Answer:
a. 121 Btu/lb
b. 211.8lb
c. 2.69/pc
Explanation:
See the attachments please
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>
Tertiary Creep has an extended creep rate and terminates when the material breaks or ruptures. It is related to each necking and formation of grain boundary voids. The wide variety of possible stress-temperature- time combos is infinite.
Therefore, 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. Option C is correct.
Learn more about creep:
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Answer:
a) 4160 V
b) 12 kW and 81 kVAR
c) 54 kW and 477 kVAR
Explanation:
1) The phase voltage is given as:

The complex power S is given as:


The line current I is given as:

The phase voltage at the sending end is:

The magnitude of the line voltage at the source end of the line (
b) The Total real and reactive power loss in the line is:

The real power loss is 12000 W = 12 kW
The reactive power loss is 81000 kVAR = 81 kVAR
c) The sending power is:

The Real power delivered by the supply = 54000 W = 54 kW
The Reactive power delivered by the supply = 477000 VAR = 477 kVAR
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 -
...........1
put here value and we get
0.5 = 1 - e^{(-k(114)^{1.1})}
solve it we get
k = 0.003786 = 37.86 ×
so here
y = 1 -
1 - y =
take ln both side
ln(1-y) = -k ×
so
t =
.............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}}}](https://tex.z-dn.net/?f=%5Csqrt%5B1.1%5D%7B-%5Cfrac%7Bln%281-0.87%29%7D%7B37.86%2A10%5E%7B-4%7D%7D%7D)
total time = 304.21 s