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What can your employer do to protect you from overhead power lines?

agasfer [191]

Answer:

Have the power company install insulated sleeves (also known as “eels”) over power lines.

Wearing PPE is the only way to prevent being electrocuted

Explanation:

To prevent electrocution at workplace, employers can ensure that the power company install insulated sleeves (also known as “eels”) over power lines. Additionally, the employees should wear PPEs which are insulators to prevent electrocution.

5 0

2 years ago

Air at 400kPa, 970 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occu

Sonja [21]

Answer:

a

The rate of work developed is $\frac{\r W}{\r m}= 300kJ/kg$

b

The rate of entropy produced within the turbine is $\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

Explanation:

From the question we are told

The rate at which heat is transferred is $\frac{\r Q}{\r m } = - 30KJ/kg$

the negative sign because the heat is transferred from the turbine

The specific heat capacity of air is $c_p = 1.1KJ/kg \cdot K$

The inlet temperature is $T_1 = 970K$

The outlet temperature is $T_2 = 670K$

The pressure at the inlet of the turbine is $p_1 = 400 kPa$

The pressure at the exist of the turbine is $p_2 = 100kPa$

The temperature at outer surface is $T_s = 315K$

The individual gas constant of air R with a constant value $R = 0.287kJ/kg \cdot K$

The general equation for the turbine operating at steady state is \

$\r Q - \r W + \r m (h_1 - h_2) = 0$

h is the enthalpy of the turbine and it is mathematically represented as

$h = c_p T$

The above equation becomes

$\r Q - \r W + \r m c_p(T_1 - T_2) = 0$

$\frac{\r W}{\r m} = \frac{\r Q}{\r m} + c_p (T_1 -T_2)$

Where $\r Q$ is the heat transfer from the turbine

$\r W$ is the work output from the turbine

$\r m$ is the mass flow rate of air

$\frac{\r W}{\r m}$ is the rate of work developed

Substituting values

$\frac{\r W}{\r m} = (-30)+1.1(970-670)$

$\frac{\r W}{\r m}= 300kJ/kg$

The general balance equation for an entropy rate is represented mathematically as

$\frac{\r Q}{T_s} + \r m (s_1 -s_2) + \sigma = 0$

=> $\frac{\sigma}{\r m} = - \frac{\r Q}{\r m T_s} + (s_1 -s_2)$

generally $(s_1 -s_2) = \Delta s = c_p\ ln[\frac{T_2}{T_1} ] + R \ ln[\frac{v_2}{v_1} ]$

substituting for $(s_1 -s_2)$

$\frac{\sigma}{\r m} = \frac{-\r Q}{\r m} * \frac{1}{T_s} + c_p\ ln[\frac{T_2}{T_1} ] - R \ ln[\frac{p_2}{p_1} ]$

Where $\frac{\sigma}{\r m}$ is the rate of entropy produced within the turbine

substituting values

$\frac{\sigma}{\r m} = - (-30) * \frac{1}{315} + 1.1 * ln\frac{670}{970} - 0.287 * ln [\frac{100kPa}{400kPa} ]$

$\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

5 0

3 years ago

Flow and Pressure Drop of Gases in Packed Bed. Air at 394.3 K flows through a packed bed of cylinders having a diameter of 0.012

devlian [24]

The pressure drop of air in the bed is 14.5 kPa.

<u>Explanation:</u>

To calculate Re:

$R e=\frac{1}{1-\varepsilon} \frac{\rho q d_{p}}{\mu}$

From the tables air property

$\mu_{394 k}=2.27 \times 10^{-5}$

Ideal gas law is used to calculate the density:

ρ = $\frac{2.2}{2.83 \times 10^{-3} \times 394.3}$

ρ = 1.97 Kg / $m^{3}$

ρ = $\frac{P}{RT}$

R = $\frac{R_{c} }{M}$ = 8.2 × $10^{-5}$ / 28.97× $10^{-3}$

R = 2.83 × $10^{-3}$ $m^{3}$ atm / K Kg

q is expressed in the unit m/s

$q=\frac{2.45}{1.97}$

q = 1.24 m/s

Re = $\frac{1}{1-0.4} \frac{1.97 \times 1.24 \times 0.0127}{2.27 \times 10^{-5}}$

Re = 2278

The Ergun equation is used when Re > 10,

$\frac{\Delta P}{L}=\frac{180 \mu}{d_{p}^{2}} \frac{(1-\varepsilon)^{2}}{\varepsilon^{3}} q+\frac{7}{4} \frac{\rho}{d_{p}} \frac{(1-\varepsilon)}{\varepsilon^{3}} q^{2}$

$\frac{\Delta P}{L}=\frac{180 \times 2.27 \times 10^{-5}}{0.0127^{2}} \frac{(1-0.4)^{2}}{0.4^{3}} 1.24$ $+\frac{7}{4} \frac{1.97}{0.0127} \frac{(1-0.4)}{0.4^{3}} 1.24^{2}$

= 4089.748 Pa/m

ΔP = 4089.748 × 3.66

ΔP = 14.5 kPa

4 0

2 years ago

A joining process in which a filler metal is melted and distributed by capillary action between faying surfaces, the base metals

PolarNik [594]

Answer:

A soldering process

Explanation:

Given that ,The filler metal's melting point temperature is less than 450 ° C.

Usually, the brazing material has the liquid temperature of the melting point, the full melting point of the filler material approaching 450 degrees centigrade, while the filler material is less than 450 degrees centigrade in the case of soldering.

Therefore the answer is "A soldering process".

4 0

2 years ago

Raul doesn’t feel like he needs to write down events that will happen months from now. Explain to him why it is important to use

MAXImum [283]

The primary reason for having a calendar is to organize the days, weeks, months and years. It keeps a track of which day of the week events fall and when special events are going to happen. Historically speaking calendars were often uses to preserve religious holidays and events.

4 0

2 years ago

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