To find final temperature when an evacuated bottle is opened

[Monsterboy]

Homework Statement

The internal energy of air is given, at ordinary temperatures, by

$u= u_o + 0.718t$
where $u$ is in kJ/kg, $u_o$ is any arbitrary value of $u$ at 0^oC, kJ/kg, and t is temperature in ^oC.

Also for air,
$pv= 0.287(t+273)$

where p is in kPa and $v$ is in m³/kg.
(a) An evacuated bottle is fitted with a valve through which air from the atmosphere, at 760mm of Hg and 25^oC, is allowed to flow slowly to fill the bottle. If no heat is transferred to or from the air in the bottle, what will it's temperature be when the pressure in the bottle reaches 760mm of Hg ?

Ans 144.2^oC

(b) If the bottle initially contains 0.03 m³ of air at 400mm Hg and 25^oC, what will the temperature be when the pressure in the bottle reaches 760mm Hg ?

Ans 71.6^oC

Homework Equations

[/B]
$u= u_o + 0.718t$

$pv= 0.287(t+273)$

$PV= mRT$

$Q-W= \Delta u$

Q- heat transfer
W- work transfer
$\Delta u$ = Internal energy change.

The Attempt at a Solution

[/B]
(a)

$P_1 = 13600*9.81*0.760 = 101396.16 Pa$
$T_1$ = 25^oC

$u_1 = u_o +0.718(25) = u_o +17.95 kJ/kg$

To find $T_2$

$Q-W= \Delta u$

Q= 0 as no heat is transferred , W=0 no work is done in free expansion ??

So $\Delta u$ has to be zero and T₁ =T₂ ??

I don't know how they got an increase in temperature.

(b)
$V_1 = 0.03 m^3$
$P_1 = 13600*9.81*0.400 = 53366.4 Pa$
$T_1 = 298K$
$P_2= 13600*9.81*0.760 = 101396.16 Pa$

To find $T_2$

Since volume of the bottle is constant $V_1= V_2$ ??

Initial mass = $m_1 = \frac{P_1 V_1} { RT_1} = \frac{53366.4*0.03} {287*298} = 0.01871936$ kg

Final mass = $m_2 =\frac {P_2 V_2} {RT_2} = \frac{101396.16*0.03} {287*T_2} = \frac {10.5890} {T_2}$ kg ...is this useful ?
Can i use $\frac{P_1 V_1} {T_1} = \frac{ P_2 V_2} {T_2}$ ??

I get T_2 = 566.2 K ...clearly wrong , how to continue ?

 

[Chestermiller]

This is an example of a problem in which you are not dealing with a closed system, because mass is entering the system. So you can't use the closed-system version of the first law of thermodynamics to solve it. Are you familiar with the open-system (control volume) version of the first law of thermodynamics? If so, please write the equation.

 

[I like Serena]

Q= 0 as no heat is transferred , W=0 no work is done in free expansion ??

Hi Monsterboy!

It's not free expansion, which is happens when the fluid is already inside the bottle.
Instead the outer atmosphere does flow work on the fluid as it moves inside the bottle.
On one unit of mass, the flow work is the pressure outside $P_0$ times the volume displaced, which is $v$.
So the final internal energy is $u_{final} = u_{initial} + P_0 v$.

 

[Monsterboy]

This is an example of a problem in which you are not dealing with a closed system, because mass is entering the system. So you can't use the closed-system version of the first law of thermodynamics to solve it. Are you familiar with the open-system (control volume) version of the first law of thermodynamics? If so, please write the equation.

You mean $\dot m_1 (h_1 + \frac{V_1^2} {2} +Z_1 g) + \frac{dQ} {dT} = \dot m_2 ( h_2 + \frac{V_2^2} {2} +Z_2 g) + \frac {dW} {dT}$ ??

(a)
As heat transfer is absent

$\frac {dQ} {dT} = 0$

$h_1 =h_2$ ??

 

[Monsterboy]

It's not free expansion as it happens when the fluid is already inside the bottle.

No , in problem (a) the bottle is initially evacuated and air is slowly allowed to flow inside.

 

[Chestermiller]

You mean $m_1 (h_1 + \frac{V_1^2} {2} +Z_1 g) + \frac{dQ} {dT} = m_2 ( h_2 + \frac{V_2^2} {2} +Z_2 g) + \frac {dW} {dT}$ ??

No. This is not a steady state problem. You omitted a key term from this equation.

As heat transfer is absent

$\frac {dQ} {dT} = 0$

$h_1 =h_2$ ??

No. There is no stream leaving the bottle, so $m_2=0$

 

[Monsterboy]

No. This is not a steady state problem. You omitted a key term from this equation.

I am not familiar with the unsteady state equation , is the term i omitted related to pressure energy ? like $\frac{1}{2} \rho v^2$ or $\frac{P}{\rho}$ something?

 

[I like Serena]

No , in problem (a) the bottle is initially evacuated and air is slowly allowed to flow inside.

Indeed. It's not free expansion. Instead...

I am not familiar with the unsteady state equation , is the term i omitted related to pressure energy ? like $\frac{1}{2} \rho v^2$ or $\frac{P}{\rho}$ something?

The first law states that energy is conserved.
That is:
$$U_{initial} + E_{in} = U_{final} + E_{out}$$
where
$$E_{in} = m_{in}(h_{in} + \frac{\mathscr V_{in}^2}{2} + Z_{in} g) + Q$$
and
$$E_{out} = m_{out}(h_{out} + \frac{\mathscr V_{out}^2}{2} + Z_{out} g) + W_{shaft}$$

 

[Andrew Mason]

Think of the process in a) in two parts starting with two bottles, one full at P = 760 mm Hg; T = 298K; V = V₀. and the other empty. The bottles are connected and air is allowed to flow between them. The first part is a free expansion which results in both bottles at P = 380 mm Hg and T = 298K; V = 2V₀. Then you adiabatically compress the air into one of the bottles. Use the adiabatic relation to determine the temperature when the pressure reaches 760.

AM

 

[Chestermiller]

I am not familiar with the unsteady state equation , is the term i omitted related to pressure energy ? like $\frac{1}{2} \rho v^2$ or $\frac{P}{\rho}$ something?

In your control volume first law equation, you omitted the rate of accumulation of internal energy within the bottle. The open-system version of the first law for part (a) reduces to:
$$h_{in}\frac{d(m_{in})}{dt}=\frac{d(mu)}{dt}$$where m is the mass of air in the bottle at time t, u is the internal energy per unit mass in the bottle at time t, $h_{in}$ is the enthalpy per unit mass of the air entering the bottle, and $d(m_{in})/dt$ represents the rate at which mass of air is flowing into the bottle. For part (a), if we integrate this equation with respect to time, we get:
$$m_{in}h_{in}=mu$$where $m_{in}$ is the cumulative mass of air that has flowed into the bottle up to time t and m is the mass of air in the bottle at time t. How are $m_{in}$ and m related? What does that leave you with?

 

[Chestermiller]

Think of the process in a) in two parts starting with two bottles, one full at P = 760 mm Hg; T = 298K; V = V₀. and the other empty. The bottles are connected and air is allowed to flow between them. The first part is a free expansion which results in both bottles at P = 380 mm Hg and T = 298K; V = 2V₀. Then you adiabatically compress the air into one of the bottles. Use the adiabatic relation to determine the temperature when the pressure reaches 760.

AM

In my judgment, this is not correct. Certainly the adiabatic compression will require work to be done, and that work should not be considered reversible (as required by the adiabatic relation). The amount of work that the surrounding air does in forcing air into the bottle is at constant pressure, and, in passing through the valve, the air undergoes an irreversible Joule-Thompson pressure reduction. It would be interesting to see what this approximation provides in comparison with the correct control volume (open system) application of the first law of thermodynamics.

This student is in the process of learning about the open-system (control volume) version of the first law of thermodynamics, and he should be applying that specific approach to this problem.

 

[Monsterboy]

In your control volume first law equation, you omitted the rate of accumulation of internal energy within the bottle. The open-system version of the first law for part (a) reduces to:
$$h_{in}\frac{d(m_{in})}{dt}=\frac{d(mu)}{dt}$$where m is the mass of air in the bottle at time t, u is the internal energy per unit mass in the bottle at time t, $h_{in}$ is the enthalpy per unit mass of the air entering the bottle, and $d(m_{in})/dt$ represents the rate at which mass of air is flowing into the bottle. For part (a), if we integrate this equation with respect to time, we get:
$$m_{in}h_{in}=mu$$where $m_{in}$ is the cumulative mass of air that has flowed into the bottle up to time t and m is the mass of air in the bottle at time t. How are $m_{in}$ and m related? What does that leave you with?

$m_{in} h_{in} =mu$

$m_{in}[ u_{in} +p_1v_1] = mu$

Since $P_1v_1 = RT_1$
$m_{in}[u_o +718(25) + 287(298) ] = mu$

$m_{in}[ u_o +103476] =mu = m[u_o + 718(25)]$

am i going the right way ?

 

[Chestermiller]

$m_{in} h_{in} =mu$

$m_{in}[ u_{in} +p_1v_1] = mu$

Since $P_1v_1 = RT_1$
$m_{in}[u_o +718(25) + 287(298) ] = mu$

$m_{in}[ u_o +103476] =mu = m[u_o + 718(25)]$

am i going the right way ?

Almost. Since the amount of air that enters is equal to the amount of air in the bottle, $m=m_{in}$. I guess you are assuming that the m's are in grams, rather than kg, correct? Finally, the term in brackets on the right hand side should have a T instead of a 25, where T is the final temperature in C.

 

[Monsterboy]

Almost. Since the amount of air that enters is equal to the amount of air in the bottle, $m=m_{in}$. I guess you are assuming that the m's are in grams, rather than kg, correct? Finally, the term in brackets on the right hand side should have a T instead of a 25, where T is the final temperature in C.

Thanks , got the first problem

$T = \frac{103476} {718}$ = 144.116^oC

 

[Monsterboy]

(b)

Initial mass = $m_1 = \frac{P_1 V_1} { RT_1} = \frac{53366.4*0.03} {287*298} = 0.01871936$ kg

Final mass = $m_{final} =\frac {P_2 V_2} {RT_2} = \frac{101396.16*0.03} {287*T_2} = \frac {10.5890} {T_2}$

$$ m_{in}h_{in} + m_1u_1 = m_{final}u_{final}\tag{1} $$ correct ?

$m_{in}[u_{in} + 287(298)] + 0.01871936[ u_o + 0.718(25)] = \frac {10.5890} {T_2}(u_o +0.718T_2) \tag{2}$ correct ?

 

[Chestermiller]

(b)

Initial mass = $m_1 = \frac{P_1 V_1} { RT_1} = \frac{53366.4*0.03} {287*298} = 0.01871936$ kg

Final mass = $m_{final} =\frac {P_2 V_2} {RT_2} = \frac{101396.16*0.03} {287*T_2} = \frac {10.5890} {T_2}$

$$ m_{in}h_{in} + m_1u_1 = m_{final}u_{final}\tag{1} $$ correct ?

Yes.

$m_{in}[u_{in} + 287(298)] + 0.01871936[ u_o + 0.718(25)] = \frac {10.5890} {T_2}(u_o +0.718T_2) \tag{2}$ correct ?

Yes, but a couple of corrections. In the first term on the left hand side, replace $u_{in}$ with $u_o$. For the term on the rhs, $T_2$ from the ideal gas law should be replaced by $(T_2+273)$. From a mass balance, how are $m_{in}$, $m_1$, and $m_{final}$ reltated?

 

[Monsterboy]

Yes, but a couple of corrections. In the first term on the left hand side, replace $u_{in}$ with $u_o$. For the term on the rhs, $T_2$ from the ideal gas law should be replaced by $(T_2+273)$. From a mass balance, how are $m_{in}$, $m_1$, and $m_{final}$ reltated?

$m_{in}=m_{final}-m_1$ ??

$( \frac{10.589}{T_2 + 273} - 0.01871936)(u_o +85526) + 0.01871936[u_o + 17.95] = \frac{10.589}{T_2+273} (u_o +0.718T_2)$ ??

$u_0$ terms cancel out ,I am getting $T_2$ =291.4^oC , something is wrong.

 

[Chestermiller]

Shouldn't $h_{in}$ be given by $$h_{in}=u_o+(0.718)(25)+(0.287)(298)$$?

 

[Monsterboy]

Shouldn't $h_{in}$ be given by $$h_{in}=u_o+(0.718)(25)+(0.287)(298)$$?

Yea , i got it ,thanks for the help.

I want to find the work done by the gas in both (a) and (b) , "I like Serena" said it is not zero in the case (a) as well ,i don't how, i thought when a gas freely expands into an evacuated chamber zero work is done.

In (b) work is done against the pressure of 400mm Hg ? so is it $(m_2- m_1)(p_1v_1)$ ?

where $p_1$ and $v_1$ are initial values inside the bottle.

In the first case $m_1 =0$ so it is $m_{in}(0) =0 ?$

 

[Chestermiller]

Yea , i got it ,thanks for the help.

I want to find the work done by the gas in both (a) and (b) , "I like Serena" said it is not zero in the case (a) as well ,i don't how, i thought when a gas freely expands into an evacuated chamber zero work is done.

In (b) work is done against the pressure of 400mm Hg ? so is it $(m_2- m_1)(p_1v_1)$ ?

where $p_1$ and $v_1$ are initial values inside the bottle.

In the first case $m_1 =0$ so it is $m_{in}(0) =0 ?$

We need to be more precise here. You are aware that this problem is an exercise in applying the open-system (control volume) of the first law, correct? You are aware that, in the open-system version, the work is subdivided into two distinct parts (a) the work done to push fluid into and out of the control volume and (b) all other work (typically called shaft work), correct? You are aware that the former is included in the enthalpy of the entering and exit streams, correct?

Are you trying to also interpret this problem in terms of the ordinary closed system version of the first law (which is mathematically equivalent to the open system version)? If so, you need to precisely define the constant mass included in your system in order to calculate the work done on the system. The work changes when you change what you call the system.

 

[I like Serena]

"I like Serena" said it is not zero in the case (a) as well ,i don't how, i thought when a gas freely expands into an evacuated chamber zero work is done.

In free expansion, also known as the Joule expansion, the fluid is already inside the chamber, and a membrane (or equivalent) that separates it from the vacuum is removed. In that case no work is done.
In our case the fluid is not inside yet, but is "pushed" inside by the atmosphere, which exerts work on the fluid. This work is captured in the enthalpy part: $h_{in} = u_{in} + p_{in}v_{in}$. That is, it's the $pv$ part, which is also known as "flow work". As a consequence we get a higher internal energy inside the container than we would otherwise have, resulting in a (much) higher temperature.
Btw, in practice the temperature does not get that high, since it won't be adiabatic.

 

[Monsterboy]

Sorry for being imprecise.

(a) the work done to push fluid into and out of the control volume

Without any internal friction , this is the only work done in problem (a) right ?

$m_{in}h_{in}$ itself is the work done ?

(b) all other work (typically called shaft work), correct?

In problem (b) both the parts are included right ? work done = $m_{in}h_{in} + (m_2-m_1)p_1v_1$ ??

Are you trying to also interpret this problem in terms of the ordinary closed system version of the first law (which is mathematically equivalent to the open system version)? If so, you need to precisely define the constant mass included in your system in order to calculate the work done on the system. The work changes when you change what you call the system.

In (b)
If i consider the bottle and everything inside it as the system ,is the formula $(m_2-m_1)p_1v_1$ for work done on the system correct ?

 

[Chestermiller]

Sorry for being imprecise.

Without any internal friction , this is the only work done in problem (a) right ?

$m_{in}h_{in}$ itself is the work done ?

No. Just $m_{in}(P_{atm}v_{in}$), where $v_{in}$ is the specific volume of the air outside the bottle. This is the work done by the surrounding air to push the air into the bottle. And this is the term included as part of the entering enthalpy.

In problem (b) both the parts are included right ? work done = $m_{in}h_{in} + (m_2-m_1)p_1v_1$ ??

No. Again, just $m_{in}(P_{atm}v_{in})$.

In (b)
If i consider the bottle and everything inside it as the system ,is the formula $(m_2-m_1)p_1v_1$ for work done on the system correct ?

Are you talking about everything inside the bottle initially or everything inside the bottle finally? Either way, this relationship is incorrect.

 

[Monsterboy]

Are you talking about everything inside the bottle initially or everything inside the bottle finally? Either way, this relationship is incorrect.

I was talking about everything in the bottle initially.

So it is $\int pdV = P_{atm}V$ where V=0.03m³ for problem (a)??

For (b) $-\int Vdp = 0.03(P_{atm}-P_{in})$ ?

 

[Chestermiller]

I was talking about everything in the bottle initially.

So it is $\int pdV = P_{atm}V$ where V=0.03m³ for problem (a)??

For (b) $-\int Vdp = 0.03(P_{atm}-P_{in})$ ?

You seem to be moving in the wrong direction, so I'm going to try to get you back on the right track.

So far we have solved this problem using the open-system (control volume) version of the first law of thermodynamics, and have (a) found this approach pretty easy to apply and (b) gotten the right answer. Now, would it be correct to say that you are trying to understand how to solve this same problem using the (more familiar) closed system version of the first law (in which the mass of material within the selected system is constant)?

 

[Monsterboy]

Now, would it be correct to say that you are trying to understand how to solve this same problem using the (more familiar) closed system version of the first law (in which the mass of material within the selected system is constant)?

Yes , the original problem is solved since i am not very familiar with unsteady state open system version of the first law , i wanted to extend the original problem by trying to include the the work done in both (a) and (b), this is also done.

You said using the closed system version of the first law will be mathematically equivalent when finding the work done. I want to use the closed system version of the first law to find the work done on the system(the evacuated bottle as in prob (a) and the bottle with 400mm Hg pressure in (b) ) .

 

[Chestermiller]

Yes , the original problem is solved since i am not very familiar with unsteady state open system version of the first law , i wanted to extend the original problem by trying to include the the work done in both (a) and (b), this is also done.

You said using the closed system version of the first law will be mathematically equivalent when finding the work done. I want to use the closed system version of the first law to find the work done on the system(the evacuated bottle as in prob (a) and the bottle with 400mm Hg pressure in (b) ) .

OK. But it is important to choose the simplest closed system to apply the first law to for this problem.

Picture an imaginary membrane lining the interior of the bottle and passing through the neck of the bottle to the outside. In the initial state of the system, the remainder of this (closed) membrane is situated outside the bottle. The membrane (in the initial state) also encloses the mass of air that will eventually enter the bottle from the outside. In the final state of the system, the boundary of the membrane is entirely contained within the bottle, up to the neck. So, all the way between initial state and the final state, no mass passes across the boundary of the membrane, and the membrane encloses a constant mass of air. (This all involves treating the air as a continuum, as we usually do in thermodynamics. We can also treat the air as individual molecules, but you get the same answer). This approach applies to both case (a) and case (b).

So, if $m_1$ represents the initial mass of air inside the bottle and $m_2$ represents the final mass of air inside the bottle, the amount of air that enters the bottle is $m_2-m_1$. The change in volume enclosed in our membrane between the initial and final states of our system is $-(m_2-m_1)v_{atm}$, where $v_{atm}$ is the specific volume of the air outside the bottle. All this volume change is driven by an external pressure $P_{ext}$ of $P_{ext}=p_{atm}$. So the work done by the air enclosed within our membrane (on its surroundings outside the membrane) between the initial and final states of the system is $$W=-(m_2-m_1)p_{atm}v_{atm}$$