Does gas flow from low to high pressure?

  • #71
Given the change in height, is there a simple way to characterize it?
53-0=53 mm of water or 519 Pa.

This is a little bit of a brain teaser/not enough information problem. At first glance, the lack of height change of C implies it is atmospheric pressure, but I think the reality is that the pressure is substantially higher than atmospheric. Based on the system shape the largest loss should be at the outlet and the loss through the rest shouldnt be much unless the velocity is high.
[Edit: beaten by @Baluncore ]

The air source is probably lab bench compressed air at 7 ATM or so (common value = 100 psi) throttled down with the valve. The flow is probably high subsonic/compressible and the air is cold. There might be a way to untangle this and figure out the velocities, but I don't see the value in it. It isn't worth the effort to me, especially given how the thread is going.
 
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  • #72
True.
Since the manometers here only show differential pressure, we are missing the absolute pressure of the system, which makes pressure ratios difficult to assess.
Yeah, as @russ_watters mentions, we are going to need to pull a lot of figures out of thin air…quite literally for a computation.

In general if we assume adiabatic process between A and C then this is a constant enthalpy process:

$$ U_A + p_A v_A = U_C+ p_C v_C$$

I believe is the basic idea.
 
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  • #73
I don't see the value in it. It isn't worth the effort to me, especially given how the thread is going.
I am sorry that the thread went downhill, but I am really interested to see if this can be solved using gas laws, calorimetry and hydrostatics rather than fluid dynamics.
 
  • #74
I am sorry that the thread went downhill, but I am really interested to see if this can be solved using gas laws, calorimetry and hydrostatics rather than fluid dynamics.
What exactly do you want to solve? There is just not enough data to do much of anything (as far as I can tell).
 
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  • #75
What exactly do you want to solve? There is just not enough data to do much of anything (as far as I can tell).
I am happy for us to make intelligent estimates for the missing data.
What I am trying to find out is if we can calculate the flow rate in Kg/sec in the system using gas laws, hydrostatics and calorimetry rather than fluid dynamics.
Which numbers would you need?
 
  • #76
I am sorry that the thread went downhill, but I am really interested to see if this can be solved using gas laws, calorimetry and hydrostatics rather than fluid dynamics.
Given that this is a pointlessly bad and difficult problem and you both have a poor grasp of the basics and don't even believe in the real science of it, it is too big of a leap and not productive. Personally I think it's a wild goose chase. If you were serious about learning this stuff you'd pick a basic problem to work through.
 
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  • #77
Which numbers would you need?
We've already said, but how in the world do you expect to be able to pull them out of the air when you have no idea what they could be?

Ok, I'm going to try to focus this thread, because I don't want to be on a wild goose chase down a rabbit hole. You've repeatedly said things like this from the OP and a later post:

Sailor Al said:
Classical Physics states that a gas will only flow from a region of high to one of low pressure...

"Most references claim that a gas only flows from high to low...."

These statements are wrong as stated, as many of us have pointed out, and you've demanded references from us to prove they are wrong. That's backwards: you need to provide references to these statements so we can see what they actually say. Because the reality here is almost certainly that either you're using bad sources or you've misunderstood them/missed context. That's where the thread should have started. We need to fix that before we can do anything else useful here.
 
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  • #78
These statements are wrong as stated, as many of us have pointed out
Yes, and as I stated in posts #23 and #39, I accept that statement was incorrect. I hoped that we had moved on to a more interesting approach to the problem.
Given that this is a pointlessly bad and difficult problem
I am not sure why this is a "pointlessly bad ... problem". It is one that is of great interest to me and, by participating, other forum members, including yourself, appear to have shown some interest.
As to it being a "difficult problem": yes, It is difficult for me, which is why I posed it. I was hoping that some members might be able to suggest a solution. I don't think the difficulty of a problem should bar it from being posed.
We've already said, but how in the world do you expect to be able to pull them out of the air when you have no idea what they could be?
Well, actually, I don't think anyone has yet indicated which data are missing from the problem.
@erobz suggested "we need to pull a lot of figures out of the air", I am simply asking which data would be required for a solution.
You yourself suggested some data about the pressure in the air supply but didn't specify which data would be useful.
@erobz in #69 also requested of the mass flow rate and the inlet temperature. Well, the mass flow rate is actually the answer I am seeking, and I'm open to suggestions about the temperature at the inlet.
@Baluncore in #70 asks for the absolute pressure of the system, which I don't quite understand.

I think it's a problem worth exploring, and look forward to hearing the views of other members.
 
  • #79
These statements are wrong as stated, as many of us have pointed out
Yes, and as I stated in posts #23 and #39, I accept that statement was incorrect. I hoped that we had moved on to a more interesting approach to the problem.
Given that this is a pointlessly bad and difficult problem
I am not sure why this is a "pointlessly bad ... problem". It is one that is of great interest to me and, by participating, other forum members, including yourself, appear to have shown some interest.
As to it being a "difficult problem": yes, It is difficult for me, which is why I posed it. I was hoping that some members might be able to suggest a solution. I don't think the difficulty of a problem should bar it from being posed.
We've already said, but how in the world do you expect to be able to pull them out of the air when you have no idea what they could be?
Well, actually, I don't think anyone has yet indicated which data are missing from the problem.
@erobz suggested "we need to pull a lot of figures out of the air", I am simply asking which data would be required for a solution. You yourself suggested some data about the pressure in the air supply.
 
  • #80
Baluncore in #70 asks for the absolute pressure of the system, which I don't quite understand.

I think it's a problem worth exploring, and look forward to hearing the views of other members.
I think you should explore it by repeating the experiment with three manometers, open to the atmosphere. Then the ratios of the pressures can be known. At present, because the manometers are isolated from the atmosphere, only the differential pressures are known.
 
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  • #81
I think you should explore it by repeating the experiment with three manometers, open to the atmosphere. Then the ratios of the pressures can be known. At present, because the manometers are isolated from the atmosphere, only the differential pressures are known.
Yes, indeed, that would be interesting. Unfortunately we have what we have.
It is the calculation of the differential pressures that pose a most interesting problem. Can hydrostatics resolve it?
Does the fact that the pressure at the base of each column must be equal provide a starting point?
 
  • #82
Yes, indeed, that would be interesting. Unfortunately we have what we have.
And what we now have, is insufficient to proceed.
You need more data. Do the experiment.
 
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  • #83
I am sorry that the thread went downhill,

Good, that's a start. Also you're more specific.

but I am really interested to see if this can be solved using gas laws, calorimetry and hydrostatics rather than fluid dynamics.

This is an interesting remark, since gas laws, to some extent calorimetry and hydrostatics are all part of fluid dynamics. This is to say that all of these equations (calorimetry to some extent) can be derived from the Navier-Stokes equations, the Continuity equation and the energy equation (often they are colloquially referred to as the Navier-Stokes equations, even though originally they only refer to the momentum equations).

[Edit] I made a mistake here. You cannot derive the gas-law from Navier-Stokes. But it is a necessary part (or rather: thermodynamics is a necessary part) when you want to solve the Navier-Stokes equation, so for me it is still an intrinsic part of Fluid Dynamics[/Edit]

But, lets ignore that and look at the equations you are willing to use:

The ideal gas law states ##P = \rho R_{specific} T## (we're going to do the ideal gas assumption here, I'm not willing to make it more complex than necessary). We don't have the pressure ##P## only relative differences, we don't have the density ##\rho## and we don't have the temperature ##T##. You need to assume at least two of the three variables.

A simple and to my best guess: valid assumption is that the temperature can be considered constant everywhere. Yes, kinetic energy is transferred to heat energy through friction, but it will not have a significant effect on the pressure and temperature I think (that's my educated guess based on my experience as fluid dynamics engineer, that's how I make my living...).

The density is another thing I think will hardly change in this experiment. The rule of thumb here is that for velocities lower than Mach 0.3 density can be considered constant. Mach is the speed of sound, Mach 0.3 in air at sea-level conditions is roughly 100m/s or 360 km/h. I think we are well below that... This is actually for external flow, but since the exit of this device is connected to the atmosphere, I would say this is valid, valid enough for our purposes anyway.

So, ##\rho## nor ##T## will change, does that mean the pressure is everywhere equal, since that would then be what the gas equation tells you right? Well no! What I'm saying is that the changes in temperature and density are small enough that it doesn't have a significant influence on the pressure, you will not learn anything new about the experiment by taking those into account.

If you would now insist to be 'more accurate' and take the changes of temperature and density into account anyway, you're not one single step closer to the solution. This is because the pressure differences and velocities are dominated, swamped really, by other effects than temperature and density differences.

This means the gas-equation will not help us. Since calorimetry is just heat transfer, and I've just told you that this is not important, we can ignore that as well.

Lastly hydrostatics. Hydrostatics tells you the pressure in a fluid that is not moving at all. The only way this becomes interesting is if gravity plays a role. In this case gravity does not play a role. Gravity is a potential force, long story short: you can take the Navier-Stokes equation for a single incompressible fluid and take out any potential force and you are still left with the same equation to solve, only the pressure is not an absolute pressure but relative to the pressure at some point. This means gravity has no influence over the dynamical behavior of the flow for this experiment.

So... what would solve this problem? The Bernoulli equation is a good start. But applying that would tell you that the pressure at A and C should be equal (Bernoulli is an energy conservation statement, it assumes no energy is lost), as pointed out by others earlier. Since this is clearly not the case energy is indeed lost. This is due to friction and turbulence. To compute how much friction there is and how much turbulence is generated you need to solve Navier-Stokes somehow. You can do assumptions here as well (you need a 'turbulence model') and maybe even derive some analytic relation, but that's way to complex and mathematically heavy to do on a PF thread (for me anyway).
 
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  • #84
Yes, indeed, that would be interesting. Unfortunately we have what we have.
It is the calculation of the differential pressures that pose a most interesting problem. Can hydrostatics resolve it?
Does the fact that the pressure at the base of each column must be equal provide a starting point?
Let’s take an (exploratory) walk and see where we hit a wall. At this point you can relate the pressures in each chamber to each other using hydrostatics. The base of each tap shares a common pressure. Find the relationships.
 
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  • #85
Yes, and as I stated in posts #23 and #39, I accept that statement was incorrect. I hoped that we had moved on to a more interesting approach to the problem.
No, that's not sufficient/not the problem I described. What it appears you said in #23 and #39 is you believe the existing physics explanation is wrong and you're still searching for the correct explanation. The reality here is almost certainly that you've misunderstood the explanation you read, not that what you read was incorrect. Frankly, I think that was an intentional strawman, but either way, the only way to sort it out is to see what you actually read.

And then, when it comes to problems you should walk before you run. See this link:
https://www.engineeringtoolbox.com/pitot-tubes-d_612.html

Take hd = 100 mm and calculate the velocity. Let us know if you don't understand how (but put an effort into it first).
 
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  • #86
I am working up a solution to the above venturi problem using :
a) The density version of the Ideal Gas Laws: PM = dRT.
where P = Pressure in Atmospheres
M = Molar mass of the air is 28.9
R = 0.821 atm litres/°Kelvin
T = temperature in °K
d = density in grams/litre.
b) The relative hydrostatic pressures in the manometer tubes using the density of water as 1000 g/Litre.
c) The flow rate of the gas φ, in grams/per second, which remains constant through the system.

And some assumptions about the system:
a) The air behaves as an Ideal Gas.
b) There is no heat transfer between the gas and the surrounding glass tubes.
c)The temperature of the gas at chamber A is 0°C or 273°K
d)The inside diameter of the cylindrical chambers A and C is 1.6 cm, giving a cross-section area 2.01 cm^2
e)The inside diameter of cylindrical chamber B is 0.5 cm: cross-section area 0.196 cm^2
f) The ambient air has a temperature of 20°C and pressure of 1033 cm water.
g) the heights of the water columns in the manometer tubes A, B and C are 0, 9.5 and 5.3 cm resp.

I am analysing the density and temperature of the air as it passes through the circular cross-sections of the chambers.
 
  • #87
I am working up a solution to the above venturi problem using :
a) The density version of the Ideal Gas Laws: PM = dRT.
where P = Pressure in Atmospheres
M = Molar mass of the air is 28.9
R = 0.821 atm litres/°Kelvin
T = temperature in °K
d = density in grams/litre.
b) The relative hydrostatic pressures in the manometer tubes using the density of water as 1000 g/Litre.
c) The flow rate of the gas φ, in grams/per second, which remains constant through the system.

And some assumptions about the system:
a) The air behaves as an Ideal Gas.
b) There is no heat transfer between the gas and the surrounding glass tubes.
c)The temperature of the gas at chamber A is 0°C or 273°K
d)The inside diameter of the cylindrical chambers A and C is 1.6 cm, giving a cross-section area 2.01 cm^2
e)The inside diameter of cylindrical chamber B is 0.5 cm: cross-section area 0.196 cm^2
f) The ambient air has a temperature of 20°C and pressure of 1033 cm water.
g) the heights of the water columns in the manometer tubes A, B and C are 0, 9.5 and 5.3 cm resp.

I am analysing the density and temperature of the air as it passes through the circular cross-sections of the chambers.
Specifying the chamber A temp and pressure should be sufficient (those are the variables I need to define to get a numerical solution). My assumption is adiabatic expansion ( ##\gamma = 1.4## )of the gas between chamber A and B. I have a general approach worked out using the first law of thermodynamics... we'll see if they are in agreement once you've completed it.

I'm using Engineering thermodynamic tables to evaluate the enthalpies. Deriving them from first principles is not something I have any experience with.
 
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  • #88
Specifying the chamber A temp and pressure should be sufficient (those are the variables I need to define to get a numerical solution). My assumption is adiabatic expansion of the gas between chamber A and B. I have a general approach worked out using the first law of thermodynamics... we'll see if they are in agreement once you've completed it.

I'm using Engineering thermodynamic tables to evaluate the enthalpies. Deriving them from first principles is not something I have any experience with.
I am deeply confused about the application of enthalpy to gas calculations, as I thought enthalpy was only applicable when there is a state change or chemical reaction. My reading indicates it was introduces in the 19th century when analysing steam engines - where water changes states between liquid and gas.
I look forward to comparing results. Will your results provide temperature and density of the air in the downstream sections of the system?
 
  • #89
I am deeply confused about the application of enthalpy to gas calculations, as I thought enthalpy was only applicable when there is a state change or chemical reaction. My reading indicates it was introduces in the 19th century when analysing steam engines - where water changes states between liquid and gas.
I look forward to comparing results. Will your results provide temperature and density of the air in the downstream sections of the system?
I solved for the mass flow rate as a function of Chamber A pressure and Temperature ##\dot m( P_A,T_A)##, these are the independent variables. The specific enthalpies of the flow in chamber A and B for an ideal gas are fully specified by the temperature in a particular chamber. Which is specified downstream by the adiabatic assumption.

1685231599678.png


$$\dot m = \frac{P_A}{R T_A} A \sqrt{ \frac{ 2 \left( h_B(T_B) - h_A(T_A) \right) }{1- \left( \frac{P_A}{P_A+ \rho_w g ( l_A -l_b)} \right)^{ \frac{2}{ \gamma} } \left( \frac{A}{a} \right)^2}} $$

Where

##T_B= T_A \left( \frac{P_A+ \rho_w g ( l_A -l_b)}{P_A} \right)^{\frac{ \gamma - 1}{ \gamma}}##
 
  • #90
I solved for the mass flow rate as a function of Chamber A pressure and Temperature ##\dot m( P_A,T_A)##, these are the independent variables. The specific enthalpies of the flow in chamber A and B for an ideal gas are fully specified by the temperature in a particular chamber. Which is specified downstream by the adiabatic assumption.

View attachment 327145

$$\dot m = \frac{P_A}{R T_A} A \sqrt{ \frac{ 2 \left( h_B(T_B) - h_A(T_A) \right) }{1- \left( \frac{P_A}{P_A+ \rho_w g ( l_A -l_b)} \right)^{ \frac{2}{ \gamma} } \left( \frac{A}{a} \right)^2}} $$

Where

##T_B= T_A \left( \frac{P_A+ \rho_w g ( l_A -l_b)}{P_A} \right)^{\frac{ \gamma - 1}{ \gamma}}##
But have you answered the problem?
What is the flow rate in g/sec of the air in the system?
I am on my third whiteboard and getting close to an answer.

(And BTW, how do you embed images in your replies? I'm using Chrome on a Mac and the toolbar - for text styles, formatting, link embedding, image pasting etc. -is displayed in red and inactive. )
 
  • #91
But have you answered the problem: What is the flow rate in g/sec of the air in the system?
(And BTW, how do you embed images in your replies? I'm using Chrome on a Mac and the toolbar - for text styles, formatting, link embedding, image pasting etc. -is displayed in red and inactive. )
Yeah...its right there in variable form...Plug in whatever values for ##P_A,T_A## you desire and the mass flow rate can be computed (with the aid of thermodynamic tables for the specific enthalpy of air)? So, No...I did not bother to evaluate anything yet. For starters, you don't have chamber A pressure specified. So I can't.

I make the graphics in PowerPoint, the copy paste them into the body of the reply.
 
  • #92
(And BTW, how do you embed images in your replies? I'm using Chrome on a Mac and the toolbar - for text styles, formatting, link embedding, image pasting etc. -is displayed in red and inactive. )
When you are creating a post, there's a button at the lower left corner - Attach files. Use it to attach whatever image you want to embed in the post. Once the image is attached, a small image will be shown. At the top left, it says Insert... Click on it to choose from Thumbnail or Full image. I normally choose Full image.
 
  • #93
When you are creating a post, there's a button at the lower left corner...
Ah, thanks - that works. I guess the text editing menu is still not available like when stating a thread.
 
  • #94
So, No...I did not bother to evaluate anything yet. For starters, you don't have chamber A pressure specified. So I can't.
I am working to a solution that deduces both the pressure in chamber and the flow rate from the water levels in the manometers. I think, with the assumptions I have provided about the state of the system, (dimensions, temp, pressures etc) that both the flow rate and the pressures in A, B and C can be calculated as well as the the temperatures and densities in the three chambers.
I was interested that I unthinkingly chose φ to symbolise the flow rate as it turns out to be the symbol for flux in electro/magnetic field theory and has very similar properties. I'm currently using the flux density B to quantify the flow rate per area (gm/sec/cm^2) to calculate the pressure in chamber B.
 
  • #95
I am working to a solution that deduces both the pressure in chamber and the flow rate from the water levels in the manometers. I think, with the assumptions I have provided about the state of the system, (dimensions, temp, pressures etc) that both the flow rate and the pressures in A, B and C can be calculated as well as the the temperatures and densities in the three chambers.
I was interested that I unthinkingly chose φ to symbolise the flow rate as it turns out to be the symbol for flux in electro/magnetic field theory and has very similar properties. I'm currently using the flux density B to quantify the flow rate per area (gm/sec/cm^2) to calculate the pressure in chamber B.
Great. Before you post those workings please take a moment to learn how to typeset mathematics in this space See: LaTeX Guide
 
  • #96
Great. Before you post those workings please take a moment to learn how to typeset mathematics in this space See: LaTeX Guide
I have recently become aware of Latex and will certainly do that.
 
  • #97
I think the following statements are a bit wooly and can be tidied up, but I think they are important.
In a venturi, a low pressure occurs in the narrow section when the fluid is either liquid or gas.
With liquid, the pressure change is due to the "trading" of potential energy and kinetic energy by a change in momentum of the fluid. Since liquid is incompressible, i.e its density is constant, and so the mass per unit volume is constant, the momentum change arises from a change in the speed of the fluid.
With gas, the low pressure is due to the "trading" of the components of the gas energy, i.e. PV and nRT, and occurs through a change in density, pressure and temperature . This "trading" can be handled mathematically using the density version of the Ideal Gas Law : PM = dRT .
That's the theory I'm using to work up the answer.
 
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  • #98
I am working up a solution to the above venturi problem using :
a) The density version of the Ideal Gas Laws: PM = dRT.
where P = Pressure in Atmospheres
M = Molar mass of the air is 28.9
R = 0.821 atm litres/°Kelvin
T = temperature in °K
d = density in grams/litre.
Dammit! I have been confused by pressure units.
R, the gas constant is not 0.821, but 0.0821 atm litres/°K
No wonder my numbers don't make sense.
At STP (273°K, 1 atm) the density of air, d, is 1.29 grams/litre
So ##PM = 1 * 28.9 = 28.9##
and ##dRT = 1.29 * 0.0821 * 273 = 28.9##
So PM = dRT finally balances for atmospheric air at STP.
 
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  • #99
I guess the text editing menu is still not available like when stating a thread.
The text-editing menu is not available when the BB code mode is off. This control is a pair of brackets [ ] almost all the way to the right in the menu bar at the top of the input pane. This control is a toggle.
 
  • #100
1685281539812.png


I'll explain my result while you are working it out.

Applying hydrostatics in the manometers we have that (neglecting the weight of the gas):

$$ P_{ref} = P_A + \rho_w g l_A = P_B + \rho_w g l_B $$

$$ \implies P_B = P_A + \rho_w g \left( l_A - l_B\right) \tag{1} $$

This relates the pressures in chamber A to chamber B, put that aside for now.

Next, apply the First Law of Thermodynamics to the control volume(flow properties assumed uniform over section A and B):

$$ \dot E_{cv} = \dot Q_{cv} - \dot W_{cv} + \dot m_A \left( u_A + \frac{P_A}{ \rho_A } + \frac{ V^2_A }{2} + gz_A \right) - \dot m_B \left( u_B + \frac{P_B}{ \rho_B } + \frac{ V^2_B }{2} + gz_B \right)$$

Assumptions:
  1. Ideal Gas
  2. Polytropic Process ##PV^{\gamma} = \text{constant}## where for Air ##\gamma = 1.4##
  3. Steady State ##\dot E = 0## and ## \dot m_A=\dot m_B = \dot m##
  4. The gas in passing from chamber A to B expands adiabatically (without heat transfer with the surroundings) ##\dot Q_{cv} = 0##
  5. There is no work the gas is doing on the surroundings nor are the surroundings doing work on the gas in the control volume ## \dot W_{cv} = 0 ##
  6. The terms ## u + \frac{P}{ \rho }## are combined for convenience into the term referred to as the enthalpy ##h##
  7. Change in gravitational potential ##\Delta PE = g ( z_A - z_B) = 0 ##

Applying the assumptions to the 1st Law we are left with:

$$ 0 = h_A - h_B + \frac{V^2_A - V^2_B}{2} \tag{2}$$

The enthalpy terms for an ideal gas are purely functions of temperature and are to be evaluated by engineering thermodynamic tables.

Otherwise, they are defined as:

$$\Delta h = \int c_p dT$$

Where knowledge of how ##c_p## varies with temperature is required to evaluate the integral for a particular gas.

In order to evaluate the change in enthalpy, we are using assumption 2 for the adiabatic expansion of an ideal gas:

$$ T_B = T_A \left( \frac{P_B}{P_A} \right)^{ \frac{ 1- \gamma}{ \gamma } } \tag{3}$$

Subbing 1 into 3 for ##P_B##, we see that the temperature of the flow in chamber B is fully constrained by the initial state ##P_A,T_A## under the adiabatic assumption.

$$ T_B = T_A \left( \frac{ P_A + \rho_w g \left( l_A - l_B\right)}{P_A} \right)^{ \frac{ 1- \gamma}{ \gamma } } \tag{3'}$$

Using this result we will say that:

$$ h_B( T_B) - h_A(T_A)= \frac{V^2_A - V^2_B}{2} \tag{2'}$$

Next the velocity of B as a function of A via continuity (conservation of mass):

$$ \dot m = \rho_A A V_A = \rho_B a V_B $$

Where ##A## and ##a## refer to the cross sectional areas of section A and B respectively.

$$ \implies V_B = \frac{\rho_A}{\rho_B}\frac{A}{a} V_A \tag{4} $$

Subbing ##4## into ##2'##:

$$ h_B( T_B) - h_A(T_A)= \frac{V^2_A - \left( \frac{\rho_A}{\rho_B}\frac{A}{a} \right)^2 V_A^2}{2} \tag{5}$$

$$ \implies h_B( T_B) - h_A(T_A) = \frac{V^2_A}{2} \left( 1 - \left( \frac{\rho_A}{\rho_B} \right)^2 \left( \frac{A}{a} \right)^2 \right) \tag{5} $$

Next we would like to eliminate the density ratio by using the Ideal Gas Law:

$$ \frac{P_A}{P_B} = \frac{ \rho_A}{\rho_B} \frac{T_A}{T_B} $$

$$ \implies \frac{\rho_A}{\rho_B} = \frac{P_A}{P_B} \frac{T_B}{T_A} \tag{6} $$

Plugging ##6## into ##5##:

$$ \implies h_B( T_B) - h_A(T_A) = \frac{V^2_A}{2} \left( 1 - \left( \frac{P_A}{P_B} \frac{T_B}{T_A}\right)^2 \left( \frac{A}{a} \right)^2 \right) \tag{7} $$

Using the adiabatic assumption we can eliminate ## \frac{T_B}{T_A}## in ##7## with ##3## :

$$ \frac{P_A}{P_B} \frac{T_B}{T_A} = \frac{P_A}{P_B} \left( \frac{P_B}{P_A} \right)^{ \frac{ 1- \gamma}{ \gamma } } = \frac{P_B}{P_A}^{(-1)} \left( \frac{P_B}{P_A} \right)^{ \frac{ 1- \gamma}{ \gamma } } = \left( \frac{P_A}{P_B} \right)^{\frac{1}{\gamma}}$$

Subbing that into ##7##:

$$ \implies h_B(T_B) - h_A(T_A) = \frac{V^2_A}{2} \left( 1 - \left( \frac{P_A}{P_B} \right)^{ \frac{2}{\gamma}} \left( \frac{A}{a} \right)^2 \right) \tag{8} $$

##8## can now be rearranged to solve for ##V_A##:

$$ V_A = \sqrt{ \frac{ 2 \left( h_B(T_B) - h_A(T_A) \right) }{ 1 - \left( \frac{P_A}{P_B} \right)^{ \frac{2}{\gamma}} \left( \frac{A}{a} \right)^2 } } \tag{9}$$

Finally the mass flowrate is given by:

$$ \dot m = \rho_A A V_A = \frac{P_A}{R T_A} A \sqrt{ \frac{ 2 \left( h_B(T_B) - h_A(T_A) \right) }{ 1 - \left( \frac{P_A}{P_B} \right)^{ \frac{2}{\gamma}} \left( \frac{A}{a} \right)^2 } } \tag{10}$$

Where ##T_B## is given by:

## T_B = T_A \left( \frac{P_B}{P_A} \right)^{ \frac{ 1- \gamma}{ \gamma } }##

and ##P_B## by:

##P_B = P_A + \rho_w g \left( l_A - l_B\right)##

and the enthalpies ##h(T_B), h(T_A)## evaluated by tables commonly found in Thermodynamic texts.

Thats my best crack at it. Hope it helps.
 
Last edited:
  • #101
The text-editing menu is not available when the BB code mode is off. This control is a pair of brackets [ ] almost all the way to the right in the menu bar at the top of the input pane. This control is a toggle.
Aha! Now I see it!
Maybe a couple pf paras in the INFO section, alerting relative newbies such as me to the fact that this forum implements some pretty fancy features, and a Dummies guide to some of them. Till today, despite 40+ years of S/W development and extensive experience in web app development, I hadn't come across "BB code". Just a thought. :wink:
 
  • #102
Maybe a couple pf paras in the INFO section, alerting relative newbies such as me to the fact that this forum implements some pretty fancy features, and a Dummies guide to some of them.
Check out the Help menu item under INFO. The Help menu item links to tutorials on LaTeX (near the top) and to the BBCode tags that are supported (down quite a ways in the list).
 
  • #103
Thats my best crack at it. Hope it helps
I see a huge amount of working, but I still don't see an answer to the stated problem: to determine the mass flow rate through the system.
I still don't have an answer, but am still working on it.
I would expect the answer to me in the order of 10-2 or 10-3 grams per sec.
 
  • #104
I see a huge amount of working, but I still don't see an answer to the stated problem: to determine the mass flow rate through the system.
I still don't have an answer, but am still working on it.
I would expect the answer to me in the order of 10-2 or 10-3 grams per sec.
Give me a pressure and temperature of the gas in chamber A and you will have an answer…if you don’t accept the 1st law of thermodynamics, just say so and I’ll stop wasting my time.
 
  • #105
Give me a pressure and temperature of the gas in chamber A and you will have an answer…if you don’t accept the 1st law of thermodynamics, just say so and I’ll stop wasting my time.
I have specified the temperature in chamber A is 0°C 273 °K.
The pressure can be determined using hydrostatics from the heights of the water columns provided i the problem statement in post #65:
By deducing the scale of the image, I have done some rough estimates of the manometer pipe diameter and thus the heights of the water and wonder how we could calculate the flow rate in Kg/sec of the gas through the system?
Estimate of the diameter of the manometer tubes: 8 mm
Scale: say 1:1,
Measured heights above red line:
A = 0 mm
B = 97 mm - allowance for narrowing at the top, say 95 mm
C = 53mm
95 + 53 = 148 /3 = 47
1 Atm = 10332 mm = 101,325 Pa (x 9.8)
Each mm is 9.8 Pa

It seems obvious that the flow rate in each section has to be the same.
I think we'll need to take into account the temperature of the gas as it passes through the system, and make a guess at the ambient air conditions (20°C, 101.3kPa?) Also something about the diameters of the various chambers through which the gas flows.
My measurements of the water heights in A,B and C are:
S mm
A 0
B 95
C 53
I think the first step is to work out the pressure in A, B and C, but as they are mutually dependent, I'm not sure where to start. I think the heights would be different if the manometer tubes were independent of each other.
and #86
And some assumptions about the system:
a) The air behaves as an Ideal Gas.
b) There is no heat transfer between the gas and the surrounding glass tubes.
c)The temperature of the gas at chamber A is 0°C or 273°K
d)The inside diameter of the cylindrical chambers A and C is 1.6 cm, giving a cross-section area 2.01 cm^2
e)The inside diameter of cylindrical chamber B is 0.5 cm: cross-section area 0.196 cm^2
f) The ambient air has a temperature of 20°C and pressure of 1033 cm water.
g) the heights of the water columns in the manometer tubes A, B and C are 0, 9.5 and 5.3 cm resp.

Absolutely no problem with the first law of thermodynamics: energy cannot be created or destroyed.
 

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