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lord [1]
3 years ago
8

For ceramic-matrix composites, high interfacial strength is desirable. ( True , False )

Engineering
1 answer:
NemiM [27]3 years ago
4 0
This is true as the answer
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An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate (ϵ˙) is found to be
cupoosta [38]

Answer:

Activation energy for creep in this temperature range is Q = 252.2 kJ/mol

Explanation:

To calculate the creep rate at a particular temperature

creep rate, \zeta_{\theta} = C \exp(\frac{-Q}{R \theta} )

Creep rate at 800⁰C, \zeta_{800} = C \exp(\frac{-Q}{R (800+273)} )

\zeta_{800} = C \exp(\frac{-Q}{1073R} )\\\zeta_{800} = 1 \% per hour =0.01\\

0.01 = C \exp(\frac{-Q}{1073R} ).........................(1)

Creep rate at 700⁰C

\zeta_{700} = C \exp(\frac{-Q}{R (700+273)} )

\zeta_{800} = C \exp(\frac{-Q}{973R} )\\\zeta_{800} = 5.5 * 10^{-2}  \% per hour =5.5 * 10^{-4}

5.5 * 10^{-4}  = C \exp(\frac{-Q}{1073R} ).................(2)

Divide equation (1) by equation (2)

\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\

Take the natural log of both sides

ln 18.182= 0.0000115Q\\2.9004 = 0.0000115Q\\Q = 2.9004/0.0000115\\Q = 252211.49 J/mol\\Q = 252.2 kJ/mol

3 0
3 years ago
A steam power plant is represented as a heat engine operating between two thermal reservoirs at 800 K and 300 K. The temperature
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Tech A says that horsepower is a measurement simply of the amount of work being performed. Tech B says that horsepower can be ca
snow_tiger [21]

Answer:

Tech B

Explanation:

Horsepower (hp) refers to a unit of measurement of power in respect of the output of engines or motors.

Horsepower is the common unit of power. It indicates the rate at which work is done.

The formula \frac{rpm*T}{5252}, where rpm is the engine speed, T is the torque, and 5,252 is radians per second.

So,

Tech B is correct

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