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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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A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa √m (50 ksi √in.). If, during

astra-53 [7]

Answer:

0.024 m = 24.07 mm

Explanation:

1) Notation

$\sigma_c$ = tensile stress = 200 Mpa

$K$ = plane strain fracture toughness= 55 Mpa $\sqrt{m}$

$\lambda$ = length of a surface crack (Variable of interest)

2) Definition and Formulas

The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.

By definition we have the following formula for the tensile stress:

$\sigma_c=\frac{K}{Y\sqrt{\pi\lambda}}$ (1)

We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for $\lambda$

Multiplying both sides of equation (1) by $Y\sqrt{\pi\lambda}$

$\sigma_c Y\sqrt{\pi\lambda}=K$ (2)

Sequaring both sides of equation (2):

$(\sigma_c Y\sqrt{\pi\lambda})^2=(K)^2$

$\sigma^2_c Y^2 \pi\lambda=K^2$ (3)

Dividing both sides by $\sigma^2_c Y^2 \pi$ we got:

$\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2$ (4)

Replacing the values into equation (4) we got:

$\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m$

3) Final solution

So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.

7 0

3 years ago

An asphalt concrete mixture includes 94% aggregate by weight. The specific gravities of aggregate and asphalt are 2.65 and 1.0,

garik1379 [7]

Answer:

2.0%

Explanation:

Percentage of aggregate = 94%

Specific gravity = 2.65

Specific gravity of asphalt = 1.9

Density of mix = 147pcf = 147lb/ft³

Total weight of mix: (volume = 1ft³)

= (147lb/ft³)(1ft³)

= 147lb

Percentage weight of asphalt in<u> mix:</u>

100% - 94%

= 6%

Weight of asphalt binders

= 6% x 147lb

= 8.82lb

Weight of aggregate in mix:

= 94% x 147

= 138.18lb

Specific weight of asphalt binder:

(Gab)(Yw)

Yw = specific Weight of water

= 62.4lb

Gab = specific gravity of asphalt binder

= 1.0

(62.4lb)(1.0)

= 62.4 lb/ft³

Volume of asphalt in binder:

8.82/62.4

= 0.14ft³

Specific weight of binder in mix:

2.65 x 62.4lb/ft³

= 165.36 lb/ft³

Volume of aggregate:

= 138.18/165.36

= 0.84ft³

Volume of void in the mix:

1ft³ - 0.84ft³ - 0.14ft³

= 0.02ft³

<u>The percentage of void in total mix:</u>

VTM = (0.02ft³/1ft³)100

= 2.0%

8 0

3 years ago

A polyethylene rod exactly 10 inches long with a cross-sectional area of 0.04 in2 is used to suspend a weight of 358 lbs-f (poun

Nadya [2.5K]

Answer:

Final length of the rod = 13.90 in

Explanation:

Cross Sectional Area of the polythene rod, A = 0.04 in²

Original length of the polythene rod, l = 10 inches

Tensile modulus for the polymer, E = 25,000 psi

Viscosity, $\eta = 1*10^{9} psi -sec$

Weight = 358 lbs - f

time, t = 1 hr = 3600 sec

Stress is given by:

$\sigma = \frac{Force}{Area} \\\sigma = \frac{358}{0.04} \\\sigma = 8950 psi$

Based on Maxwell's equation, the strain is given by:

$strain = \sigma ( \frac{1}{E} + \frac{t}{\eta} )\\Strain = 8950 ( \frac{1}{25000} + \frac{3600}{10^{9} } )\\Strain = 0.39022$

Strain = Extension/(original Length)

0.39022 = Extension/10

Extension = 0.39022 * 10

Extension = 3.9022 in

Extension = Final length - Original length

3.9022 = Final length - 10

Final length = 10 + 3.9022

Final length = 13.9022 in

Final length = 13.90 in

7 0

3 years ago

Consider a pan of water being heated (a) by placing it on an electric range and (b) by placing a heating element in the water. W

Brrunno [24]

Answer:

Method B is the more efficient way of heating the water.

Explanation:

Method B is more efficient because by placing a heating element in the water as in described in method B, the heat that is lost to the surroundings is minimized which implies that more heat is supplied directly to the water. Therefore, more heating is achieved with a lesser amount of electrical energy input. Whereas placing the pan on a range means more heat losses to the surrounding and as such it will take a longer time for the water to heat up and also take more electrical energy.

7 0

3 years ago

A 5-in.-diameter pipe is supported every 9 ft by a small frame consisting of two members asshown. Knowing that the combined weig

jarptica [38.1K]

Answer:

AC: at D , M_max = 12.25 lb-ft

BC: at E , M_max = 8.75 lb-ft

Explanation:

Given:

- The diameter of the pipe d = 5-in

- The pipe is supported every L = 9 ft of pipe in length

- The weight if the pipe + contents W = 10 lb/ft

Find:

determine the magnitude and location of the maximum bending moment in members AC and BC.

Solution:

- The figure (missing) is given in the attachment.

- We will first determine the external forces acting on each member:

Section: 9-ft section of pipe.

Sum of forces perpendicular to member AC = 0

F_d - 0.8*W*L = 0

F_d = 0.8*10*9 = 72 lb

Sum of forces perpendicular to member BC = 0

F_e - 0.6*W*L = 0

F_e = 0.6*10*9 = 54 lb

F_d = 72 lb , F_e = 54 lb

- Then we will determine the support reactions for each member AC point A and BC point B.

Section: Entire Frame.

Sum of moments about point B = 0

-A_y*(18.75/12) + F_d*(d /2*12) + F_e*((11.25-2.5)/12) = 0

-A_y*(1.5625) + 15 + 39.375 = 0

A_y = 34.8 lb

Sum of forces in vertical direction = 0

A_y + B_y - 0.8*F_d - 0.6*F_e = 0

B_y = 0.8*(72) + 0.6*(54) - 34.8

B_y = 55.2 lb

Sum of forces in horizontal direction = 0

A_x + B_x - 0.6*F_d + 0.8*F_e = 0

A_x + B_x = 0

Section: Member AC

Sum of moments about point C = 0

F_d*(2.5/12) - A_y*(12/12) - A_x*(9/12) = 0

72*2.5 - 34.8*12 - 9*A_x = 0

A_x = -237.6 / 9 = - 26.4 lb

B_x = - A_x = 26.4 lb

A_x = -26.4 lb , B_x = 26.4 lb

- Now we will calculate bending moment for each member at different sections.

Member AC:

From point A till just before point D

-0.6*A_x*x - A_y*0.8*x + M = 0

15.84*x - 27.84*x + M = 0

M = 12*x ..... max value at D, x = 12.25 in

M_max = 12*12.25/12 = 12.25 lb-ft

Member BC:

From point B till just before point E

-0.8*B_x*x + B_y*0.6*x + M = 0

-21.12*x + 33.12*x + M = 0

M = -12*x ..... max value at E, x = 11.25 - 2.5 = 8.75 in

M_max = -12*8.75/12 = -8.75 lb-ft

- The maximum bending moments and their locations are:

AC: at D , M_max = 12.25 lb-ft

BC: at E , M_max = 8.75 lb-ft

5 0

3 years ago