Q:Hydrostatic Pressure vs. Energy Conservation Equation

In summary: So if you take k_m, A_m > 0 you shouldn't find Q_b \to 0, unless Q = 0, or in the limit as k_b \to \infty, A_b \to 0...In summary, the conversation discusses two methods for determining ##P2## in order to determine ##h_{LM}##. The first method uses the steady flow assumption and the energy conservation equation, while the second method uses the continuity equation. The conversation also discusses the implications of a zero differential pressure across 1-2 and the contradiction that may arise in assuming no flow in the side branch. It is concluded that the contradiction may be hidden in the problem statement and further analysis is needed to determine the correct
  • #1
tracker890 Source h
90
11
Homework Statement
To determine P2
Relevant Equations
hydrostatic pressure and energy conservation equation.
Please help me to understand which ans is correct.
1671103143471.png
To determine the ##P2##.
$$
h_{LM}\ne 0
$$

Method 1:
$$dP=\frac{\partial P}{\partial x}dx+\frac{\partial P}{\partial y}dy+\frac{\partial P}{\partial z}dz$$$$\phantom{\rule{0ex}{0ex}}\rho \overset\rightharpoonup{a}=-\triangledown p+\rho \overset\rightharpoonup{g}\phantom{\rule{0ex}{0ex}}$$$$\triangledown P=\rho (\overset\rightharpoonup{g}-\overset\rightharpoonup{a})$$$$\because steady flow \therefore \overset\rightharpoonup{a}=0$$$$\therefore \triangledown P=\rho \left(\overset\rightharpoonup{g}\right)$$$$\therefore\triangledown P=\rho\left(\overset\rightharpoonup g\right)=\left\langle0,0,-g\right\rangle=\frac{\partial P}{\partial x}dx+\frac{\partial P}{\partial y}dy+\frac{\partial P}{\partial z}dz$$$$dp=-\rho gdz=-\gamma _wdz$$
$$P2=P1+\gamma _wh...........\left( ans1 \right)$$
$$////////////////////////////////////$$
Mtheod 2:
Energy Conservation Equation.$$\frac{{P}_{1}}{\gamma }+\frac{{{V}_{1}}^{2}}{2g}+{Z}_{1}=\frac{{P}_{2}}{\gamma }+\frac{{{V}_{2}}^{2}}{2g}+{Z}_{2}+{h}_{LM}$$$$
Q=AV_1=AV_2\
$$
$$
\therefore V_1=V_2
$$
$$
\therefore \frac{P_1}{\gamma _w}+Z_1=\frac{P_2}{\gamma _w}+Z_2+h_{LM}
$$
$$
\frac{P_2}{\gamma _w}=\frac{P_1}{\gamma _w}+Z_1-Z_2-h_{LM}
$$
$$
P_2=P_1+\gamma _w\left( Z_1-Z_2 \right) -\gamma _w\left( h_{LM} \right)
$$
$$
\ \ =P_1+\gamma _w\left( h \right) -\gamma _w\left( h_{LM} \right) ................\text{(ans2)}
$$
 

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  • #3
haruspex said:
What is hLM?
energy loss
ref
 
  • #4
If water isn’t flowing in the side branch, the differential pressure across 1-2 is 0. So what does that imply about the head loss given the change in elevation?
 
  • #5
erobz said:
If water isn’t flowing in the side branch, the differential pressure across 1-2 is 0. So what does that imply about the head loss given the change in elevation?
Energy loss is the loss caused by friction in the intermediate pipeline as the fluid flows through the pipe.
 
  • #6
tracker890 Source h said:
Energy loss is the loss caused by friction in the intermediate pipeline as the fluid flows through the pipe.
Yeah, I get that. If there is viscous loss in the pipe, then for steady flow method 2 is correct. What is the question? I suspect you are supposed to determine ##h_{LM}##?

If that is the case, like I said. The differential pressure across 1-2 is 0. That has implications for the head loss from 1-2, given the change in elevation from 1-2.
 
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  • #7
erobz said:
Yeah, I get that. If there is viscous loss in the pipe, then for steady flow method 2 is correct. What is the question? I suspect you are supposed to determine ##h_{LM}##?

If that is the case, like I said. The differential pressure across 1-2 is 0. That has implications for the head loss from 1-2, given the change in elevation from 1-2.
So, if there is frictional loss in the pipe flow, then
##
\rho \overset{\rightharpoonup}{a}=-\triangledown p+\rho \overset{\rightharpoonup}{g}
##
is not true?
 
  • #8
tracker890 Source h said:
So, if there is frictional loss in the pipe flow, then
##
\rho \overset{\rightharpoonup}{a}=-\triangledown p+\rho \overset{\rightharpoonup}{g}
##
is not true?
When you are diving into this kind of analysis, I'm on very thin ice. But, if there is viscous loss, there is shear stress acting on the fluid element.

There is no shear "force term" in what you have presented?
 
  • #9
tracker890 Source h said:
So, if there is frictional loss in the pipe flow, then
##
\rho \overset{\rightharpoonup}{a}=-\triangledown p+\rho \overset{\rightharpoonup}{g}
##
is not true?
this equation omits the viscous friction term.
 
  • #10
erobz said:
If water isn’t flowing in the side branch, the differential pressure across 1-2 is 0.
This is not correct. $$\Delta P=\rho g\Delta z$$
 
  • #11
Chestermiller said:
This is not correct. $$\Delta P=\rho g\Delta z$$
But the following formula is established.
why?
1671112155040.png
 
  • #13
Chestermiller said:
##V_z## is a function of r.
so Vz profile is,
1671112773582.png
 
  • #15
Chestermiller said:
This is not correct. $$\Delta P=\rho g\Delta z$$
We are getting contradiction somehow?

In the main applying COE:

##v_1 = v_2##
##z_2 = 0## elevation datum

$$ \frac{P_2}{\gamma} = \frac{P_1}{\gamma} + z_1 - \sum_{1 \to 2} h_L$$

Where, ##\sum_{1 \to 2} h_L > 0##

Applying COE across the branch:

##v_1 = v_2 = 0##
## \sum_{1 \to 2} h_L = 0 ##

$$ \frac{P_2}{\gamma} = \frac{P_1}{\gamma} + z_1 $$

?

For parallel sections the loss ## h_{main} = h_{branch}##

We also have to satisfy continuity.
 
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  • #16
erobz said:
We are getting contradiction somehow?

In the main applying COE:

##v_1 = v_2##
##z_2 = 0## elevation datum

$$ \frac{P_2}{\gamma} = \frac{P_1}{\gamma} + z_1 - \sum_{1 \to 2} h_L$$

Where, ##\sum_{1 \to 2} h_L > 0##

Applying COE across the branch:

##v_1 = v_2 = 0##
## \sum_{1 \to 2} h_L = 0 ##

$$ \frac{P_2}{\gamma} = \frac{P_1}{\gamma} + z_1 $$

?

For parallel sections the loss ## h_{main} = h_{branch}##

We also have to satisfy continuity.
$$
I\ think\ h_{LM}\ne 0\ is\ not\ correct\ in\ this\ example,\ h_{LM}\ne 0\ must\ be\ changed\ to\ h_{LM}=0.
$$
 
  • #17
My belief is the contradiction is hidden in the assumption of the problem statement, that the flow in the side branch is 0.

If you apply COE (using what I shown above) without invoking the "no flow" condition in the branch the following is the system of equations:

$$ \frac{P_2}{\gamma} = \frac{P_1}{\gamma} + z_1 - \sum_{1 \to 2} h_{L_m}$$

$$ \frac{P_2}{\gamma} = \frac{P_1}{\gamma} + z_1 - \sum_{1 \to 2} h_{L_{b}}$$

That system reduces to the established result:

$$ \sum_{1 \to 2} h_{L_m} = \sum_{1 \to 2} h_{L_b}$$

$$ k_{m} \frac{Q_{m}^2}{A_{m}^2 2g } = k_{b} \frac{Q_{b}^2}{A_{b}^2 2g }$$

Let the total flow into the node be known and equal to ##Q##. It follows that:

$$ k_{m} \frac{(Q - Q_b)^2}{A_{m}^2} = k_{b} \frac{Q_{b}^2}{A_{b}^2}$$

$$ \implies Q^2 - 2 Q Q_b + \left( 1 - \left( \frac{A_m}{A_b} \right)^2 \frac{k_b}{k_m} \right) Q_b^2 = 0 $$

I get the following solution:

$$ Q_b = Q \left( \frac{1+ \frac{A_m}{A_b} \sqrt{ \frac{k_b}{k_m} } }{ 1 - \left( \frac{A_m}{A_b} \right)^2 \frac{k_b}{k_m} } \right)$$

So if you take ##k_m, A_m > 0## you shouldn't find ## Q_b \to 0 ##, unless ##Q = 0##, or in the limit as ## k_b \to \infty, A_b \to 0 ##.
 
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  • #18
For laminar flow, the force balance ;on the fluid between points 1 and 2 reads:
$$P_1\left(\frac{\pi D^2}{4}\right)+\rho\left(\frac{\pi D^2}{4}\Delta z \right)g=P_2\left(\frac{\pi D^2}{4}\right)+\tau_w(\pi D \Delta z) $$where ##\tau_w## is the viscous shear stress at the wall: $$\tau_w=\left(\frac{32Q}{\pi D^3}\right)\eta$$where Q is the downward volumetric flow rate and ##\eta## is the viscosity. What does this give you when you divide both sides of the first equation by ##\left(\frac{\pi D^2}{4}\right)## and substitute the 2nd equation?
 
  • #19
Chestermiller said:
For laminar flow, the force balance ;on the fluid between points 1 and 2 reads:
$$P_1\left(\frac{\pi D^2}{4}\right)+\rho\left(\frac{\pi D^2}{4}\Delta z \right)g=P_2\left(\frac{\pi D^2}{4}\right)+\tau_w(\pi D \Delta z) $$where ##\tau_w## is the viscous shear stress at the wall: $$\tau_w=\left(\frac{32Q}{\pi D^3}\right)\eta$$where Q is the downward volumetric flow rate and ##\eta## is the viscosity. What does this give you when you divide both sides of the first equation by ##\left(\frac{\pi D^2}{4}\right)## and substitute the 2nd equation?
It gives me two different results for ##P_2## depending on which branch I take in going from ##1 \to 2##.

In the OP ##Q_m>0## in the main and in the branch ##Q_b = 0##. I don't care what flow regime we are working in, that is an issue.

The faulty assumption in the problem statement leading to the contradiction is that ##Q_b = 0##.
 
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  • #20
erobz said:
It gives me two different results for ##P_2## depending on which branch I take in going from ##1 \to 2##.

In the OP ##Q_m>0## in the main and in the branch ##Q_b = 0##. I don't care what flow regime we are working in, that is an issue.

The faulty assumption in the problem statement leading to the contradiction is that ##Q_b = 0##.
Yes. That's right. It only works out if the fluid is inviscid.
 
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  • #21
tracker890 Source h said:
I think ##h_{LM}\ne 0## is not correct in this example,##h_{LM}\ne 0## must be changed to ##h_{LM}=0##.
I fixed your Latex a bit. ( don't put back slash unless its introducing a special function like "\sin" or "\frac{}{}" and the delimiters "##" put code inline with text "$$" give it its own line centered justified)

The problem is, if you make the fluid inviscid then then the flow is split evenly between each branch (regardless of branch dimension)?

Inviscid flow and multi branch analysis don't mix.
 
  • #22
erobz said:
I fixed your Latex a bit. ( don't put back slash unless its introducing a special function like "\sin" or "\frac{}{}" and the delimiters "##" put code inline with text "$$" give it its own line centered justified)

The problem is, if you make the fluid inviscid then then the flow is split evenly between each branch (regardless of branch dimension)?

Inviscid flow and multi branch analysis don't mix.
For viscous flow, the flow in the branch is related to the flow in the main channel by $$Q_B=\left(\frac{D_B}{D}\right)^4Q$$So, for a diameter ratio of 0.1, the flow ratio is 0.0001.
 
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Related to Q:Hydrostatic Pressure vs. Energy Conservation Equation

1. What is the hydrostatic pressure equation?

The hydrostatic pressure equation is a fundamental equation in fluid mechanics that relates the pressure at a point in a fluid to the density, acceleration due to gravity, and height of the fluid column above that point. It is given by P = ρgh, where P is the pressure, ρ is the density, g is the acceleration due to gravity, and h is the height of the fluid column.

2. How does hydrostatic pressure affect energy conservation?

Hydrostatic pressure plays a crucial role in the energy conservation equation for fluids. It is a form of potential energy that is stored in the fluid due to its position in a gravitational field. As the fluid moves, this potential energy is converted into kinetic energy. Therefore, any changes in hydrostatic pressure will affect the total energy of the fluid system.

3. What is the relationship between hydrostatic pressure and energy conservation?

The relationship between hydrostatic pressure and energy conservation is that hydrostatic pressure is a form of potential energy that is included in the energy conservation equation for fluids. This means that changes in hydrostatic pressure will affect the total energy of the fluid system, and vice versa.

4. Can the hydrostatic pressure equation be used for all types of fluids?

Yes, the hydrostatic pressure equation can be used for all types of fluids, including liquids and gases. However, it is important to note that this equation assumes the fluid is incompressible and at rest. For compressible fluids or fluids in motion, additional equations and considerations may be necessary.

5. How is hydrostatic pressure measured?

Hydrostatic pressure is typically measured using a device called a manometer, which consists of a U-shaped tube filled with a liquid, such as mercury or water. The height difference between the two sides of the tube is directly related to the hydrostatic pressure of the fluid being measured. Digital pressure sensors can also be used to measure hydrostatic pressure.

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