Thermodynamics Question Regarding assuming Irreversibilities

[ScareCrow271828]

Homework Statement

There exists a tank filled with air with a given volume, temperature, and pressure. The tank exists in a room at a given temperature and pressure.
That is:
For the tank: P=1MPa, T=700k, V=1m^3
Outside: T=295K, P=100kPa

Homework Equations

\psi ₂-\psi ₁=(1-T₀/T_b)*₁Q₂-[W=P[SUB]0[/SUB]*(V₂-V₁)-T₀*δ

E₂-E₁=Q-W
PV=mRT

3. Attempt at solution
Will the transfer happen slowly enough that it is safe to assume zero reversabilities? Meaning sigma is zero? In this case I am a little confused between the definitions of internal and external reversibilities.

With the assumptions that work and entropy production are zero.

\psi ₂-\psi ₁=(1-T₀/T_b)*₁Q₂

Then we can get Q from first law with the assumption that the tank will eventually reach the environmental temperatures. So we can get u2 and u1 from the tables at the tank conditions and the environment conditions.

U2-U1=Q
m(u2-u1)=Q
Q/m=u2-u1
(210.49-512.33)Kj/kg*k=-301.84 Kj/kg*k=Q/m

So (multiplying (Q/m)*m)
m=PV/RT=171.823kg
\psi₂-\psi₁=(1-700k/295k)*(-301.84 Kj/kg*k)*(171.823)
=71,201.8302 Kj*K

That is a very large number. Is that correct?

 

[Chestermiller]

`. Problem Statement
There exists a tank filled with air with a given volume, temperature, and pressure. The tank exists in a room at a given temperature and pressure.
That is:
For the tank: P=1MPa, T=700k, V=1m^3
Outside: T=295K, P=100kPa
2. Homework Equations

\psi ₂-\psi ₁=(1-T₀/T_b)*₁Q₂-[W=P[SUB]0[/SUB]*(V₂-V₁)-T₀*δ

E₂-E₁=Q-W
PV=mRT

3. Attempt at solution
Will the transfer happen slowly enough that it is safe to assume zero reversabilities? Meaning sigma is zero? In this case I am a little confused between the definitions of internal and external reversibilities.

With the assumptions that work and entropy production are zero.

\psi ₂-\psi ₁=(1-T₀/T_b)*₁Q₂

Then we can get Q from first law with the assumption that the tank will eventually reach the environmental temperatures. So we can get u2 and u1 from the tables at the tank conditions and the environment conditions.

U2-U1=Q
m(u2-u1)=Q
Q/m=u2-u1
(210.49-512.33)Kj/kg*k=-301.84 Kj/kg*k=Q/m

So (multiplying (Q/m)*m)
m=PV/RT=171.823kg
\psi₂-\psi₁=(1-700k/295k)*(-301.84 Kj/kg*k)*(171.823)
=71,201.8302 Kj*K

That is a very large number. Is that correct?

I'm confused about the problem statement and your approach to the problem. First regarding the problem statement:

1. Are you assuming that the tank is not insulated, so that the room air temperature and the tank air temperatures are the same in the final state?
2. Are you assuming that the pressures inside the tank and outside the tank are equal in the final state/
3. Are you assuming that the room is sealed and isolated from the environment, and that the initial mass of air in the room is much larger than the initial mass of air in the tank?

Let's temporarily wait on discussing reversibilities, internal and external reversibilities, and sigma for the moment. I would like to first like to focus on establishing the initial and final states.

Is that OK with you?

 

[ScareCrow271828]

I'm confused about the problem statement and your approach to the problem. First regarding the problem statement:

1. Are you assuming that the tank is not insulated, so that the room air temperature and the tank air temperatures are the same in the final state?
2. Are you assuming that the pressures inside the tank and outside the tank are equal in the final state/
3. Are you assuming that the room is sealed and isolated from the environment, and that the initial mass of air in the room is much larger than the initial mass of air in the tank?

Let's temporarily wait on discussing reversibilities, internal and external reversibilities, and sigma for the moment. I would like to first like to focus on establishing the initial and final states.

Is that OK with you?

Great thank you. I am assuming 1 and 3. I did not take into account 2 because I only used the pressure inside of the tank to calculate the mass in the tank. Since volume stays constant in the rigid tank (another assumption) the P₀ term goes to zero. So no I did not assume #2 since I did not think I needed to.

Thanks for your time,
Otto

 

[Chestermiller]

Great thank you. I am assuming 1 and 3. I did not take into account 2 because I only used the pressure inside of the tank to calculate the mass in the tank. Since volume stays constant in the rigid tank (another assumption) the P₀ term goes to zero. So no I did not assume #2 since I did not think I needed to.

Thanks for your time,
Otto

Still confused. Are you saying that the final pressure in the tank does not equilibrate with the pressure in the room? Do you shut the valve before the two pressures equilibrate?

 

[ScareCrow271828]

My exact problem statement is:

Still confused. Are you saying that the final pressure in the tank does not equilibrate with the pressure in the room? Do you shut the valve before the two pressures equilibrate?

I am allowed reasonable assumptions. So I guess I would be assuming that pressures would also equilibrate. Sorry for the confusion!
Thanks for your time.

 

[Chestermiller]

My exact problem statement is:
I am allowed reasonable assumptions. So I guess I would be assuming that pressures would also equilibrate. Sorry for the confusion!
Thanks for your time.

OK. So, in the final state, the temperature and pressure of all the air that was originally inside the tank (some of this air remains in the thank and some is outside in the final state) are 295 K and 1 bar, correct?

Are you trying to determine the entropy change of just the air that was originally inside the tank, or the change in entropy for the combination of the air inside the tank and the room air?

 

[ScareCrow271828]

OK. So, in the final state, the temperature and pressure of all the air that was originally inside the tank (some of this air remains in the thank and some is outside in the final state) are 295 K and 1 bar, correct?

Are you trying to determine the entropy change of just the air that was originally inside the tank, or the change in entropy for the combination of the air inside the tank and the room air?

A one cubic meter tank containts a substance at 700k and 1Mpa. The tank is in a room at 295k and 100kPa. I am asked to determine the exergy in kJ. So I would imagine it would be the exergy in the tank.

 

[Chestermiller]

A one cubic meter tank containts a substance at 700k and 1Mpa. The tank is in a room at 295k and 100kPa. I am asked to determine the exergy in kJ. So I would imagine it would be the exergy in the tank.

I don't know too much about exergy, but can't it be calculated from the straightforward equation:
$$E=(U-U_0)+p_0(V-V_0)-T_0(S-S_0)$$

 

[ScareCrow271828]

I don't know too much about exergy, but can't it be calculated from the straightforward equation:
$$E=(U-U_0)+p_0(V-V_0)-T_0(S-S_0)$$

Yes it can be. I am using the net change form for a closed system. My question is: is it okay to assume that entropy production is zero, and there are no reversibilities because the system cools down slowly to the environmental temperature?

 

[Chestermiller]

Yes it can be. I am using the net change form for a closed system. My question is: is it okay to assume that entropy production is zero, and there are no reversibilities because the system cools down slowly to the environmental temperature?

Why don't you try it both ways, and compare the results?