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zloy xaker [14]
3 years ago
6

**Please Help, ASAP**

Engineering
1 answer:
algol133 years ago
5 0

Explanation:

1) x= y/(1+y)

2)x= -y/(1-y)

3) x=y/(1-y)

4)square root (2E/m-v^2)

5) (( x^2b^2- a^2b^2)/a^2) square root

6) square root( x^3/a)

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Kindly check the attached image below to see the step by step explanation to the above question.

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Who's your favorite singer and WHT your favorite song​
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Which of the following sensors is used to provide suspension control module with feedback regarding vehicle cornering​ forces?
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2 years ago
Ammonia enters the expansion valve of a refrigeration system at a pressure of 1.4 MPa and a temperature of 32degreeC and exits a
AveGali [126]

Answer:

the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %

Explanation:

given data

pressure p1 = 1.4 MPa = 14 bar

temperature t1 = 32°C

exit pressure = 0.08 MPa = 0.8 bar

to find out

the quality of the refrigerant exiting the expansion valve

solution

we know here refrigerant undergoes at throtting process so

h1 = h2

so by table A 14 at p1 = 14 bar

t1 ≤ Tsat

so we use equation here that is

h1 = hf(t1) = 332.17 kJ/kg

this value we get from table A13

so as h1 = h2

h1 = h(f2)  + x(2) * h(fg2)

so

exit quality  = \frac{h1 - h(f2)}{h(fg2)}

exit quality  = \frac{332.17- 9.04}{1382.73)}

so exit quality = 0.2337 = 23.37 %

the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %

5 0
3 years ago
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
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