How engineer can find problems and solutions Give example
1 answer:
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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
Latent heat for ice H=336 KJ/kg
Specific heat for ice
We know that sensible heat given as
Heat for -15F to 32 F:
Q=858.69 KJ
Heat for 32 Fto 200 F:
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.
Answer:
the ratio of the etched to the original crack tip radius is 30.24
Explanation:
Given the data in the question;
we determine the initial fracture stress using the following expression;
(σf)₁ = 2(σ₀)₁ α₁/(
)₁
----- let this be equation 1
where; (σ₀)₁ is the initial fracture strength
()₁ is the original crack tip radius
α₁ is the original crack length.
first, we determine the final crack length;
α₂ = α₁ - 16% of α₁
α₂ = α₁ - ( 0.16 × α₁)
α₂ = α₁ - 0.16α₁
α₂ = 0.84α₁
next, we calculate the final fracture stress;
the fracture strength is increased by a factor of 6;
(σ₀)₂ = 6( σ₀ )₁
Now, expression for the final fracture stress
(σf)₂ = 2(σ₀)₂ α₂/(
)₂
------- let this be equation 2
where ()₂ is the etched crack tip radius
value of fracture stress of glass is constant
Now, we substitute 2(σ₀)₁ α₁/(
)₁
from equation for (σf)₂ in equation 2.
0.84α₁ for α₂.
6( σ₀ )₁ for (σ₀)₂.
∴
2(σ₀)₁ α₁/(
)₁
= 2(6( σ₀ )₁)
0.84α₁/(
)₂
divide both sides by 2(σ₀)₁
α₁/(
)₁
= 6
0.84α₁/(
)₂
1/(
)₁
= 6
0.84/(
)₂
1/(
)₁
= 36
0.84/(
)₂
1 / ()₁ = 30.24 / (
)₂
()₂ = 30.24(
)₁
()₂/(
)₁ = 30.24
Therefore, the ratio of the etched to the original crack tip radius is 30.24