The Joule Thompson Coefficient -- Throttling effect

[Riverbirdy]

Say I have a line of steam with temperature 800 deg C, at 10bars connected with it is a partially evacuated tank 0.01 bar pressure, room temperature. The connection was built with a long capillary and a valve. Does throttling effect experience when the valve suddenly opened? Why and why not?

 

[Chestermiller]

What is your assessment of this?

 

[Riverbirdy]

What is your assessment of this?

I guess, momentarily throttling effect might be experienced. But a soon as pressure build up, the effect diminished and tank and source come into the same pressure. I am not sure about this. Do you have insights? Perhaps, the actual effect might be different. I never have practical experience on this yet

 

[Chestermiller]

Is this a homework problem?

When you are talking about throttling, you are referreing exclusively to what is happening between the inlet and outlet of the capillary.

My understanding is that you want to know how the temperatures and pressures in the two tanks vary as a function of time. Is that correct?

 

[Riverbirdy]

Is this a homework problem?

When you are talking about throttling, you are referring exclusively to what is happening between the inlet and outlet of the capillary.

My understanding is that you want to know how the temperatures and pressures in the two tanks vary as a function of time. Is that correct?

Yes, Chet. A sort of things I am troubled to ponder. My concern is only at the outlet of the capillary at low pressure side where the tank can be considered in adiabatic condition. I really wonder what's the final temperature. Some references says Joule Thompson can be negative or positive. I do not know if that is indeed applicable to such as this uniform state uniform flow problems.

 

[Chestermiller]

Yes, Chet. A sort of things I am troubled to ponder. My concern is only at the outlet of the capillary at low pressure side where the tank can be considered in adiabatic condition. I really wonder what's the final temperature. Some references says Joule Thompson can be negative or positive. I do not know if that is indeed applicable to such as this uniform state uniform flow problems.

The Joule Thompson effect is applicable exclusively to the capillary, where the change in enthalpy (per unit mass flowing through) is zero. If you want to analyze the problem of steam equilibrating between the two adiabatic tanks (by flowing through the capillary), you can do that. Would you like to solve that problem ?

 

[Riverbirdy]

The Joule Thompson effect is applicable exclusively to the capillary, where the change in enthalpy (per unit mass flowing through) is zero. If you want to analyze the problem of steam equilibrating between the two adiabatic tanks (by flowing through the capillary), you can do that. Would you like to solve that problem ?

Yap, Chet. How do I start?

 

[Chestermiller]

Yap, Chet. How do I start?

OK. Let's start by specifying the volume of each of the two tanks and determining the number of moles of water (or the mass of water) in each tank. The ball is in your court for doing this now.

 

[Riverbirdy]

OK. Let's start by specifying the volume of each of the two tanks and determining the number of moles of water (or the mass of water) in each tank. The ball is in your court for doing this now.

Say, I've got only one tank (adiabatically-insulated so that no heat crosses the inside to the outside) with 6 kg steam at 138kPa, 200⁰C. This is connected via long capillary and a valve to hold the cross main pipe of superheated steam having a pressure of 680kPa, 350⁰C to enter the tanks.

Then how to do it?
What's next Chet?

 

[Chestermiller]

Say, I've got only one tank (adiabatically-insulated so that no heat crosses the inside to the outside) with 6 kg steam at 138kPa, 200⁰C. This is connected via long capillary and a valve to hold the cross main pipe of superheated steam having a pressure of 680kPa, 350⁰C to enter the tanks.

Then how to do it?
What's next Chet?

I don't understand your description. Can you please provide a diagram? Thanks.

Chet

 

[Riverbirdy]

I don't understand your description. Can you please provide a diagram? Thanks.

Chet

 

[Chestermiller]

View attachment 109606

Is it safe to assume that the capillary is insulated?

 

[Riverbirdy]

Is it safe to assume that the capillary is insulated?

Yes, Chet.

 

[Chestermiller]

OK. We are going to assume that, at any given time, there is negligible mass holdup in the capillary (compared to the tank). Our open system is going to be the combination of the tank and capillary. I assume you are currently studying the open system version of the first law of thermodynamics, correct? Are you allowed to use the steam tables to solve this problem?

 

[Riverbirdy]

OK. We are going to assume that, at any given time, there is negligible mass holdup in the capillary (compared to the tank). Our open system is going to be the combination of the tank and capillary. I assume you are currently studying the open system version of the first law of thermodynamics, correct? Are you allowed to use the steam tables to solve this problem?

Yes, Chet I know how to get the properties from the tables. Yes, Open and close systems. I more interested on the equation formulation.

 

[Chestermiller]

Yes, Chet I know how to get the properties from the tables. Yes, Open and close systems. I more interested on the equation formulation.

Let $m_0$ represent the initial mass of steam in the tank
Let $v_0$ represent the initial specific volume of steam in the tank
Let $u_0$ represent the initial internal energy per unit mass of steam in the tank
Let $\Delta m$ represent the mass of steam added from the pipe to the tank (through the capillary)
Let $u_f$ represent the final internal energy per unit mass of steam in the tank (after $\Delta m$ has been added)
let $v_f$ represent the final specific volume of the steam in the tank (after $\Delta m$ has been added)
Let $h_p$ represent the enthalpy per unit mass of the steam in the pipe

Using the open system version of the first law, derive an equation for $u_f$ in terms of $\Delta m$, $h_p$, $u_0$, and $m_0$. Derive an equation for $v_f$ in terms of $m_0$, $v_0$, and $\Delta m$.

For $\Delta m=0.5 kg$, what are the values of $u_f$ and $v_f$?

 

[Riverbirdy]

Let $m_0$ represent the initial mass of steam in the tank
Let $v_0$ represent the initial specific volume of steam in the tank
Let $u_0$ represent the initial internal energy per unit mass of steam in the tank
Let $\Delta m$ represent the mass of steam added from the pipe to the tank (through the capillary)
Let $u_f$ represent the final internal energy per unit mass of steam in the tank (after $\Delta m$ has been added)
let $v_f$ represent the final specific volume of the steam in the tank (after $\Delta m$ has been added)
Let $h_p$ represent the enthalpy per unit mass of the steam in the pipe

Using the open system version of the first law, derive an equation for $u_f$ in terms of $\Delta m$, $h_p$, $h_p$, $u_0$, and $m_0$. Derive an equation for $v_f$ in terms of $m_0$, $v_0$, and $\Delta m$.

For $\Delta m=0.5 kg$, what are the values of $u_f$ and $v_f$?

I see, I guess if we use Steady State Steady Flow analysis, there will be a throttling effect only that, pressure builds up on the tank as well, so the joule-thompson coeffecient also changed with time, but this is a good approach. Uniform State Uniform Flow analysis would tend to neglect the Joule thomson effect. I am stilll confused how to derive the rate eqaution.

 

[Chestermiller]

I see, I guess if we use Steady State Steady Flow analysis, there will be a throttling effect only that, pressure builds up on the tank as well, so the joule-thompson coeffecient also changed with time, but this is a good approach. Uniform State Uniform Flow analysis would tend to neglect the Joule thomson effect. I am stilll confused how to derive the rate eqaution.

To get the rate equation, you need to include the fluid dynamic pressure-drop/flow-rate relationship for the capillary.

Have you decided not to pursue the derivation of the equations that I asked for?

 

[Riverbirdy]

To get the rate equation, you need to include the fluid dynamic pressure-drop/flow-rate relationship for the capillary.

Have you decided not to pursue the derivation of the equations that I asked for?

You know, Chet. It makes me dizzy, now i am entangled really

 

[Chestermiller]

You know, Chet. It makes me dizzy, now i am entangled really

Sorry to hear that. Do you want me to write down the equations, so that you can then plug numbers in?

 

[Riverbirdy]

Sorry to hear that. Do you want me to write down the equations, so that you can then plug numbers in?

yes, please so you might enlighten me

 

[Chestermiller]

yes, please so you might enlighten me

First Law Open System Energy Balance on Tank:$$\Delta U=(\Delta m) h_p$$or equivalently:$$(m_0+\Delta m)u_f-m_0u_0=(\Delta m) h_p$$Final Specific Volume:
$$v_f=\frac{m_0v_0}{(m_0+\Delta m)}$$

 

[Chestermiller]

The equation I gave in my previous post was the integrated relationship (integrated with respect to time). It just struck me that you might have been interested in the differential version of the relationship:

$$\frac{dU}{dt}=\frac{(mu)}{dt}=\dot{m}h_p$$

And, yes, since we are assuming there is no mass holdup in the capillary, the enthalpy of the steam entering the tank from the capillary is the same as the enthalpy of the steam in the pipe. This assumption permits us to regard the capillary as operating at an instantaneous steady state.