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3 years ago

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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