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3 years ago

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Large quantities of liquefied natural gas (LNG) are shipped by ocean tanker. At the unloading port, provision is made for vapori

zation of the LNG so that it may be delivered to pipelines as gas. The LNG arrives in the tanker at atmospheric pressure and 113.7 K and represents a possible heat sink for use as the cold reservoir of a heat engine. Assuming unloading of LNG as a vapor at the rate of 8,000 m3s-1, as measured at 25 C and 1.0133 bar, and assuming the availability of an adequate heat source at 35 C, what is the maximum amount of work that could be generated and what is the rate of heat transfer from the heat source

1 answer:

kodGreya [7K]3 years ago

6 0

Answer:

yfyuhvckydfxcvkjyfvgrjfvhkhgfyhtfhchghjgyfvyghvhygvkuh

Explanation:

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Two routes connect an origin and a destination. Routes 1 and 2 have performance functions t1 = 2 + X1 and t2 = 1 + X2, where the

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Solution :

Given

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$t_2=1+x_2$

Now,

$$P(h$

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Therefore, $x_1=368 \ veh/h$

$=\frac{368}{1000} = 0.368$

Given, $t_1=2+x_1$

= 2 + 0.368

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∴ $t_2$ = 2.368 min

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$2.368=1+x_2$

$x_2 = 1.368$

$x_2 = 1.368 \times 1000$

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A farmer has 12 hectares of land on which he grows corn, wheat, and soybeans. It costs $4500 per hectare to grow corn, $6000 to

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The number of hectares of each crop he should plant are; 250 hectares of Corn, 500 hectares of Wheat and 450 hectares of soybeans

<h3>How to solve algebra word problem?</h3>

He grows corn, wheat and soya beans on the farm of 1200 hectares. Thus;

C + W + S = 12 ----(1)

It costs $45 per hectare to grow corn, $60 to grow wheat, and $50 to grow soybeans. Thus;

45C + 60W + 50S = 63750 -----(2)

He will grow twice as many hectares of wheat as corn. Thus;

W = 2C ------(3)

Put 2C for W in eq 1 and eq 2 to get;

C + 2C + S = 1200

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45C + 60(2C) + 50S = 63750

45C + 120C + 50S = 63750

165C + 50S = 63750 ------(5)

Solving eq 4 and 5 simultaneosly gives;

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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