All categories

3 years ago

This acronym is a reminder of the most common types of hazards or injuries caused by electricity.

Engineering

1 answer:

yulyashka [42]3 years ago

3 0

Answer:

BE SAFE

Explanation:

it stands for Burns, Electrocution, Shock, Arch Flash/Arc Blast, Fire, Explosions

You might be interested in

A cylindrical specimen of cold-worked steel has a Brinell hardness of 250.(a) Estimate its ductility in percent elongation.(b) I

koban [17]

Answer:

A) Ductility = 11% EL

B) Radius after deformation = 4.27 mm

Explanation:

A) From equations in steel test,

Tensile Strength (Ts) = 3.45 x HB

Where HB is brinell hardness;

Thus, Ts = 3.45 x 250 = 862MPa

From image 1 attached below, for steel at Tensile strength of 862 MPa, %CW = 27%.

Also, from image 2,at CW of 27%,

Ductility is approximately, 11% EL

B) Now we know that formula for %CW is;

%CW = (Ao - Ad)/(Ao)

Where Ao is area with initial radius and Ad is area deformation.

Thus;

%CW = [[π(ro)² - π(rd)²] /π(ro)²] x 100

%CW = [1 - (rd)²/(ro)²]

1 - (%CW/100) = (rd)²/(ro)²

So;

(rd)²[1 - (%CW/100)] = (ro)²

So putting the values as gotten initially ;

(ro)² = 5²([1 - (27/100)]

(ro)² = 25 - 6.75

(ro) ² = 18.25

ro = √18.25

So ro = 4.27 mm

6 0

4 years ago

A well-insulated rigid tank contains 5 kg of a saturated liquid–vapor mixture of water at 200 kPa. Initially, three-quarters of

vlabodo [156]

Answer:

$S_{gen} = 18.519\,\frac{kJ}{K}$

Explanation:

Given that rigid tank is a closed system, the following model is constructed after the First Law of Thermodynamics:

$W_{heater} + U_{sys,1} - U_{sys,2} = 0$

$W_{heater} = U_{sys,2} - U_{sys,1}$

$W_{heater} = m\cdot (u_{2}-u_{1})$

The entropy generation inside the rigid tank is determined by appropriate application of the Second Law of Thermodynamics:

$S_{sys,1} - S_{sys,2} + S_{gen} = 0$

$S_{gen} = S_{sys,2} - S_{sys,1}$

$S_{gen} = m\cdot (s_{2}-s_{1})$

The properties of the steam are obtained from steam tables:

Intial State

$P = 200\,kPa$

$T = 120.21\,^{\textdegree}C$

$\nu = 0.2222\,\frac{m^{3}}{kg}$

$u = 1010.7\,\frac{kJ}{kg}$

$s = 2.9294\,\frac{kJ}{kg\cdot K}$

$x = 0.25$

Final State

$P = 869.567\,kPa$

$T = 173.88\,^{\textdegree} C$

$\nu = 0.2222\,\frac{m^{3}}{kg}$

$u = 2578.6\,\frac{kJ}{kg}$

$s = 6.6332\,\frac{kJ}{kg\cdot K}$

$x = 1.00$

The entropy change of the steam during the process is:

$S_{gen} = (5\,kg)\cdot \left(6.6332\,\frac{kJ}{kg\cdot K} - 2.9294\,\frac{kJ}{kg\cdot K} \right)$

$S_{gen} = 18.519\,\frac{kJ}{K}$

8 0

3 years ago

A piston-cylinder assembly contains 2 lb of air at a temperature of 540 degrees R and a pressure of 1 atm. The air is compressed

attashe74 [19]

Answer:

$W=-123.8Btu$

Explanation:

Hello,

In this problem, the piston-cylinder assembly make us state the energy balance as:

$Q-W=\Delta U$

Thus, we must now compute $\Delta U$ in terms of Cv as follows:

$\Delta U=mCv(T_2-T_1)\\\Delta U=2lb*0.173\frac{Btu}{lb*R}*(840R-540R)\\ \Delta U=103.8Btu$

Now, since heat is given off, its sign is negative, thus, the work is computed as:

$W=Q-\Delta U\\W=-20Btu-103.8Btu=-123.8Btu$

This work means that work was done over the system in order to allow the compression.

The suppositions were:

- The change in the internal energy is a function of the temperature.

- Air is an ideal gas.

Best regards.

8 0

4 years ago

Automotive class

sp2606 [1]

Answer:

what are you in

Explanation:

7 0

3 years ago

Which of the following influenced the hair of the 1980s?

geniusboy [140]

Answer:

OB. the styles of popular musicians.

Explanation:

6 0

3 years ago