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List irreversibilities

natali 33 [55]

Answer:

Some of the irreversibilities are listed below:

Plastic deformation of solids
Transfer of heat over finite difference of temperature
When two fluids are mixed together the process is irreversible
Combustion of a gas
Current flowing through a finite resistor
Diffusion and free compression or expansion of gas
Relative motion of body with force of friction
Processes involving chemical reactions(spontaneous)

6 0

3 years ago

#198. Moment of inertia about center of a segmented bar A bar of width is formed of three uniform segments with lengths and area

zaharov [31]

Complete Complete

The complete question is shown on the first uploaded image

Answer:

The moment of inertia of the bar about the center of mass is

$I_r = 1888.80 \ kg m^2$

Explanation:

The free body diagram is shown on the second uploaded image

From the diagram we see that is

The mass of each segment is

$m_1 = \rho_1 l_1 w = 1 * 6 * 2 = 12$

$m_1 = \rho_2 l_2 w = 8 * 6 * 2 = 96$

$m_1 = \rho_2 l_2 w = 5 * 5 * 2 = 50$

The distance from the origin to the center of the segments i.e the center of masses for the individual segments

$x_2 = \frac{6}{2} + 6 = 9 m$

$x_3 = \frac{4}{2} + 12 = 14 m$

The resultant center of mass is mathematically evaluated as

$x_r = \frac{m_1 * x_1 + m_2 *x_2 + m_3 *x_3}{m_1 + m_2 + m_3}$

$= \frac{12 * 3 + 96 *9 + 50 *14}{12+ 96 + 50}$

$x_r = 10.13m$

The moment of Inertia of each segment of the bar is mathematically evaluated

$I_1 =\frac{m_1}{12}(l_1^2 + w^2)$ = $\frac{12}{12}(1^2 + 2^2)$

$I_1 = 4 \ kgm^2$

$I_2 =\frac{m_2}{12}(l_2^2 + w^2)$ = $\frac{96}{12}(6^2 + 2^2)$

$I_2 = 320 \ kgm^2$

$I_3 =\frac{m_3}{12}(l_3^2 + w^2)$ = $\frac{50}{12}(4^2 + 2^2)$

$I_2 = 83.334 \ kgm^2$

According to parallel axis theorem the moment of inertia about the center ( $x_r$ ) is mathematically evaluated as

$I_r = (I_1 + m_1 r_1^2) + (I_2 + m_2 r_2^2) +(I_3 + m_3 r_3^2)$

$I_r = (I_1 + m_1 |x_r - x_1|^2) + (I_2 + m_2 |x_r - x_2|^2) +(I_3 + m_3 |x_r - x_3|^2)$

$I_r = (4 + 12 |10.13 - 3|^2) + (320 + 96 |10.13 - 9|^2) +(83.334 + 50 |10.13 - 14|^2)$

$I_r = 1888.80 \ kg m^2$

6 0

3 years ago

Water is contained in a rigid vessel of 5 m3 at a quality of 0.8 and a pressure of 1 MPa. If the pressure is reduced to 270.3 kP

professor190 [17]

Answer:

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Explanation:

4 0

4 years ago

Pipe (2) is supported by a pin at bracket C and by tie rod (1). The structure supports a load P at pin B. Tie rod (1) has a diam

Galina-37 [17]

Answer:

P_max = 25204 N

Explanation:

Given:

- Rod 1 : Diameter D = 12 mm , stress_1 = 110 MPa

- Rod 2: OD = 48 mm , thickness t = 5 mm , stress_2 = 65 MPa

- x_1 = 3.5 mm ; x_2 = 2.1 m ; y_1 = 3.7 m

Find:

- Maximum Force P_max that this structure can support.

Solution:

- We will investigate the maximum load that each Rod can bear by computing the normal stress due to applied force and the geometry of the structure.

- The two components of force P normal to rods are:

Rod 1 : P*cos(Q)

Rod 2: - P*sin(Q)

where Q: angle subtended between x_1 and Rod 1 @ A. Hence,

Q = arctan ( y_1 / x_1)

Q = arctan (3.7 / 2.1 ) = 60.422 degrees.

- The normal stress in each Rod due to normal force P are:

Rod 1 : stress_1 = P*cos(Q) / A_1

Rod 2: stress_2 = - P*sin(Q) / A_2

- The cross sectional Area of both rods are A_1 and A_2:

A_1 = pi*D^2 / 4

A_2 = pi*(OD^2 - ID^2) / 4

- The maximum force for the given allowable stresses are:

Rod 1: P_max = stress_1 * A_1 / cos(Q)

P_max = (110*10^6)*pi*0.012^2 / 4*cos(60.422)

P_max = 25203.61848 N

Rod 2: P_max = stress_2 * A_2 / sin(Q)

P_max = (65*10^6)*pi*(0.048^2 - 0.038^2) / 4*sin(60.422)

P_max = 50483.4 N

- The maximum force that the structure can with-stand is governed by the member of the structure that fails first. In our case Rod 1 with P_max = 25204 N.

8 0

3 years ago

Showing or hiding records in a database is called “filtering.”<br> True<br> False

agasfer [191]

Answer:

TRUE

Explanation:

4 0

3 years ago

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