160V

b) The Total real and reactive power loss in the line is:

S_l=3|I|^2(R+jX)=3|100|^2(0.4+j2.7)=12000+j81000

The real power loss is 12000 W = 12 kW

The reactive power loss is 81000 kVAR = 81 kVAR

c) The sending power is:

S_s=3V_sI^*=3(2401.7\angle 4.58)(100\angle 36.87)=54000+j477000

The Real power delivered by the supply = 54000 W = 54 kW

The Reactive power delivered by the supply = 477000 VAR = 477 kVAR

5 0
4 years ago
For some transformation having kinetics that obey the Avrami equation , the parameter n is known to have a value of 1.1. If, aft
ivanzaharov [21]

Answer:

total time  = 304.21 s

Explanation:

given data

y = 50% = 0.5

n = 1.1

t = 114 s

y = 1 - exp(-kt^n)

solution

first we get here k value by given equation

y = 1 - e^{(-kt^n)}   ...........1    

put here value and we get

0.5 = 1 - e^{(-k(114)^{1.1})}    

solve it we get

k = 0.003786  = 37.86 × 10^{4}

so here

y = 1 - e^{(-kt^n)}

1 - y  =  e^{(-kt^n)}

take ln both side

ln(1-y) = -k × t^n  

so

t = \sqrt[n]{-\frac{ln(1-y)}{k}}    .............2

now we will put the value of y = 87% in equation  with k and find out t

t = \sqrt[1.1]{-\frac{ln(1-0.87)}{37.86*10^{-4}}}

total time  = 304.21 s

7 0
3 years ago
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