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2 years ago

Component of earthing and reasons why each material is being used

Engineering

1 answer:

timofeeve [1]2 years ago

3 0

Answer:

The components of earthing are Earth electrode, Main earthing terminals/bars, Earthing conductors, Protective conductors Equipontential binding conductors and Electrically independent electrodes.

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A milling operation was used to remove a portion of a solid bar of square cross section. Forces of magnitude P = 18 kN are appli

monitta

The smallest allowable depth is $d=16.04 \mathrm{mm}$ for the milled portion of bar.

<u>Explanation:</u>

Given,

Magnitude of force, $\mathbf{p}=18 \mathrm{kN}$

$a=30 \mathrm{mm}$

$=0.03 \mathrm{m}$

Allowable stress, $\sigma_{a l l}=135 \mathrm{MPa}$

cross sectional area of bar,

$A=a \times d$

$A=a d$

e - eccentricity

$e=\frac{a}{2}-\frac{d}{2}$

The internal forces in the cross section are equivalent to a centric force P and a bending couple M.

$M=P e$

$=P\left(\frac{a}{2}-\frac{d}{2}\right)$

$=\frac{P(a-d)}{2}$

Allowable stress

$\sigma=\frac{P}{A}+\frac{M c}{I}$

$c=\frac{d}{2}$

Moment of Inertia,

$I=\frac{b d^{3}}{12}$

$=\frac{a d^{3}}{12}$

$\therefore \sigma=\frac{P}{a d}+\frac{\frac{P(a-d)}{2} \times \frac{d}{2}}{\frac{a d^{3}}{12}}$

$\sigma=\frac{P}{a d}+\frac{3 P(a-d)}{a d^{2}}\\$

$\sqrt{x} \sigma\left(a d^{2}\right)=P d+3 P(a-d)$

$\sigma\left(a d^{2}\right)=P d+3 P a-3 P d$

$\sigma\left(a d^{2}\right)=(P-3 P) d+3 P a$

$\left(\sigma a d^{2}\right)=-2 P d+3 P a$

$\sigma d^{2}=-\frac{2 P}{a} d+3 P$

By substituting values we get,

$\left(135 \times 10^{6}\right) d^{2}+\frac{2 \times 18 \times 10^{3}}{0.03} d-3\left(18 \times 10^{3}\right)=0$

$\left(135 \times 10^{6}\right) d^{2}+\left(12 \times 10^{5}\right) d-54 \times 10^{3}=0$

On solving above equation we get, $d=0.01604 \mathrm{m}\\$

$d=16.04 \mathrm{mm}$

3 0

3 years ago

# 17

vazorg [7]

Answer:

FALSE

Explanation:

It is best practice to ALWAYS change the password of your router to something other than the default.

_____

Leaving the password as the default leaves the router open to exploitation by hostiles.

6 0

3 years ago

Read 2 more answers

Liquid water is fed to a boiler at 24°C and 10 bar is converted at a constant pressure to saturated steam.

zepelin [54]

We can find the change in the enthalpy through the tables A5 for Saturated water, pressure table.

For 1bar=1000kPa:

$T_{sat}=179.88\°c$

$H_{fg} = 2014.6kJ/kg$

$c_p=4.18 kJkg^{-1}{K^{-1}$

$\nu_g = 0.19436m^3/kg$

Replacing,

$\Delta h = h_{fg}+c_p(T_{sat}-T_{inlet})$

$\Delta h = 2014.6+4.18(179.88-24)$

$\Delta h=2666.17kJ/kg$

With the specific volume we know can calculate the mass flow, that is

$\dot{m}=\frac{\frac{15000}{3600}}{0.19436}$

$\dot{m} = 21.4378kg/s$

Then the heat required in input is,

$Q=\dot{m}\Delta h$

$Q=21.4378*2666.17$

$Q=57157.036kW$

With the same value required of 15000m^3/h, we can calculate the velocity of the water, that is given by,

$V= \frac{\dotV}{A}$

$V = \frac{\frac{15000}{3600}}{\pi /4 *(0.15)^2}$

$V=235.79m/s$

Finally we can apply the steady flow energy equation, that is

$\dot{m}(h_1+\frac{V^2}{2000})+Q = \dot{m}h_2$

Re-arrange for Q,

$Q=\dot{m}(h_2-h_1-\frac{V^2}{2000})$

$Q=\dot{m}(\Delta h-\frac{V^2}{2000})$

$Q= (21.4378)(2666.17-\frac{235.79^2}{2000})$

$Q= 56560.88kW$

We can note that consider the Kinetic Energy will decrease the heat input.

4 0

3 years ago

The van der Waals equation is a modification of the ideal gas equation. What two factors does this equation account for? A. (1)

Liula [17]

Answer:

Explanation:

The vander waals equation

$(P+n^2a/V^2)(V-nb)$ =nRT

The factor n^2a/V^2 accounts for the attractive forces between the gases molecules whereas nb accounts for the decrease in volume occupied by the gas. Therefore, correct option would be E. None of the above have BOTH of the two factors accurately stated

6 0

3 years ago

A flame ionization detector, which is often used in gas chromatography, responds to a change in

Lena [83]

Answer:

Option A

Explanation:

We know that ions are present in hydrogen-air flame and when the burning of an organic compound takes place in this flame more ions are produced in the flame.

Thus when we apply a voltage across this flame, the ion collector plate attracts the all the ions in the flame.

The presence of organic compounds increases the voltage across the hydrogen ion flame produced at the ion collector increases and as the voltage increases, the detection of the organic compound can be made in turn.

Thus flame ionization detector clearly responds to the variation in the collection of ions or electrons in a flame.

3 0

3 years ago

Component of earthing and reasons why each material is being used​

Component of earthing and reasons why each material is being used