Thermodynamics: Pressure Drop over a Valve

[TimoD]

Hey Guys! Anyone able to help out here?

I have already happily solved for T₂ = 369.91K and m₂ = 11.492kg

However, for question 2.3, I'm terribly stuck. I'm not even sure what to make my control surface.
How can I relate pressure outside the valve with the two thermodynamic states I have inside the tank, as well as with the entropy generated?

Many thanks in advance for any advice!

PS α₄ = 6

 

[Chestermiller]

What is $\alpha_4$?

 

[TimoD]

What is $\alpha_4$?

My apologies, in my case it is 6, so V = 6.1m³. Just a randomiser to prevent students copying directly.

 

[Chestermiller]

What is the equation for the change in entropy per mole of an ideal gas in terms of the initial and final temperatures and pressures?
If you choose the inlet plane and the outlet plane to the valve as the boundaries of your control surface, for any parcel of air passing through this control volume, what is the change in enthalpy? What is the change in temperature of the parcel for an ideal gas? Do all parcels of gas enter the valve at the same temperature and pressure?

Chet

 

[TimoD]

What is the equation for the change in entropy per mole of an ideal gas in terms of the initial and final temperatures and pressures?
If you choose the inlet plane and the outlet plane to the valve as the boundaries of your control surface, for any parcel of air passing through this control volume, what is the change in enthalpy? What is the change in temperature of the parcel for an ideal gas? Do all parcels of gas enter the valve at the same temperature and pressure?

Chet

Ok:
I know how to find change in specific entropy using a Gibbs equation manipulation ( Δs = ∫c_vdT/T + Rln(v₂/v₁) ).

Then, using the first law for a single flow system, I get that h_in = h_out. I think it is correct to assume an adiabatic process here? So I guess change in enthalpy is zero.

As to the change in temperature, T_out - T_in = (P_outv_out - P_inv_in)/R... I think??

Finally no, since the state inside the tank in constantly changing.

Am I doing anything right?

 

[Chestermiller]

Ok:
I know how to find change in specific entropy using a Gibbs equation manipulation ( Δs = ∫c_vdT/T + Rln(v₂/v₁) ).

The equation I had in mind was $\Delta s=\int{C_p}\frac{dT}{T}-R\ln \frac{p_2}{p_1}$

Then, using the first law for a single flow system, I get that h_in = h_out. I think it is correct to assume an adiabatic process here? So I guess change in enthalpy is zero.

Correct.

As to the change in temperature, T_out - T_in = (P_outv_out - P_inv_in)/R... I think??

Not quite. If the change in enthalpy for an ideal gas is zero, and the enthalpy is a unique function of temperature, what is the change in temperature through the valve?

Given the answer to this question, what is the change in entropy per mole in passing through the valve (using p_in to represent the pressure in the tank upstream of the valve and p_out to represent the pressure coming out of the valve (assumed constant according to the problem statement)?

Finally no, since the state inside the tank in constantly changing.

Correct. The pressure and temperature of the gas entering the valve change with time. We will take this into account.

Am I doing anything right?

Yes!

 

[TimoD]

The equation I had in mind was $\Delta s=\int{C_p}\frac{dT}{T}-R\ln \frac{p_2}{p_1}$

Correct.

Not quite. If the change in enthalpy for an ideal gas is zero, and the enthalpy is a unique function of temperature, what is the change in temperature through the valve?

Given the answer to this question, what is the change in entropy per mole in passing through the valve (using p_in to represent the pressure in the tank upstream of the valve and p_out to represent the pressure coming out of the valve (assumed constant according to the problem statement)?

Correct. The pressure and temperature of the gas entering the valve change with time. We will take this into account.

Yes!

Well if enthalpy is strictly a function of temperature then it seems temperature is constant through the valve also. Is assuming this a bit of an approximation, since h = u + Pv, clearly pressure and volume have an influence too?

In that case though, $\Delta s=-R\ln \frac{p_{out}}{p_{in}}$

Where to from here though? How do I involve the changing inlet pressure and temperature?

 

[Chestermiller]

Well if enthalpy is strictly a function of temperature then it seems temperature is constant through the valve also. Is assuming this a bit of an approximation, since h = u + Pv, clearly pressure and volume have an influence too?

No. For an ideal gas, there is no approximation involved. This is exact. So Δ(Pv)=0.

In that case though, $\Delta s=-R\ln \frac{p_{out}}{p_{in}}$

Where to from here though? How do I involve the changing inlet pressure and temperature?

Suppose that the number of moles of gas remaining in the tank at any instant of time is m. Then the number of moles of gas passing through the valve over any short time interval -dm. So, during the time that -dm passes through the valve, the amount of entropy generated is:
$$dS=-dmΔs=R\ln\left(\frac{p_{out}}{p_{in}}\right)dm$$
Now, from the ideal gas law for the tank, how is m related to p_in, T_in, and V?

Chet

 

[TimoD]

No. For an ideal gas, there is no approximation involved. This is exact. So Δ(Pv)=0.

Suppose that the number of moles of gas remaining in the tank at any instant of time is m. Then the number of moles of gas passing through the valve over any short time interval -dm. So, during the time that -dm passes through the valve, the amount of entropy generated is:
$$dS=-dmΔs=R\ln\left(\frac{p_{out}}{p_{in}}\right)dm$$
Now, from the ideal gas law for the tank, how is m related to p_in, T_in, and V?

Chet

$m=\frac{p_{in}V}{RT_{in}}$

 

[Chestermiller]

$m=\frac{p_{in}V}{RT_{in}}$

Good. Now before proceeding with the case where the heat capacity is a function of temperature, I recommend that we continue by first solving for the case of constant heat capacity (to see how that plays out and to make life simpler for ourselves). You need to eliminate T_in and express m exclusively in terms of p_in. You can do this from the adiabatic expansion relationship for the gas in the tank. So, what is T_in as a function of p_in, and then, what is m as a function of p_in? Then, what is dm?

Chet

 

[Chestermiller]

I can also help you get started for the case in which the temperature dependence of the heat capacity is included. In the part 2.1, you must have used the equation:
$$C_p(T)\frac{dT}{T}=R\frac{dp}{p}$$
Integrating this equation from the initial condition, you get:
$$\int_{T_0}^{T_{in}}{\frac{C_p(T')}{R}\frac{dT'}{T'}}=\ln\frac{p_{in}}{p_0}$$
where T' is a dummy variable of integration.
So, solving for p_in gives:
$$p_{in}=p_0\exp\left[{\int_{T_0}^{T_{in}}{\frac{C_p(T')}{R}\frac{dT'}{T'}}}\right]$$
See what happens when you substitute this into the equations for dS and Δs.

Chet

 

[TimoD]

Good. Now before proceeding with the case where the heat capacity is a function of temperature, I recommend that we continue by first solving for the case of constant heat capacity (to see how that plays out and to make life simpler for ourselves). You need to eliminate T_in and express m exclusively in terms of p_in. You can do this from the adiabatic expansion relationship for the gas in the tank. So, what is T_in as a function of p_in, and then, what is m as a function of p_in? Then, what is dm?

Chet

Ok I think I see where this is going, but I can't seem to find a way to write T_in in terms of p_in without using the ideal gas equation?

As for the integral of c_p with temperature dependence, we have an appendix of tabulated standard entropy values which we are encouraged to use due to their improved accuracy. I think they would apply here, so could I write $p_{in} = p_{0}exp[s_{T_{in}}-s_{T_{0}}]$?

 

[Chestermiller]

Ok I think I see where this is going, but I can't seem to find a way to write T_in in terms of p_in without using the ideal gas equation?

OK. I think you mean here "without assuming constant heat capacity," right? Let's skip the simplifying approximation implied in post # 11, and move on to post #12.

As for the integral of c_p with temperature dependence, we have an appendix of tabulated standard entropy values which we are encouraged to use due to their improved accuracy.

Oh. I wasn't aware that this was the case. So you are supposed to work with the table. So in Part 2.1, you just used s(p_final,T_final)=s(p₀,T₀) to obtain the final temperature?

I think they would apply here, so could I write $p_{in} = p_{0}exp[s_{T_{in}}-s_{T_{0}}]$?

Let's come back to this after you've answered my question above.
Chet

 

[TimoD]

OK. I think you mean here "without assuming constant heat capacity," right? Let's skip the simplifying approximation implied in post # 11, and move on to post #12.

Oh. I wasn't aware that this was the case. So you are supposed to work with the table. So in Part 2.1, you just used s(p_final,T_final)=s(p₀,T₀) to obtain the final temperature?

Let's come back to this after you've answered my question above.
Chet

In part 2.1 I used $\Delta s=\int{C_p}\frac{dT}{T}-R\ln \frac{p_2}{p_1}$, and then substituted standard entropy values for the c_p integral, following which I solved for s_T2 (because everything else is known, and $\Delta s= 0$) and then used tabulated data to interpolate and solve for T₂.

 

[Chestermiller]

In part 2.1 I used $\Delta s=\int{C_p}\frac{dT}{T}-R\ln \frac{p_2}{p_1}$, and then substituted standard entropy values for the c_p integral, following which I solved for s_T2 (because everything else is known, and $\Delta s= 0$) and then used tabulated data to interpolate and solve for T₂.

OK. I want to make sure we are on the same page. When you use the term "standard entropy," you mean the entropy at temperature T and pressure 1 atm., correct? This entropy is usually signified by using a superscript 0 to indicate that it is at 1 atm. So,
$$s^0(T)=s(1 atm, T)$$
Please confirm.

Chet

 

[TimoD]

OK. I want to make sure we are on the same page. When you use the term "standard entropy," you mean the entropy at temperature T and pressure 1 atm., correct? This entropy is usually signified by using a superscript 0 to indicate that it is at 1 atm. So,
$$s^0(T)=s(1 atm, T)$$
Please confirm.

Chet

Yes exactly that. Sorry still getting used to the syntax for equations

 

[Chestermiller]

OK. Thanks.

Let's go back to your post #12. The equation you wrote was $p_{in} = p_{0}exp[s_{T_{in}}-s_{T_{0}}]$. But you're missing an R in the denominator of the exponent. So,
$$p_{in} = p_{0}exp \left[\frac{(s_{T_{in}}-s_{T_{0}})}{R}\right]$$
Do we agree on this?

Chet

 

[TimoD]

OK. Thanks.

Let's go back to your post #12. The equation you wrote was $p_{in} = p_{0}exp[s_{T_{in}}-s_{T_{0}}]$. But you're missing an R in the denominator of the exponent. So,
$$p_{in} = p_{0}exp \left[\frac{(s_{T_{in}}-s_{T_{0}})}{R}\right]$$
Do we agree on this?

Chet

Ah yes of course! My mistake.

Ok so I have an expression for p_in, and I also know that

$\Delta s = \int\frac{Q}{T} + S_{gen} = 100J/K$

since the process is adiabatic. I'm still struggling though to find an expression for dm that I can use in our equation $$dS=-dmΔs=R\ln\left(\frac{p_{out}}{p_{in}}\right)dm$$

 

[Chestermiller]

Ah yes of course! My mistake.

Ok so I have an expression for p_in, and I also know that

$\Delta s = \int\frac{Q}{T} + S_{gen} = 100J/K$

since the process is adiabatic.

This is not the correct equation to use. But, don't worry, I'll get you there.

I'm still struggling though to find an expression for dm that I can use in our equation $$dS=-dmΔs=R\ln\left(\frac{p_{out}}{p_{in}}\right)dm$$

This is the correct equation to work with. As I said, don't worry, I'll get you there. We're almost done. We are going to integrate this equation to get ΔS which it the total amount of entropy generated in the valve, namely, 100 J/K. From this, we are going to determine p_out.

The next step is to substitute the equation for p_in from the previous post into$$dS=-dmΔs=R\ln\left(\frac{p_{out}}{p_{in}}\right)dm$$
Please show me what you get.

Chet

 

[TimoD]

This is not the correct equation to use. But, don't worry, I'll get you there.

This is the correct equation to work with. As I said, don't worry, I'll get you there. We're almost done. We are going to integrate this equation to get ΔS which it the total amount of entropy generated in the valve, namely, 100 J/K. From this, we are going to determine p_out.

The next step is to substitute the equation for p_in from the previous post into$$dS=-dmΔs=R\ln\left(\frac{p_{out}}{p_{in}}\right)dm$$
Please show me what you get.

Chet

After simplifying I get $$dS=-dmΔs=R(\ln(p_{out}) - \ln(p_{0}) - \frac{s_{T_{in}}-s_{T_{0}}}{R})dm$$

 

[Chestermiller]

After simplifying I get $$dS=-dmΔs=R(\ln(p_{out}) - \ln(p_{0}) - \frac{s_{T_{in}}-s_{T_{0}}}{R})dm$$

Nice job, except that there is a factor of R issue again. We actually get:

$$ΔS=R\ln(p_0/p_{out})(m_{0}-m_f)-\int{(s_{T_{in}}-s_{T_{0}})dm}$$
where ΔS represents the entropy generated in the valve during the process, which ends up in the air that has exited the valve.
OK so far?

Chet

 

[TimoD]

Nice job, except that there is a factor of R issue again. We actually get:

$$ΔS=R\ln(p_0/p_{out})(m_{0}-m_f)-\int{(s_{T_{in}}-s_{T_{0}})dm}$$
where ΔS represents the entropy generated in the valve during the process, which ends up in the air that has exited the valve.
OK so far?

Chet

Ok great I think I understand all that, I just didn't distribute the R into the brackets.

Where to from here? What do we do with the s_Tin in the integral?

 

[Chestermiller]

Ok great I think I understand all that, I just didn't distribute the R into the brackets.

Just to clarify our bounds of integration, why do we use initial and final mass of air in the tank when we are looking at entropy generated in the air that passed through the valve?

Let's review what's happening. Entropy is being generated as the air as it passes through the valve, and each parcel of air that leaves the value carries away its share of the entropy that was generated. The entropy change for each of the parcels of air is associated with its change in pressure (since its temperature doesn't change as it passes through the valve, i.e., constant delta h). However, since the temperature of the air that enters the valve is changing during the process, the temperatures of all the parcels that exit the value are different (even though their final pressures are the same). The total entropy generated in the valve is equal to the sum of the entropy changes of all the parcels of air that have left the valve during the process, as determined by their varying inlet pressure and (constant) exit pressure (and their individual masses).

Now, with regard to our use of the mass of air remaining in the tank, we use it to keep track of how much mass is contained in each of the parcels of air that pass through the valve. If m(t) is the amount of gas in the tank at any time, then the amount of mass that passes through the valve between time t and t + Δt is equal to m(t) - m(t+Δt). We need to sum up over all the parcels of mass that pass through the valve between the start and end of the process to get the total entropy change. We are expressing the total entropy change in terms of p_out, and, once we can do that, we can set it equal to the total entropy generated in the valve according to the problem statement, and solve for p_out.

Where to from here?

We need to figure out how to do the integral in the equation for ΔS. (That integral is independent of p_out.) I tried doing the integral analytically, but was only able to do it for the case in which the heat capacity is constant. So, instead of spending much more time on this problem, we can do it numerically, with very little effort.

Suppose I choose a value of T_in for the air entering the valve at time t during the process. This temperature is somewhere between 450 K and 370 K. You can immediately determine the value of the integrand in our equation at this temperature, since it is a function only of T_in. You can then use the equation in post #17 to get the pressure p_in at that time. Then, you can use your equation in post #9 to get the mass in the tank at that time. So you have the value of the integrand, and you have the value of m. By choosing a sequence of values of T_in, you can make a plot of the value of the integrand as a function of the integration variable m. The area under this curve will be the value of the integral. Rather than counting boxes in a graph, however, it is simple to do the integration numerically.

Let me get you started. Make a little table (say on a spreadsheet) with the column headings T_in, (s_in-s₀), p_in, and m. The first entry in the T_in column will be 450, and the last entry will be 370. We will use 10 equal increments of temperature, with the increment equal to 8 degrees, so the second entry in the column will be 442. All together there will be 11 entries in the column. Now, use your entropy table and the equations in posts #17 and #9 to fill in the rest of the columns. This should not take very long.

When you complete the table, please get back with me, and I'll tell you how to do the integration numerically.

Thanks for your diligence and your patience.

Chet

 

[TimoD]

Let's review what's happening. Entropy is being generated as the air as it passes through the valve, and each parcel of air that leaves the value carries away its share of the entropy that was generated. The entropy change for each of the parcels of air is associated with its change in pressure (since its temperature doesn't change as it passes through the valve, i.e., constant delta h). However, since the temperature of the air that enters the valve is changing during the process, the temperatures of all the parcels that exit the value are different (even though their final pressures are the same). The total entropy generated in the valve is equal to the sum of the entropy changes of all the parcels of air that have left the valve during the process, as determined by their varying inlet pressure and (constant) exit pressure (and their individual masses).

Now, with regard to our use of the mass of air remaining in the tank, we use it to keep track of how much mass is contained in each of the parcels of air that pass through the valve. If m(t) is the amount of gas in the tank at any time, then the amount of mass that passes through the valve between time t and t + Δt is equal to m(t) - m(t+Δt). We need to sum up over all the parcels of mass that pass through the valve between the start and end of the process to get the total entropy change. We are expressing the total entropy change in terms of p_out, and, once we can do that, we can set it equal to the total entropy generated in the valve according to the problem statement, and solve for p_out.

We need to figure out how to do the integral in the equation for ΔS. (That integral is independent of p_out.) I tried doing the integral analytically, but was only able to do it for the case in which the heat capacity is constant. So, instead of spending much more time on this problem, we can do it numerically, with very little effort.

Suppose I choose a value of T_in for the air entering the valve at time t during the process. This temperature is somewhere between 450 K and 370 K. You can immediately determine the value of the integrand in our equation at this temperature, since it is a function only of T_in. You can then use the equation in post #17 to get the pressure p_in at that time. Then, you can use your equation in post #9 to get the mass in the tank at that time. So you have the value of the integrand, and you have the value of m. By choosing a sequence of values of T_in, you can make a plot of the value of the integrand as a function of the integration variable m. The area under this curve will be the value of the integral. Rather than counting boxes in a graph, however, it is simple to do the integration numerically.

Let me get you started. Make a little table (say on a spreadsheet) with the column headings T_in, (s_in-s₀), p_in, and m. The first entry in the T_in column will be 450, and the last entry will be 370. We will use 10 equal increments of temperature, with the increment equal to 8 degrees, so the second entry in the column will be 442. All together there will be 11 entries in the column. Now, use your entropy table and the equations in posts #17 and #9 to fill in the rest of the columns. This should not take very long.

When you complete the table, please get back with me, and I'll tell you how to do the integration numerically.

Thanks for your diligence and your patience.

Chet

Great! I have the values in a table, and I think I see where the iterative part comes in.

Now how to numerically integrate?

No thank you!

 

[Chestermiller]

Great! I have the values in a table, and I think I see where the iterative part comes in.

Now how to numerically integrate?

No thank you!

Great. Before we start integrating, let's check to make sure that the numbers in your table are correct. Is the final pressure in your table 200 kPa? Is the difference in mass between beginning and end consistent with what you calculated in part 2.2? Also, let's be sure that the units are consistent when we do the integration.

To do the integration numerically, the integral over each interval of mass is just calculated as $\frac{I(m)+I(m+\Delta m)}{2}\Delta m$, where I(m) is the value of the integrand at mass m.

Chet

 

[TimoD]

Great. Before we start integrating, let's check to make sure that the numbers in your table are correct. Is the final pressure in your table 200 kPa? Is the difference in mass between beginning and end consistent with what you calculated in part 2.2? Also, let's be sure that the units are consistent when we do the integration.

To do the integration numerically, the integral over each interval of mass is just calculated as $\frac{I(m)+I(m+\Delta m)}{2}\Delta m$, where I(m) is the value of the integrand at mass m.

Chet

Yes I think the values are close enough.

Ok so if I use that parallelogram approximation to find the integral, then I can solve for p_in? What about the iterative method that the question speaks of?

 

[Chestermiller]

Yes I think the values are close enough.

Ok so if I use that parallelogram approximation to find the integral, then I can solve for p_in?

I think you mean the trapazoidal rule.

What about the iterative method that the question speaks of?

I don't know. It doesn't seem like an iteration is needed. Maybe they were using equal increments in pressure, rather than temperature, in which case they would have to iterate to get the temperature (and thus the mass) at each step. Or maybe they were using the word iterate to signify numerical integration. Or maybe they didn't realize that they could solve for the exit pressure explicitly. It's hard to say. All we can do is have confidence in what we did.

Chet

 

[TimoD]

I think you mean the trapazoidal rule.

I don't know. It doesn't seem like an iteration is needed. Maybe they were using equal increments in pressure, rather than temperature, in which case they would have to iterate to get the temperature (and thus the mass) at each step. Or maybe they were using the word iterate to signify numerical integration. Or maybe they didn't realize that they could solve for the exit pressure explicitly. It's hard to say. All we can do is have confidence in what we did.

Chet

Yes that's the one sorry.

Chestermiller thank you so much for your step by step help through all of this as well as your patience through all my blunders. I really appreciate it! God bless you sir

 

[Chestermiller]

Yes that's the one sorry.

Chestermiller thank you so much for your step by step help through all of this as well as your patience through all my blunders. I really appreciate it! God bless you sir

You've done very well. This was not an easy problem. I think you have a real knack for understanding thermodynamics.

Chet

 

[TimoD]

You've done very well. This was not an easy problem. I think you have a real knack for understanding thermodynamics.

Chet

Thank you!

Could you possibly just clarify for me one last time; in post #11 where you derive an expression for p_in, I see it comes from the equation $Cp(T)\frac{dT}{T}=R\frac{dp}{p}$, which comes from $ds = Cp(T)\frac{dT}{T}-R\frac{dp}{p}$ where change in entropy is zero. Why is this so? Since we use p_in and p_out, surely the change in entropy is not zero between these points?

Finally could you just confirm that you get an answer around 277kPa if you have worked it out?

 

[Chestermiller]

Thank you!

Could you possibly just clarify for me one last time; in post #11 where you derive an expression for p_in, I see it comes from the equation $Cp(T)\frac{dT}{T}=R\frac{dp}{p}$, which comes from $ds = Cp(T)\frac{dT}{T}-R\frac{dp}{p}$ where change in entropy is zero. Why is this so? Since we use p_in and p_out, surely the change in entropy is not zero between these points?

This would be the change in molar entropy for the gas in the tank. It establishes the relationship between the pressure and the temperature in the tank (and of the parcels of air that enter the valve).

Finally could you just confirm that you get an answer around 277kPa if you have worked it out?

This appears to be high. The exit pressure from the valve shouldn't be higher than the final pressure in the tank (200 kPa). Otherwise gas would be flowing back in the opposite direction. What would you get for the entropy change if you assumed that the pressure at the exit of the valve were at 200 kPa? (Maybe they gave you too small a value for the entropy change in the problem statement, not realizing that the exit pressure had to be less than the final pressure in the tank).

Chet

 

[TimoD]

This would be the change in molar entropy for the gas in the tank. It establishes the relationship between the pressure and the temperature in the tank (and of the parcels of air that enter the valve).

This appears to be high. The exit pressure from the valve shouldn't be higher than the final pressure in the tank (200 kPa). Otherwise gas would be flowing back in the opposite direction. What would you get for the entropy change if you assumed that the pressure at the exit of the valve were at 200 kPa? (Maybe they gave you too small a value for the entropy change in the problem statement, not realizing that the exit pressure had to be less than the final pressure in the tank).

Chet

Oh right I see, I misread p₀ as p_out.

Oh I thought that the average may be above even though the instantaneous is always below p_in. With a 200kPa exit pressure though, I get an entropy change of 797.7J/K. Maybe my integral value is off? For that I get quite a high 674.4J/K.

 

[Chestermiller]

Oh right I see, I misread p₀ as p_out.

Oh I thought that the average may be above even though the instantaneous is always below p_in. With a 200kPa exit pressure though, I get an entropy change of 797.7J/K. Maybe my integral value is off? For that I get quite a high 674.4J/K.

I don't think so. I'm guessing that they meant 1000 J/K in the problem statement, rather than 100 J/K. But, just in case, please check all your units. I think the 277 kPa is close to the value you would get if you took the entropy change in the valve to be close to zero. What value would you get for p_out it you used a value of 1000 J/K for the entropy generated?

Chet

 

[Chestermiller]

With your value of the integral and an assumed value of 1000 J/K for the entropy change, I get a value of about 185 kPa for p_out, which sounds about right.

Chet

 

[TimoD]

With your value of the integral and an assumed value of 1000 J/K for the entropy change, I get a value of about 185 kPa for p_out, which sounds about right.

Chet

Yup I get that as well. Ok well since our method is sound I will note this in my assignment and see what happens. Thanks once again for all your help