Thermodynamic Exercise: Adiabatic Cylinder with Double Chamber and Piston

[CVB]

Hello, my name is Roberto and I am engineering student from Spain. I am a new member in the forum and I don´t know if this is the right site to present me but I want to use this moment to say hello to the comunity.
I have a doubt with a thermodinamic exercise.I have an adiabatic cylinder with double chamber separated by a piston diatermano. In the chamber of the left there is a resistance initially turned off and the system is in thermodynamic equilibrium. When turn on the resistance the piston moves to reach a new equilibrium.My doubt is what are the right ecuations:
-Total energy balance \Delta U_ {total}=-W_ {elec}
-A chamber \Delta U_A=Q-W_{exp}-W_{elec} or \Delta U_A=-W_{exp}-W_{elec}
-B chamber \Delta U_B=Q-W_{exp} or \Delta U_B=-W_{exp}
Thank you very much and I sorry if something is wrong.

 

[Andrew Mason]

Hello, my name is Roberto and I am engineering student from Spain. I am a new member in the forum and I don´t know if this is the right site to present me but I want to use this moment to say hello to the comunity.

Welcome to PF!

I have a doubt with a thermodinamic exercise.I have an adiabatic cylinder with double chamber separated by a piston diatermano. In the chamber of the left there is a resistance initially turned off and the system is in thermodynamic equilibrium. When turn on the resistance the piston moves to reach a new equilibrium.My doubt is what are the right ecuations:
-Total energy balance \Delta U_ {total}=-W_ {elec}
-A chamber \Delta U_A=Q-W_{exp}-W_{elec} or \Delta U_A=-W_{exp}-W_{elec}
-B chamber \Delta U_B=Q-W_{exp} or \Delta U_B=-W_{exp}

I am not sure what W_exp is.

The electrical energy is all turned into heat flow. This heat flow is initially into the left side:

Q_{A} = W_{elec}

But some of the heat then flows into the right side through the piston. The question is whether there is any work done. If there is no work done, then \Delta U = Q = W_{elec} (I am not sure why you have minus signs there).

Is there any work done by the gas on either side?

AM

 

[Studiot]

But some of the heat then flows into the right side through the piston.

Don't you think the electric heater in the left chamber heats the gas inside which expands, pushing the piston into the right chamber compressing the gas in the right chamber until equilibrium is again reached?

 

[Studiot]

Hello Roberto

I think your confusion arises because of sign convention. This is a very common confusion.

The first law is normally stated

\DeltaU = q - w

q is the heat input to the system - counted positive
w is the work done by the system - counted positive

So q is energy in and w is energy out.

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Edit
Some books use

\DeltaU = q + w

Where w is negative for work done by the system (energy out) and positive for work done on the system (energy in)

This leads to the same result, but the different sign conventions can be confusing.

******************************************************************

Quite reasonably putting energy in increases the internal energy
Taking energy out decreases it.

That is why the equation is written with a minus sign.

So in your case the electrically generated heat in is positive and the expansion work done by the left chamber is also positive.

So in the left chamber

\DeltaU = q_electrical - w_expansion

In the right chamber

There is no heat input but work is done on the system so is negative (= minus work done by the system)

So \DeltaU = 0 - (-w_compression)

Hope this helps

go well

 

[CVB]

First, I want to say thank you for your quick reply, I am very grateful. I am very confused because I understand Q=W_{elec} but I am going to upload a book where W_{out}=W_{exp}+W_{elec} and the first law is \Delta U=Q-W_{out} you can see example 4.16.
Than you

 

[Studiot]

My Spanish is not that good, but I presume you mean example 4.6, PDF pages 85 to 86?

As far as I can make out they have stated the same thing I did, but they have worked out fully the equations using the fact that the work done in chamber A = - work done in chamber B.

Like me thay have assumed that no heat is transferred from chamber A to chamber B, only work.

What do do not understand about their working?

 

[CVB]

Hello,thank you for your reply. Yes the question is that in the total balance before turn on the resistance \Delta U = -W_{elec} because the system is adiabatic and Q=0. But to calculate Q in left chamber after turn on the resistance would be Q=N C_v (T_f-T_i)+W_{expansion}+W_{elec}
Thank you

 

[Andrew Mason]

Hello,thank you for your reply. Yes the question is that in the total balance before turn on the resistance \Delta U = -W_{elec} because the system is adiabatic and Q=0.

Before electricity is supplied to the resistance the system is in static equilibrium: \Delta Q = \Delta U = W = 0

But to calculate Q in left chamber after turn on the resistance would be Q=N C_v (T_f-T_i)+W_{expansion}+W_{elec}

Not quite:

Q=N C_v (T_f-T_i) + W_{expansion} = W_{elec}

if you ignore the heat flow from the left to the right side.

On the right side, without any heat flow from the left, Q = 0. What is the relationship between the work done by the right side and the work done by the left side? What does that tell you about the change in internal energy on the right side?

AM

 

[CVB]

Hello Andrew,thank you for your reply.It`s right that before electricity is supplied to the resistance the system is in static equilibrium: \Delta Q = \Delta U = W = 0
I wanted to say \Delta U_{total} = -W_{elec}.
If you don´t mind I`m going to write all exercise and like I solve it.

 

[CVB]

Have a rigid and thermally insulated tank having a diatermano piston that divides in 2 separate parts. Part A contains oxygen and inside there is a resistance connected to 220 v which supplies a electric current of 3 amperes. The other part B contains helium. Initially the system is in thermal equilibrium.Oxygen mass is 3 kg and 6 bar pressure.Helium mass is 0.5 kg, their pressure 2 bar and 100 º C. Temp. We turn on the resistance for 3 minutes while freeing the piston.In B there is a polytropic process. The gases are perfect.
Answering a) P, T, V start and end in A and B
b) Flow of heat and expansion work exchanged by both chambers
a) First I calculate moles in A (93,75 mol) and in B (124,91 mol).
Like the system is in thermal equilibrium T_A=T_B then by PV=NRT I calculate V_A=477,9 Litres and V_B= 1910.24 Litres Now I calculate

W_{elec}= P= I . v= 220 x 3 = 660 W=0,660 Kw x 180 sec = 118,8 Kj
Then \Delta U_{total} = -W_{elec}\rightarrow Then m_A Cv_A(T_f-T_i)+m_B Cv_B (T_f-T_i)= -W_{elec} I calculate T_f= 406,88ºK

 

[CVB]

To Calculate final pressure : V_A+V_B= V_T I calculate V_T=477,9+1910,24=2388,14 Litres
Like the total volume is constant

 

[CVB]

V_T= \frac{N_A . R. T_f}{P_f}+ \frac{N_B .R .T_f}{P_f}
2388,14 = 93,75 x 0,082 x 406,88/P_f + 124,91 x 0,082 x 406,88/P_f
Then P_f= 3,05 Bar and Final volume in B is V_B = 124,91 x 8,3143 x 406,88/305 = 1385,44 litres
And Final volume in A is V_A =2388,14-1385,44 = 1002,7 litres
b)To calculate heat and expansion work flow
\Delta U_B=Q-W_{expansion}
we know is a polytropic process and we need calculate polytropic index
n = \frac{Ln (P_B/P_A)}{Ln (V_B/V_A)}=1,31
After W_{exp}= \frac{P_2 . V_2-P_1 . V_1}{ 1-n } = -381,69 Kj

I`m sorry I have problems with Latex code and I don`t know why.I`m going to finish post by attachment.Thank you.\frac{}{}

 

[CVB]

Hello again, here is the exercise,do you think is ok?. I ´m sorry to send it by attachment. I will try to solve my Latex problem.Thank you for your help.