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3 years ago

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Air is compressed in a reversible, isothermal, steady- flow process from 15 psia, 100°F to 100 psia. Calculate the work of compr

ession per pound, the change of entropy, and the heat transfer per pound of air compressed.

1 answer:

mixas84 [53]3 years ago

3 0

Answer:

|W|=169.28 KJ/kg

ΔS = -0.544 KJ/Kg.K

Explanation:

Given that

T= 100°F

We know that

1 °F = 255.92 K

100°F = 310 .92 K

$P _1= 15 psia$

$P _1= 100 psia$

We know that work for isothermal process

$W=mRT\ln \dfrac{P_1}{P_2}$

Lets take mass is 1 kg.

So work per unit mass

$W=RT\ln \dfrac{P_1}{P_2}$

We know that for air R=0.287KJ/kg.K

$W=RT\ln \dfrac{P_1}{P_2}$

$W=0.287\times 310.92\ln \dfrac{15}{100}$

W= - 169.28 KJ/kg

Negative sign indicates compression

|W|=169.28 KJ/kg

We know that change in entropy at constant volume

$\Delta S=-R\ln \dfrac{P_2}{P_1}$

$\Delta S=-0.287\ln \dfrac{100}{15}$

ΔS = -0.544 KJ/Kg.K

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Answer:

t= 4.5 mm

Explanation:

Given that

P = 520 KPa ( gauge)

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The normal stress for pressure vessel given as

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We always take maximum stress for safe design.

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Now by putting the values

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Answer:

a)5.28 Å , b)3.73 Å , c)3.048 Å

Explanation:

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We know formula of distance,

d = $\frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}$

a)(100)

a=5.28 Å

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b)(110)

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c)(111)

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Answer with Explanation:

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part 1)

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part 2)

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part 3)

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