Task 1. Calculation of current passing through the body of man at unipolar and bipolar touch

To expect a current, passing through the body of man at unipolar and bipolar touch to the three phase network of alternating current by frequency 50Hz.

To consider the cases of network with earthed (380/220 v) and isolated (380v) neutral. The isolation resistance of wire Ris = 300 kΩ. The capacity of network is insignificant (C≈ 0). An apartment is moist; a man is on the, moist concrete floor. His resitance Rpeople = 1.5 kΩ.

Resistance of neutral grounding R₀ = 3 Ω.

Considering a man touching the phase wire; in this case the current through the body of man is determined on the formula.

Now we will consider a bipolar touch (figure 6.3).

At a bipolar touch two variants are also possible: a man touches a zero wire and the phase wire at the same time (B.1 - fig.6.3.a), and a man touches two phase wires at the same time (biphasic touch, B.2 - fig.6.3.b).

Consider the first variant: man touches to a zero wire and to the earthed neutral

- (a) (b)

Figure 6.3 – Bipolar touch to the network by Neutral

In this case a current through man’s body is given by a formula:

Such value of current (0.22 А) by frequency 50 Hz also is mortal for a man

6.5.3 We will consider the second variant now: man touches two phase wires (biphasic touch, figure 1.2.b).

In this case a current through the body of man is on a formula:

Where V_linear - linear voltage (in our case 380 V).

We will count up the value of current:

Such value of current by frequency 50 Hz also is mortal for a man

Electric shock occurs upon contact of a (human) body with any source of electricity that causes a sufficient current through the skin, muscles, or hair. Typically, the expression is used to denote an unwanted exposure to electricity, hence the effects are considered undesirable.

The minimum current a human can feel depends on the current type (AC or DC) and frequency. A person can feel at least 1 mA (rms) of AC at 60 Hz, while at least 5 mA for DC. The current may, if it is high enough, cause tissue damage or fibrillation which leads to cardiac arrest. 60 mA of AC (rms, 60 Hz) or 300–500 mA of DC can cause fibrillation. A sustained electric shock from AC at 120 V, 60 Hz is an especially dangerous source of ventricular fibrillation because it usually exceeds the let-go threshold, while not delivering enough initial energy to propel the person away from the source. However, the potential seriousness of the shock depends on paths through the body that the currents take. Death caused by an electric shock is called electrocution.

6.5.4 Determining the amount of reveal air on the amount of persons

To define the amount of reveal air, it is necessary to ventilate an apartment of volume V =250m³, if amount of workers in it n =8 persons

The volume of apartment for one man is less than 20m³ an amount of reveal air necessary ventilation, must be not less than G₁= 30m³ /hour on every working period. At the volume of apartment more than 20m³ per one working, am amount of reveal air for ventilation must not be less than G₁ =20m³ /hour on every working period

The volume of apartment for one man:

V₁=V/n;

V₁= 250m³ /8;

V₁= 25m³ /hour.

As V₁ > 20m³ /hour, then norm of serve of reveal air per one man G₁=20m³ /hour

Am amount of reveal air taking into account the quantity of working period (m³ /hour) settles accounts on a formula:

G =G₁ *n= 20*8= 160m³ /hour.

Task 2. Calculation of crossing of send-offs and cables for economic of density current

In accordance with recommendation of NPAOP 40.1-1.21-98 the calculation is done by a formula:

S_å = I_max/ J_ec,

where Imax is the current of line during work of network, A;

Jec- economical density of current, A/mm², which is determined depending of the material and time of usage of the maximal loading.

Select a section of bare copper conductors with the following currents of line. I_max=5, J_ec=2, 5, S_å =5/2, 5 =2.

Select a section of bare aluminium conductors with the following currents of line. I_max=6, J_ec=1.3, S_å =6/1.3 =4, 615.

Select a section of the cables with copper conductor and plastic insulation

TASK 3

Estimation of shut-down ability of devices of maximal current defense

The short circuit current, :

=

where: = is the phase tension, V;

= is an impedance of the loop a< < phase zero> >, Ω.

The device of maximal current defence is provided by the reliable disconnection of users of electric power from the network, if a condition is executed:

,

Where: is nominal current of fusible insertion of safety device. The current electromagnetic insertion of circuit breaker on a short-circuit, A;

K is coefficient of multipleness of current (According to NPAOP 40.1-1.21-98) for fuse K=3, for the electromagnetic breaker K= 1.4 or 1.25).

Selected values for the task from the table are

= 127 = 24

Using the formula: = = = 5.29A

Then into the second equation we get: = 3.79

Table 6.2- List of Objects of different settings.

I_max=7, J_ec=3, 5, S_å =7/3, 5 =2.

Select a section of the cables with aluminum conductor and plastic insulation. I_max=8, J_ec=1.9, S_å =8/1, 9 =4, 210.

Task 8. Calculation of Necessary Amount and Types of Fire-Extinguishers

The recommendation for equipment of apartments with portable fire-extinguishers

On 20 m² of floorage in such apartments: office apartments with PC, barns, electro-shield, vent chambers and other technical apartments;

On 50 m² of floorage apartments of archives, machine rooms, libraries, museums. Additionally foregoing apartments can be equipped with aerosol water from fire extinguishers with mass of charge of extinguishing matter of 400gг and more

At presence in apartment of possibility different fire classes, the amount of fire-extinguishers gets out for one of classes, for which an amount of fire-extinguishers is greater.

For apartments with PC it is necessary to equip the portable carbon-dioxide fire-extinguishers

BBK-1, 4, BBK-2, or one BBПA-400 on three PC, but not less than one fire-extinguisher.