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3 years ago

What is electrical energy being transformed into? How do you know?

Engineering

1 answer:

Olegator [25]3 years ago

7 0

Electrical Energy is converted to Kinetic Energy.

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A pump of a water distribution system at 25°C is powered by a 15 kW electric motor whose efficiency is 90 percent. The water flo

IRISSAK [1]

The friction loss in the system is 3.480 kilowatts.

<h2>Procedure - Friction loss through a pump</h2><h2 /><h3>Pump model</h3><h3 />

Let suppose that the pump within a distribution system is an open system at steady state, whose mass and energy balances are shown below:

<h3>Mass balance</h3>

$\dot m_{in}-\dot m_{out} = 0$ (1)

$\dot m_{in} = \frac{\dot V_{in}}{\nu_{in}}$ (2)

$\dot m_{out} = \frac{\dot V_{out}}{\nu_{out}}$ (3)

<h3>Energy balance</h3>

$\eta \cdot \dot W_{el} + \dot m_{in}\cdot (h_{in}-h_{out}) - \dot W_{f} = 0$ (4)

Where:

$\dot m_{in}$ - Inlet mass flow, in kilograms per second.
$\dot m_{out}$ - Outlet mass flow, in kilograms per second.
$\dot V_{in}$ - Inlet volume flow, in cubic meters per second.
$\dot V_{out}$ - Outlet volume flow, in cubic meters per second.
$\nu_{in}$ - Inlet specific volume, in cubic meters per kilogram.
$\nu_{out}$ - Outlet specific volume, in cubic meters per kilogram.
$\eta$ - Pump efficiency, no unit.
$\dot W_{el}$ - Electric motor power, in kilowatts.
$h_{in}$ - Inlet specific enthalpy, in kilojoules per kilogram.
$h_{out}$ - Outlet specific enthalpy, in kilojoules per kilogram.
$\dot W$ - Work losses due to friction, in kilowatts.

<h3>Data from steam tables</h3>

From steam tables we get the following water properties at inlet and outlet:

Inlet

$p = 100\,kPa$ , $T = 25\,^{\circ}C$ , $\nu = 0.001003\,\frac{kJ}{kg}$ , $h = 104.927\,\frac{kJ}{kg}$ , Subcooled liquid

Outlet

$p = 300\,kPa$ , $T = 25\,^{\circ}C$ , $\nu = 0.001003\,\frac{kJ}{kg}$ , $h = 105.128\,\frac{kJ}{kg}$ , Subcooled liquid

<h3>Calculation of the friction loss in the system</h3>

If we know that $\dot V_{in} = 0.05\,\frac{m^{3}}{s}$ , $\nu_{in} = 0.001003\,\frac{m^{3}}{kg}$ , $h_{in} = 104.927\,\frac{kJ}{kg}$ , $h_{out} = 105.128\,\frac{kJ}{kg}$ , $\eta = 0.90$ and $\dot W_{el} = 15\,kW$ , then the friction loss in the system is:

$\dot W_{f} = \frac{\dot V_{in}}{\nu_{in}}\cdot (h_{in} - h_{out}) + \eta \cdot \dot W_{el}$

$\dot W_{f} = \left(\frac{0.05\,\frac{m^{3}}{s} }{0.001003\,\frac{m^{3}}{kg} } \right)\cdot \left(104.927\,\frac{kJ}{kg}-105.128\,\frac{kJ}{kg}\right) + (0.90)\cdot (15\,kW)$

$\dot W_{f} = 3.480\,kW$

The friction loss in the system is 3.480 kilowatts. $\blacksquare$

To learn more on pumps, we kindly invite to check this verified question: brainly.com/question/544887

6 0

2 years ago

I NEED ANSWER'S ASAP PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

nydimaria [60]

Answer:

Sealing agents are made of eco-friendly materials that are difficult to properly position in hard-to-reach places.

Explanation:

4 0

4 years ago

Read 2 more answers

A team member who has been a good worker for many years has recently been doing poor work. You suspect that he may be tired of h

ohaa [14]

Answer:

You must follow the steps below to deal with employees who have poor performance. I hope that you will be able to handle all the issues after reading this answer.

Explanation:

Be specific with facts in hand

It is important to confront your employees about their respective actions. But to convince them about their withdrawal from lack of interest, it is imperative to have a consistent record at hand as well. For example, if the employee has been constantly delayed for a period of time, specify the precise details about the frequency and intensity of absenteeism. Be sure not to overdo your statements or use hard phrases to reduce employee self-esteem. Just be direct and precise. Reiterate the guidelines accordingly.

Consider the needs of your employees

Poor performance is not always the result of an employee's carelessness. There can be multiple genuine reasons for lack of performance and it can vary from person to person. The first is to understand the reason and judge whether they are genuine or not. Even if they aren't, don't let the other person know. Focus on your concerns and provide solutions accordingly. For example, if your employees cannot focus on their work due to some personal stress, make appointments for the counseling sessions and make sure they can get back on track.

Focus on feedback

Everyone handles comments differently. Although it is always recommended to be direct and clear in your communication, there may be certain strategies you can adopt to communicate your comments effectively. If your employee has difficulties in achieving his goals, work with him and provide him with all the necessary help to improve his performance. The best way is to provide weekly or monthly comments to your employees, so they know what they must do to achieve their goals.

Provide Performance Support Technology

When faced with existing employees who do not meet expectations, it is a smart decision to offer them training (and / or training) and different resources to help them improve. As an example, you can combine a low-performance worker with someone to act as a mentor or offer you a manual with the procedures to follow. In addition, there are performance improvement tools: WalkMe is a very valuable technology that can help a worker be more efficient and precise, within the workflow (and not take them away from their daily tasks).

Offer rewards and recognition

Whenever you see that your employees have poor performance, it is always better to adopt a carrot and stick approach for instant and consistent improvement. It is a combination of rewards and punishments that you can induce for the best and worst weekly or monthly performance. This has proven to be one of the best ways to combat low performance problems in companies of all kinds for centuries.

Facing low-performance workers for the first time may not be too encouraging for managers as well, but having an adequate system to deal with them is essential, especially during the performance of change management. It is always better to deal with such situations, instead of ignoring them, to maintain the consistency of productivity and profitability in the business.

8 0

4 years ago

Assume that in orthogonal cutting the rake angle is 15° and the coefficient of friction is 0.15. a. Determine the percentage cha

alexira [117]

Answer:

Δr=20.45 %

Explanation:

Given that

Rake angle α = 15°

coefficient of friction ,μ = 0.15

The friction angle β

tanβ = μ

tanβ = 0.15

β=8.83°

2φ + β - α = 90°

φ=Shear angle

2φ + 8.833° - 15° = 90°

φ = 48.08°

Chip thickness r given as

$r=\dfrac{tan\phi}{cos\alpha +sin\alpha\ tan\phi}$

$r=\dfrac{tan48.08^{\circ}}{cos15^{\circ} +sin15^{\circ}\ tan48.08^{\circ}}$

r=0.88

New coefficient of friction ,μ' = 0.3

tanβ' = μ'

tanβ' = 0.3

β'=16.69°

2φ' + β' - α = 90°

φ'=Shear angle

2φ' + 16.69° - 25° = 90°

φ' = 49.15°

Chip thickness r' given as

$r'=\dfrac{tan\phi'}{cos\alpha +sin\alpha\ tan\phi'}$

$r'=\dfrac{tan44.15^{\circ}}{cos49.15^{\circ} +sin49.15^{\circ}\ tan44.15^{\circ}}$

r'=0.70

Percentage change

$\Delta r=\dfrac{r-r'}{r}\times 100$

$\Delta r=\dfrac{0.88-0.70}{0.88}\times 100$

Δr=20.45 %

8 0

3 years ago

From the following numbered list of characteristics, decide which pertain to (a) precipitation hardening, and which are displaye

tia_tia [17]

Answer:

(a) Precipitation hardening - 1, 2, 4

(b) Dispersion strengthening - 1, 3, 5

Explanation:

The correct options for each are shown as follows:

Precipitation hardening

From the first statement; Dislocation movement is limited by precipitated particles. This resulted in an expansion in hardness and rigidity. Precipitates particles are separated out from the framework after heat treatment.

The aging process occurs in the second statement; because it speaks volumes on how heated solutions are treated with alloys above raised elevated temperature. As such when aging increases, there exists a decrease in the hardness of the alloy.

Also, for the third option for precipitation hardening; This cycle includes the application of heat the alloy (amalgam) to a raised temperature, maintaining such temperature for an extended period of time. This temperature relies upon alloying components. e.g. Heating of steel underneath eutectic temperature. Subsequent to heating, the alloy is extinguished and immersed in water.

Dispersion strengthening

Here: The effect of hearting is not significant to the hardness of alloys hardening by the method in statement 3.

In statement 5: The process only involves the dispersion of particles and not the application of heat.

8 0

3 years ago