Write down the understanding of these designations of the materials (a)Ck45, (b)AISI 4140

Identify SIX (6) objectives of maintenance.<br>

Rasek [7]

Answer:

to optimize the reliability of equipment and infrastructure;

- to ensure that equipment and infrastructure are always in good condition;

- to carry out prompt emergency repair of equipment and infrastructure so as to secure the best possible availability for production;

- to enhance, through modifications, extensions, or new low-cost items, the productivity of existing equipment or production capacity;

- to ensure the operation of equipment for production and for the distribution of energy and fluids;

- to improve operational safety;

- to train personnel in specific maintenance skills;

- to advise on the acquisition, installation and operation of machinery;

- to contribute to finished product quality;

- to ensure environmental protection.

Explanation:

pick whichever you want

8 0

3 years ago

An ideal reheat Rankine cycle with water as the working fluid operates the boiler at 15,000 kPa, the reheater at 2000 kPa, and t

solniwko [45]

Answer:

See the explanation below.

Explanation:

First find the enthalpies h₁, h₂, h₃, h₄, h₅, and h₆.

Find h₁:

Using Saturated Water Table and given pressure p₁ = 100 kPa

h₁ = 417.5 kJ/kg

Find h₂:

In order to find h₂, add the $w_{p}$ to h₁, where $w_{p}$ is the work done by pump and h₁ is the enthalpy computed above h₁ = 417.5 kJ/kg.

But first we need to compute $w_{p}$ To computer

Pressures:

p₁ = 100 kPa

p₂ = 15,000 kPa

and

Using saturated water pressure table, the volume of water $v_{f}$ = 1.0432

Dividing 1.0432/1000 gives us:

Volume of water = v₁ = 0.001043 m³/kg

Compute the value of h₂:

h₂ = h₁ + v₁ (p₂ - p₁)

= 417.5 kJ/kg + 0.001043 m³/kg ( 15,000 kPa - 100 kPa)

= 417.5 + 0.001043 (14900)

= 417.5 + 15.5407

= 433.04 kJ/kg

Find h₃

Using steam table:

At pressure p₃ = 15000 kPa

and Temperature = T₃ = 450°C

Then h₃ = 3159 kJ/kg

The entropy s₃ = 6.14 kJ/ kg K

Find h₄

Since entropy s₃ is equal to s₄ So

s₄ = 6.14 kJ/kgK

To compute h₄

s₄ = $s_{f}$ + $x_{4} s_{fg}$

$x_{4} = s_{4} -s_{f} /s_{fg}$

$x_{4}$ = 6.14 - 2.45 / 3.89

$x_{4}$ = 0.9497

The enthalpy h₄:

h₄ = $h_{f} +x_{4} h_{fg}$

= 908.4 + 0.9497(1889.8)

= 908.4 + 1794.7430

= 2703 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₃ = 6.14 kJ/ kg K and pressure p₄ = 2000 kPa

Find h₅

Using steam table:

At pressure p₅ = 2000 kPa

and Temperature = T₅ = 450°C

Then h₅ = 3358 kJ/kg

Find h₆:

Since the entropy s₅ = 7.286 kJ/kgK is equal s₆ to So

s₆ = 7.286 kJ/kgK = 7.29 kJ/kgK

To compute h₆

s₆ = $s_{f}$ + $x_{6} s_{fg}$

$x_{6} = s_{6} -s_{f} /s_{fg}$

$x_{6}$ = 7.29 - 1.3028 / 6.0562

$x_{6}$ = 0.988

The enthalpy h₆:

h₆ = $h_{f} +x_{6} h_{fg}$

= 417.51 + 0.988 (2257.5)

= 417.51 + 2230.41

h₆ = 2648 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₅ = 7.286 kJ/kgK and pressure p₅ = 2000 kPa

Compute power used by pump:

$P_{p}$ is found by using:

mass flow rate = m = 1.74 kg/s

Volume of water = v₁ = 0.001043 m³/kg

p₁ = 100 kPa

p₂ = 15,000 kPa

$P_{p}$ = ( m ) ( v₁ ) ( p₂ - p₁ )

= (1.74 kg/s) (0.001043 m³/kg) (15,000 kPa - 100 kPa)

= (1.74 kg/s) (0.001043 m³/kg) (14900)

= 27.04

$P_{p}$ = 27 kW

Compute heat added $q_{a}$ and heat rejected $q_{r}$ from boiler using computed enthalpies:

$q_{a}$ = ( h₃ - h₂ ) + ( h₅ - h₄ )

= ( 3159 kJ/kg - 433.04 kJ/kg ) + ( 3358 kJ/kg - 2703 kJ/kg )

= 2726 + 655

= 3381 kJ/kg

$q_{r}$ = h₆ - h₁

= 2648 kJ/kg - 417.5 kJ/kg

= 2232 kJ/kg

Compute net work

W $_{net}$ = $q_{a}$ - $q_{r}$

= 3381 kJ/kg - 2232 kJ/kg

= 1150 kJ/kg

Compute power produced by the cycle

mass flow rate = m = 1.74 kg/s

W $_{net}$ = 1150 kJ/kg

P = m * W $_{net}$

= 1.74 kg/s * 1150 kJ/kg

= 2001 kW

Compute rate of heat transfer in the reheater

Q = m * ( h₅ - h₄ )

= 1.74 kg/s * 655

= 1140 kW

Compute Thermal efficiency of this system

μ $_{t}$ = 1 - $q_{r}$ / $q_{a}$

= 1 - 2232 kJ/kg / 3381 kJ/kg

= 1 - 0.6601

= 0.34

= 34%

7 0

3 years ago

A 46.0-g meter stick is balanced at its midpoint (50.0 cm, zero point is a left end of stick). Then a 210.0-g weight is hung wit

Anna71 [15]

Clockwise torque due to 100g is 0.1029 Nm and 200g is 1.4406 Nm. Clockwise torque due to stick mass is 0.2254 Nm and Counter-clockwise torque due to normal force is 1.7689 Nm.

<h3>What is clockwise torque?</h3>

The right-hand rule for cross products determines the direction of torque, which is calculated as the cross product of force and distance. Your thumb will point in the direction of the torque if you place your palm in the direction of the applied force and extend your fingers from the pivot point in that direction.

A related right-hand rule relates the direction of the rotation to the direction of the torque. Your fingers will curl in the direction of rotation if you point your thumb in the direction of the torque.

Positive torques cause counter clockwise rotation, while negative torques cause clockwise rotation.

The sum of all torques must be zero at equilibrium since an object in equilibrium has no net torque.

When the force is applied in a direction perpendicular to the line connecting the pivot and the force, the torque is at its greatest.

You can calculate the torque's magnitude using

$\begin{displaymath}\tau =rF_{\bot }=rF\sin \theta .\end{displaymath}$

To solve problems involving torques, follow these eight steps: read the issue, create a free-body diagram, locate the pivot point, write down the expressions for all torques, For equilibrium conditions, set the sum of torques to zero, list all known variables, pick the desired variable(s), write down equations involving those variable(s), solve the equations, plug in numbers, and test your solution.

Clockwise torque due to 100 g ⇒ T1 = 0.105* 9.8* 0.1 = 0.1029 Nm

Clockwise torque due to 200 g ⇒ T2 = 0.210* 9.8* 0.7 = 1.4406 Nm

Clockwise torque due to stick mass ⇒ T3 = 0.046* 0.5* 9.8 =0.2254 Nm

Counter-clockwise torque due to normal force ⇒ T4 = (0.046 + 0.21 + 0.105)*9.8* 0.5 = 1.7689 Nm

Learn more about torque

brainly.com/question/1544595

#SPJ4

7 0

1 year ago

The lift on a spinning circular cylinder in a freestream with a velocity of 30 m/s and at standard sea level conditions is 6 N/m

Evgesh-ka [11]

Answer:

The circulation around the cylinder is 0.163 $\frac{m^{2} }{s}$

Explanation:

Given :

Velocity of spinning cylinder $v = 30$ $\frac{m}{s}$

Sea level density $\rho = 1.23$ $\frac{kg}{m^{3} }$

Sea level span $L = 6$ $\frac{N}{m}$

Lift per unit circulation is given by,

$L = \rho v c$

Where $c =$ circulation around cylinder

$c = \frac{L}{\rho v}$

$c = \frac{6}{1.23 \times 30}$

$c = 0.163$ $\frac{m^{2} }{s}$

Therefore, the circulation around the cylinder is 0.163 $\frac{m^{2} }{s}$

5 0

3 years ago

Explain the conductivity results observed for ionic compounds in the solid state and in aqueous solution.i.Solid ionic compounds

Svetach [21]

Answer:

Aqueous solution of ionic compounds conduct electricity while solid ionic compounds don't.

Explanation:

Ionic compound conduct electricity when liquid or in aqueous solution that is resolved in water because the ionic bonds of the compound become weak and the ions are free to move from place to place.

Ionic compounds don't conduct electricity while in solid state because the ionic bonds are to strong and ions cannot move around with lack of space for movement which makes the electric conductivity zero.

8 0

3 years ago