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statuscvo [17]
3 years ago
5

ladders are not required to be inspected for visible defects prior to the first use of each work shift,and after any occuurrence

that could affect their safety.. True or False?​
Engineering
1 answer:
zaharov [31]3 years ago
4 0

Answer:

The answer is False.

Explanation:

When it comes to occupational safety,<em> it is very important for ladders to be inspected by a qualified person before each use.</em> This is because ladders undergo conditions that impact their integrity while being in use. The inspection is also essential in order for the ladder to be timely replaced.

<u><em>Ladder accidents or ladder-related injuries happen every year.</em></u> Around 700 occupational deaths due to elevated fall from a ladder accounts for 15% of all occupational deaths. Misuse or damage ladders are often the reasons for this.

Thus, the answer in the above statement is False because ladders are required to be inspected for visible defects prior to the first use of each work shift and after any occurrence that could affect their safety.

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3 years ago
Air at 2.5 bar, 400 K is extracted from a main jet engine compressor for cabin cooling. The extracted air enters a heat exchange
Shkiper50 [21]

Answer:

a) Power developed by the turbine = 132.89 kW

b) magnitude of the rate of heat transfer from the air to the ambient, in kw = 251.25 kW

Explanation:

b) The process is a constant pressure process (Isobaric process)

The constant pressure specific heat of air, c_{p} = 1.005 kJ/kg -K

Specific heat ratio for air, \gamma = 1.4

The mass flow rate of air, \dot{m} = 2.5 kg/s

P₁ = 2.5 bar, T₁ = 400 K

P₂ = 2.5 bar, T₂ = 300 K

Using the steady flow energy equation:

Q_{1-2}  = \dot{m} c_{p} (T_{2} - T_{1} \\Q_{1-2}  = 2.5 * 1.005 * (300 - 400)\\Q_{1-2}  = -251.25 kW

Therefore, the magnitude of the rate of heat transfer from the air to the ambient, in kw, Q_{1-2} = 251.25 kW

a) For the isentropic process:

Power developed by the turbine is given by the relation \dot{W} = \dot{M}  c_{p} (T_{2} - T_{3})

Isentropic efficiency, \eta_{t} = 80%

P₂ = 2.5 bar, T₂ = 300 K

P₃ = 1 bar, T_{3s} = ? where T_{3s} is the isentropic temperature at 100% efficiency

The isentropic relation is given by:

\frac{T_{3s} }{T_{2} } = (\frac{P_{3} }{P_{2} }) ^{\frac{\gamma - 1}{\gamma} } \\\frac{T_{3s} }{300 } = (\frac{1 }{2.5 }) ^{\frac{1.4 - 1}{1.4 }

T_{3s} = 230.9 K

To get the temperature at 80% efficiency, we will use the relation:

\eta_{t} = \frac{T_{2} - T_{3}  }{T_{2} - T_{3s} } \\0.8= \frac{300 - T_{3}  }{300 - 230.9 }

T₃ = 244.72 K

Power developed by the turbine is given by the relation:

\dot{W} = \dot{M}  c_{p} (T_{2} - T_{3})\\ \dot{W} = 2.5 * 1.005* (300-244.72)\\ \dot{W} = 138.89 kW

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