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Daniel [21]
3 years ago
15

What are the characteristic features of stress corrosion cracks?

Engineering
1 answer:
Oduvanchick [21]3 years ago
5 0

Answer and Explanation:

The crack formation growth that takes place in an environment corrosive.

Stress corrosion cracks can be defined as the spontaneous failures of the metal alloy as a result of the combined action of corrosion and high tensile stress.

Some of the characteristic features of stress corrosion cracks are:

  • These occur at high temperatures.
  • Occurrence of failures in metals mechanically.
  • Occurrence of sudden and unexpected failures under tensile stress.
  • The rate of work hardening of the metal alloy is high.
  • Time
  • An environment that is specific for stress corrosion cracking.
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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
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

4 0
3 years ago
Read 2 more answers
According to information found in an old hydraulies book, the energy loss per unit weight of fluid flowing through a nozzle conn
nekit [7.7K]

Answer:

Yes equation is valid.

Explanation:

Given:

h = (0.04 to 0.09)(D/d)^4*V^2/2*g

Using SI units to assign dimensions to every quantity as follows:

Energy loss per unit weight h = J / N = kg m ^2 s^-2 / kg m s^-2 = [m]

Hose diameter D = [m]

Nozzle tip diameter d = [m]

Fluid velocity in the hose V = [ m s^-1 ]

Acceleration of gravity g = [ m s^-2 ]

Using the Given Equation and plug the SI units of respective quantities:

h = (0.04 to 0.09)(D/d)^4*V^2/2*g

[m] = (0.04 to 0.09)([m] / [m])^4*[ m s^-1 ]^2/2*[ m s^-2 ]

Simplify the equation above:

[m] = ( 1 )^4 * [ m^2 s^-2 ] / [ m s^-2 ]

[m] = [m]

Hence, SI units of RHS of given equation = LHS of given equation, we can say the equation has consistent dimensions.

3 0
3 years ago
What is Pressure measured from absolute zero pressure called?
7nadin3 [17]
Anything greater than total vacuum is technically a form of pressure
5 0
3 years ago
9. Calculate the total resistance and current in a parallel cir-
Taya2010 [7]

Answer:

  d. 2.3 ohms (5.3 amperes)

Explanation:

The calculator's 1/x key makes it convenient to calculate parallel resistance.

  Req = 1/(1/4 +1/8 +1/16) = 1/(7/16) = 16/7 ≈ 2.3 ohms

This corresponds to answer choice D.

__

<em>Additional comment</em>

This problem statement does not tell the applied voltage. The answer choices suggest that it is 12 V. If so, the current is 12/(16/7) = 21/4 = 5.25 amperes.

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3 years ago
Vital role of maritime english among seaferers
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Answer:

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