New answers tagged fluid-mechanics
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I think you are not presenting the fully developed boundary layer.
Because the transition to a turbulent flow occurs sooner for higher velocities you see up to 2 meters a greater boundary layer for double the speed. If you calculated further down - e.g for X>5 -then the boundary layer thickness would have stopped (somewhat) developing and in that case the ...
1
The symbol $H_{\mathsf{C}}$ here doesn't represent a head loss, it represents an "absolute" head at point $\mathsf{C}$. Assuming that the fluid velocity in the region vertically above $\mathsf{C}$ is negligible, so that the pressure variation in that region is hydrostatic, that absolute head is equal to the vertical co-ordinate of the free surface ...
1
Hi sorry I was posting as a guest before and it signed me out so I have to answer instead of comment.
It’s a duct where both ends are open and water is flowing in from converging end. Transitions are smooth and we are trying to find validity of Bernoulli’s equation.
I’ll delete this message after you see it
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Water exerts pressure on all surfaces in contact. The intensity varies, depends on the depth, the properties of the surrounding medium, and the enclosure of the cavity/reservoir. Also, the externally applied pressure, if any.
1
We do not seek to prove whether two values are the same. We seek to determine whether two values differ from each other to a defined level of confidence.
Your simulation has a limit in its precision. This may be set by something as simple as the number of significant digits in a numerical constant that you use during the calculation. For example, setting the ...
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A while ago, there was study done. They had a similar problem like yours. Namely, there was a bar, simply supported at the ends and it was in a flow with a pressure.
They gave the problem with the same parameters to about 15-20 expert researchers and professionals in the field, with the state-of-the-art hardware and software at the time. Arguably that was ...
3
Generally you would work out the precision of your measuring equipment. That'll give you a sensible +/- value either side of your measurement. You may find that your simulated result falls within this band, in which case your simulation agrees.
Also look into 'uncertainty' in measurement - this will provide you with a percentage representing how confident ...
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It's kind of a matter of semantics: if the fluid is still in a "jet" that's undergoing significant broadening or narrowing, and therefore significant acceleration that can match a pressure difference from the far-field "back pressure", then the fluid hasn't really "exited" the (influence of the) nozzle yet, and the "exit&...
1
Yes, you do have enough information. The equations you've seen that require a pressure measurement are the Colebrook-White equation (or its embodiment in the Moody chart) and the Darcy-Weisbach equation (as a definition of Darcy friction factor), right? Applying those equations three times to the situations described in your first paragraph (once to pipe A ...
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The object will flat if sum of the upward forces is more than the sum of downward forces. When the sum of the upward forces equal to the sum of the download forces, the object is in the "limiting equilibrium" condition, instability and floating will follow.
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Since there is relatively little area change and the shapes are both rectangular, I would not worry about the pressure loss that occurs transitioning between the two shapes. I would think the right angle bend assumption would account for more loss than the transition between two similar shapes. You could check the equivalent diameter between the two shapes ...
1
You can't do it with conservation of energy alone: you need another physical principle, namely conservation of momentum in the form of the Navier-Stokes momentum equation (or, if the shear stresses are negligible, the Euler momentum equation). That conservation of momentum equation will give you an expression for the surface traction density at any given ...
2
The energy equation can be simplified to the two equations you've referenced, given that in the first case there is negligible work, in the second case there is negligible heat transfer, and in both cases there is negligible viscous dissipation. For example, consider a heat exchanger with negligible inlet and outlet velocities and no elevation difference ...
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IMHO this is a matter of convention more than anything else. However, I acknowledge that there are other more experienced and knowledgeable people in this forum (e.g. alephzero), that can offer a more scientific (and less based on experience) view on the subject.
My view is that Even if the solid part's constitutive equations are elastic, the interaction ...
2
Continuum mechanics has two branches - solid mechanics, and fluid mechanics.
The subbranches of solid mechanics are Elasticity (linear) and Plasticity (nonlinear); the subbranches of fluid mechanics are Newtonian Fluids (linear) and Non-Newtonian Fluids (nonlinear). Depending on the branch of study, the nonlinear continuum mechanics involves matters with ...
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Although this is probably a gross oversimplification, I regard nonlinear continuum mechanics as a superset of linear continuum mechanics.
To my experience, the basic difference is in the form of constitutive equations. I.e in the case of linear continuum mechanics you have the stress strain relationship
$$\sigma = E \epsilon$$
where E is a constant and ...
3
Could you get it to work, sure. There are dozens of ways to turn heat into energy.
Could you get it to work at scale and cost effectively? Almost certainly not. There's just not enough energy in heated air to do much, and the capital costs of the system break the effort. This is another entry into the "Why can't Stirling engines give us almost free ...
3
First thing first, you need to get hand on the ASME Boiler and Pressure Vessel Code (BPVC Code). Starting division 8, section 1, for safety and quality of your pressure vessel.
After checking on all relevant provisions/standards, you can use the equations to derive the required wall thickness for your application.
Notation
σ_H = hoop stress, psi or MPa
D = ...
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That is certainly the hard way to do it. You could go to a local welding supplier and buy or lease a standard steel pressure bottle / tank ; good for probably 3 X your maximum pressure (designed and built in conformance to ASME Sec. 8 , Div.1 ). Stainless steel is certainly not needed for air or most gasses. You don't need a turbine to get the cooling ...
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The scaling for a simple case is easy to calculate. Assume a spherical pressure vessel of radius $r$ and wall thickness $t$ with internal pressure $P$. The stress in the walls scales as $$\frac{Pr}{t}.$$
So, if you double $r$, for the same quantity of gas you reduce $P$ by a factor of 8 (since the internal volume scales as $r^3$), which reduces $t$ by a ...
4
Taylor is straightforward:
$$
\sqrt{1+2x}
=\left.\sqrt{1+2x}\right|_0
+\left.{d \over dx}\sqrt{1+2x}\right|_0x
+O(x^2) \\
=1
+\left.{d \over dx}{1 \over \sqrt{1+2x}}\right|_0x
+O(x^2) \\
=1
+x
+O(x^2) \\
$$
Note that Taylor holds for matrices variables under some conditions.
ps.If you can handle it, you also have the Generalized Binomial Expansion for ...
0
The specification of a velocity field can help define the flow regimes that can give clues to the underlying physics, but the relevant physics depends greatly on the velocity magnitudes. There are various approaches to sound generation that are dependent upon the flow regimes, and it's impossible to focus on any particular one without knowing the governing ...
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