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Alekssandra [29.7K]
3 years ago
6

technician a uses a current output test to check ac generator output. technician b uses a voltage output test to check output. w

ho is correct?
Engineering
1 answer:
expeople1 [14]3 years ago
7 0

Answer:

both

Explanation:

Both the technician are correct, ac generator output can be tested in both ways. The two ways are  current output test to check ac generator output. and  voltage output test to check output.

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For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration will raise the carbon concentratio
diamong [38]

This question is incomplete, the complete question is;

For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration at Temperature T1 will raise the carbon concentration to 0.44 wt% at a point 1.8 mm from the surface. A separate experiment is performed at T2 that doubles the diffusion coefficient for carbon in steel.

Estimate the time necessary to achieve the same concentration at a 4.9 mm position for an identical steel and at the same carburizing temperature T2.

Answer:

the required time to achieve the same concentration at a 4.9 is 83.733 hrs

Explanation:

Given the data in the question;

treatment time t₁ = 11.3 hours

Carbon concentration = 0.444 wt%

thickness at surface x₁ = 1.8 mm = 0.0018 m

thickness at identical steel x₂ = 4.9 mm = 0.0049 m

Now, Using Fick's second law inform of diffusion

x^2 / Dt = constant

where D is constant

then

x^2 / t = constant

x^2_1 / t₁ = x^2_2 / t₂

x^2_1 t₂ = t₁x^2_2

t₂ = t₁x^2_2 / x^2_1

t₂ = (x^2_2 / x^2_1)t₁

t₂ = ( x_2 / x_1 )^2 × t₁

so we substitute

t₂ = ( 0.0049  / 0.0018  )^2 × 11.3 hrs

t₂ = 7.41 × 11.3 hrs

t₂ = 83.733 hrs

Therefore, the required time to achieve the same concentration at a 4.9 is 83.733 hrs

8 0
2 years ago
Marcus wants to pursue a career in civil engineering. He aims to work for the city council as a civil engineer. What examination
Rzqust [24]

Answer:

Marcus would have to take an exam administered by the national council of examiners for engineering and surveying.

Explanation:

Civil engineers design, construct, and maintain projects regarding infrastructure. A civil engineer also looks after the systems in the public and private sectors like roads, buildings, and systems for water supply and sewage treatment.

In order to pursue a career in civil engineering, Marcus aims to work for the city council as a civil engineer. Therefore, he would have to take an exam administered by the national council of examiners for engineering and surveying.

6 0
2 years ago
Is my paper's main idea, or thesis, clearly stated early on (within the first paragraph, ideally)?
Burka [1]
I dont know is your papers main idea stated clearly?
5 0
2 years ago
3. In order to obtain your commercial driver's license (CDL) you must first:
Murljashka [212]
A and C is the answer to the question. Be 15 years old & get a permit
8 0
2 years ago
An ideal vapor-compression refrigeration cycle operates at steady state with Refrigerant 134a as the working fluid. Saturated va
Ksju [112]

Explanation:

Note: Refer the diagram below

Obtaining data from property tables

State 1:

\left.\begin{array}{l}P_{1}=1.25 \text { bar } \\\text { Sat - vapour }\end{array}\right\} \begin{array}{l}h_{1}=234.45 \mathrm{kJ} / \mathrm{kg} \\S_{1}=0.9346 \mathrm{kJ} / \mathrm{kgk}\end{array}

State 2:

\left.\begin{array}{l}P_{2}=5 \text { bor } \\S_{2}=S_{1}\end{array}\right\} \quad h_{2}=262.78 \mathrm{kJ} / \mathrm{kg}

State 3:

\left.\begin{array}{l}P_{3}=5 \text { bar } \\\text { Sat }-4 q\end{array}\right\} h_{3}=71-33 \mathrm{kJ} / \mathrm{kg}

State 4:

Throttling process  h_{4}=h_{3}=71.33 \mathrm{kJ} / \mathrm{kg}

(a)

Magnitude of compressor power input

\dot{w}_{c}=\dot{m}\left(h_{2}-h_{1}\right)=\left(8 \cdot 5 \frac{\mathrm{kg}}{\min } \times \frac{1 \mathrm{min}}{\csc }\right)(262.78-234 \cdot 45)\frac{kj}{kg}

w_{c}=4 \cdot 013 \mathrm{kw}

(b)

Refrigerator capacity

Q_{i n}=\dot{m}\left(h_{1}-h_{4}\right)=\left(\frac{g \cdot s}{60} k_{0} / s\right) \times(234 \cdot 45-71 \cdot 33) \frac{k J}{k_{8}}

Q_{i n}=23 \cdot 108 \mathrm{kW}\\1 ton of retregiration =3.51 k \omega

\ Q_{in} =6 \cdot 583 \text { tons }

(c)

Cop:

\beta=\frac{\left(h_{1}-h_{4}\right)}{\left(h_{2}-h_{1}\right)}=\frac{Q_{i n}}{\omega_{c}}=\frac{23 \cdot 108}{4 \cdot 013}

\beta=5 \cdot 758

3 0
2 years ago
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