Ideal Gas Expansion State Properties & Exergy Balance

[Cora]

Homework Statement

Two well-insulated rigid tanks of equal volume, tank A and tank B, are connected via a valve. Tank A is initially empty. Tank B has 2 kg of Argon at 350 K and 5000 kPa. The valve is opened and the Argon fills both tanks. State 2 is the final equilibrium state. The temperature and pressure of the room in which the tanks sit are 300 K and 100 kPa, respectively. Perform a closed system analysis.
a. Determine the volume of tank B.
b. Determine the final temperature.
c. Determine the final pressure.
d. Determine the entropy produced in the process.
e. Determine the exergy (or availability) destroyed in the process.

i. Using an exergy balance.
ii. Using the entropy produced.​

f. Determine the total exergy in tank B initially.

Homework Equations

Ideal Gas law:

PV = ZmRT​

First Law of Thermo

E2-E1 = Q - W + ∑m(h+v²/2+gz)​

Second Law of Thermo

S2-S1=int(1/Tb,1,2,dQ)+σ​

The Attempt at a Solution

PART A)
using tabulated values for Tc and Pc of Argon, I found Z (compressibility factor) to be 0.9945~1, so I'm treating the Argon as behaving like an ideal gas.
given T_B1 and P_B1, specific volume is found to be v_B1=0.00137 m³/kg
V_B = v_B1*m_B1
V_B = 0.02914 m³

PART B)

My approach to find the temperatures would be to use conservation of internal energy between initial and final states.

U₁ = U₂
U₁ = U_A1 + U_B1
U₁ = u_A1*m_A1 + u_B1*m_A1
where tank A is initially evacuated, so m_A1 = 0, therefor
U₁ = u_B1*m_B1

and then for U₂

U₂ = u_A2*m_A2 + u_B2*m_B2

assuming from conservation of mass: m_B1 = m_A2 + m_B2
and that since P_A2=P_B2 and T_A2 = T_B2, m_A2 = m_B2 = m_B1/2​

U₂ = (m_B1/2)*(u_A2 + u_B2)

assuming u_A2 = u_B2 = .5*u₂​

U₂ = (m_B1/2)*(u₂)
U₂ = .5*m_B1*u₂

now equating U₁ and U₂:

u_B1*m_B1 = .5*m_B1*u₂
u_B1 = .5*U₂ = u_B2 = u_A2

so since I have u_B1, T_B1 and have equated the final specific internal energies to u_B1, I can find the final temperature and move on. Yes?

PART C)
In order to find the pressure I would equate the specific volume of the tanks, resulting in
.5*v_B1 = v_B2 = v_A2

next using final specific volume and final temperature to determine the final pressure.

Is this heading in the right direction?

 

[Chestermiller]

No. This assumes that the two tanks equilibrate thermally, such that u_A2=u_B2=u_B1, and T_A2=T_B2=T_B1. It also implies that p_A2=p_B2=p_B1/2. However, the tanks are insulated, and so, the gases in the two tanks do not thermally equilibrate.

What is your assessment of what is happening to the gas in insulated tank B during the time that gas is seeping through the valve (while the system is in the process of equilibrating)?

Chet

 

[Cora]

No. This assumes that the two tanks equilibrate thermally, such that u_A2=u_B2=u_B1, and T_A2=T_B2=T_B1. It also implies that p_A2=p_B2=p_B1/2. However, the tanks are insulated, and so, the gases in the two tanks do not thermally equilibrate.

What is your assessment of what is happening to the gas in insulated tank B during the time that gas is seeping through the valve (while the system is in the process of equilibrating)?

Chet

The gas in tank B is expanding, or decreasing in Pressure (so also in Temperature).

 

[Chestermiller]

The gas in tank B is expanding, or decreasing in Pressure (so also in Temperature).

So, are you saying that the gas remaining in the tank at any instant of time has been experiencing something very close to an adiabatic reversible expansion (particularly if the seepage rate through the valve is very slow)?

 

[Cora]

That's what it sounds like to me. So the 2nd law is where this is headed...

 

[Chestermiller]

That's what it sounds like to me. So the 2nd law is where this is headed...

No. We are going to express everything in terms of the final pressure P, and then solve for P under the constraint that the change in internal energy of the overall system is zero. So, let's get started.

Given that the initial temperature and pressure in tank B is 350 K and 5000 kPa, in terms of P, what is the final temperature T in tank B?
Given the final volume in tank B is 0.02914 cubic meters, in terms of P, what is the final mass of argon in tank B?

 

[Cora]

for T₂:
So if I'm not using the second law, the other equation I see T and P in is the ideal gas law. Can I use this even though I can't check the validity of the second state behaving as an ideal gas?

P₂V₂/T₂=P₁T₁/V₁
is it appropriate here to use V2 = 2*V1, since I've been told the tanks are equal in size? Then it would become:
T₂=2*P₂T₁/P₁

for m₂:
the ideal gas law also seems appropriate here because it has P, V and m in it. Yeah?

m_B2=P₂V₁/(R*T₂)

 

[Chestermiller]

None of this is correct. Our focus is on the gas that remains in tank B during and after the system equilibrates. This gas is experiencing and adiabatic reversible expansion. Do you know the relationship between temperature and pressure for an ideal gas experiencing an adiabatic reversible expansion? If not, what about the relationship between molar volume and pressure for an ideal gas experiencing and adiabatic reversible expansion?

Chet

 

[Sarooj94]

Hey

So I followed similar approach to yours and I believe we get the same answer
my approach:

assumption 1: Argon is an ideal gas
mAB = mB (COM)

1st law:
E2- E1 =Q -W
since it is well insulated Q = 0 (Adiabatic)
and because it is rigid body I assumed there is boundary work so W = 0

From this I went on and
equated E2-E1=0
so U2-U1 +change in KE and PE =0 (assumption neglect PE and KE)

mAB * uAB = mA*uA + mBuB
since mA is 0 and mAB = mB
uAB = uB
so TempB = T2

for final pressure
we know that v2 = 2*vB
so P2 = (m*R_bar*TempB)/(2*vB)

which is approximately half PressureB

I got stuck in the last part of the question which is finding the total energy in tank B initially! :(

 

[Chestermiller]

Hey

So I followed similar approach to yours and I believe we get the same answer
my approach:

assumption 1: Argon is an ideal gas
mAB = mB (COM)

1st law:
E2- E1 =Q -W
since it is well insulated Q = 0 (Adiabatic)
and because it is rigid body I assumed there is boundary work so W = 0

From this I went on and
equated E2-E1=0
so U2-U1 +change in KE and PE =0 (assumption neglect PE and KE)

mAB * uAB = mA*uA + mBuB
since mA is 0 and mAB = mB
uAB = uB
so TempB = T2

for final pressure
we know that v2 = 2*vB
so P2 = (m*R_bar*TempB)/(2*vB)

which is approximately half PressureB

I got stuck in the last part of the question which is finding the total energy in tank B initially! :(

As I said earlier, this is only correct if the two tanks can reach thermal equilibrium with one another, but, according to the problem statement, they cannot.

 

[Chestermiller]

For a nice analysis of the change in tank B between the initial and final states of the system, see Example 6.10 in Fundamentals of Engineering Thermodynamics by Moran, Shapiro, Boettner, and Bailey.