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3 years ago

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Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 3 m3, is stirred

until its temperature is 600 K. Assuming the ideal gas model for the air, and ignoring kinetic and potential energy, determine : (a) the final pressure, in bar. (b) the work, in kJ. (c) the amount of entropy produced, in kJ/K.

1 answer:

madam [21]3 years ago

5 0

Answer:

a)P₂ =4 bar

b)W= - 1482.48 KJ

It means that work done on the system.

c)S₂ - S₁ = 3.42 KJ/K

Explanation:

Given that

T₁ = 300 K ,V₁ = 3 m³ ,P₁=2 bar

T₂ = 600 K ,V₂=V₁ 3 m³

Given that tank is rigid and insulated.It means that volume of the gas will remain constant.

Lets take the final pressure = P₂

For ideal gas P V = m R T

$\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}$

$P_2=\dfrac{T_2}{T_1}\times P_1$

$P_2=\dfrac{600}{300}\times 2$

P₂ =4 bar

Internal energy

ΔU = m Cv ΔT

Cv=0.71 KJ/kg.k for air

$m=\dfrac{PV}{RT}$

$m=\dfrac{200\times 3}{0.287\times 300}\ kg$

m= 6.96 kg

ΔU= 6.96 x 0.71 x (600 - 300)

ΔU=1482.48 KJ

From first law

Q= ΔU + W

Q= 0 Insulated

W = - ΔU

W= - 1482.48 KJ

It means that work done on the system.

Change in the entropy

$S_2-S_1=mC_v\ \ln\dfrac{T_2}{T_1}$

$S_2-S_1=6.96\times 0.71\ \ln\dfrac{600}{300}$

S₂ - S₁ = 3.42 KJ/K

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Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

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$A_o=1m^2$ is the outlet area

$P_o=100Kpa$ pressure at the outlet area

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Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

$h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}$

$h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}$

Our interest value would be given using interpolation like this:

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Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that $\dot{m}=\frac{\dot{V}}{\upsilon}$ , but since $\dot{V}=Av$ we have this:

$\dot{m}=\frac{Av}{\upsilon}$

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$v=\frac{\upsilon \dot{m}}{A}$ (*)

And now we just need to replace the values into equation (*)

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$v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}$

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$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

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