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ipn [44]
3 years ago
9

Explain why the following acts lead to hazardous safety conditions when working with electrical equipmenta. Wearing metal ring o

r braceletsb. Being barefootc. Working on a damp concrete floord. Touching grounded conductors while working on electrical equipmente. Working on electrical equipment with sweaty hands
Engineering
1 answer:
Rus_ich [418]3 years ago
4 0

Answer:

Please find the answer in the explanation

Explanation:

In Engineering, safety is very essential and very important for all engineers to stick to.

A.) Wearing metal ring or bracelets.

When their is discharg of electrical charges, wearing of metal rings or braceletsb can allow charges to pass through them into the body which can eventually lead to electrical shock.

B.) Being barefoot

Being bare footed is very dangerous because someone can mistakenly step on naked wire which can lead to electrical shock.

C. Working on a damp concrete floor.

It is very hazardous to be working on a damp concrete floor because of the water moisture. It is very unsafe for any electrical job to be done on any wet area because water can conduct electricity which can lead to electrical shock.

d. Touching grounded conductors while working on electrical equipment

Grounded conductor can allow charges to flow through them. So, it is very unsafe to have them have contact with the body because of electrical charges.

e. Working on electrical equipment with sweaty hands

A sweaty hands contain some content of water which can conduct electricity and lead to electrical shock.

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KVL holds for the supermesh, so we can write a KVL equation to generate the second equation we need to solve for the two unknown
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Answer:

The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V

Explanation:

As the complete question is not given the complete question is found online and is attached herewith.

By applying KCL at node 1

i_x+50mA=i_y\\i_x-i_y=0.05A

Also

V_{\Delta}=1K*i_y

Now applying KVL on loop 1 as indicated in the attached figure

1K*i_y+5K(i_y-i_z)+3K*i_x=0\\3i_x+6 i_y-5i_z=0

Similarly for loop 2

2V_{\Delta}+5K(i_z-i_y)=0\\2*1K*i_y+5K(i_z-i_y)=0\\2K*i_y+5K(i_z-i_y)=0\\3i_y-5i_z=0

So the system of equations become

i_x-i_y+0i_z=0.05\\3i_x+6i_y-5i_z=0\\0i_x-3i_y+5i_z=0

Solving these give the values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA. Also the value of voltage is given as

V_{\Delta}=1K*i_y\\V_{\Delta}=1K*-25 mA\\V_{\Delta}=-25 V

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4 years ago
A student proposes a complex design for a steam power plant with a high efficiency. The power plant has several turbines, pumps,
masha68 [24]

Answer:

(a). max possible efficiency = 55.62%

(b). max power output = = 133.5 MW

Explanation:

From the question we were given the Maximum temperature in the system as

Tmax = 500°C

Minimum temperature in the system Tmin = 70°C

the Heat supplied to the boiler Qb = 240000 KJ/s

we use the temperature conversion factor from °C to K

given T(K) = T (°C) + 273

⇒ Tmax = 500 + 273 = 773 K

⇒ Tmin = 70 + 273 = 343 K

(a). we are to determine the maximum possible thermal efficiency;

(Πth)max = 1 - Tmin/Tmax

(Πth)max = 1 - 343/773  = 0.5562

(Πth)max = 55.62%

(b). to determine the maximum possible power output for the plant we have;

(Πth)max = Wmax/Qb

where Wmax rep the maximum power output

(Πth)max = 0.5562

Qb = 240000

∴ Wmax = 0.5562 × 240000 = 133505.6 Kw

Wmax = 133.5 MW

cheers i hope this helps

3 0
3 years ago
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