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3 years ago

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Air is compressed in the compressor of a turbojet engine. Air enters the compressor at 270 K and 58 kPa and exits the compressor

at 465 K and 350 kPa. Take the average specific heat at constant pressure (cp,avg) of air as 1.015 kJ/ kg.K and Rair = 0.287 kJ/ kg.K. Find the mass specific entropy change associated with the compression process

1 answer:

tia_tia [17]3 years ago

8 0

Answer:

Recall the Entropy Balance (Second Law of Thermodynamics) equations for the system below

And for ideal gas, we know that

Change in Entropy of the air kJ/Kg.K, ΔS = S₁ - S ₂ = Cp Ln (T₂/T₁) - R Ln(P₂/P₁)

where R is gas constant =0.287kJ/kg.K and given

Cp=1.015 kJ/ kg.K

T₂ = 465k T₁=270k, P1=58kPa, P2=350kPa

substituting these values into the eqn above, we have

S = (1.015x 0.544) - (0.287x1.798)

S = 0.0363kJ/Kg.K

Hence, the mass specific entropy change associated with the compression process = 0.0363kJ/Kg.K

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Explanation:

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a. On the attached document, you can see a brief scheme of the process. Thus, to know the degrees of freedom, we state the following unknowns:

- $\xi_1$ and $\xi_2$ : extent of the reactions (2).

- $F_{O_2}^2$ , $F_{CH_4}^2$ , $F_{H_2O}^2$ , $F_{HCHO}^2$ and $F_{CO_2}^2$ : Molar flows at the second stream (5).

On the other hand, we've got the following equations:

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- $F_{CH_4}^2=50mol/s-\xi_1-\xi_2$ : methane mole balance.

- $F_{H_2O}^2=\xi_1+2\xi_2$ : water mole balance.

- $F_{HCHO}^2=\xi_1$ : formaldehyde mole balance.

- $F_{CO_2}^2=\xi_2$ : carbon dioxide mole balance.

Thus, the degrees of freedom are:

$DF=7unknowns-5equations=2$

It means that we need two additional equations or data to solve the problem.

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$0.900=\frac{\xi_1+\xi_2}{50mol/s}$

And for the fractional yield of formaldehyde we can set it in terms of methane as the reagents are equimolar:

$0.860=\frac{F_{HCHO}^2}{50mol/s}$

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$F_{HCHO}^2=0.860*50mol/s=43mol/s$

Which is equal to the first reaction extent $\xi_1$ , therefore, one computes the second one from the fractional conversion of methane as:

$\xi_2=0.900*50mol/s-\xi_1\\\xi_2=0.900*50mol/s-43mol/s\\\xi_2=2mol/s$

Now, one computes the rest of the output flows via:

- $F_{O_2}^2=50mol/s-43mol/s-2*2mol/s=3mol/s$

- $F_{CH_4}^2=50mol/s-43mol/s-2mol/s=5mol/s$

- $F_{H_2O}^2=43mol/s+2*2mol/s=47mol/s$

- $F_{HCHO}^2=43mol/s$

- $F_{CO_2}^2=2mol/s$

The total output molar flow is:

$F_{O_2}+F_{CH_4}+F_{H_2O}+F_{HCHO}+F_{CO_2}=100mol/s$

Therefore the output stream composition turns out into:

$y_{CH_4}^2=\frac{5mol/s}{100mol/s}=0.05\\y_{O_2}^2=\frac{3mol/s}{100mol/s}=0.03\\y_{H_2O}^2=\frac{47mol/s}{100mol/s}=0.47\\y_{HCHO}^2=\frac{43mol/s}{100mol/s}=0.43\\y_{CO_2}^2=\frac{2mol/s}{100mol/s}=0.02$

Best regards.

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Explanation:

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Solution

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The First Law of Thermodynamics states that heat is a form of energy, subject to the principle of conservation of energy, that heat energy cannot be created or destroyed. It can be transferred from one location to another location. i.e.

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