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Naya [18.7K]
3 years ago
6

Work-producing devices that operate on reversible processes deliver the most work, and work-consuming devices that operate on re

versible processes consume the least work. a)- True b)- False
Engineering
1 answer:
Margarita [4]3 years ago
4 0

Answer: True

Explanation:

Yes, it is true that work producing device that can operate on reversible process deliver the most work as, the process must be quasi equilibrium to be reversible and work consuming device that operate on reversible process consume the least work as, energy released by the reversible process can do maximum amount of work as, less energy lost as heat and energy loss is very less in reversible energy and less work require to operate as consuming device.

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An ideal gas initially occupying 0.020 m3 at 1.0 MPa is quasistatically expanded inside a piston-cylinder device at a constant p
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An ideal gas initially occupying 0.020 m3 at 1.0 MPa is quasistatically expanded inside a piston-cylinder device at a constant pressure until its volume doubles. Next the expansion is continued at constant volume till the pressure reaches half of the initial pressure. Finally it is brought back to the initial state in a polytropic process with exponent n=1.6

a. Draw the processes on a P-v diagram and calculate the total work.
b. Calculate the total heat transfered, what is the difference between the initial and final temperature?an answer is to present a question of how you are not able to join the world and how you can help please answer
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3 years ago
Consider a cylinder of height h, diameter d, and wall thickness t pressurized to an internal pressure P_0 (gauge pressure, relat
Serggg [28]

Explanation:

Note: For equations refer the attached document!

The net upward pressure force per unit height p*D must be balanced by the downward tensile force per unit height 2T, a force that can also be expressed as a stress, σhoop, times area 2t. Equating and solving for σh gives:

 Eq 1

Similarly, the axial stress σaxial can be calculated by dividing the total force on the end of the can, pA=pπ(D/2)2 by the cross sectional area of the wall, πDt, giving:

Eq 2

For a flat sheet in biaxial tension, the strain in a given direction such as the ‘hoop’ tangential direction is given by the following constitutive relation - with Young’s modulus E and Poisson’s ratio ν:

 Eq 3

 

Finally, solving for unknown pressure as a function of hoop strain:

 Eq 4

 

Resistance of a conductor of length L, cross-sectional area A, and resistivity ρ is

 Eq 5

Consequently, a small differential change in ΔR/R can be expressed as

 Eq 6

Where ΔL/L is longitudinal strain ε, and ΔA/A is –2νε where ν is the Poisson’s ratio of the resistive material. Substitution and factoring out ε from the right hand side leaves

 Eq 7

Where Δρ/ρε can be considered nearly constant, and thus the parenthetical term effectively becomes a single constant, the gage factor, GF

 Eq 8

For Wheat stone bridge:

 Eq 9

Given that R1=R3=R4=Ro, and R2 (the strain gage) = Ro + ΔR, substituting into equation above:

Eq9

Substituting e with respective stress-strain relation

Eq 10

 

part b

Since, axial strain(1-2v) < hoop strain (2-v). V out axial < V out hoop.

Hence, dV hoop < dV axial.

part c  

Given data:

P = 253313 Pa

D = d + 2t = 0.09013 m

t = 65 um

GF = 2

E = 75 GPa

v = 0.33

Use the data above and compute Vout using Eq3

Eq 11  

Download docx
3 0
3 years ago
What is engineering
kirill [66]

Answer:

Engineering is the use of scientific principles to design and build machines, structures, and other items, including bridges, tunnels, roads, vehicles, and buildings. In other words, Engineering is the use of science and math to design or make things.

Explanation:

7 0
3 years ago
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The supplement file* that enclosed to this homework consists Time Versus Force data. The first column in the file stands for tim
AysviL [449]

Answer:

A.) 1mv = 2000N

B.) Impulse = 60Ns

C.) Acceleration = 66.67 m/s^2

Velocity = 4 m/s

Displacement = 0.075 metre

Absorbed energy = 60 J

Explanation:

A.) Using a mathematical linear equation,

Y = MX + C

Where M = (2000 - 0)/( 898 - 0 )

M = 2000/898

M = 2.23

Let Y = 2000 and X = 898

2000 = 2.23(898) + C

2000 = 2000 + C

C = 0

We can therefore conclude that

1 mV = 2000N

B.) Impulse is the product of force and time.

Also, impulse = momentum

Given that

Mass M = 30kg

Velocity V = 2 m/s

Impulse = M × V = momentum

Impulse = 30 × 2 = 60 Ns

C.) Force = mass × acceleration

F = ma

Substitute force and mass into the formula

2000 = 30a

Make a the subject of formula

a = 2000/30

acceleration a = 66.67 m/s^2

Since impulse = 60 Ns

From Newton 2nd law,

Force = rate of change in momentum

Where

change in momentum = -MV - (- MU)

Impulse = -MV + MU

Where U = initial velocity

60 = -60 + MU

30U = 120

U = 120/30

U = 4 m/s

Force = 2000N

Impulse = Ft

Substitute force and impulse to get time

60 = 2000t

t = 60/2000

t = 0.03 second

Using third equation of motion

V^2 = U^2 + 2as

Where S = displacement

4^2 = 2^2 + 2 × 66.67S

16 = 4 + 133.4S

133.4S = 10

S = 10/133.4

S = 0.075 metre

D.) Energy = 1/2 mV^2

Energy = 0.5 × 30 × 2^2

Energy = 15 × 4 = 60J

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4 years ago
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DedPeter [7]

Answer:

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Explanation:

4 0
3 years ago
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