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MArishka [77]
3 years ago
15

A metal shear can be used to cut flat stock , round stock , channel iron and which of the following?

Engineering
1 answer:
Charra [1.4K]3 years ago
3 0

Answer:

Angle Iron

Explanation:

Mark brainliest please.

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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
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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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Im pretty sure it is true !
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