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3 years ago

What is the area enclosed by the cycle area of the Carnot cycle illustrating on a P-V diagram?

Engineering

1 answer:

Inga [223]3 years ago

7 0

Answer:

The work of the cycle.

Explanation:

The area enclosed by the cycle of the Pressure-Volume diagram of a Carnot engine represents the net work performed by the cycle.

The expansions yield work, and this is represented by the area under the curve all the way to the p=0 line. But the compressions consume work (or add negative work) and this is substracted fro the total work. Therefore the areas under the compressions are eliminated and you are left with only the enclosed area.

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Answer:

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3 years ago

A masonry facade consisting of 3,800 square feet is to be constructed for a building. The total cost per worker hour is estimate

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2 years ago

The state of plane strain on an element is:

balu736 [363]

Answer:

a. ε₁=-0.000317

ε₂=0.000017

θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain =3.335 *10^-4

Associated average normal strain ε(avg) =150 *10^-6

θ = 31.71 or -58.29

Explanation:

$\epsilon _{1,2} =\frac{\epsilon_x + \epsilon_y}{2} \pm \sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\epsilon _{1,2} =\frac{-300 \times 10^{-6} + 0}{2} \pm \sqrt{(\frac{-300 \times 10^{-6}+ 0}{2}) ^2 + (\frac{150 \times 10^-6}{2})^2}\\\\\epsilon _{1,2} = -150 \times 10^{-6} \pm 1.67 \times 10^{-4}$

ε₁=-0.000317

ε₂=0.000017

To determine the orientation of ε₁ and ε₂

$tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5$

θ= -13.28° and 76.72°

To determine the direction of ε₁ and ε₂

$\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2} + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2} + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56) + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\$

=-0.000284 -0.0000335 = -0.000317 =ε₁

Therefore θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain

$\gamma_{max \ in \ plane} =2\sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\gamma_{max \ in \ plane} = 2\sqrt{(\frac{-300 *10^{-6} + 0}{2} )^2 + (\frac{150 *10^{-6}}{2})^2}$

=3.335 *10^-4

$\epsilon_{avg} =(\frac{\epsilon_x + \epsilon_y}{2} )$

ε(avg) =150 *10^-6

orientation of γmax

$tan 2 \theta_s = \frac{-(\epsilon_x - \epsilon_y)}{\gamma_xy} \\\\tan 2 \theta_s = \frac{-(-300*10^{-6} - 0)}{150*10^{-6}}$

θ = 31.71 or -58.29

To determine the direction of γmax

$\gamma _{x'y' }= - \frac{\epsilon_x -\epsilon_y}{2} sin2\theta + \frac{\gamma_xy}{2}cos2\theta \\\\\gamma _{x'y' }= - \frac{-300*10^{-6} - \ 0}{2} sin(63.42) + \frac{150*10^{-6}}{2}cos(63.42)$

= 1.67 *10^-4

4 0

3 years ago

What are the success factors for mechanical engineering?

Eva8 [605]

Answer:

-effective technical skills.

-the ability to work under pressure.

-problem-solving skills.

-creativity.

-interpersonal skills.

-verbal and written communication skills.

-commercial awareness.

-teamworking skills.

Explanation:

is this what ur looking for? if so there ya go lol

7 0

2 years ago

Read 2 more answers

The velocity of a particle which moves along the s-axis is given by = 40 − 3 2/ , ℎ t is in seconds. Calculate the displacement

scoundrel [369]

The displacement ∆S of the particle during the interval from t = 2sec to 4sec is; 210 sec

<h3>How to find the displacement?</h3>

We are given the velocity equation as;

s' = 40 - 3t²

Thus, the speed equation will be gotten by integration of the velocity equation to get;

s = ∫40 - 3t²

s = 40t - ¹/₂t³

Thus, the displacement between times of t = 2 sec and t = 4 sec is;

∆S = [40(4) - ¹/₂(4)³] - [40(2) - ¹/₂(2)³]

∆S = 210 m

Read more about Displacement at; brainly.com/question/4931057

#SPJ1

8 0

2 years ago