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Ronch [10]
2 years ago
10

Does anyone know what this is​

Engineering
2 answers:
attashe74 [19]2 years ago
6 0

Answer:

it looked like their something for whatever pressure or and electrical thingy

sammy [17]2 years ago
4 0

Answer:

Looks like mold that got frosted over

Explanation:

You might be interested in
1. A pipeline constructed of carbon steel failed after 3 years of operation. On examination it was found that the wall thickness
jek_recluse [69]

Answer:

check the explanation

Explanation:

1.

Thickness Loss = t =\frac{t_{o}-t_{i}}{2} = \frac{114.3-102.3}{2} = 2mm

t_{f} = \frac{1}{2}*6 = 3mm

Hence Rate of Corrosion = 6*\frac{1-0.5}{3} = 1mm/year = 0.03 inches per year

2.

As the expected future life is 7 years,

40 carbon steel pipe has to be replaced every 3 years as given in the question,

Cost per unit length is the sum of material cost and installation cost.

Cost of 40 carbon steel = (5 dollars + 16.5 dollars) * 3 = 64.5 dollars

For 80 carbon steel pipe, first calculate the thickness loss,

\frac{114.3-97.2}{2} = 8.55mm

The critical thickness is given to be 3mm, Hence change in thickness is 8.55-3 = 5.5mm

This 80 carbon steel pipe has to be replaced one more time

Hence, Cost per unit length is the sum of material cost and installation cost.

Cost of 80 carbon steel = (8.3 dollars + 16.5 dollars) * 2 = 49.6 dollars

The best is of stainless steel which does not undergo corrosion at all and thus it needs to be replaced only once throughout the plant operation. Its cost = 24.8 dollars + 16.5 dollars = 41.3 dollars

Hence, stainless steel is the recommended pipe to be used.

3 0
3 years ago
What is an air mass?​
kotegsom [21]

Answer:

An air mass is a body of air with horizontally uniform temperature, humidity, and pressure.

Explanation:

Because it is

8 0
3 years ago
Read 2 more answers
Twenty-five wooden beams were ordered or a construction project. The sample mean and he sample standard deviation were measured
aksik [14]

Answer:

Correct option: B. 90%

Explanation:

The confidence interval is given by:

CI = [\bar{x} - z\sigma_{\bar{x}} , \bar{x}+z\sigma_{\bar{x}} ]

If \bar{x} is 190, we can find the value of z\sigma_{\bar{x}}:

\bar{x} - z\sigma_{\bar{x}}  = 188.29

190 - z\sigma_{\bar{x}}  = 188.29

z\sigma_{\bar{x}}  = 1.71

Now we need to find the value of \sigma_{\bar{x}}:

\sigma_{\bar{x}} = s / \sqrt{n}

\sigma_{\bar{x}} = 5/ \sqrt{25}

\sigma_{\bar{x}} = 1

So the value of z is 1.71.

Looking at the z-table, the z value that gives a z-score of 1.71 is 0.0436

This value will occur in both sides of the normal curve, so the confidence level is:

CI = 1 - 2*0.0436 = 0.9128 = 91.28\%

The nearest CI in the options is 90%, so the correct option is B.

4 0
3 years ago
A cylindrical specimen of some metal alloy having an elastic modulus of 124 GPa and an original cross-sectional diameter of 4.2
IrinaVladis [17]

Answer:

the maximum length of the specimen before deformation is 0.4366 m

Explanation:

Given the data in the question;

Elastic modulus E = 124 GPa = 124 × 10⁹ Nm⁻²

cross-sectional diameter D = 4.2 mm = 4.2 × 10⁻³ m

tensile load F = 1810 N

maximum allowable elongation Δl = 0.46 mm = 0.46 × 10⁻³ m

Now to calculate the maximum length l for the deformation, we use the following relation;

l = [ Δl × E × π × D² ] / 4F

so we substitute our values into the formula

l = [ (0.46 × 10⁻³) × (124 × 10⁹) × π × (4.2 × 10⁻³)² ] / ( 4 × 1810 )

l = 3161.025289 / 7240

l = 0.4366 m

Therefore, the maximum length of the specimen before deformation is 0.4366 m

5 0
3 years ago
Carnot heat engine A operates between 20ºC and 520ºC. Carnot heat engine B operates between 20ºC and 820ºC. Which Carnot heat en
nikklg [1K]

Answer:

engine B is more efficient.

Explanation:

We know that Carnot cycle is an ideal cycle for all working heat engine.In Carnot cycle there are four processes in which two are constant temperature processes and others two are isentropic process.

We also kn ow that the efficiency of Carnot cycle given as follows  

\eta =1-\dfrac{T_1}{T_2}

Here temperature should be in Kelvin.

For engine A

\eta =1-\dfrac{T_1}{T_2}

\eta =1-\dfrac{273+20}{520+273}

\eta =0.63

For engine B

\eta =1-\dfrac{T_1}{T_2}

\eta =1-\dfrac{273+20}{820+273}

\eta =0.73

So from above we can say that engine B is more efficient.

4 0
3 years ago
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