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dmitriy555 [2]
2 years ago
13

Reason fo I.EE regulations in electrical installations​

Engineering
1 answer:
klasskru [66]2 years ago
7 0

Answer:

The IEE Wiring Regulations Explained and Illustrated, Second Edition discusses the recommendations of the IEE Regulations for the Electrical Equipment of Buildings for the safe selection or erection of wiring installations. The book emphasizes earthing, bonding, protection, and circuit design of electrical wirings.

Explanation:

i hope it help you please make me brainlist and answer my question

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svet-max [94.6K]
OA bloom is smaller than a bar
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3 years ago
Disconnecting a circuit while in operation can create a voltage blank
zlopas [31]

Answer:

what is the question

Explanation:

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3 years ago
An air conditioning system operating on reversed carnot cycle is required to remove heat from the house at a rate of 32kj/s to m
Brilliant_brown [7]

Answer:

(e) 1.64 kW

Explanation:

The Coefficient of Performance of the Reverse Carnot's Cycle is:

COP = \frac{T_{L}}{T_{H}-T_{L}}

COP = \frac{293.15\,K}{308.15\,K-293.15\,K}

COP = 19.543

Lastly, the power required to operate the air conditioning system is:

\dot W = \frac{\dot Q_{L}}{COP}

\dot W = \frac{32\,kW}{19.543}

\dot W = 1.637\,kW

Hence, the answer is E.

3 0
3 years ago
A train consists of a 50 Mg engine and three cars, each having a mass of 30 Mg . If it takes 75 s for the train to increase its
ohaa [14]

Answer:

T = 15 kN

F = 23.33 kN

Explanation:

Given the data in the question,

We apply the impulse momentum principle on the total system,

mv₁ + ∑\int\limits^{t2}_{t1} {Fx} \, dt = mv₂

we substitute

[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ ×  ( 45 × 1000 / 3600 )  

F( 75 - 0 ) =  1.75 × 10⁶

The resultant frictional tractive force F is will then be;

F =  1.75 × 10⁶ / 75

F = 23333.33 N

F = 23.33 kN

Applying the impulse momentum principle on the three cars;

mv₁ + ∑\int\limits^{t2}_{t1} {Fx} \, dt = mv₂

[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ ×  ( 45 × 1000 / 3600 )  

F(75-0) = 1.125 × 10⁶

The force T developed is then;

T =  1.125 × 10⁶ / 75

T = 15000 N

T = 15 kN

7 0
3 years ago
A rectangular block of material with shear modulus G= 620 MPa is fixed to rigid plates at its top and bottom surfaces. Thelower
PIT_PIT [208]

Answer:

γ_{xy} =0.01, P=248 kN

Explanation:

Given Data:

displacement = 2mm ;

height = 200mm ;

l = 400mm ;

w = 100 ;

G = 620 MPa = 620 N//mm²;    1MPa = 1N//mm²

a. Average Shear Strain:

The average shear strain can be determined by dividing the total displacement of plate by height

γ_{xy} = displacement / total height

     = 2/200 = 0.01

b. Force P on upper plate:

Now, as we know that force per unit area equals to stress

τ = P/A

Also,  τ = Gγ_{xy}

By comapring both equations, we get

P/A = Gγ_{xy}   ------------ eq(1)

First we need to calculate total area,

A = l*w = 400 * 100= 4*10^4mm²

By putting the values in equation 1, we get

P/40000 = 620 * 0.01

P = 248000 N or 2.48 *10^5 N or 248 kN

6 0
3 years ago
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