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Damm [24]
3 years ago
11

A glass plate is subjected to a tensile stress of 40 MPa. If the specific surface energy is 0.3 J/m2 and the modulus of elastici

ty is 69 GPa, determine the maximum length of a surface flaw that is possible without fracture.
Engineering
1 answer:
Pavlova-9 [17]3 years ago
4 0

Answer:

8.24μm

Explanation:

The theory of brittle fracture was used to solve this problem.

And if you follow through with the attachment made a the subject of the formula

Such that,

a = 2x(69x10⁹)x0.3/pi(40x10⁶)²

= 4.14x10¹⁰/5.024x10¹⁵

= 8.24x10^-06

= 8.24μm

This is the the maximum length of the surface flaw

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What are the characteristic features of stress corrosion cracks?
Oduvanchick [21]

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.
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5 0
3 years ago
A block of ice weighing 20 lb is taken from the freezer where it was stored at -15"F. How many Btu of heat will be required to c
Rus_ich [418]

Answer:

Heat required =7126.58 Btu.

Explanation:

Given that

Mass m=20 lb

We know that

1 lb =0.45 kg

So 20 lb=9 kg

m=9 kg

Ice at -15° F and we have to covert it at 200° F.

First ice will take sensible heat at up to 32 F then it will take latent heat at constant temperature and temperature will remain 32 F.After that it will convert in water and water will take sensible heat and reach at 200 F.

We know that

Specific heat for ice C_p=2.03\ KJ/kg.K

Latent heat for ice H=336 KJ/kg

Specific heat for ice C_p=4.187\ KJ/kg.K

We know that sensible heat given as

Q=mC_p\Delta T

Heat for -15F to 32 F:

Q=mC_p\Delta T

Q=9\times 2.03(32+15) KJ

Q=858.69 KJ

Heat for 32 Fto 200 F:

Q=mC_p\Delta T

Q=9\times 4.187(200-32) KJ

Q=6330.74 KJ

Total heat=858.69 + 336 +6330.74 KJ

Total heat=7525.43 KJ

We know that 1 KJ=0.947 Btu

So   7525.43 KJ=7126.58 Btu

So heat required to covert ice into water is 7126.58 Btu.

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3 years ago
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A room is kept at −5°C by a vapor-compression refrigeration cycle with R-134a as the refrigerant. Heat is rejected to cooling wa
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Answer:

note:

<u>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</u>

Download docx
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3 years ago
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