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# Get help with fluid mechanics questions

Recent questions in Fluid Mechanics

### Matching Bernoulli equation and Hagen Poiseuille law for viscous fluid motionI thought that Bernoulli equation could be used only in the case of non viscous fluid. But doing exercises on 2500 Solved Problems In Fluid Mechanics and Hydraulics (Schaum's Solved Problems Series) I found that this procedure is followed.In the case of viscous laminar flow, Bernoulli equation is written as$\begin{array}{}\text{(1)}& {z}_{1}+\frac{{v}_{1}^{2}}{2g}+\frac{{p}_{1}}{\rho g}={z}_{2}+\frac{{v}_{2}^{2}}{2g}+\frac{{p}_{2}}{\rho g}+{h}_{L}\end{array}$Where ${h}_{L}$ is the head loss (due to viscosity) calculated using Hagen Poiseuille law.$\begin{array}{}\text{(2)}& {h}_{L}=\rho g\frac{8\eta L\overline{v}}{{R}^{2}}\end{array}$Is this a correct way to solve exercises involving viscosity?Furthermore are there limitation to this use of Bernoulli equation (in case of viscosity)?In particularIf the flow is not laminar, I cannot use $\left(2\right)$, but can I still write $\left(1\right)$ in that way?Is $\left(1\right)$ valid only along the singular streamline or between different ones (assuming the fluid irrotational)?

Amappyaccon22j7e 2022-05-09

### Deriving the scaling law for the Reynolds numberI'm trying to derive a scaling law for the Reynolds number, to get a better understanding of how it changes for microsystem applications, but I'm getting stuck. From textbook tables I should end up with ${l}^{2}$, but can't get there. The goal is to find a simplistic relation of Reynolds number and the system dimension.This is my reasoning so far:Considering a tube$Re=\frac{\rho v2R}{\mu }$$v=\frac{Q}{A}=\frac{Q}{\pi {R}^{2}}$From Hagen-Poiseuille:$Q=\frac{\delta P\pi {R}^{4}}{8\mu L},$therefore:$v=\frac{\frac{\delta P\pi {R}^{4}}{8\mu L}}{\pi {R}^{2}}=\frac{\delta P\pi {R}^{4}}{8\mu L\pi {R}^{2}}=\frac{\delta P{R}^{2}}{8\mu L}$Replacing $v$ on $Re$$Re=\frac{\frac{\rho \delta P{R}^{2}2R}{8\mu L}}{\mu }=\frac{\rho \delta P{R}^{3}2}{8{\mu }^{2}L}$From this, if I consider the $\delta P$ as constant with the size scaling, $\rho$ and $\mu$ are material properties that are constant at different scales, I get:$Re\sim \frac{{R}^{3}}{L}$I can only think that I could consider it as $\frac{{l}^{3}}{l}={l}^{2}$ but doesn't seem right.What am I missing here?

tuehanhyd8ml 2022-05-09

### Intuitive explanation of Poiseuille's law -- why ${r}^{4}$?According to Poiseuille's law, the effective resistance of a tube is inversely proportional to the fourth power of its radius (as given by the following equation).$R=\frac{8\eta \mathrm{\Delta }x}{\pi {r}^{4}}$I can go through the derivation of the law but is there an intuitive explanation for why it is the fourth power of radius, or a thought experiment that might make it more obvious?

Aedan Gonzales 2022-05-08

### Steady isothermal flow of an ideal gasSo I have a steady isothermal flow of an ideal gas through a smooth duct (no frictional losses) and need to compute the mass flow rate (per unit area) as a function of pressures at any two different arbitrary points, say 1 and 2. I have the following momentum equation in differential form:$\rho vdv+dP=0$where $v$ is the gas the flow velocity and $P$ is static pressure. The mass flow rate per unit cross section $G$, can be calculated by integrating this equation between points 1 and 2. This is where it gets confusing. I do the integration by two ways:1) Use the ideal gas equation $P=\rho RT$ right away and restructure the momentum equation:$vdv+RT\frac{dP}{P}=0$, integrate it between points 1 and 2 and arrive at:${v}_{1}^{2}-{v}_{2}^{2}+2RTln\frac{{P}_{1}}{{P}_{2}}=0$Since the flow is steady, I can write $G={\rho }_{1}{v}_{1}={\rho }_{2}{v}_{2}$, again use the ideal gas law to write density in terms of gas pressure and finally arrive at the mass flow rate expression:${G}^{2}=\frac{2ln\frac{{P}_{2}}{{P}_{1}}}{RT\left(\frac{1}{{P}_{1}^{2}}-\frac{1}{{P}_{2}^{2}}\right)}$2) In another way of integrating (which is mathematically correct), I start by multiplying the original momentum equation by $\rho$ to get${\rho }^{2}vdv+\frac{1}{RT}PdP=0$write ${\rho }^{2}v={G}^{2}/v$, integrate between points 1 and 2 to arrive at${G}^{2}ln\frac{{v}_{2}}{{v}_{1}}=\frac{{P}_{1}^{2}-{P}_{2}^{2}}{2RT}$Using the ideal gas law the velocity ratio can be written as the pressure ratio to finally arrive at the mass flow rate equation${G}^{2}=\frac{{P}_{1}^{2}-{P}_{2}^{2}}{2RTln\frac{{P}_{1}}{{P}_{2}}}$Both the expressions are dimensionaly sound and I know that the second expression is the correct one. My question is, whats wrong with first expression.

Azzalictpdv 2022-05-08

### What determines the dominant pressure-flow relationship for a gas across a flow restriction?If one measures the pressure drop across any gas flow restriction you can generally fit the relationship to$\mathrm{\Delta }P={K}_{2}{Q}^{2}+{K}_{1}Q$where $\mathrm{\Delta }P$ is the pressure drop and $Q$ is the volumetric flowand what I've observed is that if the restriction is orifice-like, ${K}_{2}>>{K}_{1}$ and if the restriction is somewhat more of a complex, tortuous path, ${K}_{1}>>{K}_{2}$ and ${K}_{2}$ tends towards zero.I get that the Bernoulli equation will dominate when velocities are large and so the square relationship component. But what's determining the ${K}_{1}$ component behavior? Is this due to viscosity effrects becoming dominant? Does the Pouiselle relationship become dominant?

Yaritza Oneill 2022-05-08

### Pneumatic flow formulaI'm looking for a pneumatic formula in order to have the flow. I found many formula, but only for hydraulic :$\mathrm{\Delta }P={R}_{p}Q$ where ${R}_{p}$ is the resistance, $Q$ the flow, and $\mathrm{\Delta }P$ the pressure potential:${R}_{p}=\frac{8\eta L}{\pi {R}^{4}}$with $\eta =$ dynamic viscosity of the liquid ; $L=$ Length of the pipe ; $R=$ radius of the pipeI don't know if I can use them, since I'm in pneumatic ! After doing research on the internet, I found some other variables like : sonic conductance, critical pressure coefficient, but no formula...I think that I have all the information in order the calculate the flow : Pipe length = 1m ; Pipe diameter = 10mm ; $\mathrm{\Delta }P=$ 2 barsThanks !

Marissa Singh 2022-05-08

format: plain


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# RNN not matching expected output

I am trying to build an RNN to predict a time-series signal based on knowledge of 3 others that I believe to be related to the output. I am using the rnn Lua Torch package, but I am quite new to the area and the platform. I admit to be struggling a bit since much of the documentation for Torch rnn appears to be for modules (such as recurrent) that have been deprecated and removed from the current version (checked out latest commit 83a5f17 dated last change Aug 23, 2017 in https://github.com/torch/rnn), and many of the examples in the repo seem too complex for my needs.

What I currently have is the following:

opt = {
rho = 10,-- maximum number of time steps for back-propagate through time (BPTT)
inputSize = 3,
outputSize = 1,
hiddenSize = 10,-- number of hidden units used at output of the recurrent layer
batchSize = 10,-- number of training samples per batch
nIterations = 2000,-- max number of training iterations
learningRate = 0.001-- learning rate
}
local rm = nn.Sequential()

rnn = nn.Sequencer(rm)
criterion = nn.SequencerCriterion(nn.MSECriterion())

And the batch training follows the rnn module examples:

minErr = 100000 -- report min error
minK = 0
avgErrs = torch.Tensor(opt.nIterations):fill(0)
for k = 1, opt.nIterations do
---- 1. create a sequence of rho time-steps-----
local inputs, targets = {}, {}
for step = 1, opt.rho do
-- batch of inputs
inputs[step] = inputs[step] or sequence.new()
inputs[step]:index(sequence, 1, offsets[step])
-- batch of targets
offsets:add(1) -- increase indices by 1
offsets[offsets:gt(opt.dataSize)] = 1
targets[step] = targets[step] or output.new()
targets[step]:index(output, 1, offsets[step])
end

---- 2. forward sequence through rnn-------------
local outputs = rnn:forward(inputs)
local err = criterion:forward(outputs, targets)

if k % 100 == 0 then
print('Iter: ' .. k .. '   Err: ' .. err)
end

avgErrs[k] = err
if avgErrs[k] < minErr then
minErr = avgErrs[k]
minK = k
end

---- 3. backward sequence through rnn -------------

----------------------------------------------------

rnn:updateParameters(opt.learningRate)
----------------------------------------------------
end

The plot below shows the thee input channels at the top, followed by the predicted output from the RNN, and the expected output at the bottom. FOr the sake of simplicity training and testing have used the data dataset, though I get similar results when I use disjoint datasets. As you can see, the expected output is quite a simple sawtooth-wave, and I intuitively expected the important phase details to emerge from the three input values and readily determine the right structure of the output, but actually the predicted output is much more like a square-wave signal.

I have tried varying the number of hidden nodes, adding a second RNN layer, ramping up the number of iterations, and varying the batch length and number of back-propagation steps, all to little avail. I wonder whether I have simply designed the RNN incorrectly. Are there some obvious things that I should try to improve things?

• rnn:forget() needs to be called at the end of each training iteration to clear out the previous state.


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# Seminarios

## Javiera Barrera (UAI)

### Título:

Sharp Bounds for the Reliability of a k-out-of-n Sys- tem Under Dependent Failures Using Cutoff Phenomenon Techniques

### Abstract:

In this work we consider the reliability of a network where link failures are correlated. We define the reliability as the probability of the network to be working at a given time instant. Our main contribution is a collection of results giving a detailed analysis of a non-trivial scaling regime for the probability of the network being working at a certain time, as the time and size of network scales. Here we consider that the network fails when there are no links working, or more generally when less than k-out-of-n edges are working (with k close to n) like in [2] and [5]. Our results allow to study the common-cause failure models describe in [3] on networks in a realistic, relevant, yet practical, fashion: it allows to capture correlated components in the network; it allows to estimate and give error bounds for the failure probabilities of the system; and at same time only needs to specify a reduced family of parameters. Moreover, our results for the k-out-of-n failure model allow to give new scaling regimes for the probabilistic behavior of the last-ordinals in the theory of extreme values for dependent tuples. The techniques are similar to those used to estimate the asymptotic convergence profile for ergodic Markov chains [1] or [4].

1. [1]  J. Barrera and B. Ycart Bounds for Left and Right Window Cutoffs. ALEA, Lat. Am. J. Probab. Math. Stat., 11 (2): 445–458, 2014.

2. [2]  I. Bayramoglu and M. Ozkut. The Reliability of Coherent Systems Subjected to Marshall?Olkin Type Shocks. EEE Transactions on Reliability,, 64 (1): 435–443, 2015.

3. [3]  U. Cherubini, F. Durante, and S. Mulinacci. Marshall – Olkin Distributions-Advances in Theory and Applications: Bologna, Italy, October 2013, volume 141. Springer, 2015.

4. [4]  B. Lachaud and B. Ycart Convergence Times for Parallel Markov Chains. Positive systems, 169–176, 2006.

5. [5]  T. Yuge, M. Maruyama, and S. Yanagi Reliability of a k-out-of-n Systemwith Common-Cause Failures Using Multivariate Exponential Distribution. Procedia Computer Sci- ence, 96: 968?976, 2016.

may / 2018 15

## Rangel Baldasso (Bar Ilan University)

### Título:

Noise sensitivity for Voronoi Percolation

### Abstract:

Noise sensitivity is a concept that measures if the outcome of Boolean function can be predicted when one is given its value for a perturbation of the configuration. A sequence of functions is noise sensitive when this is asymptotically not possible. A non-trivial example of a sequence that is noise sensitive is the crossing functions in critical two-dimensional Bernoulli bond percolation. In this setting, noise sensitivity can be understood via the study of randomized algorithms. Together with a discretization argument, these techniques can be extended to the continuum setting. In this talk, we prove noise sensitivity for critical Voronoi percolation in dimension two, and derive some consequences of it.
Based on a joint work with D. Ahlberg.
may / 2018 03

## Jean-Dominique Deuschel (T.U. Berlin )

### Título:

Harnack inequality for degenerate balanced random walks.

### Abstract:

We consider an i.i.d. balanced environment  $\omega(x,e)=\omega(x,-e)$,genuinely d dimensional on the lattice and show that there exist a
positive constant $C$ and a random radius $R(\omega)$ with streched
exponential tail such that every non negative
$\omega$ harmonic function $u$ on the ball  $B_{2r}$ of radius
$2r>R(\omega)$,
we have $\max_{B_r} u <= C \min_{B_r} u$. Our proof relies on a quantitative quenched invariance principle for the corresponding random walk in  balanced random environment and a careful analysis of the directed percolation cluster. This result extends Martins Barlow’s Harnack’s inequality for i.i.d. bond percolation to the directed case.

This is joint work with N.Berger  M. Cohen and X. Guo.

Sala 3, Facultad de Matemáticas, PUC

abr / 2018 02

## Luis Fredes (Bordeaux)

### Título:

“Invariant measures of discrete interacting particle systems: Algebraic aspects”

### Abstract:

We consider a continuous time particle system on a graph L being either Z,  Z_n, a segment {1,…, n}, or Z^d, with state space Ek={0,…,k-1} for some k belonging to {infinity, 2, 3, …}. We also assume that the Markovian evolution is driven by some translation invariant local dynamics with bounded width dependence, encoded by a rate matrix T. These are standard settings, satisfied by many studied particle systems. We provide some sufficient and/or necessary conditions on the matrix T, so that this Markov process admits some simple invariant distribution, as a product measure, as the distribution of a Markov process indexed by Z or {1,…, n} (if L=Z or {1,…,n}), or as a Gibbs measure (if L=Z_n). These results are mainly obtained with some manipulations of finite words, with alphabet Ek, representing subconfigurations of the systems. For the case L=Z, we give a procedure to find the set of invariant i.i.d and Markov measures.

Sala John Von Neumann, 7mo piso, CMM, 15:00 hrs.
ene / 2018 09

## Nikolas Tapia, Universidad de Chile

### Título:

Construction of geometric rough paths

### Abstract:

This talk is based on a joint work in progress with L. Zambotti (UPMC). First, I will give a brief introduction to the theory of rough paths focusing on the case of Hölder regularity between 1/3 and 1/2. After this, I will address the basic problem of construction of a geometric rough path over a given ɑ-Hölder path in a finite-dimensional vector space. Although this problem was already solved by Lyons and Victoir in 2007, their method relies on the axiom of choice and thus is not explicit; in exchange the proof is simpler. In an upcoming paper, we provide an explicit construction clarifying the connection between rough paths theory and free (nilpotent) Lie algebras. In particular, we use an explicit form of the Baker–Campbell–Hausdorff formula due to Loday in order to provide explicit expressions and bounds to achieve such a construction.

Sala de seminarios del 5to piso, CMM, 16:00 hrs.

nov / 2017 27
12345...10...
##### Departamento de Matemáticas

Pontificia Universidad Católica de Chile (PUC-Chile)

Av. Vicuña Mackenna 4860, Macul,

Santiago – Chile

(+56 2) 2354 5779

##### Centro de Modelamiento Matemático (CMM)

Facultad de Ciencias Físicas y Matemáticas (FCFM)


text:
# Tag Info

After a bit of searching in Aladin. I found that the other star is V* V2170 Ori. By comparing their respective Simbad pages one can see the following: LL Ori is actually slightly further away. V2170 Ori has a GAIA parallax of 2.84 while LL Ori has one of 2.55. Simply using $D=1/p$, this translates to a distance of 0.35 kpc and 0.39 kpc LL Ori and V2170 Ori ...


text:
# Klein-Gordon Equation

118articles on
the Psi Epsilon Wikia
Quantum Field Theory
... no spooky action at a distance (Einstein)
Early Results
Relativistic Quantum Mechanics Klein-Gordon Equation
Dirac Equation
The Dawn of QFT Spinors
Spin
Feynman Slash Notation
Antimatter
Klein-Gordon Field
Dirac Field
Renormalisation
Grassman Variable
Conformal Field Theory
Countdown to the Standard Model
From a framework to a model Yang-Mills Theory
Quantum Electrodynamics
Quantum Chromodynamics
Electroweak Theory
Higgs Mechanism
Standard Model
Semi-Classical Gravity and the Dark Age Hawking Radiation
Chandrashekhar Limit
Inflation
Problems with the Standard Model
Outlook
Beyond the Standard Model Beyond the Standard Model
Quantum Gravity
Theory of Everything
Related
Related De Donder-Weyl Theory

The Klein-Gordon Equation is a relativistic generalisation of the Schrordinger Equation to Relativistic Quantum Mechanics. In Quantum Field Theory, it is a relativistic field equation for a spin-0 field (Klein-Gordon Field).

## HistoryEdit this section

This equation was actually discovered by Erwin Schrodinger before he discovered the Schrodinger Equation, that bears his name. However, for reasons which we will be discussing later, he soon discarded the equation and decided to use the non-relativistic energy instead, which resulted in the intrinsically not Lorentz-Invariant Schrodinger Equation.  ;

This equation was named after Oskar Klein and Walter Gordon, who proposed (1926) that it describes relativistic electrons. Vladimir Fock, Johann Kudar, Theophile de Donder and Frans H van den Dungen, and Louis de Broglie also made these claims.

We now know that the Dirac Equation is the actual field equation for these spin-1/2 electrons. The Klein-Gordon Equation however, does describe spin-0 particles, such as the Higgs Field/ .

## StatementEdit this section

 Klein-Gordon Equation $\left(\Box+\mu^2\right)\Psi=0$
Derivation of Klein-Gordon Equation from Schrodinger's Wave Equation

In the Schrodinger formulation of Quantum Mechanics, quantisation is as follows:

$E=\frac{m}{2}{{\left( \frac{\text{d}x}{\text{d}t} \right)}^{2}}+U=\frac{{{p}^{2}}}{2m}+U$

Then, quantise as follows:

\begin{align} & E\Psi =i\hbar \frac{\partial \Psi }{\partial t} \\ & \left( \frac{{{p}^{2}}}{2m}+U \right)\Psi =i\hbar \frac{\partial \Psi }{\partial t} \\ \end{align}

$E=\sqrt{{{p}^{2}}c_{0}^{2}+m_{0}^{2}c_{0}^{4}}+U$

\begin{align} & E\Psi =i\hbar \frac{\partial \Psi }{\partial t} \\ & \left( \sqrt{{{p}^{2}}c_{0}^{2}+m_{0}^{2}c_{0}^{4}}+U \right)\Psi =i\hbar \frac{\partial \Psi }{\partial t} \\ \end{align}

$\left( \sqrt{m_{0}^{2}c_{0}^{4}-{{\hbar }^{2}}c_{0}^{2}{{\nabla }^{2}}}+U \right)\Psi =i\hbar \frac{\partial \Psi }{\partial t}$

The Laplacian cannot be calculated under the square root, so instead, we use

${{E}^{2}}={{p}^{2}}c_{0}^{2}+m_{0}^{2}c_{0}^{4}+2U\sqrt{{{p}^{2}}c_{0}^{2}+m_{0}^{2}c_{0}^{4}}+{{U}^{2}}$

We realise that this cannot be properly calculated (the same problems as before, the square root), so, at least for now, we let $U = 0$, i.e. only free particles.

${{E}^{2}}={{p}^{2}}c_{0}^{2}+m_{0}^{2}c_{0}^{4}$

\begin{align} & {{E}^{2}}\Psi =-{{\hbar }^{2}}\frac{{{\partial }^{2}}\Psi }{\partial {{t}^{2}}} \\ & \left( {{p}^{2}}c_{0}^{2}+m_{0}^{2}c_{0}^{4} \right)\Psi =-{{\hbar }^{2}}\frac{{{\partial }^{2}}\Psi }{\partial {{t}^{2}}} \\ & -{{\hbar }^{2}}c_{0}^{2}{{\nabla }^{2}}\Psi +m_{0}^{2}c_{0}^{4}\Psi =-{{\hbar }^{2}}\frac{{{\partial }^{2}}\Psi }{\partial {{t}^{2}}} \\ & {{\hbar }^{2}}\frac{{{\partial }^{2}}\Psi }{\partial {{t}^{2}}}-{{\hbar }^{2}}c_{0}^{2}{{\nabla }^{2}}\Psi +m_{0}^{2}c_{0}^{4}\Psi =0 \\ & {{\hbar }^{2}}c_{0}^{2}\left( \frac{1}{c_{0}^{2}}\frac{{{\partial }^{2}}\Psi }{\partial {{t}^{2}}}-{{\nabla }^{2}}\Psi \right)+m_{0}^{2}c_{0}^{4}\Psi =0 \\ & -{{\hbar }^{2}}c_{0}^{2}\left( -\frac{{{\partial }^{2}}\Psi }{\partial {{\left( {{c}_{0}}t \right)}^{2}}}+\frac{{{\partial }^{2}}\Psi }{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}\Psi }{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}\Psi }{\partial {{z}^{2}}} \right)+m_{0}^{2}c_{0}^{4}\Psi =0 \\ & -{{\hbar }^{2}}{{\eta }^{\mu \nu }}{{\partial }_{\mu }}{{\partial }_{\nu }}\Psi +m_{0}^{2}c_{0}^{2}\Psi =0 \\ & \left( \square +{{\mu }^{2}} \right)\Psi =0 \\ \end{align}

## In a PotentialEdit this section

In a potential, the Klein-Gordon Equation obviously gets modified as follows:

$\left(\Box+\mu^2\right)\Psi+\frac{\partial V}{\partial\psi}=0$

## Free Particle SolutionEdit this section

For the free particle solution described earlier, the solution for the wavefunction remains the same as in the Non-Relativistic case:

$\psi(\mathbf{r}, t) = e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}$

This is because the additional energy $m_0c_0^2$ only induces a very fast phase rotation of the state.

However, an additional constraint, called the Dispersion Relation, arises:

$-k^2+\frac{\omega^2}{c^2}=\mu^2$ \

For massless particles, this becomes $k=\pm\frac{\omega}{c_0}$.

## Gravitational InteractionEdit this section

To see the effect of gravitational interaction, use the form of the Klein-Gordon Equation:

$-{{\hbar }^{2}}{{\eta }^{\mu \nu }}{{\partial }_{\mu }}{{\partial }_{\nu }}\Psi +m_{0}^{2}c_{0}^{2}\Psi =0$

It is obvious to anyone with any knowledge about General Relativity that when considering gravitational interaction, one does transformations such as $\eta_{\mu\nu}\to g_{\mu\nu},\partial\to\nabla$. Therefore, in the presence of a gravitational field, the Klein-Gordon Equation becomes:

$-{{\hbar }^{2}}{{g }^{\mu \nu }}{{\nabla }_{\mu }}{{\nabla }_{\nu }}\Psi +m_{0}^{2}c_{0}^{2}\Psi =0$

Or, with a potential,

$-{{\hbar }^{2}}{{g }^{\mu \nu }}{{\nabla }_{\mu }}{{\nabla }_{\nu }}\Psi +m_{0}^{2}c_{0}^{2}\Psi + \frac{\partial V}{\partial\psi}=0$

## IssuesEdit this section

However, the Klein-Gordon Equation has issues when interpreted as a standard wave equation.

• Since it is second-order in time (like the classical wave equation from Classical Lorentz-Invariant EM), probability density is not explicitly always conserved.
• The energy is not forced to be positive.

The Dirac Equation was an attempt to solve this issue/.


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# What is the value of x? (1) 2x + 1 = 0 (2) (x + 1)^2 = x^2

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What is the value of x? (1) 2x + 1 = 0 (2) (x + 1)^2 = x^2  [#permalink]

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23 Sep 2018, 21:12
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What is the value of x?

(1) $$2x + 1 = 0$$

(2) $$(x + 1)^2 = x^2$$

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Re: What is the value of x? (1) 2x + 1 = 0 (2) (x + 1)^2 = x^2  [#permalink]

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28 Sep 2018, 23:10
1
Time needed: 31s

Statement 1:
$$2x+1=0$$
We have an equation with one variable, which will give us only one solution --> Sufficient

Statement 2:
$$(x+1)^2 = x^2$$
The two $$x^2$$ will cancel out, leaving us with an equation with one variable. Again, there's no need for calculations since they would make us waste time. --> Sufficient

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Re: What is the value of x? (1) 2x + 1 = 0 (2) (x + 1)^2 = x^2  [#permalink]

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29 Sep 2018, 05:47
Bunuel wrote:
What is the value of x?

(1) $$2x + 1 = 0$$

(2) $$(x + 1)^2 = x^2$$

Question: What is the value of x?

Statement 1: $$2x + 1 = 0$$
$$i.e. x = -1/2$$
SUFFICIENT

Statement 2: $$(x + 1)^2 = x^2$$

i.e. $$x+1 = -x$$
i.e. $$i.e. x = -1/2$$
SUFFICIENT

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Re: What is the value of x? (1) 2x + 1 = 0 (2) (x + 1)^2 = x^2   [#permalink] 29 Sep 2018, 05:47
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text:
What temperature in Celsius corresponds to 20° F?

Jul 16, 2017

$- 6.6666666$
℉ - 32/1.8, is the formula we use to convert °F to °C.
$= 20 - \frac{32}{1.8}$
°C=-6.6666666


text:
Research website of Vyacheslav Gorchilin
2016-07-29
All articles
Parametric variation of the resistance, capacitance and inductance in RLC-circuits
It is known that mathematical modeling and analysis of RLC-circuits give a very accurate coincidence with practice. In this work we use this achievement of civilization, and on the example of two different chains will try to prove or disprove some hypotheses more energy by means of parametric changes in resistance, capacitance or inductance.
1. Parametric variation of the resistance in the RC and RL-circuit
Here we consider one approach to solving the search problem of an increase of energy in the RC or RL-circuits, where the resistance $$R$$ can depend on voltage or current in the circuit, or from time to time. This approach is more General than the one here considered previously. To begin with, we generalize the parametric dependence, and present it as: $$R=R(t)$$. As such, it includes all other possible dependencies, because ultimately the whole process runs in time $$t$$.
Our electrical circuit consists of a voltage source $$U$$ (with a current source, all can be proved similarly), parametric resistance $$R$$ and the characteristic of $$Z$$, which in reality may be a constant capacitance or inductance. We believe that for the first time in the capacitor (inductor) energy is missing.
Capacitive circuit
Write the differential equation for this circuit [1]: $Z\,R(t)\,\dot Y_t + Y = U(t), \quad Y=Y(t) \qquad (1.1)$ where: $$Y$$ is the voltage, and $$Z$$ — capacity. We are also aware of a law that changes the current in the tank. Recall it: $\Phi(t) = Z\,\dot Y_t \qquad (1.2)$ where: $$\Phi$$ is current. Substitute the formula (1.2) in (1.1) we get the characteristic equation: $R(t)\,\Phi(t) + Y = U(t) \qquad (1.3)$ Now domnain right and left parts of $$\Phi$$ and printeriem: $\int_{0}^{T} R(t)\,\Phi(t)^2\, dt + \int_{0}^{T} Y\,\Phi(t)\, dt = \int_{0}^{T} U(t)\,\Phi(t)\, dt \qquad (1.4)$ where: $$T$$ — time of the studied process. In the formula (1.4) the rst term is nothing but the energy dissipated in the resistance $$R$$, the second — potential energy in the capacitor, and to the right of the equal sign is the energy that takes the power source for the whole process.
If the first and the third term is clear, then we need to clarify. Substituting back to the formula (1.2), we obtain this integral: $\int_{0}^{T} Y\,Z\,\dot Y_t\, dt = Z \int_{0}^{T} Y\, dY = \frac{Z}{2} Y^2\, \bigg |_{0}^{T} \qquad (1.5)$ But $$Y(0)$$ is the voltage on the tank for the first time which can be only zero, and hence the desired second term is: $\int_{0}^{T} Y\,\Phi(t)\, dt = \frac{Z}{2} Y(T)^2 \qquad (1.6)$ note that this term can only be positive. Move him to the right side of (1.4) and get our proof: $\int_{0}^{T} R(t)\,\Phi(t)^2\, dt = \int_{0}^{T} U(t)\,\Phi(t)\, dt - \frac{Z}{2} Y(T)^2 \qquad (1.7)$ Now the left — the energy that we receive on resistance, and the right — expended by the power source minus the residual — capacity. It is obvious that the resulting energy can be only less or equal to the energy of the power source.
Inductive circuit
The proof for the inductive circuit is carried out in a similar way. The differential equation is: $R(t)\,Y + Z\,\dot Y_t = U(t), \quad Y=Y(t) \qquad (1.8)$ where: $$Y$$ is the current in the circuit, and $$Z$$ — inductance. The voltage dependence of the current on the record form the same: $\Phi(t) = Z\,\dot Y_t \qquad (1.9)$ here $$\Phi$$ is the voltage across the inductor. The characteristic equation in this case looks like this: $R(t)\,Y + \Phi(t) = U(t) \qquad (1.10)$ Damnosa all its members $$Y$$ and integrating we get: $\int_{0}^{T} R(t)\,Y^2\, dt + \int_{0}^{T} Y\,\Phi(t)\, dt = \int_{0}^{T} U(t)\,Y\, dt \qquad (1.11)$ the second member of the equation do the same as in a capacitive circuit, which leads us to the final equation: $\int_{0}^{T} R(t)\,Y^2\, dt = \int_{0}^{T} U(t)\,Y\, dt - \frac{Z}{2} Y(T)^2 \qquad (1.12)$ Output from it is similar to the previous case: we get the resistance of the energy can only be less than or equal to the energy of the power source.
Insights
The proof shows that in an RC or RL circuit, where capacitance and inductance are constant, and the resistance is parametric, the gain of energy can not be. Moreover, it does not depend on the nature of parametric dependencies, nor from the selected time range.
Next we will describe a much more complex process with parametric capacity and we find a condition for obtaining a gain of energy in this circuit.

The materials used


text:
Problem 1-12: Using Voltage Sources to Determine Node Voltages

Determine the power of $R_1, R_2$ and $Vs_1$ . (Hint: there is no need to use nodal analysis; voltages between nodes can be easily found by the voltage sources.)

Solution

$V_{R_1}= Vs_1 = 10v \rightarrow P_{R_1}=\frac{V_{R_1}^2}{R_1}=50 W$

Problem 1-11: Solving a Circuit with Three Nodes by the Nodal Analysis

Solve the circuit by nodal analysis and find the power of $Is_1$ .

Solution
a) Choose a reference node, label node voltages:

Problem 1-10: Solving by Nodal Analysis - Circuit with Four Nodes

Solve the circuit by nodal analysis and find $V_a$ .

Solution
a) Choose a reference node, label node voltages:


text:
# Van der Waals equation of state and Maxwell construction

1. Aug 20, 2009

### TheDestroyer

Hello people :),

I have a question because I couldn't understand something written in Schwabl's book of Statistical mechanics.

Can you please explain why we in the Van der Waals equation of state choose a constant pressure under the critical temperature. I understood the part that we want to have some plausible compressibility, therefore the derivative of the free energy with respect to volume should be >= 0, but I don't understand:

1) Why we choose the Maxwell construction to be a line, not some curve.
2) why we have to have equal areas above and below the line.

What would happen if we don't follow those rules?

Find attached the corresponding graphs.

Thank you

#### Attached Files:

• ###### Statistical Mechanics, Schwabl, Springer 2006.pdf
File size:
53.9 KB
Views:
202
2. Aug 20, 2009

I'm not sure if this is what you want but experiment shows that below the critical temperature the PV graph has a a flat portion(constant pressure)where liquid and vapour are in equilibrium.Van der Waals improved upon the simple kinetic model ideal gas equation by taking into account the fact that the molecules have a finite size and that they exert forces on eachother.When his results are plotted they do not give the flat portion but they give the peculiar wiggly line which goes above and below the flat portion
Clearly V de Ws equation is not perfect but it is an improvement on the equation it replaced.

3. Aug 20, 2009

### TheDestroyer

Is it that simple? just because the experiments have shown that? isn't there some good theoretical reason?

Thanks

4. Aug 20, 2009

Again I am not sure if this is what you want but there is a good theoretical reason for the flat portion of the graph.To liquify a gas you have to both compress it and cool it below its critical temperature.If it wasn't cooled enough you would end up with a highly compressed gas where the molecules are moving too fast to get continually, momentarily captured by other molecules, as is the case with liquids.If a gas is compressed below its critical point a pressure is reached where it starts to liquify and a meniscus appears.The gas can now be described as a saturated vapour where the gas and liquid are in equilibrium with eachother.With a saturated vapour the volume can be changed without a change in pressure,reducing the volume results in more liquid and less vapour and vice versa.As the volume is reduced futher a point is reached where all of the vapour has changed to liquid and since liquids are highly incompressible a massive increase in pressure is needed for a small reduction in volume.Of course different gases have different characteristics.A good equation should explain experimental results and although V de Ws doesn't do this exactly it is a step up from the ideal gas equation.

5. Aug 20, 2009

### TheDestroyer

Thank you, but what about the equal areas?

6. Aug 20, 2009

Are you referring to the area of the V de W curve above the flat portion as compared to the area below the flat portion?If so I wasn't aware that the areas were equal but when I recall the curves they certainly seem to be.Interesting if they are equal but I think that it just pops out of the equation.Perhaps you could most easily prove the equality by integration.

7. Aug 20, 2009

### TheDestroyer

Yes, the areas should be equal, this is the deal! This line should be constructed in a way giving equal areas according to the integral, it's something related to the free energy I couldn't understand. It's not a mathematical issue, it's a physical issue, there is some kind of a condition or something with the free energy, homogeneity.

Any ideas guys?

8. Aug 20, 2009

The work done on or by a gas is equal to the integral of PdV which is equal to the area under the PV graph.Assume that the work done in the V de Ws wiggly portion of the graph is the same as the work done during the experimentally found flat portion of the graph,if so the areas under both graphs must be the same.With this in mind we can take V de Ws graph and draw a line through through the wiggly portion, the position of the line being such that the area of the V de Ws graph above the line must be exactly equal to the area of the V de Ws graph below the line.Imagine the gas was being compressed,when it reaches the flat portion, V de Ws graph rises above the line showing that extra work is done.Later,when V de Ws graph goes below the line less work is done the total work done being the same as that which is done when P remains constant.I hope others can come in and justify more fully why V de Ws graph, which is not exactly correct, can be manipulated like this.

9. Aug 20, 2009

### TheDestroyer

Thank you for your tries and efforts :)

More explanation guys, come on!

10. Aug 21, 2009

### TheDestroyer

Oh my god! this is really disappointing :(

11. Aug 23, 2009

### kanato

Ok, so you have to draw a line somewhere. In the P-V curve, there is a maximum, and to the left of that dP/dV < 0. Where dP/dV < 0, you have negative compressibility which means the system will spontaneously compress to a liquid, so the system is thermodynamically unstable at that point. This means you could draw a line at the top so in any case you would have a line drawn at the point where dP/dV = 0.

Then you need to minimize the free energy. Pick any constant pressure line that intersects the P-V isotherm at three points. The left point is the volume of the liquid, V_L and the right point is the volume of the gas V_G. At a given volume on that line, that line, number of particles in gas or liquid phases are

$$N_L + N_G = N$$
$$n_LV_L + n_g V_G = V$$

where n_L(n_G) is the number fraction of particles in the liquid(gas), ie. N_L/N.

The free energy along that line will be

$$A = n_G A(N,V_G,T) + n_L A(N,V_L,T)$$
$$A = n_G A(N,V_G,T) + (1 - n_G) A(N,V_L,T)$$

will be a straight line, since N, V_G, V_L and T are fixed. Picking different constant pressure lines will shift the line in the free energy up and down. Now, what constant pressure can you pick that gives the lowest line in the free energy? The Maxwell construction gives you that, and guarantees convexity in the free energy. A more detailed explanation is avilable here:
http://www.pma.caltech.edu/~mcc/Ph127/b/Lecture3.pdf [Broken]

Last edited by a moderator: May 4, 2017
12. Aug 26, 2009

### TheDestroyer

Thanks pal :)


text:
# Conditional Probability Problems about Die Rolling

## Problem 728

A fair six-sided die is rolled.

(1) What is the conditional probability that the die lands on a prime number given the die lands on an odd number?

(2) What is the conditional probability that the die lands on 1 given the die lands on a prime number?

## Solution.

Let $E$ and $F$ be events. Then the conditional probability that $E$ occurs given $F$ has occurred, which is denoted by $P(E\mid F)$, is defined as
$P(E \mid F) = \frac{P(E \cap F)}{P(F)}.$

For uniform space (such as fair die rolling), this can be rewritten as
$P(E \mid F) = \frac{|E \cap F|}{|F|}.$ Here the notation $|A|$ means the number of elements in $A$.

### Solution (1)

Let $E = \{2, 3, 5\}$ be the set of prime number faces of a die. Let $F = \{1, 3, 5\}$ be the set of odd faces of a die. Then the required conditional probability can be calculated as follows:
\begin{align*}
P(E \mid F) &= \frac{|E \cap F|}{|F|}\\
&= \frac{|\{2, 3, 5\} \cap \{1, 3, 5\}|}{|\{1, 3, 5\}|}\\
&= \frac{|\{3, 5\}|}{\{1,3,5\}|}\\
&= \frac{2}{3}
\end{align*}

### Solution (2)

The required conditional probability is
\begin{align*}
P(\{1\} \mid E) &= \frac{|\{1\} \cap E|}{|E|}\\
&= \frac{|\{1\} \cap \{2, 3, 5\}|}{|\{2, 3, 5\}|}\\
&= \frac{|\emptyset|}{|\{2, 3, 5\}|}\\
&= \frac{0}{3} = 0
\end{align*}

### More from my site

• Probabilities of An Infinite Sequence of Die Rolling Consider an infinite series of events of rolling a fair six-sided die. Assume that each event is independent of each other. For each of the below, determine its probability. (1) At least one die lands on the face 5 in the first $n$ rolls. (2) Exactly $k$ dice land on the face 5 […]
• Probability Problems about Two Dice Two fair and distinguishable six-sided dice are rolled. (1) What is the probability that the sum of the upturned faces will equal $5$? (2) What is the probability that the outcome of the second die is strictly greater than the first die? Solution. The sample space $S$ is […]
• Overall Fraction of Defective Smartphones of Three Factories A certain model of smartphone is manufactured by three factories A, B, and C. Factories A, B, and C produce $60\%$, $25\%$, and $15\%$ of the smartphones, respectively. Suppose that their defective rates are $5\%$, $2\%$, and $7\%$, respectively. Determine the overall fraction of […]
• Lower and Upper Bounds of the Probability of the Intersection of Two Events Let $A, B$ be events with probabilities $P(A)=2/5$, $P(B)=5/6$, respectively. Find the best lower and upper bound of the probability $P(A \cap B)$ of the intersection $A \cap B$. Namely, find real numbers $a, b$ such that $a \leq P(A \cap B) \leq b$ and $P(A \cap B)$ could […]
• What is the Probability that All Coins Land Heads When Four Coins are Tossed If…? Four fair coins are tossed. (1) What is the probability that all coins land heads? (2) What is the probability that all coins land heads if the first coin is heads? (3) What is the probability that all coins land heads if at least one coin lands […]
• What is the Probability that Selected Coin was Two-Headed? There are three coins in a box. The first coin is two-headed. The second one is a fair coin. The third one is a biased coin that comes up heads $75\%$ of the time. When one of the three coins was picked at random from the box and tossed, it landed heads. What is the probability […]
• If At Least One of Two Coins Lands Heads, What is the Conditional Probability that the First Coin Lands Heads? Two fair coins are tossed. Given that at least one of them lands heads, what is the conditional probability that the first coin lands heads? We give two proofs. The first one uses Bays' theorem and the second one simply uses the definition of the conditional […]
• Independent Events of Playing Cards A card is chosen randomly from a deck of the standard 52 playing cards. Let $E$ be the event that the selected card is a king and let $F$ be the event that it is a heart. Prove or disprove that the events $E$ and $F$ are independent. Definition of Independence Events […]

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##### Probability Problems about Two Dice

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Hydrogen fuel cells combine hydrogen and oxygen to produce water and energy. Assume 1 kg of fuel is used, and the mass of the water produced is 1.10 x

Question:

Hydrogen fuel cells combine hydrogen and oxygen to produce water and energy. Assume 1 kg of fuel is used, and the mass of the water produced is 1.10 x 10-11 kg. How much energy is produced by this fuel cell? (Hint: Use E = mc2.)

What are the powers and functions of the Supreme Court? Explain at least 3 ofthem.

What are the powers and functions of the Supreme Court? Explain at least 3 ofthem....

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The radius of a basketball is 4.7 inches what is the voloume of the basketball round to the nearest tenth

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# White Balls, Black Balls

There are two bags $A$ and $B$. Bag $A$ contains $n$ white balls and $2$ black balls while bag $B$ contains $2$ white balls and $n$ black balls . One of the two bags is selected at random and two balls are drawn from it without replacement . We know that if both the balls drawn are white balls, the probability of bag $A$ being selected to draw the balls is $\frac{6}{7}$. Find the value of $n$.

×


text:
# Study on hybrid au-underfill resin bonding method with lock-and-key structure for 3-D integration

Masatsugu Nimura, Jun Mizuno, Akitsu Shigetou, Katsuyuki Sakuma, Hiroshi Ogino, Tomoyuki Enomoto, Shuichi Shoji

Research output: Contribution to journalArticle

8 Citations (Scopus)

This paper describes a hybrid Au-underfill resin bonding method with lock-and-key structure for 3-D integration. In 3-D large scale integration (LSI), the gap between stacked chips becomes narrower because the bump dimension and pitch are smaller than those encountered in 2-D LSI. Therefore, the filling of gaps less than 10 $\mu{\rm m}$ using capillary forces often becomes insufficient because of the surface condition. To address this challenge, we study a hybrid bonding method in which the metal-metal and resin-resin bonding are carried out simultaneously with a chip resin applied previously only around the bump. To realize hybrid bonding on the entire chip, we fabricate indent and protrusion structures, which are called lock-and-key structures. The key structure is fabricated by a process that can remove the resin on the bumps by ${\rm O}2 plasma irradiation. The lock structure is fabricated by conventional photolithography and dry etching. By means of hybrid bonding with the lock-and-key structure, we have achieved the Au bump bonding and the filling of 4-$\mu{\rm m}$gaps between the stacked chips, concurrently. The cross-sectional transmission electron microscopy image of the bonded sample demonstrated that no significant gap exists at both the Au-Au and resin-resin interfaces. In addition, the shear strength of the sample bonded with resin is 10 times higher than that without the resin. The electrical continuity of the Au bump connections after hybrid bonding has also been determined. Original language English 6449300 558-565 8 IEEE Transactions on Components, Packaging and Manufacturing Technology 3 4 https://doi.org/10.1109/TCPMT.2013.2240566 Published - 2013 ### Fingerprint Resins LSI circuits Metals Dry etching Photolithography Shear strength Irradiation Transmission electron microscopy Plasmas ### Keywords • 3-D integration • Au bump • bonding • flip chip • micro-bump • underfill ### ASJC Scopus subject areas • Electrical and Electronic Engineering • Electronic, Optical and Magnetic Materials • Industrial and Manufacturing Engineering ### Cite this Study on hybrid au-underfill resin bonding method with lock-and-key structure for 3-D integration. / Nimura, Masatsugu; Mizuno, Jun; Shigetou, Akitsu; Sakuma, Katsuyuki; Ogino, Hiroshi; Enomoto, Tomoyuki; Shoji, Shuichi. In: IEEE Transactions on Components, Packaging and Manufacturing Technology, Vol. 3, No. 4, 6449300, 2013, p. 558-565. Research output: Contribution to journalArticle Nimura, Masatsugu ; Mizuno, Jun ; Shigetou, Akitsu ; Sakuma, Katsuyuki ; Ogino, Hiroshi ; Enomoto, Tomoyuki ; Shoji, Shuichi. / Study on hybrid au-underfill resin bonding method with lock-and-key structure for 3-D integration. In: IEEE Transactions on Components, Packaging and Manufacturing Technology. 2013 ; Vol. 3, No. 4. pp. 558-565. @article{eedf25bd04a74defa9abaf5655e83b0b, title = "Study on hybrid au-underfill resin bonding method with lock-and-key structure for 3-D integration", abstract = "This paper describes a hybrid Au-underfill resin bonding method with lock-and-key structure for 3-D integration. In 3-D large scale integration (LSI), the gap between stacked chips becomes narrower because the bump dimension and pitch are smaller than those encountered in 2-D LSI. Therefore, the filling of gaps less than 10$\mu{\rm m}$using capillary forces often becomes insufficient because of the surface condition. To address this challenge, we study a hybrid bonding method in which the metal-metal and resin-resin bonding are carried out simultaneously with a chip resin applied previously only around the bump. To realize hybrid bonding on the entire chip, we fabricate indent and protrusion structures, which are called lock-and-key structures. The key structure is fabricated by a process that can remove the resin on the bumps by${\rm O}2 plasma irradiation. The lock structure is fabricated by conventional photolithography and dry etching. By means of hybrid bonding with the lock-and-key structure, we have achieved the Au bump bonding and the filling of 4-$\mu{\rm m}$ gaps between the stacked chips, concurrently. The cross-sectional transmission electron microscopy image of the bonded sample demonstrated that no significant gap exists at both the Au-Au and resin-resin interfaces. In addition, the shear strength of the sample bonded with resin is 10 times higher than that without the resin. The electrical continuity of the Au bump connections after hybrid bonding has also been determined.",
keywords = "3-D integration, Au bump, bonding, flip chip, micro-bump, underfill",
author = "Masatsugu Nimura and Jun Mizuno and Akitsu Shigetou and Katsuyuki Sakuma and Hiroshi Ogino and Tomoyuki Enomoto and Shuichi Shoji",
year = "2013",
doi = "10.1109/TCPMT.2013.2240566",
language = "English",
volume = "3",
pages = "558--565",
journal = "IEEE Transactions on Components, Packaging and Manufacturing Technology",
issn = "2156-3950",
publisher = "Institute of Electrical and Electronics Engineers Inc.",
number = "4",

}

TY - JOUR

T1 - Study on hybrid au-underfill resin bonding method with lock-and-key structure for 3-D integration

AU - Nimura, Masatsugu

AU - Mizuno, Jun

AU - Shigetou, Akitsu

AU - Sakuma, Katsuyuki

AU - Ogino, Hiroshi

AU - Enomoto, Tomoyuki

AU - Shoji, Shuichi

PY - 2013

Y1 - 2013

N2 - This paper describes a hybrid Au-underfill resin bonding method with lock-and-key structure for 3-D integration. In 3-D large scale integration (LSI), the gap between stacked chips becomes narrower because the bump dimension and pitch are smaller than those encountered in 2-D LSI. Therefore, the filling of gaps less than 10 $\mu{\rm m}$ using capillary forces often becomes insufficient because of the surface condition. To address this challenge, we study a hybrid bonding method in which the metal-metal and resin-resin bonding are carried out simultaneously with a chip resin applied previously only around the bump. To realize hybrid bonding on the entire chip, we fabricate indent and protrusion structures, which are called lock-and-key structures. The key structure is fabricated by a process that can remove the resin on the bumps by ${\rm O}2 plasma irradiation. The lock structure is fabricated by conventional photolithography and dry etching. By means of hybrid bonding with the lock-and-key structure, we have achieved the Au bump bonding and the filling of 4-$\mu{\rm m}$gaps between the stacked chips, concurrently. The cross-sectional transmission electron microscopy image of the bonded sample demonstrated that no significant gap exists at both the Au-Au and resin-resin interfaces. In addition, the shear strength of the sample bonded with resin is 10 times higher than that without the resin. The electrical continuity of the Au bump connections after hybrid bonding has also been determined. AB - This paper describes a hybrid Au-underfill resin bonding method with lock-and-key structure for 3-D integration. In 3-D large scale integration (LSI), the gap between stacked chips becomes narrower because the bump dimension and pitch are smaller than those encountered in 2-D LSI. Therefore, the filling of gaps less than 10$\mu{\rm m}$using capillary forces often becomes insufficient because of the surface condition. To address this challenge, we study a hybrid bonding method in which the metal-metal and resin-resin bonding are carried out simultaneously with a chip resin applied previously only around the bump. To realize hybrid bonding on the entire chip, we fabricate indent and protrusion structures, which are called lock-and-key structures. The key structure is fabricated by a process that can remove the resin on the bumps by${\rm O}2 plasma irradiation. The lock structure is fabricated by conventional photolithography and dry etching. By means of hybrid bonding with the lock-and-key structure, we have achieved the Au bump bonding and the filling of 4-$\mu{\rm m}$ gaps between the stacked chips, concurrently. The cross-sectional transmission electron microscopy image of the bonded sample demonstrated that no significant gap exists at both the Au-Au and resin-resin interfaces. In addition, the shear strength of the sample bonded with resin is 10 times higher than that without the resin. The electrical continuity of the Au bump connections after hybrid bonding has also been determined.

KW - 3-D integration

KW - Au bump

KW - bonding

KW - flip chip

KW - micro-bump

KW - underfill

UR - http://www.scopus.com/inward/record.url?scp=84875907152&partnerID=8YFLogxK

UR - http://www.scopus.com/inward/citedby.url?scp=84875907152&partnerID=8YFLogxK

U2 - 10.1109/TCPMT.2013.2240566

DO - 10.1109/TCPMT.2013.2240566

M3 - Article

AN - SCOPUS:84875907152

VL - 3

SP - 558

EP - 565

JO - IEEE Transactions on Components, Packaging and Manufacturing Technology

JF - IEEE Transactions on Components, Packaging and Manufacturing Technology

SN - 2156-3950

IS - 4

M1 - 6449300

ER -


text:
Windows – Stopping Microsoft Search from eating the hard drive

uninstallwindowswindows-vista

I have Windows Vista on a machine and I noticed quite a bit of hard drive was disappearing. I ran a utility to show me where it all went. I found the following directory consumes over 2GB of space:
C:\ProgramData\Microsoft\Search\Data\Applications\Windows.

The Microsoft Search service is now disabled, but I want it removed completely. I see nothing on Add/Remove Programs. Also, will I get blue screens if I delete this directory?

Here's what I had to do:

• open services.msc
• Stop the Windows Search service (I was this far already).
• Rename C:\ProgramData\Microsoft\Search\Data\Applications\Windows\Windows.edb
• Start the Windows Search service
• The directory rebuilds Windows.edb (34 MB initial size in my case)
• Open Control Panel / Indexing Options
• Wait a long time for buttons to become enabled
• Click Modify
• Uncheck Users directory or whatever else you don't want indexed (I left Start Menu enabled because it's so small)
• Click Ok
• Stand on your head and count to three while drinking water with a straw. (not sure if this part is necessary but I wanted to try everything I could)

It now says indexing complete with only 800 or so items. Total size of the directory is now just over 48 MB. Pretty large index if you ask me, but better than 2 GB.


text:
## Month: December, 2012

### Computational Combinatorics Roundup: November-December

We continue our monthly update of collecting research papers (or blog posts!) that were posted in the previous month+$\varepsilon$ (or were missed by previous lists). So far, I’ve just been watching the arXiv updates and my Google Scholar updates. Please comment or send me an email if you have an addition. (You can even tell me about papers when you see them, and I’ll add them to the next one!)

I’ve added my own short synopsis in italics based on a cursory glance through the papers. These are not reviews which verify their correctness or quality.
Read the rest of this entry »


text:
# Directory `systems/win32/latexmng`

```LaTexMng
Version 2.0

LaTeX is a high-quality typesetting system, with features designed for the production of technical and scientific documentation. LaTeX is the de facto standard for the communication and publication of scientific
documents.

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LaTexMng requires Windows 2000 or XP, or later editions of MS Windows.

We do not warrant that the software is error free.

This is not free software. Subject to the terms below, you are hereby.

Licensed to use Shareware Version of this software for evaluation purposes without charge for a period of 30 days. The free evaluation period starts at the moment of installation of the software.

New features in version 2.0:
* Support of work with BibTeX files;
* Insertion of TeX-objects (commands, symbols) with new toolbox;
* Preparation \bibitem elements for thebibliography environments of your LaTeX document with Bibliography template.

Installation

LaTexMng is distributed as LaTexMngSetup.exe. Run LaTexMngSetup.exe and simply follow the on-screen instructions to install LaTexMng.

LaTexMng is copyright (C) 2004-2006, by VNP Comp.

Contact: