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3 years ago

When must an Assured Equipment Grounding Conductor Program (AEGCP) be in place?

Engineering

1 answer:

sasho [114]3 years ago

8 0

Answer:

Only when working around power lines

Explanation:

Assured Equipment Grounding Conductor Program (AEGCP) is applicable mainly in sites dealing with power. Therefore, the most appropriate worksite to have Assured Equipment Grounding Conductor Program is when working around power lines.

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A gas in a piston–cylinder assembly undergoes a compression process for which the relation between pressure and volume is given

viktelen [127]

Answer:

A.) P = 2bar, W = - 12kJ

B.) P = 0.8 bar, W = - 7.3 kJ

C.) P = 0.608 bar, W = - 6.4kJ

Explanation: Given that the relation between pressure and volume is

PV^n = constant.

That is, P1V1^n = P2V2^n

P1 = P2 × ( V2/V1 )^n

If the initial volume V1 = 0.1 m3,

the final volume V2 = 0.04 m3, and

the final pressure P2 = 2 bar.

A.) When n = 0

Substitute all the parameters into the formula

(V2/V1)^0 = 1

Therefore, P2 = P1 = 2 bar

Work = ∫ PdV = constant × dV

Work = 2 × 10^5 × [ 0.04 - 0.1 ]

Work = 200000 × - 0.06

Work = - 12000J

Work = - 12 kJ

B.) When n = 1

P1 = 2 × (0.04/0.1)^1

P1 = 2 × 0.4 = 0.8 bar

Work = ∫ PdV = constant × ∫dV/V

Work = P1V1 × ln ( V2/V1 )

Work = 0.8 ×10^5 × 0.1 × ln 0.4

Work = - 7330.3J

Work = -7.33 kJ

C.) When n = 1.3

P1 = 2 × (0.04/0.1)^1.3

P1 = 0.6077 bar

Work = ∫ PdV

Work = (P2V2 - P1V1)/ ( 1 - 1.3 )

Work = (2×10^5×0.04) - (0.608 10^5×0.1)/ ( 1 - 1.3 )

Work = (8000 - 6080)/ -0.3

Work = -1920/0.3

Work = -6400 J

Work = -6.4 kJ

5 0

3 years ago

Anna makes arrangements to reuse waste water that has been used in sinks and showers. Which term refers to the waste water that

victus00 [196]

Answer:

Greywater.

Explanation:

Greywater is also known as sullage and it can be defined as any form of gently used wastewater derived from sources within a residential or office building such as showers, washing machines, bathroom sinks, bathroom tub, etc.

Generally, greywater or sullage is completely free of fecal materials (faeces) because it is independent from all toilet activities. However, greywater is not clean for direct use because it usually contains food particles, dirt, oil from dishes, hair, etc.

In this scenario, Anna makes arrangements to reuse waste water that has been used in sinks and showers. Greywater is a term which refers to the waste water that Anna reuses to conserve resources.

Therefore, Anna reuses greywater to conserve resources.

8 0

3 years ago

In a certain chemical plant, a closed tank contains ethyl alcohol to a depth of 71 ft. Air at a pressure of 17 psi fills the gap

Yuliya22 [10]

Answer:

the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

Explanation:

Given that;

depth 1 = 71 ft

depth 2 = 10 ft

pressure p = 17 psi = 2448 lb/ft²

depth h = 71 ft - 10 ft = 61 ft

we know that;

p = P_air + yh

where y is the specific weight of ethyl alcohol ( 49.3 lb/ft³ )

so we substitute;

p = 2448 + ( 49.3 × 61 )

= 2448 + 3007.3

= 5455.3 lb/ft³

= 37.88 psi

Therefore, the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

5 0

3 years ago

zhenek [66]

The answer is b i did the same thing and i got it right

5 0

2 years ago

Read 2 more answers

An asbestos pad is square in cross section, measuring 5 cm on a side at its small end increasing linearly to 10 cm on a side at

polet [3.4K]

Answer:

q = 1.73 W

Explanation:

given data

small end = 5 cm

large end = 10 cm

high = 15 cm

small end is held = 600 K

large end at = 300 K

thermal conductivity of asbestos = 0.173 W/mK

solution

first we will get here side of cross section that is express as

$S = S1 + \frac{S2-S1}{L} x$ ...............1

here x is distance from small end and S1 is side of square at small end

and S2 is side of square of large end and L is length

put here value and we get

S = 5 + $\frac{10-5}{15} x$

S = $\frac{0.15 + x}{3}$ m

and

now we get here Area of section at distance x is

area A = S² ...............2

area A = $(\frac{0.15 + x}{3})^2$ m²

and

now we take here small length dx and temperature difference is dt

so as per fourier law

heat conduction is express as

heat conduction q = $\frac{-k\times A\ dt}{dx}$ ...............3

put here value and we get

heat conduction q = $-k\times (\frac{0.15 + x}{3})^2 \ \frac{dt}{dx}$

it will be express as

$q \times \frac{dx}{(\frac{0.15 + x}{3})^2} = -k (dt)$

now we intergrate it with limit 0 to 0.15 and take temp 600 to 300 K

$q \int\limits^{0.15}_0 {\frac{dx}{(\frac{0.15 + x}{3})^2 } = -0.173 \int\limits^{300}_{600} {dt}$

solve it and we get

q (30) = (0.173) × (600 - 300)

q = 1.73 W

5 0

3 years ago