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2 years ago

Explain the problems and their possible solution for electricity problems ?

Engineering

1 answer:

Temka [501]2 years ago

5 0

Answer:

1) Electrical surges

It can be occurred due to poor wiring in the house or lightning strikes or faulty appliances or damaged power lines. Surges are common and last for a microsecond but if you experience frequent surges lead to equipment damage that degrade life expectancy particularly.

Check the device that connects to the home grid or the wiring and try disconnecting the poor quality powerboards or devices from the outlet. If the surges don’t occur again, your problem is solved. If it is not, you must call an electrician.

2) Overloading

Sometimes your light fixture has a bulb or other fitting with high watts than the designed fixture. This is a code violation and the risk level is quite high. The high heat from the bulb can melt the socket and insulation present in wires of the fixture. This results in sparks from one wire to another and causes electrical fires. Even after the bulb is removed, the socket and wires will still be under damage.

It is always better to fit a bulb or any other fittings by staying within the wattage. If the fixtures are not marked with wattage, it is advisable to use a 60-watt bulb or even smaller ones.

3) Power sags and dips

Sags are dips usually occur when the power grip is faulty and electrical appliances are connected to it. It also occurs when the grid is made of low-quality materials. When this is the case, it draws more power when switched on.

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What is the line called that has the red arrow pointing to it in the attached picture?

Simora [160]

Answer:

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Explanation:

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3 years ago

The cross-section of a rough, rectangular, concrete() channel measures . The channel slope is 0.02ft/ft. Using the Darcy-Weisbac

bazaltina [42]

Answer:

The following are the answer to this question:

Explanation:

In point a, Calculating the are of flow:

$\bold{Area =B \times D_f}$

$=6\times 5\\\\=30 \ ft^2$

In point b, Calculating the wetter perimeter.

$\bold{P_w =B+2\times D_f}$

$= 6 +2\times (5)\\\\= 6 +10 \\\\=16 \ ft$

In point c, Calculating the hydraulic radius:

$\bold{R=\frac{A}{P_w}}$

$=\frac{30}{16}\\\\= 1.875 \ ft$

In point d, Calculating the value of Reynolds's number.

$\bold{Re =\frac{4VR}{v}}$

$=\frac{4V \times 1.875}{1 \times 10^{-5} \frac{ft^2}{s}}\\\\$

$=750,000 V$

Calculating the velocity:

$V= \sqrt{\frac{8gRS}{f}}$

$= \sqrt{\frac{8\times 32.2 \times 1.875 \times 0.02}{f}}\\\\=\frac{3.108}{\sqrt{f}}\\\\$

$\sqrt{f}=\frac{3.108}{V}\\\\$

calculating the Cole-brook-White value:

$\frac{1}{\sqrt{f}}= -2 \log (\frac{K}{12 R} +\frac{2.51}{R_e \sqrt{f}})\\\\ \frac{1}{\frac{3.108}{V}}= -2 \log (\frac{2 \times 10^{-2}}{12 \times 1.875} +\frac{2.51}{750,000V\sqrt{f}})\\$

$\frac{V}{3.108} =-2\log(8.88 \times 10^{-5} + \frac{3.346 \times 10^{-6}}{750,000(3.108)})$

After calculating the value of V it will give:

$V= 25.18 \ \frac{ft}{s^2}\\$

In point a, Calculating the value of Froude:

$F= \frac{V}{\sqrt{gD}}$

$= \frac{V}{\sqrt{g\frac{A}{\text{Width flow}}}}\\$

$= \frac{25.18}{\sqrt{32.2\frac{30}{6}}}\\\\= \frac{25.18}{\sqrt{32.2 \times 5}}\\\\= \frac{25.18}{\sqrt{161}}\\\\= \frac{25.18}{12.68}\\\\= 1.98$

The flow is supercritical because the amount of Froude is greater than 1.

Calculating the channel flow rate.

$Q= AV$

$=30x 25.18\\\\= 755.4 \ \frac{ft^3}{s}\\$

4 0

3 years ago

Water is pumped from a lower reservoir to a higher reservoir by a pump that provides 20 kW of shaft power. The free surface of t

ahrayia [7]

Answer:

note:

<u><em>solution is attached due to error in mathematical equation. please find the attachment</em></u>

8 0

3 years ago

A plate clutch is used to connect a motor shaft running at 1500rpm to shaft 1. The motor is rated at 4 hp. Using a service facto

vazorg [7]

Answer:

$(M_t)_{rated}=61.11lb-in$

Explanation:

speed of motor (N)=1500 rpm

power=4 hp = $4 \times 0.7457$ =2.9828 KW

service factor(k)= 2.75

now,

$KW=\frac{2\pi n M_t}{60 \times 10^6} \\2.9828=\frac{2\pi \times 1500 M_t}{60 \times 10^6}\\M_t=\frac{2.9828\times 60 \times 10^6}{2\pi \times 1500 }$