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jeyben [28]
3 years ago
8

Explain the problems and their possible solution for electricity problems ?​

Engineering
1 answer:
Temka [501]3 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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A large heat pump should upgrade 5 MW of heat at 85°C to be delivered as heat at 150°C. Suppose the actual heat pump has a COP o
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Answer:

W=2 MW

Explanation:

Given that

COP= 2.5

Heat extracted from 85°C  

Qa= 5 MW

Lets heat supplied at 150°C   = Qr

The power input to heat pump = W

From first law of thermodynamics

Qr= Qa+ W

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COP=\dfrac{Qr}{W}

2.5=\dfrac{5}{W}

2.5=\dfrac{5}{W}

W=2 MW

For Carnot heat pump

COP=\dfrac{T_2}{T_2-T_1}

2.5=\dfrac{T_2}{T_2-(273+85)}

2.5 T₂ -  895= T₂

T₂=596.66 K

T₂=323.6 °C

7 0
3 years ago
A segment of four-lane freeway (two lanes in each direction) has a 3% upgrade that is 1500 ft long followed by a 1000-ft 4% upgr
Dennis_Churaev [7]

Answer:

The level of service of of compound grade freeway is LOSB.

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3 years ago
Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa
Helen [10]

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air,  R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes  2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

(c) The next step is to find the total work needed in kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

4 0
3 years ago
All of the following safety tips are true EXCEPT Select one: a. It is not acceptable to handle broken glass with your bare hands
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Answer:

Explanation:

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