Thermodynamics: Pressure Drop over a Valve

  • #1
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Hey Guys! Anyone able to help out here?

upload_2015-10-22_10-12-24.png


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
 
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Answers and Replies

  • #3
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What is ##\alpha_4##?
My apologies, in my case it is 6, so V = 6.1m3. Just a randomiser to prevent students copying directly.
 
  • #4
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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
 
  • #5
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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 = ∫cvdT/T + Rln(v2/v1) ).

Then, using the first law for a single flow system, I get that hin = hout. 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, Tout - Tin = (Poutvout - Pinvin)/R... I think??

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

Am I doing anything right?
 
  • #6
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Ok:
I know how to find change in specific entropy using a Gibbs equation manipulation ( Δs = ∫cvdT/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 hin = hout. 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, Tout - Tin = (Poutvout - Pinvin)/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 pin to represent the pressure in the tank upstream of the valve and pout 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!!!
 
  • #7
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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 pin to represent the pressure in the tank upstream of the valve and pout 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?
 
  • #8
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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 pin, Tin, and V?

Chet
 
  • #9
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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 pin, Tin, and V?

Chet

##m=\frac{p_{in}V}{RT_{in}}##
 
  • #10
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##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 Tin and express m exclusively in terms of pin. You can do this from the adiabatic expansion relationship for the gas in the tank. So, what is Tin as a function of pin, and then, what is m as a function of pin? Then, what is dm?

Chet
 
  • #11
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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 pin 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
 
  • #12
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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 Tin and express m exclusively in terms of pin. You can do this from the adiabatic expansion relationship for the gas in the tank. So, what is Tin as a function of pin, and then, what is m as a function of pin? 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 Tin in terms of pin without using the ideal gas equation?

As for the integral of cp 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}}]##?
 
  • #13
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Ok I think I see where this is going, but I can't seem to find a way to write Tin in terms of pin 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 cp 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(pfinal,Tfinal)=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
 
  • #14
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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(pfinal,Tfinal)=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 cp integral, following which I solved for sT2 (because everything else is known, and ##\Delta s= 0##) and then used tabulated data to interpolate and solve for T2.
 
  • #15
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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 cp integral, following which I solved for sT2 (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
 
  • #16
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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
 
  • #17
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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
 
  • #18
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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 pin, 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 $$dS=-dmΔs=R\ln\left(\frac{p_{out}}{p_{in}}\right)dm$$
 
  • #19
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Ah yes of course! My mistake.

Ok so I have an expression for pin, 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 pout.

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

Chet
 
  • #20
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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 pout.

The next step is to substitute the equation for pin 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 $$dS=-dmΔs=R(\ln(p_{out}) - \ln(p_{0}) - \frac{s_{T_{in}}-s_{T_{0}}}{R})dm$$
 
  • #21
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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
 
  • #22
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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 sTin in the integral?
 
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  • #23
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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 pout, 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 pout.
Where to from here?
We need to figure out how to do the integral in the equation for ΔS. (That integral is independent of pout.) 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 Tin 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 Tin. You can then use the equation in post #17 to get the pressure pin 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 Tin, 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 Tin, (sin-s0), pin, and m. The first entry in the Tin 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
 
  • #24
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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 pout, 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 pout.

We need to figure out how to do the integral in the equation for ΔS. (That integral is independent of pout.) 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 Tin 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 Tin. You can then use the equation in post #17 to get the pressure pin 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 Tin, 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 Tin, (sin-s0), pin, and m. The first entry in the Tin 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!
 
  • #25
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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
 

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