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Bingel [31]
2 years ago
12

On the generalized enthalpy departure chart, the normalized enthalpy departure values seem to approach zero as the reduced press

ure PR approaches zero. How do you explain this behavior?
Engineering
1 answer:
Alekssandra [29.7K]2 years ago
4 0

Answer:

Enthalpy is a function of pressure hence normalized enthalpy departure values will approach zero with reduced pressure approaching zero

Explanation:

On the generalized enthalpy departure chart, the normalized enthalpy departure values seem to approach zero as the reduced pressure PR approaches zero. this is because enthalpy is a function of pressure therefore as the Pressure is reducing towards the zero value, the gas associated with the pressure tends to behave more like an Ideal gas.

For an Ideal gas the Normalized enthalpy departure value will be approaching the zero value.

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The fatigue data for a brass alloy are given as follows: Stress Amplitude (MPa) Cycles to Failure 170 3.7 × 104 148 1.0 × 105 13
prohojiy [21]

Answer:

i) S–N plot is attached

ii) fatigue strength = 100 MPa

iii) fatigue life = 5.62 x 10^(5) cycles

Explanation:

i) I have attached the S–N plot (stress amplitude versus logarithm of cycles to failure)

ii) The question says we should find the fatigue strength at 4 × 10^(6) cycles.

So let's find the log of this and trace it on the graph attached.

Log(4 × 10^(6)) = 6.6

From the graph attached, at log of cycle value of 6.6, the fatigue strength is approximately 100 MPa

iii) The question says we should find the fatigue life for 120 MPa.

Thus, from the graph, at stress amplitude of 120 MPa, the log of cycles is approximately 5.75.

Thus,the fatigue life will be the inverse log of 5.75.

Thus, fatigue life = 10^(5.75)

Fatigue life = 5.62 x 10^(5)

8 0
2 years ago
Six years ago, an 80-kW diesel electric set cost $160,000. The cost index for this class of equipment six years ago was 187 and
exis [7]

Answer:

new boiler total cost = $229706.825

new boiler total cost = $127512

Explanation:

given data

power p1 = 80 kW

cost C = $160000

cost index CI 1 = 187

cost index CI 2= 194

cost capacity factor f = 0.6

power p2 = 120 kW

current cost = $18000

to find out

total cost and cost of 40 kW

solution

we consider here CN cost for new boiler and CO cost for old boiler

and x is capacity of new boiler

first we find old boiler current cost that is

current cost CO = C × \frac{CI 1 }{CI 2 }   .............1

put here value

current cost = 160000 × \frac{194 }{187 }

new current cost = $165989.304

and

use here power sizing technique for 124 kW

CN/CO = (\frac{p2}{p1} )^{f}    ...............2

put here value and find CN

CN/CO = (\frac{p2}{p1} )^{f}  

CN / 165989.304 = (\frac{120}{80} )^{0.6}  

CN = 211706.825

so new cost = $211706.825

so

total cost for new boiler is

total cost = new cost + current cost

total cost = 211706.825 + 18000

new boiler total cost = $229706.825

and

for 40 kW new cost will be

use equation 2

CN/CO = (\frac{p2}{p1} )^{f}

CN / 165989.304 = (\frac{40}{80} )^{0.6}  

CN = 109512

so new cost is $109512

so

total cost for new boiler is

total cost = new cost + current cost

total cost = 109512 + 18000

new boiler total cost = $127512

7 0
2 years ago
Consider insulation on a circular pipe For the same thickness and type of insulation, the thermal resistance of the insulation i
leonid [27]

Answer:

b). The same for all pipes independent of the diameter

Explanation:

We know,

R_{conduction}=\frac{ln(\frac{r_{2}}{r_{1}})}{2\pi LK}

R_{convection}=\frac{1}{h(2\pi r_{2}L)}

From the above formulas we can conclude that the thermal resistance of a substance mainly depends upon heat transfer coefficient,whereas radius has negligible effects on heat transfer coefficient.

We also know,

Factors on which thermal resistance of insulation depends are :

1. Thickness of the insulation

2. Thermal conductivity of the insulating material.

Therefore from above observation we can conclude that the thermal resistance of the insulation is same for all pipes independent of diameter.

5 0
2 years ago
Identify the four engineering economy symbols and their values from the following problem statement. Use a question mark with th
Vilka [71]

Answer:

The company will have $311,424 in its investment set-aside account.

Explanation:

To determine the amount of money that the company will have after 7 years with an interest rate of 11% per year, we must calculate the price to start with an increase of 11%, and then repeat the operation until reaching seven years:

Year 0: 150,000

Year 1: 150,000 x 1.11 = 166,500

Year 2: 166,000 x 1.11 = 184,815

Year 3: 184,815 x 1.11 = 205,144.65

Year 4: 205,144.65 x 1.11 = 227,710.56

Year 5: 227,710.56 x 1.11 = 252,758.72

Year 6: 252,758.72 x 1.11 = 280,562.18

Year 7: 280,562.18 x 1.11 = 311,424

3 0
2 years ago
In which of the following branches of engineering is the practice not restricted?
fgiga [73]

Answer:

a) civil engineering.

Explanation:

Civil engineering is a professional engineering program that deals with the construction, design, and maintenance of all the natural and man-made environments including dams, buildings, railways, and roads.

Civil engineering is the branch of engineering that is the practice not restricted because civil engineer is not restricted to academic profession but practice in designing and construction can make someone a professional civil engineer.

Hence, the correct answer is "a)."

6 0
2 years ago
Read 2 more answers
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