Struggling Applying General Equation of Energy to Gas Flows 2

[MumtazDogan]

Hello everyone,

I am really struggling when I try to deal with gaseous flows. For liquids, everything is really straightforward for me, the terms in the energy equation, applying Bernoulli's Equation or Energy Equation etc.

But when it comes to gases, I really start to struggle and can't describe the system and what is happening. I can approximate some gaseous systems with Bernoulli, but when it comes to General Equation of Energy, they seem to contradict and Energy Equation seem not intuitively to me.

The equation I am trying to use is this

I always thought that the Bernoulli's Equation and head form of it was coming from the General Equation of Energy. But, the results seem not being interpreted the same.

Below, I have two questions, one is related with an atmospheric tank vent, the other is related with a gas flowing in a very long pipe whose radius is increasing to keep the linear velocity constant. I really do not understand how the terms in the General Equation of Energy changes in these situations.

1)

2)

Thanks in advance.

 

[Snickster]

In the first situation the change in stagnation enthalpy inside the tank and the flowing enthalpy in the vent line, assuming no velocity inside tank and gas is at stagnation conditions, equals the increase in velocity as follow:

Hi = Ho + V[sup]2[/sup]/2g
or
Cp(Ti-To) = V[sup]2[/sup]/(2g(778))

where Cp is specific heat in Btu/lb-F
T is deg F
V is velocity in ft/sec
g is gravity constant 32 ft/sec[sup]2[/sup]
778 is conversion from ft-lbs to Btu

So basically the change in enthaply is converted into velocity and also provides energy for frictional flow losses. However any frictional losses does not change the enthalpy since any energy loss to friction is put back into the fluid as heat so temperature or enthalpy does not change with friction, what changes is an increase in entropy and a decrease in stagnation pressure. Therefore the above equation is true with or without friction.

The flow is actually not constant temperature unless there is enough heat transfer into the fluid when it flows into the pipe to offset the energy required, that comes from the decrease in enthalpy, for the increase in kinetic energy needed to increase in velocity. In such case if the temperature wre to remain constant the the Q-in term will not be equal to zerobut would need to equal the kinetic energy term on the right side of the above equation. Although you can consider the flow to be isothermal and apply the isothermal frictional flow equation, true pipe flow is closer to adiabatic where not much heat enters the flowing fluid from the extenal environment. For adiabatic flow with friction more accurate flow equation is the Fanno flow equation. In true fanno flow temperature always decrease because of the reduction in enthaply reqired to provide the energy needed for the increase in velocity.

 

[pierreick]

HI,
You have a nice set of videos (tutorials) by Dr Biddle on YouTube, based on text book from F.White Fluid mechanics -section 9
Take a look:

https://www.youtube.com/results?search_query=fluid+mechanics+compressible+flow

Pierre

 

[Snickster]

In the increasing diameter situation the energy from friction loss goes back into the fluid so even though there is energy loss due to friction the enthalpy does not change, what does change is the entropy, ie. loss of usefull work, and stagnation pressure. The stagnation pressure is if you brought the flowing fluid to a stop isentropically so the velocity is re-converted to enthalpy. Upstream the stagnation pressue is higher while after friction downstream the stagnation pressure is lower signifying a loss of available energy and increase in entropy.

You can think of this flow with friction as having two tanks connected to each other with one tank pressurized and a valve is open to equalize pressue in the tanks, it follows the same path on a T-H-S/P-V diagrams. No work was done and no heat was added but the volume occupied increase and pressure decreased changing the pressure and entropy and available energy in exactly the same way on the diagrams as it changes for the flow case you described.

 

[MumtazDogan]

Hello Snickster, thank you so much for your answer. I really appreciate you for giving a very helpful answer to my detailed questions. Thanks a lot.

About The Tank Vent Flow Question:

I think I understood your explanation about the question in general. But to clear my mind more, please consider the gas (air) ideal and we talk about the General Equation of Energy not in enthalpy form, but in internal energy form.

So for ideal gases, "P/rho" = "R*T/M" and both the internal energy (U) and enthalpy (H) are only dependent on Temperature. So, basically, when the air flows from inside the tank to atmosphere through an atmospheric vent, the temperature decreases and this accounts for the linear velocity at the outlet of the vent, right ? Because when the temperature decreases, both "U" and "P/rho" values decreases as much as the "(v2^2)/2" term increases. I understood this situation thanks to you, but this is still a little counter-intuitive for me because it is like the pressure is not the reason why the flow occurs here, it is like the reduction in temperature. For example, with temperature decrease, if the pressure in the tank was lower than the atmospheric, say 900 mbars, it would be still possible to flow from inside to tank to atmosphere through the vent, right ? I think maybe the best way to look it is like that, for hydraulic calculations, we can interpret the situation by Euler's (if deltaP is low then Bernoulli's)Equation and for Thermodynamic purposes, by the General Equation of Energy. What do you think about these ? Especially, how can we interpret from the General Equation of Energy to which direction the fluid will flow and to what degree the temperature will decrease to increase the linear velocity ?



And lastly, I think the "Pressure/density" term means two different things in the General Equation of Energy and in Bernoulli's Equation. In Bernoulli's Equation, it can be considered the driving force for the pressure difference. It is like the mechanical energy available to obtain a flow and the linear velocity of the flow depends the pressure difference between two points. Yet, in the General Equation of Energy, for compressible flows, the "Pressure/density" terms means not much about flow because it is constant for ideal gases and nearly constant for real gases. Even a gas is going from 30 bars to 1 bars, this term is near equal which actually will cause an extremely high speed flow.

I still think about your answers about my second question, I will also write the parts that I have still questions about.

Thank you very much again.

 

[Snickster]

So for ideal gases, "P/rho" = "R*T/M" and both the internal energy (U) and enthalpy (H) are only dependent on Temperature. So, basically, when the air flows from inside the tank to atmosphere through an atmospheric vent, the temperature decreases and this accounts for the linear velocity at the outlet of the vent, right ? Because when the temperature decreases, both "U" and "P/rho" values decreases as much as the "(v2^2)/2" term increases.

Yes for flow with or without friction dH = dU + d(pv) = Cp (dT) = Cp (T2-T1) = V[sup]2[/sup][sub]2[/sub]/2g(778) - V[sup]2[/sup][sub]1[/sub]/2g(778)

Where Cp is specific heat in BTU/# deg F
T is in deg F
Velocity V is in ft/sec
g is gravity constant 32.2 ft/sec[sup]2[/sup]
778 is conversion from ft-lb units to BTU

Any increase in velocity has to come from the energy contained within the gas itself if there is no heat or work input externally. Without friction there is no frictional pressure drop so the process is reversible. In other words if you reverse the process the pressure would go back to the pressure in the tank when velocity goes back to zero. With friction there is irreversible friction loss but any frictional energy loss goes back into the gas so that the enthalpy does not change although the pressure is reduced more than the reversible case because of not only pressure decreasing due to velocity but pressure decreasing also due to friction loss. Therefore if you then reverse the process with friction and bring fluid backwards to zero then the pressure at zero velocity will be less than what you started with in the tank. This is the effect of irreversible process with increase in entropy. However in both the reversible and irreversible case the enthalpy remains constant so the above equation applies. For instance if the gas entered the pipe through a reversible perfect nozzle then there would ideally be no frictional pressure drop in the nozzle so the fluid on the downstream side of the nozzle will be at a lower pressure due to isentropic expansion only. Then once in the pipe the velocity would further decrease due to pressure drop due to friction. In reality there is a non-ideal expansion of the gas in the tank at zero velocity to the gas in the pipe since the opening in the tank is not a perfect nozzle. In any case the enthalpy is the same for the reversible or irreversible expansion of the non-flowing fluid to flowing fluid in the pipe. Then further pressure drop due to friction occurs and velocity further increases in accordance with the above equation.

I understood this situation thanks to you, but this is still a little counter-intuitive for me because it is like the pressure is not the reason why the flow occurs here, it is like the reduction in temperature. For example, with temperature decrease, if the pressure in the tank was lower than the atmospheric, say 900 mbars, it would be still possible to flow from inside to tank to atmosphere through the vent, right ?

No, the pressure differential is the driving force in accordance with Newtons law F = M x A. Pressure difference is the only reason why flow exists. If the fluid in the tank were a higher temperature but same pressure as the outside of the tank there would be no flow.

I think maybe the best way to look it is like that, for hydraulic calculations, we can interpret the situation by Euler's (if deltaP is low then Bernoulli's)Equation and for Thermodynamic purposes, by the General Equation of Energy. What do you think about these ?

No Eulers analysis of Sum of Forces in a pipe along the streamline is same for liquid or gas.

Especially, how can we interpret from the General Equation of Energy to which direction the fluid will flow and to what degree the temperature will decrease to increase the linear velocity?

High pressure to low pressure based on Sum F = M x A always, and the temperature will decrease in accordance with the above equation.

To be continued.

 

[Snickster]

And lastly, I think the "Pressure/density" term means two different things in the General Equation of Energy and in Bernoulli's Equation. In Bernoulli's Equation, it can be considered the driving force for the pressure difference. It is like the mechanical energy available to obtain a flow and the linear velocity of the flow depends the pressure difference between two points. Yet, in the General Equation of Energy, for compressible flows, the "Pressure/density" terms means not much about flow because it is constant for ideal gases and nearly constant for real gases. Even a gas is going from 30 bars to 1 bars, this term is near equal which actually will cause an extremely high speed flow.

No it means the same thing for both cases of liquid flow or gas flow. It is always the driving force for flow. The flow between two points is always dependent on the pressure differential. The greater the differential the greater the driving force and greater the flow.

For an ideal gas although with friction but no velocity increase such as your increasing diameter pipe of your original question 2, the pressure differential created the flow. So, the pressure at the downstream end of the pipe was lower due to friction loss and this pressure differential created the flow. But in an adiabatic process (no heat transferred to or from the system) the frictional losses went back into the fluid and the enthalpy did not change because you forced the velocity not to change by increasing the area of the pipe. In this case although the enthalpy remained constant, there was energy consumption due to friction and the stagnation pressure of the fluid decreased and entropy increased signifying a loss of available energy because you used energy in the fluid flow due to the friction. So if you brought this flow back to zero velocity to stagnation pressure and temperature (at zero velocity) the stagnation temperature will be same as in the tank but the stagnation pressure will be lower than what was originally in the tank signifying a loss in available energy due to friction. You can check the actual change is entropy by using the formula for entropy change R ln (p2/p1)where p is stagnation pressures with and without friction.

 

[MumtazDogan]

Hello @Snickster,

Thank you a lot for your answers, again. Thank you very much. Due to a business trip, I could just reply to your answers.

To start, I understand fully the situation on my question 1, thanks to you. In summary, in terms of the General Equation of energy, assuming ideal gas and frictionless environment, mainly the decrease in temperature reduces both "Uo" and "Po/Density0" terms (or as total, the Enthalpy) which accounts for the increase in the velocity. These terms only are dependent on "Temperature" because we already assumed the gas is ideal. I do not have problems until now, I only think that the way to calculate Vo is by Euler's Equation. Even though the situation in General Equation of Energy is fully understood by me, there are two unknowns as the outlet velocity and outlet temperature in it.

And again about Question 1, but with friction this time, I understand that the friction losses of ideal gases in adiabatic systems go back into gases "Internal Energy" or thus "Enthalpy" term. So, no change in "Temperature" and thus no change in "Internal Energy","P/rho", or "Enthalpy". It is intuitive because the energy created due to friction in the form of heat does not have anywere to go due to the adiabatic system, so it goes back to the fluid. However, about the Question 1 - Frictional Case, it can be clearly seen that by using head loss form of Bernoulli's Equation, when there is relatively high amount of friction, the outlet velocity (v2) gets decreased a lot, as you can see from my approximation below:

So, the friction affects the difference between "Internal Energy", "P/rho" and "Enthalpy" terms in the tank and at the outlet of the vent a lot, especially when the friction is relatively high. It decreases the velocity v2, so, there is a smaller need of enthalpy change to account for the velocity creation, right ?

If I am right until now, my question is that is the General Equation of Energy useful in flow calculations of gases ? Because the existence of friction affected all the terms, especially the outlet velocity. Yet, again, this seems not calculatable by the General Equation of Energy. On the other hand, for liquids, the general equation of energy can directly turn to head form of Bernoulli's equation I guess.

About the Question 2, I understand from your explanation that the all the terms are constant in the General Equation of Energy, right ? (The temperature, the p/rho term, internal energy and enthalpy). The reason is that the energy converted to heat due to friction goes back to the fluid since it is an adiabatic system. So, the total energy of the system remains constant.

So my question is, does the General Equation of Energy give anything about the situation to us, in terms of fluid dynamics? For example, if the system in Question 2 was for liquid flow and constant diameter, the pressure loss would be clearly represented in the "P/rho" term since the rho is constant, and the pressure loss term ((P2-P1)/rho) would be equal to U2-U1. And for liquid case, it is totally in agreement with the Bernoulli's Equation and head form of it.

Lastly, again about Question 2, what would happen when the diameter was not increasing, but constant ? It would be that the temperature would decrease again to account for the increase in velocity which stems from the specific volume expansion from 10 bar to 5 bar, right ?

Thanks in advance.

 

[Snickster]

Mumtaz, I need a little time to digest what you are saying and will get back to you.

 

[Snickster]

RESPONSE TO QUESTION 1 (Part 1 of 2)

The amount of friction in a pipe is what limits the flowrate and velocity in a pipe for a given upstream and downstream pressure for flow with friction. There is nothing in the General Energy Equation (GEE) that will predict the flowrate for flow with friction. The GEE can predict the flow for frictionless reversible flow. However, to use the GEE to predict the flow for ideal gas flow without friction you need to know some other conditions such as what type of process it is (isentropic, adiabatic, constant temperature, constant pressure, etc.).

For instance, if you know what type of process it is then you know some other things, for instance if you know it is an isentropic adiabatic process you know that PV[sup]k[/sup] = Constant. So for instance if there were a complete frictionless flow of gas from inside your tank to outside of tank through your vent pipe it would actually follow and adiabatic isentropic expansion process. This assumes the pipe is completely insulated so no heat enters the fluid from the environment as it flows through the pipe to the discharge. Any process where a fluid starts from zero velocity and increases to a final a velocity reversibly without friction is an isentropic expansion process, and the reverse is also true that when a fluid is at a velocity and comes to a complete rest then it follows an isentropic compression path to what called stagnation condition (stagnation pressure, stagnation temperature, stagnation enthalpy, etc.). Then for an isentropic expansion from zero velocity in your tank to a final velocity in your vent pipe without friction then you know P1 (pressure in tank) and P2 (atmospheric pressure at pipe discharge). You also know T1 since that is a given temperature inside the tank. Then you can calculate T2 at the pipe exit since T2 depends on the ratio of P2/P1 and the ratio of specific heats k and there is a formular to find T2 for and isentropic expansion in textbooks. So from the GEE, work W and heat Q are zero and PE is negligible, then the GEE reduces to change in enthalpy:

dH = dKE = V[sup]2[/sup]/2g Where V is velocity in pipe

And by dH = M Cp dT you can now calculate V (using consistent units) where the total dH went into the velocity of the fluid since there was no friction losses. So the General Energy Equation is useful for ideal flow and it also can be used in combination with frictional flow to determine some unknows that would otherwise be unable to be determined.

So for your vent example there is friction in the piping. You have an upstream pressure in the tank and a downstream pressure of atmospheric at the exit of your pipe (assuming the pipe discharges to the atmosphere). The only thing that can make something move in this universe per Newtons laws is a net force on the object. A net force on an object is the only thing that will make it move. Temperature or anything in the energy equation only keeps track of the energy and how it is transferred between U, PV, KE and PE once it starts moving.

So without any flow at time = 0, you have a force of P1*A in the tank acting on the gas in the pipe where A is the crossectional area of the pipe and at the exit of the pipe you have a force of P2*A where P2 is the atmospheric pressure of 0 psig or 14.7 psia. Therefore you have a net force on that fluid in the pipe which causes the fluid to accelerate in accordance with Sum F = M*Acc. Now at T=0 there is a maximum net force on the fluid since you have P1*A acting on one side and P2*A acting on the other side since without velocity there is no friction force. Therefore you have a maximum acceleration of the fluid at T=0. As the fluid accelerates and picks up velocity then the frictional shear forces also increase which act in the same direction of P2*A. So as the velocity increases the net forces on the fluid decrease and so the acceleration decreases until there reaches a point where the velocity reaches a value such that the friction force plus P2*A equals P1*A and there is no more net force and no more acceleration so the fluid is now at steady flow velocity. See attached photo.

And that is how the velocity and flowrate is determined in a pipe. When the velocity reaches a point such that the frictional losses plus P2*A equals the upstream pressure P1*A less the energy required for the KE of the velocity of the fluid. This is the same equation as the Bernoulli equation which you show in your latest post at the top and which can also be derived by the Euler Equation by summing the forces on the fluid streamline.

CONTINUED NEXT POST

 

[Snickster]

RESPONSE TO QUESTION 1 (Part 2 of 2)

So in a real life calculation to determine what size pipe you need for a given flowrate and friction loss (or how much flow you would get with a given pipe size and differential pressure) then you would simply do a frictional pressure drop calculation to determine at what flowrate and velocity will give you the pressure drop in the piping between what pressure you have available between upstream versus what pressure you have downstream. Therefore the flowrate will be such that the pressure downstream plus friction losses will equal your available pressure upstream. That is how to size piping and how you determine the resulting flowrate.

If you assume isothermal flow with friction (this assumes that enough heat enters the pipe from the environment to the pipe to keep the flowing temperature constant which is close enough assumption for most cases) then you get the Isothermal Friction Flow Equation for a gas which assumes a constant temperature and density. This equation looks a lot like the Bernoulli equation you have in the top part of your recent post. See attached photo.

There are other friction flow equations for a gas such as the Fanno Flow equations which consider flow to be completely adiabatic with friction, or Weymouth equation for long pipelines which I believe only considers frictional pressure losses but no losses due to increase in KE (velocity of gas), or Rayliegh Flow calcs which considers frictional flow with heat transfer to/from pipe. Unless you are doing a very precise calculation any of these equations are good to use for piping pressure drop and give answers that are close to each other.

So in regards to your questions in your post:

So, the friction affects the difference between "Internal Energy", "P/rho" and "Enthalpy" terms in the tank and at the outlet of the vent a lot, especially when the friction is relatively high. It decreases the velocity v2, so, there is a smaller need of enthalpy change to account for the velocity creation, right ?

As I described, friction only determines the maximum flowrate and velocity in a pipe based on the balance of forces and how much energy is available for frictional losses and KE. Friction does not affect any of the terms in the GEE directly but since it limits the velocity then will indirectly affect the terms in the GEE. The higher the pipe friction the lower the resulting flow and velocity and therefore the lower the resulting change in enthalpy. If there is a lower velocity then yes the enthalpy has to decrease less to provide the KE for that velocity change.

If I am right until now, my question is that is the General Equation of Energy useful in flow calculations of gases ? Because the existence of friction affected all the terms, especially the outlet velocity. Yet, again, this seems not calculatable by the General Equation of Energy. On the other hand, for liquids, the general equation of energy can directly turn to head form of Bernoulli's equation I guess.

The GEE only keeps track of how energy is converted from one form to another and can be used to predict flow for ideal non-frictional flow. It is also useful in determining unknowns through the use of energy relations. To predict flow and velocity for frictional flow one of the frictional flow equations previously mentioned must be used.

I need to read over your Question 2 and will reply in a separate post.

 

[Snickster]

So my question is, does the General Equation of Energy give anything about the situation to us, in terms of fluid dynamics? For example, if the system in Question 2 was for liquid flow and constant diameter, the pressure loss would be clearly represented in the "P/rho" term since the rho is constant, and the pressure loss term ((P2-P1)/rho) would be equal to U2-U1. And for liquid case, it is totally in agreement with the Bernoulli's Equation and head form of it.

Yes you are correct. For liquid flow you would see the change since density is constant so P1V1 greater than P2V2. However the GEE only keeps track of energy and since H1 = H2 there is no indication in the energy terms that the fluid has undergone a change. And the change with friction is a increase in entropy and and a decrease in the available energy. In other words the pressure drop has created an irreversibel energy loss in the system that has reduced the available energy to do more work. The way you know this is that the stagnation pressure has been reduced although the stagnation temperature and stagnation enthapy is the same, and a reduction of stagnation pressure is an indication of increase in entropy and the loss of available energy. Stagnation conditions are when a flowing fluid comes to a complete stop at zero velocity and all velocity head is converted back to enthalpy. Since enthalpy and temperature don't change in friction flow the stagnation enthalpy and temperature remains same as before or after friction, but since pressure drops with friction the stagnation pressure is reduced. This is basically a throttling process where pressure drop due to friction is same as pressure drop across a throttling valve. This is the same process on a TdS or PV diagram as if you had a pressurized volume in a tank that was attached to another tank with another empty volume and would suddenly open the valve to let the pressurized volume occupy both tanks. The pressure would decrease, but temperature and enthapy would remain constant, and the entropy and stagnation pressure would increase/decrease respectively. See attached paper. Also for more information you can search for "available energy loss in throttling process".

Lastly, again about Question 2, what would happen when the diameter was not increasing, but constant ? It would be that the temperature would decrease again to account for the increase in velocity which stems from the specific volume expansion from 10 bar to 5 bar, right ?

Yes correct. This is called Fanno Flow which considers adiabatic flow in a constant area duct with friction. If you do a search you will find alot of info on Fanno Flow calculations. A good book that I learned compressible flow from and Fanno Flow equations is "Compressible Fluid Flow", by Michel A. Saad.

 

[Snickster]

The HTML file attached above did not print correctly - attached is a PDF.

 

[MumtazDogan]

Thank you a lot Snickster. You are the best. Appreciate for your all efforts.

See you.

 

[Snickster]