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34kurt
1 year ago
11

What part of the scope pattern show the duration of the spark?

Engineering
2 answers:
malfutka [58]1 year ago
6 0

Answer:

This spark energy trigger ignition and combustion in the compressed air-fuel mixture. This discharge is of extremely brief duration (about 1/1000 of a second) and is extraordinarily complex.

dezoksy [38]1 year ago
3 0

Answer: This spark energy trigger ignition and combustion in the compressed air-fuel mixture. This discharge is of extremely brief duration (about 1/1000 of a second) and is extraordinarily complex!

-Your Welcome-

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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
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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)

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\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

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