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3 years ago

To protect their employees from caught-in and -between hazards, what must employers do?

Engineering

1 answer:

marusya05 [52]3 years ago

5 0

Answer:

Employers must comply with OSHA's regulations to safeguard workers from caught-in or -between dangers, which include, but are not limited to, the following:

• Protect power tools and other equipment with moving parts with guards.

• Support, secure, or otherwise make safe any equipment that has pieces that workers could become entangled in.

· Take precautions to avoid workers being crushed by tipped-over heavy machinery.

• Take precautions to avoid pinning workers between equipment and a solid object.

• Provide workers with protection when trenching and excavating.

• Provide ways to prevent constructions, such as scaffolds, from collapsing.

Explanation:

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You and your design group are competing for a multidisciplinary concept project. Your firm is the lead group in the design profe

sasho [114]

Answer: the answer for the question is a

Explanation:

4 0

3 years ago

Air modeled as an ideal gas enters a turbine operating at steady state at 1040 K, 278 kPa and exits at 120 kPa. The mass flow ra

gladu [14]

Answer:

a) $T_{2}=837.2K$

b) $e=91.3$ %

Explanation:

A) First, let's write the energy balance:

$W=m*(h_{2}-h_{1})\\W=m*Cp*(T_{2}-T_{1})$ (The enthalpy of an ideal gas is just function of the temperature, not the pressure).

The Cp of air is: 1.004 $\frac{kJ}{kgK}$ And its specific R constant is 0.287 $\frac{kJ}{kgK}$ .

The only unknown from the energy balance is $T_{2}$ , so it is possible to calculate it. The power must be negative because the work is done by the fluid, so the energy is going out from it.

$T_{2}=T_{1}+\frac{W}{mCp}=1040K-\frac{1120kW}{5.5\frac{kg}{s}*1.004\frac{kJ}{kgk}} \\T_{2}=837.2K$

B) The isentropic efficiency (e) is defined as:

$e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}$

Where ${h_{2s}$ is the isentropic enthalpy at the exit of the turbine for the isentropic process. The only missing in the last equation is that variable, because $h_{2}-h_{1}$ can be obtained from the energy balance $\frac{W}{m}=h_{2}-h_{1}$

$h_{2}-h_{1}=\frac{-1120kW}{5.5\frac{kg}{s}}=-203.64\frac{kJ}{kg}$

An entropy change for an ideal gas with constant Cp is given by:

$s_{2}-s_{1}=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})$

You can review its deduction on van Wylen 6 Edition, section 8.10.

For the isentropic process the equation is:

$0=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})\\Rln(\frac{P_{2}}{P_{1}})=Cpln(\frac{T_{2}}{T_{1}})$

Applying logarithm properties:

$ln((\frac{P_{2}}{P_{1}})^{R} )=ln((\frac{T_{2}}{T_{1}})^{Cp} )\\(\frac{P_{2}}{P_{1}})^{R}=(\frac{T_{2}}{T_{1}})^{Cp}\\(\frac{P_{2}}{P_{1}})^{R/Cp}=(\frac{T_{2}}{T_{1}})\\T_{2}=T_{1}(\frac{P_{2}}{P_{1}})^{R/Cp}$

Then,

$T_{2}=1040K(\frac{120kPa}{278kPa})^{0.287/1.004}=817.96K$

So, now it is possible to calculate $h_{2s}-h_{1}$ :

$h_{2s}-h_{1}}=Cp(T_{2s}-T_{1}})=1.004\frac{kJ}{kgK}*(817.96K-1040K)=-222.92\frac{kJ}{kg}$

Finally, the efficiency can be calculated:

$e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}=\frac{-203.64\frac{kJ}{kg}}{-222.92\frac{kJ}{kg}}\\e=0.913=91.3$ %

4 0

4 years ago

Giải giúp mình câu này được không ạ.em cảm ơn

sp2606 [1]

Answer:

çâdßèöñbvjsjaushdhshs

4 0

3 years ago

In a steam power plant, 1 MW is added in the boiler, 0.58 MW is taken out in the condenser and the pump work is 0.02 MW.

Free_Kalibri [48]

Answer:

a) $\eta = 42\,\%$ , b) $COP_{R} = 29$

Explanation:

a) The thermal efficiency is:

$\eta = \frac{\dot Q_{in} - \dot Q_{out}}{\dot Q_{in}}\times 100\,\%$

$\eta = \frac{1\,MW-0.58\,MW}{1\,MW} \,\times 100\,\%$

$\eta = 42\,\%$

b) The coefficient of performance is:

$COP_{R} = \frac{\dot Q_{L}}{\dot W}$

$COP_{R} = \frac{0.58\,MW}{0.02\,MW}$

$COP_{R} = 29$

3 0

3 years ago

Stone has been used as a building material since ancient times because of its compressive strength, which is the __________.

Doss [256]

Complete Question:

Stone has been used as a building material since ancient times because of its compressive strength, which is the?

Group of answer choices

A. ability to support pressure without breaking.

B. ability to press down and become solidly fixed in the ground.

C. relative weight of a block of stone to its size.

D. force a block exerts on the blocks around it.

Answer:

A. ability to support pressure without breaking.

Explanation:

Stone has been used as a building material since ancient times because of its compressive strength, which is the ability to support pressure without breaking.

Compressive strength can be defined as the ability of a structural element or particular material to withstand an applied, which is aimed at reducing the size of the structural element.

Simply stated, it is the ability of a structural element or material to withstand an applied load without deflections, fracture or having any crack.

In this context, a stone possesses the ability to resist or withstand compression loads.

Some examples of other materials or structural elements having good compressive strength are steel, bones, concrete etc. The standard unit of measurement of the compressive strength of a material is Mega Pascal (MPa) or pound-force per square inch (psi) in the United States of America.

7 0

4 years ago