# The Joule Thompson Coefficient -- Throttling effect

• Riverbirdy
In summary, the conversation discusses a scenario where a line of steam with a temperature of 800 deg C and a pressure of 10 bars is connected to a partially evacuated tank with a pressure of 0.01 bar and room temperature. The connection between the two is made using a long capillary and a valve. The question is whether throttling effect is experienced when the valve is suddenly opened, and what the final temperature of the tank would be. The conversation also touches on the Joule Thompson effect and solving the problem using the open system version of the first law of thermodynamics and steam tables.
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?

What is your assessment of this?

Chestermiller said:
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

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?

Chestermiller said:
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.

Riverbirdy said:
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 ?

Chestermiller said:
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?

Riverbirdy said:
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.

Chestermiller said:
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?

Riverbirdy said:
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

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

Chet

Riverbirdy said:
Is it safe to assume that the capillary is insulated?

Chestermiller said:
Is it safe to assume that the capillary is insulated?
Yes, Chet.

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?

Chestermiller said:
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.

Riverbirdy said:
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##?

Chestermiller said:
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.

Riverbirdy said:
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?

Chestermiller said:
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

Riverbirdy said:
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?

Chestermiller said:
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

Riverbirdy said:
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:
$$v_f=\frac{m_0v_0}{(m_0+\Delta m)}$$

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.

## 1. What is the Joule-Thompson coefficient?

The Joule-Thompson coefficient, also known as the throttling coefficient, is a thermodynamic property that describes the change in temperature of a gas when it is allowed to expand or contract without performing work.

## 2. How is the Joule-Thompson coefficient calculated?

The Joule-Thompson coefficient is calculated by taking the partial derivative of the enthalpy of the gas with respect to its temperature, while keeping the pressure constant. It is often denoted by the symbol mu (μ).

## 3. What is the significance of the Joule-Thompson coefficient?

The Joule-Thompson coefficient is an important parameter in understanding the behavior of gases, particularly in refrigeration and liquefaction processes. It also plays a role in determining the conditions for the onset of condensation or evaporation.

## 4. How does the Joule-Thompson coefficient affect the temperature of a gas?

The sign of the Joule-Thompson coefficient determines whether the gas will experience a temperature increase or decrease when it undergoes throttling. A positive coefficient results in a decrease in temperature, while a negative coefficient results in an increase.

## 5. Are there any factors that can influence the Joule-Thompson coefficient?

Yes, the Joule-Thompson coefficient can be influenced by factors such as the composition of the gas, the pressure and temperature at which the throttling occurs, and the presence of impurities or non-ideal behavior in the gas. It is also dependent on the specific gas being studied.

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