Thermodynamics: Constant Pressure Tank Heating Problem

In summary: I'll try, but I haven't done these types of problems before so don't know if I'm going about it correctly.
  • #1

Chestermiller

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Homework Statement
Given a volume V linked to the outdoors through a small hole, with pressure outside P, determine the energy required to heat from 𝑇1 to 𝑇2.
Relevant Equations
Open system (control volume) version of first law of thermodynamics
I found this interesting thermodynamics problem on another site. I thought PF members might find it challenging to attack. I'm not asking for help since I've already solved it. So pease feel free to present your entire solution if you desire.

Chet
 
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  • #2
Just to clarify, T1 and T2 are the initial and final temperatures of the air in volume V, correct?
 
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  • #3
kuruman said:
Just to clarify, T1 and T2 are the initial and final temperatures of the air in volume V, correct?
Yes.
 
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  • #4
Since this problem has not received any significant activity, I am offering the following hints to its solution. The problem can be analyzed using either the open system version of the first law of thermodynamics, or the open system version of the 2nd law of thermodynamics. Applying the open system version of the first law of thermodynamics to the problem leads to the following equation:
$$d(nu)=dQ-h(-dn)$$where u is the molar internal energy of the volume contents, n is the number of moles of gas in the volume, dQ is the differential heat being added during the process, and h(-dn) is the differential enthalpy of the gas leaving the control volume through the hole during the process; h includes the molar internal energy of the exit stream, plus the work per mole of gas forced out through the hole. So, ##h =u+Pv##, where v is the molar volume. This is a standard approach to applying the open system version of the first law.

The second law of thermodynamics applied to this system reads: $$d(ns)=\frac{dQ}{T}-s(-dn)$$where s is the entropy per mole.

Both these approaches give the exact same answer for the heat load.
 
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  • #5
Chestermiller said:
Since this problem has not received any significant activity, I am offering the following hints to its solution.
This problem may not have received any significant activity in the form of posts on this thread, but it generated significant activity in my brain. I explored the application of the first law to open systems in preparation for providing a solution to your problem. My undergraduate education decades ago included enthalpy only in passing, but gave me the necessary foundational tools to pick up the subject now. So thank you, @Chestermiller, for teaching this old dog a new trick.
 
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  • #6
kuruman said:
This problem may not have received any significant activity in the form of posts on this thread, but it generated significant activity in my brain. I explored the application of the first law to open systems in preparation for providing a solution to your problem. My undergraduate education decades ago included enthalpy only in passing, but gave me the necessary foundational tools to pick up the subject now. So thank you, @Chestermiller, for teaching this old dog a new trick.
You're very welcome. Thank you for this very impressive discussion on the value of continuing to learn new things and re-mastering old things. I feel the same way.

Are you inclined to flesh out the remainder of the solution, or are you in favor of allowing other members to take a shot at it?
 
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  • #7
I'll try, but I haven't done these types of problems before so don't know if I'm going about it correctly. Denote the initial and final states by ##A## and ##B## respectively. The total internal energies of these states are ##U_A = n_A u_A## and ##U_B = n_B u_B## respectively [I will neglect the velocity of the gas entering the inlet, and also the gravitational potential energy of the gas inside the box]. We then have that$$\Delta U \equiv U_B - U_A = Q_i + \int_{t_A}^{t_B} \dot{n} h_i dt$$where ##h_i = u_i + Pv_i## is the molar enthalpy at the inlet. The ##h_i = h_i(t)## is a function of time, but as an approximation we might be able to use a constant value ##h_i = \frac{1}{2} (h_A + h_B)##, in which case$$\begin{align*}

Q_i &= n_B u_B - n_A u_A - \frac{1}{2} (u_A + Pv_A + u_B + Pv_B)(n_B - n_A) \\

&= \frac{1}{2} n_B u_B - \frac{1}{2} n_A u_A - \frac{1}{2} n_B u_A + \frac{1}{2} n_A u_B

- \frac{1}{2} \left( P v_A (n_B - n_A) + Pv_B (n_B - n_A) \right) \\

&= \frac{1}{2} n_B (u_B - u_A) + \frac{1}{2} n_A (u_B - u_A)- \frac{1}{2} \left( P v_A (n_B - n_A) + Pv_B (n_B - n_A) \right) \\

&= \frac{1}{2}n_B C_v (T_B - T_A) + \frac{1}{2} n_A C_v (T_B - T_A) - \frac{1}{2} \left( RT_A (n_B - n_A) + RT_B (n_B -n_A) \right)

\end{align*}$$Which is the same as$$Q_i = \frac{1}{2} C_v (T_B - T_A) (n_B + n_A) - \frac{1}{2} R(n_B - n_A)(T_A + T_B)$$Somewhere, we might need to use ##PV = n_A R T_A = n_B R T_B \iff n_A T_A = n_B T_B## which gives that ##n_B = (T_A/ T_B) n_A##, i.e.$$Q_i = \frac{1}{2} C_v n_A (T_B - T_A) (1+ \frac{T_A}{T_B} ) - \frac{1}{2} R n_A (\frac{T_A}{T_B} - 1)(T_A + T_B)$$We'd need to eliminate ##n_A## from this, which would be possible if we know ##V##, i.e. we can write$$\begin{align*}

Q_i &= \frac{C_v}{2} \frac{PV}{RT_A} (T_B - T_A) (1+ \frac{T_A}{T_B}) - \frac{R}{2} \frac{PV}{RT_A} (\frac{T_A}{T_B} - 1)(T_A + T_B) \\ \\

&= \frac{PV}{2R} \left[ C_v (T_B - T_A) (\frac{1}{T_A} + \frac{1}{T_B}) - R(\frac{1}{T_B} - \frac{1}{T_A})(T_A + T_B)\right] \\ \\

&= \frac{PV}{2RT_A T_B} \left[ C_v(T_B - T_A)(T_B + T_A) + R(T_B - T_A)(T_A + T_B)\right] \\ \\

&= \frac{PV(T_B - T_A)(T_B + T_A)}{2RT_A T_B} \left[ C_v + R \right] \\ \\

&= \frac{C_p PV(T_B - T_A)(T_B + T_A)}{2RT_A T_B} = \frac{C_p PV (T_B^2 - T_A^2)}{2RT_A T_B}

\end{align*}$$is this, along the right lines? Or, did I go wrong assuming that ##h_i \sim \frac{1}{2} (h_A + h_B)##?
 
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  • #8
Chestermiller said:
Are you inclined to flesh out the remainder of the solution, or are you in favor of allowing other members to take a shot at it?
Both. I will provide just the bottom line I got which is unlike the one provided by @etotheipi.
$$\Delta Q=C_v(T_2-T_1)+PV\ln\left(\frac{T_2}{T_1}\right).$$On edit:
40 minutes after posting I changed the sign of the log term from negative to positive.
 
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  • #9
My analysis agrees with neither of these. I held off integrating until later so that I would not have to make any approximation about the integral of the exiting enthalpy stream.

First Law analysis: $$d(nu)=dQ+hdn$$
$$ndu+udn=dQ+(u+Pv)dn=dQ+udn+Pvdn=dQ+udn+RTdn$$
So, $$ndu=nC_vdT=dQ+RTdn$$
From the ideal gas law, the number of moles of gas in the tank at any time is $$n=\frac{PV}{RT}$$where P is the (constant) pressure, V is the (constant) tank volume, and T is the varying temperature. Substituting this into the differential equation gives:
$$\frac{PV}{R}C_v\frac{dT}{T}=dQ-PV\frac{dT}{T}$$or$$PV\left(\frac{C_v}{R}+1\right)\frac{dT}{T}=dQ$$or$$PV\frac{C_p}{R}\frac{dT}{T}=dQ$$
Therefore, integrating gives:
$$Q=PV\frac{C_p}{R}\ln{(T_B/T_A)}$$
Second Law Analysis:
$$d(ns)=\frac{dQ}{T}+sdn$$
or $$nds=\frac{dQ}{T}$$Substituting the ideal gas law then gives:
$$\frac{PV}{R}ds=dQ$$or, integrating,
$$Q=\frac{PV}{R}\Delta s$$
For an ideal gas at constant pressure, $$\Delta s=C_p\ln{(T_B/T_A)}$$So, here again,
$$Q=PV\frac{C_p}{R}\ln{(T_B/T_A)}$$
 
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  • #10
I think my approximation ##h_i \sim \frac{1}{2} (h_A + h_B)## must have been invalid, because what I got fits quite closely your answer for very small ##T_B / T_A##, but diverges after.
 
  • #11
etotheipi said:
I think my approximation ##h_i \sim \frac{1}{2} (h_A + h_B)## must have been invalid, because what I got fits quite closely your answer for very small ##T_B / T_A##, but diverges after.
In general, we can express internal energy and enthalpy of an ideal gas having constant heat capacity as ##u=C_v(T-T_R)## and ##h=C_p(T-T_R)+RT_R## where ##T_R## is an arbitrary reference temperature at which u is taken as zero. The solution to any problem, including the present one, the is independent of the choice of the reference temperature. If we choose ##T_R## as absolute zero, we ##u=C_vT## and ##h=C_pT##. In this problem, that would mean that nu is constant, and the change in internal energy of the tank contents would be zero (the temperature rise is offset by the decrease in the number of moles).

Here is a supplementary problem for you. For this choice of reference temperature, what would the average exit enthalpy and gas temperature have to be to give the right answer?
 
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  • #12
I have nothing to add but it always amazes me how people can be so fluent in thermodynamics. I feel like I have a pretty good grasp on E&M and to a lesser extent classical and quantum but thermo is a whole other language for me so I tip my hat to anyone that can “speak it”.
 
  • #13
Chestermiller said:
Here is a supplementary problem for you. For this choice of reference temperature, what would the average exit enthalpy and gas temperature have to be to give the right answer?

Okay, so let's suppose the correct average exit enthalpy is ##h_a##. The heat transferred to the system during the process is$$\begin{align*}

Q &= \Delta U - h_a \Delta n = \left[C_v(T_B - T_R) - C_v(T_A - T_R) \right] - h_a \left[ \left(\frac{PV}{RT_B} \right) - \left(\frac{PV}{RT_A} \right) \right] \\

Q &= C_v(T_B - T_A) - \frac{h_a PV}{R} \left( \frac{1}{T_B} - \frac{1}{T_A} \right) \overset{!}{=} PV\frac{C_p}{R}\ln{\frac{T_B}{T_A}}

\end{align*}$$Which we can rearrange for ##h_a##,$$C_v(T_B - T_A) - \frac{PV C_p}{R}\ln{\frac{T_B}{T_A}} = \frac{h_a PV}{R} \frac{T_A - T_B}{T_A T_B}$$ $$h_a = \frac{R T_A T_B}{PV(T_A - T_B)} \left[ C_v(T_B - T_A) - \frac{PV C_p}{R}\ln{\frac{T_B}{T_A}}\right]$$ $$h_a = - T_A T_B \left[\frac{R C_v}{PV} + \frac{C_p}{(T_A - T_B)} \ln{\frac{T_B}{T_A}} \right]$$And I presume that to find the required average gas temperature, we could do almost essentially exactly the same procedure except just using the entropy approach :smile:
 
  • #14
Here is my solution

Start with the expression provided in post #4 after substituting the enthalpy, $$d(nu)=dQ-(u+Pv)(-dn)~\Rightarrow~ndu=dQ+Pvdn.$$ We now isolate ##dQ## and use ##ndu=dU## and ##v=\dfrac{V}{n}## to get $$dQ=dU-PV\frac{dn}{n}.$$From the ideal gas law at constant ##P## and ##V## we get ##\dfrac{dn}{n}=-\dfrac{dT}{T}## so that $$dQ=dU+PV\frac{dT}{T}=\left(C_v+\frac{PV}{T}\right)dT.$$Then $$Q=\int_A^B\left(C_v+\frac{PV}{T}\right)dT=C_v(T_B-T_A)+PV\ln\left(\frac{T_B}{T_A}\right).$$I read the caveat
Chestermiller said:
I held off integrating until later so that I would not have to make any approximation about the integral of the exiting enthalpy stream.
but all this is new to me. I am not savvy enough to figure out where, in my derivation, I made an approximation in what I thought is obvious math. Can you please explain?

(I know I will not master this until I have to teach it which will not happen.)
 
  • #15
etotheipi said:
Okay, so let's suppose the correct average exit enthalpy is ##h_a##. The heat transferred to the system during the process is$$\begin{align*}

Q &= \Delta U - h_a \Delta n = \left[C_v(T_B - T_R) - C_v(T_A - T_R) \right] - h_a \left[ \left(\frac{PV}{RT_B} \right) - \left(\frac{PV}{RT_A} \right) \right] \\

Q &= C_v(T_B - T_A) - \frac{h_a PV}{R} \left( \frac{1}{T_B} - \frac{1}{T_A} \right) \overset{!}{=} PV\frac{C_p}{R}\ln{\frac{T_B}{T_A}}

\end{align*}$$Which we can rearrange for ##h_a##,$$C_v(T_B - T_A) - \frac{PV C_p}{R}\ln{\frac{T_B}{T_A}} = \frac{h_a PV}{R} \frac{T_A - T_B}{T_A T_B}$$ $$h_a = \frac{R T_A T_B}{PV(T_A - T_B)} \left[ C_v(T_B - T_A) - \frac{PV C_p}{R}\ln{\frac{T_B}{T_A}}\right]$$ $$h_a = - T_A T_B \left[\frac{R C_v}{PV} + \frac{C_p}{(T_A - T_B)} \ln{\frac{T_B}{T_A}} \right]$$And I presume that to find the required average gas temperature, we could do almost essentially exactly the same procedure except just using the entropy approach :smile:
This is the results of taking TA as the reference temperature. I was asking what the average enthalpy and average temperature of the exit stream should be if the reference temperature is taken as absolute zero.

Also, in getting ##\Delta U##, you forgot to take into account that the number of moles change. ##U = nu=nC_v(T-T_R)##

Your result is actually correct, except for the Cv term which should not be in there.
 
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  • #16
PhDeezNutz said:
I have nothing to add but it always amazes me how people can be so fluent in thermodynamics. I feel like I have a pretty good grasp on E&M and to a lesser extent classical and quantum but thermo is a whole other language for me so I tip my hat to anyone that can “speak it”.
Yes, in my mind @Chestermiller is the undisputed guru of PF thermodynamics.
 
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  • #17
kuruman said:
Here is my solution

Start with the expression provided in post #4 after substituting the enthalpy, $$d(nu)=dQ-(u+Pv)(-dn)~\Rightarrow~ndu=dQ+Pvdn.$$ We now isolate ##dQ## and use ##ndu=dU## and ##v=\dfrac{V}{n}## to get $$dQ=dU-PV\frac{dn}{n}.$$From the ideal gas law at constant ##P## and ##V## we get ##\dfrac{dn}{n}=-\dfrac{dT}{T}## so that $$dQ=dU+PV\frac{dT}{T}=\left(C_v+\frac{PV}{T}\right)dT.$$Then $$Q=\int_A^B\left(C_v+\frac{PV}{T}\right)dT=C_v(T_B-T_A)+PV\ln\left(\frac{T_B}{T_A}\right).$$I read the caveat

but all this is new to me. I am not savvy enough to figure out where, in my derivation, I made an approximation and in what I thought is obvious math. Can you please explain?

(I know I will not master this until I have to teach it which will not happen.)
$$dU=d(nu)=udn+nC_vdT$$
 
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  • #18
Chestermiller said:
$$dU=d(nu)=udn+nC_vdT$$
Yesss! I need some time to get used to thinking outside the box of the closed system and adopt the open system frame of mind. Thank you.
 
  • #19
kuruman said:
Yes, in my mind @Chestermiller is the undisputed guru of PF thermodynamics.

I absolutely agree!
 
  • #20
You guys are making me blush. In my judgment, the reason thermodynamics often seems so difficult is that the books out there are typically awful. I sometimes feel like the authors themselves don't really understand the fundamentals. The two books that I like the best are Fundamentals of Engineering Thermodynamics by Moran et al (available as a PDF online) and Introduction to Chemical Engineering Thermodynamics by Smith and van Ness.
 
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  • #21
Chestermiller said:
You guys are making me blush. In my judgment, the reason thermodynamics often seems so difficult is that the books out there are typically awful. I sometimes feel like the authors themselves don't really understand the fundamentals. The two books that I like the best are Fundamentals of Engineering Thermodynamics by Moran et al (available as a PDF online) and Introduction to Chemical Engineering Thermodynamics by Smith and van Ness.

I will definitely check them out! I’m a physics grad student and a lot of the time I feel like our exposure to thermo is so theoretical that we don’t really get a feeling for what these various quantities really mean/their utility. I really think an engineering perspective would help to rectify that situation so thank you for the recommendation.
 
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1. What is the purpose of a constant pressure tank in thermodynamics?

The purpose of a constant pressure tank in thermodynamics is to maintain a constant pressure throughout a system, even as the temperature changes. This is important in many industrial and scientific applications where a specific pressure must be maintained for the system to function properly.

2. How does heating a constant pressure tank affect the temperature inside the tank?

Heating a constant pressure tank will cause the temperature inside the tank to increase. This is because the pressure inside the tank remains constant, but the volume of the gas inside the tank will increase as it is heated, resulting in an increase in temperature according to the ideal gas law.

3. What factors affect the rate of heating in a constant pressure tank?

The rate of heating in a constant pressure tank is affected by several factors, including the heat source, the volume and composition of the gas inside the tank, and the thermal conductivity of the tank material. The rate of heating can also be affected by external factors such as ambient temperature and pressure.

4. How can the temperature inside a constant pressure tank be controlled?

The temperature inside a constant pressure tank can be controlled by adjusting the heat input, changing the volume or composition of the gas inside the tank, or using an external cooling system. The rate of heating can also be controlled by adjusting the pressure inside the tank.

5. What are some real-world applications of the constant pressure tank heating problem?

The constant pressure tank heating problem has many practical applications, including in chemical and industrial processes, such as distillation and chemical reactions. It is also used in laboratory experiments to control and maintain a specific temperature for reactions or to study the behavior of gases under different conditions.

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