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AP Physics C: Mechanics : Electricity

Example Questions

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Example Question #1 : Electricity And Magnetism Exam

A hollow metal sphere with a diameter of 10cm has a net charge of  distributed uniformly across its surface. What is the magnitude of the field a distance 2.0m from the center of the sphere?

Explanation:

Relevant equations:

(electric field of point charge)

Anywhere outside the metal sphere, the electric field is the same as it would be for a point charge of the same magnitude, located at the center of the sphere. So, calculate the electric field of a point charge given:

Plugging in gives:

Example Question #1 : Electricity And Magnetism Exam

Two infinite parallel conducting sheets each have positive charge density . What is the magnitude and direction of the electric field to the right of the right sheet?

, to the right

, to the left

, to the right

, to the right

, to the right

Explanation:

Relevant equations:

(field due to single infinite plane)

Electric field is additive; in other words, the total electric field from the two planes is the sum of their individual fields:

The direction of the electric field is away from positive source charges. Thus, to the right of these positively charged planes, the field points away to the right.

Example Question #1 : Electricity And Magnetism Exam

Four particles, each of charge , make up the four corners of a square with equal side lengths of . For the charge in the top left corner of the square, in what direction is the net force that it experiences due to its interactions with the other three particles?

downwards of right

upwards of left

Directly to the left

Directly to the right

upwards of left

Explanation:

The correct answer is 45 degrees upwards of left. Since all particles have charge , all forces will be repulsive (there will be no attracting forces). The particle in the top-right corner creates a repulsive force directly to the left, and the particle in the bottom-left corner creates a repulsive force directly upwards. These are equal in magnitude, since they are both at distance  from the top left corner. The bottom-right corner also creates a repulsive force, but acts along the same direction as the vector sum of a leftwards and upwards force.

Example Question #1 : Electricity And Magnetism Exam

Consider a spherical capacitor made of two nested spheres. The smaller sphere has a radius of  and a charge of , and lies within a larger sphere with radius  and a charge of

Which of the following equations accurately describes the capacitance of this spherical capacitor?

Due to symmetry, this scenario would not produce capacitance

Explanation:

To solve this problem, we will need to derive an equation.

We know that:

We can use Gauss's law to derive the electric field between the two circles yielding:

Doing our integration with respect to  from  to , we get:

We can plug this back into our equation for capacitance to get:

Example Question #1 : Electricity And Magnetism Exam

We have a point charge of . Determine the electric field at a distance of  away from that charge.

Explanation:

Coulomb's law for the electric field from point charges is , where we know the values of the following variables.

Using these values, we can solve for the electric field.

Example Question #2 : Electricity And Magnetism Exam

Two positive point charges of  and  are place at a distance  away from each other, as shown below. If a positive test charge, , is placed in between, at what distance away from  will this test charge experience zero net force?

Explanation:

To find the location at which the test charge experience zero net force, write the net force equation as , where  is the force on the test charge from , and  is the force on the same test charge from . Using Coulomb's law, we can rewrite the force equation and set it equal to zero.

In this equation, the distance, , is how far away the test charge is from , while  represents how far away the test charge is from .  Now, we simplify and solve for .

Cross-multiply.

We can cancel and . We do not need to know these values in order to solve the question.

Now that we have isolated , we can plug in the values given in the question and solve.

Example Question #2 : Electricity And Magnetism Exam

You are standing on top of a very large positively charged metal plate with a surface charge of .

Assuming that the plate is infinitely large and your mass is , how much charge does your body need to have in order for you to float?

Explanation:

Consider the forces that are acting on you. There is the downward (negative direction) force of gravity, . In order for you to float, there has to be an upward (positive direction) force, and that upward force is coming from the metal plate, . To show that you would float, the net force equation is written as , where  is the charge on you.

For plates that are charged, know that .

Knowing this, the force equation becomes .

Solve for .

Now we can plug in our given values, and solve for the charge.

Example Question #1 : Electricity

A point charge of  exerts a force of  on another charge with . How far apart are the two charges?

Explanation:

To find the distance between the two charges, use Coulomb's Law.

Since we want to find distance, , we solve for .

We know the values of the force and the two charges.

We can plug in these values and solve for the distance.

Example Question #1 : Electricity

What is the electric force between two charges, and , located apart?

Explanation:

The equation for finding the electric force between two charges is , where . Using this, we can rewrite the force equation.

Now, we can use the values given in the question to solve for the electric force between the two particles.

Example Question #1 : Using Coulomb's Law

What is the magnitude of the electric field at a field point  from a point charge of ?


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# KCL

P@K
Probability at King's

# Probability Seminar at King's

The Probability Seminar at King's is dedicated to presenting the latest advances in Pure and Applied Probability. If only a room number is given for a talk, then it is taking place in King's College London, Strand Campus.

Subscribe to the P@K mailing list here.

## Talks

Forthcoming Past

### February 2020

03/02/2020 - Alexander Gnedin (Queen Mary University of London)
16:00-17:00
In 1981 Samuels and Steele showed that the maximum expected length of subsequence which can be selected from $n$ iid observations by a nonanticipating online strategy is asymptotic to $\sqrt{2n}$. This can be compared with the classic $2\sqrt{n}$ asymptotics for the (offline) longest increasing subsequence in the Ulam-Hammersley problem. The talk will survey history of the online selection problem and then focus on recent results on its poissonised version. The latter include a renewal-style approximation leading to an asymptotic expansion of the moments, as well as functional limits for the transversal and longitudinal fluctuations of the shape of the increasing subsequence under the optimal or near-optimal selection strategies. The talk is based on a joint work with Amirlan Seksenbayev.
10/02/2020 - Zorana Grbac (Université Paris Diderot, France)
16:00-17:00
10/02/2020 - Gechun Liang (Warwick University)
17:00-18:00
17/02/2020 - Marie Kratz
16:00-17:00 S4.29
24/02/2020 - Pierre Le Doussal
16:00-17:00 S4.29

### March 2020

02/03/2020 - Martin Herdegen (University of Warwick)
15:00-16:00
09/03/2020 - Pierre-François Rodriguez
16:00-17:00 S4.29
16/03/2020 - Eveliina Peltola
16:00-17:00 S4.29

format: plain


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### Home > AC > Chapter 12 > Lesson 12.2.1 > Problem12-41

12-41.

Solve each equation or inequality below, if possible. Check your solution.

1. $−(x − 3)(x + 1) = 4$

2. $10 | x − 3 | > 40$

3. $(x − 5)^3 = 8$

Use a generic rectangle to multiply the factors together then distribute the negative and solve.

Set the equation equal to $0$. Then use a generic rectangle to factor and solve for $x$ using the Zero Product Property.

Solve for the absolute value then make two equations to remove the absolute value.

$x − 3 = 4$ or $x − 3 = −4$
$x = 7$ or $x = −1$

Test numbers to determine which region is the solution.

What number cubed equals $8$?


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Dynamics of agents on random networks.

Overview
What are the necessary conditions in order to have consensus among an assembly of people? Why, sometimes, do all people share the same opinion in a short time? Is consensus more difficult to achieve in our complex and connected society than in the less connected society of the beginning of the 20th century?

Consider for example, a school class of students who want to make an important travel. They have to choose a destination among two possible choices, Alaska or Rio de Janeiro, and every student has naturally a strong preference for one of the destinations. If students are isolated then nobody changes his opinion, and no consensus will be found. When students take advice from one or two friends in the class, they may change their opinion quickly and a consensus will be achieved. Now if each student has many friends, he will always find a small group among his friends that share his opinion and support his choice. Hence students will keep their opinion and no consensus is found although every student is connected to a large number of friends.

A new model of basic dynamics is introduced where agents have a three state opinion: yes, no or no opinion. Each agent has a group of advisors taken at random among other agents. When the group of advisors is large, opinions are no more correlated and are random. A phase transition occures between the two states. Many real systems like social networks, neural networks, gene networks and supply chains have an optimal evolution at the maximum of activity, precisely at the critical connectivity depending on their intrinsic noise.
 In this figure, the average opinion (order parameter) is plotted for different values of the noise r showing two transitions. Following a line defined by a constant order parameter: when p is very small, most agents have zero advisor, and information cannot percolate through the network. When p increases, there is a critical probability p where agents have enough connections in order to establish consensus among the entire network. If we increase again p, a second transition occurs where groups of advisors are so large that agents cannot change their opinion anymore.

The Java Applet
Try to vary the noise r, i.e. the number of people who are changing their opinion randomly: if the noise is too large opinions of agents are different and =0. If r is small, then opinions are suddenly correlated below a certain critical value of r. This is similar to a ferromagnetic phase transition. Observe the number of advisor in the group at the transition.

Opinion of agents are represented on a circle (positions are arbitrary). Colors mean: Green = -1, Red = +1, White = 0.
For clarity, arrows represent active advisors only for one agent.

Observations
• If advisors groups are small, then consensus is present: since agent a more likely to change their opinion when they have a small number of advisors, information can percolate through the network and consensus can be established.
• If advisors groups are large, then consensus disappears: too many advisors block the decision processus since all advisors have rarely the same and opposite opinion as the agent.
• At the transition, the number of advisors is independent of the system size (fix noise).
 The critical connectivity (points) marks the transition between stable consensus and stable disorder phases. The transition coincides with the maximum of the standard deviation or activity $\sigma$ of the order parameter. Computer simulations have been done for different noise $r$ and probability $p$. The standard deviation, and therefore the activity or evolution, is maximum in the red region between the two phases where it is lower (blue). The vertical dashed line is where the majority rule applies (this corresponds to occidental democratic model!). Surprisingly the line cross the maximum of the critical connectivty curve. This means that the majority rule is the most tolerant rule to noise where activity is still possible.

References:

Optimal evolution of random networks: from social to airports networks.
Ph. Curty
pre-print physics/0509074 (arXiv.org)


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# How do you solve -7( 1- 4x ) + 8( 1- 6x ) \leq 41?

$x \ge - 2$
$- 7 + 28 x + 8 - 48 x \le 41$
$- 20 x \le 41 + 7 - 8$
$20 x \ge - 40$


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# Isomorphic Gaussian Integers

Can we show that the ring of Gaussian integers $$\mathbb{Z}[\sqrt{17}]:=\{a+b\sqrt{17}:a,b\in\mathbb{Z}\}$$ $$\mathbb{Z}[\sqrt{11}]:=\{a+b\sqrt{11}:a,b\in\mathbb{Z}\}$$

equipped with standard addition and multiplication are not isomorphic?

-
If they were isomorphic, then wouldn't they both have the same number of solutions of the equation $x^2=11$? –  Old John Oct 27 '13 at 18:16

An isomorphism of rings would map $1$ to $1$. Hence it would fix all integers. In $\mathbb Z[\sqrt{17}]$ we have $(\sqrt{17})^2 = 17$. It is easy to check that no element $x$ of $\mathbb Z[\sqrt{11}]$ satisfies $x^2 = 17$.
Hint: Both rings can be equipped with multiplicative norms. Use this to show that $11$ is the product of nontrivial divisors in one ring, but not the other.


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The growing electrification of transportation systems is dramatically increasing the waste heat that must be dissipated from high-density power electronics. Transformative embedded heat spreading technologies must be developed in next-generation systems to enable air cooling of power semiconductors with heat fluxes exceeding 500 W/cm2 over large hotspot areas up to 1 cm2. In this study, vapor chamber heat spreaders, or thermal ground planes (TGPs), with customized wick structures are investigated as one possibility. A 10 cm × 10 cm TGP with hybrid wick, which is a blend of a biporous wick with a standard monoporous wick, was designed. The TGP was tested in combination with a straight pin fin heat sink under air jet impingement and a 1 cm2 size heat source. The experimental performance of the hybrid wick TGP was compared under the same air-cooled conditions with an off-the-shelf TGP of the same size from a commercial vendor and a TGP with a biporous wick only. The customized hybrid wick TGP exhibits ∼28% lower thermal resistance compared with a traditional commercial TGP, and the capillary limit heat flux is measured as 450 W/cm2. Technical challenges in extending this capillary limit heat flux value and TGP integration into packaged electronics are described.

## Introduction

A core theme in the development of next generation compact power electronics for electrified vehicles is increased power density. Active liquid cooling is conventionally employed to effectively dissipate waste heat from high power density systems. Forced liquid cooling exhibits excellent and reliable heat transfer performance in such high heat flux (>250 W/cm2) applications. However, the cooling loop should be leak-proof, may require filtering, and involves the connection of auxiliary components, which in combination increases overall system complexity. While air cooling is considered a relatively more simple and reliable approach, it is limited to low heat transfer coefficients because of the inferior thermophysical properties of air as a fluid. Thus, to apply air cooling in high volumetric power density electronics packaging, a high-performance embedded heat spreading device as a high-to-low heat flux converter is required.

Typical heat spreading technologies include solid material heat spreaders such as copper (Cu), graphite, or diamond. Another option is a vapor chamber thermal ground plane (TGP). A vapor chamber is a sealed two-phase heat transfer device enclosing a working fluid that is evaporated upon localized heat input. This vapor is then condensed over a broader heat rejection, or condenser, surface area. The liquid condensates return to the evaporator side of the chamber by the capillary force of an inner wick structure. By absorbing energy in the form of latent heat of vaporization, a vapor chamber exhibits advantages in terms of superior heat transport (i.e., spreading) capabilities.

The importance of heat spreading for air cooling and the feasibility of air cooling for a large-area (>1 cm2) high heat flux (>1 kW/cm2) device are highlighted by considering a conduction heat transfer finite element analysis (FEA) as follows [1]. In Fig. 1(a), a center cross section image of a 100 mm × 100 mm × 3 mm thick thermally conductive heat spreader is shown. On the bottom of the heat spreader, a uniform heat flux of 1 kW/cm2 is applied over a 1 cm × 1 cm center region. A convective heat transfer coefficient, h, is then applied to the top surface of the heat spreader. For simplicity, the remaining surfaces of the heat spreader are assumed to be adiabatic. The heat spreader thermal conductivity, k, is defined as a parameter in the model. Thermal conductivity values of 400, 1000, 2000, and 5000 W/(m K) are selected to represent, respectively, Cu, graphite, and diamond solid state materials plus a high value for a TGP, per Bar-Cohen et al. [2]. At the same time, the heat transfer coefficient, h, is set as a parameter to range through moderate to aggressive forced air convection values of 62.5–1000 W/(m2 K).

Figure 1(b) shows the numerical results obtained using commercial FEA software [3]. The maximum temperature at the center of the bottom of the heat spreader is plotted versus the convective heat transfer coefficient applied to the top side of the heat spreader. To achieve maximum heat spreader temperatures, 150–200 °C, that are compatible with next generation silicon carbide power devices, aggressive air cooling should be used in combination with a heat spreader that has a thermal conductivity value approaching that of a TGP. Figure 2 shows the heat flux profile along the centerline of the top of the heat spreader for the four different materials. In this case, the heat transfer coefficient, h, is set to 750 W/(m2 K). As expected, an average heat flux value of 10 W/cm2 is obtained in all four cases. However, effective thermal conductivity is critical to decreasing the peak heat flux value, which is vital for air cooling. Observe that the TGP provides the most uniform heat flux profile with a maximum value that is within a feasible range for forced convection air cooling.

From the above introduction and numerical analysis, a need exists for low-thermal-resistance and ultrathin heat spreaders that are capable of dissipating heat fluxes larger than 500 W/cm2 from relatively large electronic device areas (∼1 cm2) [1]. Thus, the major gap for air cooling of next generation wide band-gap electronics is extremely high-performance heat spreading technology such as the TGP. As explained in Ref. [4], various research groups have been investigating the underlying wick structure necessary for the proper operation of a TGP. Table 1 provides a summary of the existing wick technologies. Very high (>1.2 kW/cm2) heat fluxes that are applied over very small areas (∼0.6 mm2) with very low superheat (<10 K) have recently emerged; see Ref. [5]. Additionally, as described in Ref. [10], moderate heat flux, ∼580 W/cm2, over a large area, ∼100 mm2, with large superheat, ∼72 K, has been studied. However, a technology gap exists in Table 1, where the cooling of both large size (∼100 mm2) and extreme heat flux (>1 kW/cm2) heat sources at low superheat (<35 K) still needs to be addressed. The supply of an adequate amount of liquid coolant to avoid dryout of the wick at the evaporator of a TGP via a 3D liquid feed structure should be considered a primary research direction [5,12]. Thus, the development of high performance two-phase thermal ground plane vapor chambers remains a critical field for further research in the context of future compact air-cooled electronics.

The current work herein applies the biporous sintered wick structure developed and tested by Semenic et al. [13] at wick level. In this work, they demonstrated the capability of the wick to spread a heat flux of 990 W/cm2 applied over a relatively large area of 0.32 cm2. Here, the same strategy is applied to a vapor chamber TGP with a spreading area of ∼10 × 10 cm2 and a heated area of 1 × 1 cm2. The heat spreading performance of the TGP is tested with an air jet impingement cooled heat sink, as a benchmark. Besides the biporous wick, a hybrid wick reported later by the same group in Ref. [14], which is a combination of a biporous wick and a standard monoporous wick, is applied to the same size TGP for experimental evaluation. The performance of the two TGPs is compared with an off-the-shelf TGP of the same size from a commercial vendor2 under the same air-cooled conditions. Most often, the performance of a TGP is tested with liquid cooling on the condenser side, and only a few publications [1518] that utilize an air-cooled system, albeit under low heat flux conditions, can be found. Hence, in this investigation, the TGP performance is tested under high heat flux conditions (produced by a large heater area) with air cooling, which is challenging. In the end, a novel two-layer wick structure, which theoretically should have improved performance, is proposed and will be applied to a TGP for future evaluation.

## Vapor Chamber Wick Structures

Depending on porosity, a wick structure can be categorized as either a monoporous or a biporous structure. Low porosity monoporous structures offer larger capillary driving force for more efficient return of condensate [4]. However, the low permeability of the small-particle features prevents the vapor bubbles from penetrating through the wick readily [4]. Hence, high porosity and permeability biporous wicks with large pores are an alternative, despite their shortcoming of higher thermal resistance. A hybrid porous wick design combining monoporous and biporous features thus naturally emerged as another compromise solution. In the hybrid wick structure, a monoporous interface layer is sandwiched between the biporous material and the Cu substrate; refer to Ref. [14]. Vapor travels easily through the wick structure due to the upper biporous structure. Meanwhile, working fluid resupply benefits from the embedded monoporous layer, and at high heat flux, the top biporous wick can help local liquid supply as well.

Figure 3 shows a schematic of the monoporous wick (left), biporous wick (middle), and hybrid wick (right). The TGP with monoporous wick is a standard product from a commercial vendor, Celsia Inc. (Santa Clara, CA)2 with 60 μm Cu particles being sintered to form a $θmono=$ 1500 μm thick wick layer. Same wick structures were applied to both the evaporator side and the condenser side of all the three TGPs.

For fair comparison, the other two TGPs have the same wick thickness and were constructed based on the specifications from this research investigation. The biporous wick was fabricated by sintering 60 μm Cu particles to form a 300 μm thick wick sheet with about five particles through the thickness. The wick sheet then was broken into pieces (clusters) and sieved with screen mesh to obtain a specific cluster size (300 μm). The combination of 60 μm Cu particles with the 300 μm cluster size originated from the results reported by Semenic et al. [9,13], who experimentally studied monoporous and biporous wicks. They found that the 60 μm Cu particles have a distinct performance characteristic in monoporous wicks when compared with other size particles, and the 302/58, 302/72 copper cluster-to-particle ratios had the lowest wall superheat under the same heat flux in biporous wicks. The concept of the hybrid wick was reported by Reilly and Catton in Ref. [14], but no test results were publicly reported. In the current work, the hybrid wick was designed as a 300 μm thick monoporous layer attached to the Cu substrate with a 1200 μm thick biporous layer on top for a total wick thickness of 1500 μm (same as the monoporous standard product). The overall size of each vapor chamber is 107 mm × 99 mm × 5.5 mm with a 1 mm thick Cu substrate thickness, leaving a 0.5 mm thick remaining vapor space.

The scanning electron microscope (SEM) images of the monoporous wick and biporous wick are shown in Figs. 4(a) and 4(b), respectively, each at 150 times magnification. It was confirmed by measurement that the monoporous wick has final sintered particles in the range of 60–100 μm size as expected, with some particle deformation due to the sintering process. From the SEM image of biporous wick, clusters of ∼300 μm size were observed and larger pores between these clusters were visualized, as well. Qualitatively, due to the existence of these larger pores, the biporous wick has relatively larger porosity compared to the monoporous wick of the same particle size. This is further qualitatively confirmed from the SEM images of each wick at a lower magnification (50 times); see Fig. 5.

These larger pores in the biporous wick are expected to provide for good vapor escape, while the clusters are anticipated to provide adequate liquid supply. However, these larger pores may also lead to larger wall superheat when compared to the performance of a monoporous wick, as seen in Table 1 [9]. Therefore, adding a monoporous layer between the biporous wick layer and the Cu substrate should counterbalance the negative effect of the biporous wick, while exploiting its advantages of easy vapor flow and good liquid supply.

## Experimental Setup

### Heater Design.

The interest of the current work is to evaluate the performance of the three described TGPs at a large heat flux (up to 1 kW/cm2) applied over a large heating area (1 cm2). Therefore, at the center of each TGP, a 10 mm × 10 mm commercial film heater (Mesoscribe™, Cental Islip, NY [19]) with an embedded K-type thermocouple (TC) is printed directly on the center of the Cu substrate with high-temperature solder for the electrical wire connections, as shown in Fig. 6.

The heater is nominally capable of supplying ∼500 W of power (500 W/cm2 heat flux) per the manufacturer specifications with possibility of reaching a 1000 W/cm2 heat flux. To confirm if the heater has the capability to supply higher amounts of power during the TGP thermal performance testing, a pretest of heater performance is carried out using a liquid-cooled test setup. In experimental setup shown in Fig. 7, a Cu substrate of the same footprint area size as the TGPs is clamped between a cold plate and an aluminum (Al) frame. The heater is installed at the center of the Cu substrate. A chiller is connected to the cold plate and the inlet temperature of the circulated water–ethylene glycol coolant is set to 5 °C. A calibrated thermal camera, FLIR SC7650 (accuracy ±1 °C and pixel resolution 640 × 512), is positioned over the top of the heater, Cu substrate, and Al frame. On the upper surface area of the Al frame (including the exposed portion of the heater plus Cu substrate), a 50 mm × 50 mm area is uniformly coated with flat black paint for accurate temperature measurement by infrared (IR) camera. The power supplied to the heater is increased upward from 100 W to confirm the maximum heat flux capability. Each power level is maintained until the temperature reaches steady-state. The heater temperature reading from the embedded TC is monitored by a data acquisition system.

As shown in Fig. 7, there is no temperature fluctuation as the heater power input level is increased from 100 W to 500 W, which indicates that the heater is operationally stable. However, as the heat flux passed the 500 W/cm2 threshold, i.e., at 622.5 W/cm2, the heater temperature did not reach steady-state and continuously increased. When the power input reached ∼700 W, the heater temperature increased at a faster rate making it difficult to control and thus ending the experiment.

In addition to measuring the heater temperature via the embedded TC, the heater temperature was also monitored by the IR camera. It was found that, at 500 W/cm2 heat flux, the maximum temperature within a peripheral portion of the scribed heater trace had already exceeded 300 °C, while the center TC reading was only at ∼140 °C indicating a nonuniform distribution of the applied heat flux. This was not unexpected considering the intrinsic characteristics of the serpentine shaped heater design. At even higher heat flux, it is plausible that hot spots within the heater serpentine trace had a temperature approaching or even exceeding the maximum allowable temperature of the heater of 500 °C. Due to a limitation on the calibrated temperature range of the IR camera, any temperature higher than 400 °C could not be adequately resolved, which is another reason that the heater evaluation test was concluded at 700 W/cm2.

### Test Setup.

An air jet impingement test setup is constructed for the vapor chamber thermal performance evaluation test, as shown in Fig. 8. A regenerative blower (Airtech PB1300, Rutherford, NJ) with speed control supplies pressurized air that is passed through two flowmeters (King instrument 7510/7511 Series, Garden Grove, CA), which are arranged in parallel with ranges of 0 ∼ 20 CFM and 0 ∼ 40 CFM, respectively. As shown in Fig. 8, after arriving at the inlet manifold section, the air flows through a series of air straighteners to reduce turbulence. A calibrated K-type TC (±2.2 °C) is mounted in-line with the jet orifice to measure the inlet air temperature. Cool air flows through the 38.1 mm diameter (d) circular orifice and reaches the main test-section, which is fixed on an adjustable positioning stand; refer to Ref. [20] for additional details. In the main test-section, the vapor chamber sample is sandwiched between a straight pin-fin Al heat sink and a polyetheretherketone (PEEK) cap for minimizing heat loss from the evaporator side to the surrounding environment. The heat sink has a 30 × 30 array of 2 mm × 2 mm square pin fins. The fin height is 34 mm, and the fin pitch is 4 mm. Stainless steel bolts are used for clamping the overall stack-up. A 0.5 mm thick graphite thermal pad is placed between vapor chamber condenser side upper surface and the heat sink to reduce the interfacial thermal resistance. A preliminary test is performed to study the effect of the height between the jet orifice and the heat sink fin top surface on cooling performance. The height is tuned from 2.5 cm to 20 cm with a power input up to 60 W. It was found that the TGP effective thermal resistance is nearly independent of the height, H, between the jet orifice and heat sink, which is expected. However, this distance affects the heater temperature due to height dependency of the air side heat transfer coefficient for jet impingement. While the H/d value may logically be further tuned for maximum air side heat transfer, for simplicity, a H/d = 0.33 value was selected based on the prior study [20].

Five calibrated K-type thermocouples (±0.5 K accuracy) are attached on the TGP condenser side surface, see Fig. 8, with four TCs on the four corners, Tb1 ∼ Tb4, and one at the center, Tc. The evaporator side center temperature, Th, is measured using the embedded heater TC. Aluminum tape is applied to affix the thermocouples and overlapped by another thin layer of thermal insulation tape to ensure that the TC readings represent local TGP temperatures. The overall thermal resistance of the TGP is calculated as
$Rtot=Th−TwQh−Qloss=Th−(Tc+Tb1+Tb2+Tb3+Tb4)/5Qh−Qloss$
(1)
where Tw is the average temperature of the condenser wall, Qh is the heater input power, and Qloss is the system heat loss. Note that the heater input power is directly obtained from the DC power supply via the measured current and a remote-sense line connection that is used to accurately measure the heater voltage drop (to within 1%) directly.

For the current test setup, the bottom side of the heater is insulated by 6.35 mm thick low thermal conductivity, 0.25 W/(m K), PEEK material, while the top side of the heater is a highly thermally conducting TGP plus heat sink. A steady-state conduction-based finite element model of heat loss was developed, where the 100 mm × 100 mm × 5.5 mm TGP was modeled to be a high thermal conductivity, 5000 W/(m K), material. A 1 kW/cm2 heat flux was applied to the 10 mm × 10 mm heated area, and the PEEK insulation material was modeled on the lower surface of the spreader. A representative, 500 W/(m2 K), heat transfer coefficient for air jet impingement was applied to the top surface of the TGP, while a free convection heat transfer coefficient, 2 W/(m2 K), was applied to the exterior surfaces of the PEEK insulation. A mesh refinement study resulted in the use of 2.37 × 105 tetrahedral elements for the analysis domain. Based on this numerical analysis, more than 99% of the applied heat flows upward through the TGP plus heat sink and into the environment. Considering that the system utilizes external air convective cooling, nearly all the heat lost through the heat sink (either by free or forced convection) is logically carried away by the coolant, air. Therefore, in the current work, Qloss was assumed to be negligible, and the total power input was effectively counted toward heating.

## Results and Discussion

The experiments were carried out starting from low power with an increment of approximately 10∼20 W until wick dry out occurred. The inlet air temperature was a stable 30.5 °C during testing. The influence of the air volumetric flow rate on the TGP thermal resistance was explored by changing the flow rate from 15.0 CFM to 31.5 CFM under different heater power levels. Although the TGP evaporator side temperature is lower at large air flow rate due to better heat dissipation, the TGP thermal resistance is independent of the air flow rate. This behavior is analogous to the independency of the thermal resistance in relation to varying the H/d value. Therefore, the inlet air flow rate was maximized at 31.5 CFM and kept constant for the experiments. Experiments were carried out multiple times to establish repeatability with no found degradation in performance over repeated tests.

Figure 9(a) shows the heater temperature as a function of time as the heat input is gradually increased from 20 W to 450 W for the biporous wick TGP. Figure 9(b) provides schematics that describe the heat transfer operational stages within the biporous wick. At an early stage, the heat is mainly transferred by evaporation. The boiling incipience started at a power input of ∼60 W, when Th suddenly drops, as shown in Fig. 9(a). This temperature drop is an indication of boiling onset in the wick [21]. Phase change during nucleate boiling absorbs considerable heat, which leads to a sudden drop of evaporator side temperature at the same power input level. As the power level is increased, the TGP operates stably within the boiling stage until the heat flux reaches ∼420 W/cm2. At this point, dry out occurs, which is indicated by the dramatic increase in heater temperature; see Fig. 9(a). At this high-power level, the boiling is so severe that the capillary force provided by the wick cannot supply enough liquid to balance with the vaporization process. At the end of the test, the TGP vapor chamber physically deformed due to overheating and excessive inner vapor pressure at the high power input. The temperature profile of the monoporous wick TGP and hybrid wick TGP followed a similar trend to that shown for the biporous wick TGP in Fig. 9(a).

Figure 10 shows the thermal resistance of the three different TGPs as a function of applied heat flux. All three types of TGPs started with a high thermal resistance at low heat flux, and the thermal resistance gradually decreased as the heat flux increased. When the applied heat flux reached ∼60 W/cm2, there was a sudden drop of thermal resistance for the TGPs with monoporous wick and biporous wick, again due to the onset of boiling in each vapor chamber. The heat flux value at which this sudden drop was observed for the TGP with hybrid wick was around 70 W/cm2 indicating that the hybrid wick produced a delayed onset of boiling. After the boiling onset, the thermal resistance of the hybrid wick TGP decreased more gradually until reaching an approximately constant value after passing 200 W/cm2.

Observe that the commercial monoporous wick TGP exhibits the highest thermal resistance, 0.21 K/W at a heat flux higher than 200 W/cm2, followed by the biporous wicked TGP, 0.18 K/W. The TGP with hybrid wick displays an even lower thermal resistance, 0.15 K/W, which represents an approximate 28% decrease in thermal resistance when compared with the commercial monoporous TGP. Additionally, the dry-out heat flux of the three TGPs is also different at 360 W/cm2, 420 W/cm2, and 450 W/cm2, respectively, for the monoporous wick, biporous wick, and hybrid wick TGPs.

Figure 11 shows the relationship between the heat flux and the temperature differential across each vapor chamber TGP, ThTw. Each curve can be divided into three sections. The switch from the first section to the second section occurs around a ∼20 °C TGP temperature differential for mono/biporous TGPs, while the switch for hybrid wicked TGP takes place at a temperature differential of ∼40 °C. This is due to the delayed boiling onset as discussed in relationship to Fig. 10. There is limited fundamental understanding and experimental evidence of the factors that influence the onset of boiling in vapor chambers. Weibel et al. [22] found that carbon nanotube coatings clearly have the ability to reduce the incipience superheat, although the exact mechanism is not known. This demonstrates that different wick designs can indeed have a significant effect on the behavior.

The superheat at which incipience occurs during capillary-fed conditions can be very high, much higher than is observed in pool boiling. One theory is that there is a sharp temperature gradient in the thin liquid films formed in the wick that increases the required surface superheat per Hsu's seminal theory for bubble nucleation [23]. This is analogous to the increase in wall superheat required for incipience during flow boiling, due to the thinned boundary layer. We speculate that nucleation can be delayed by any structures that promote a thin liquid film near the heated surface during evaporation, based on a body of evidence in the literature on this topic. It is consistent that the hybrid wick structure, with the thin base layer of particles, might increase the superheat required for incipience.

The curves in Fig. 11 have a steeper slope in the second section when compared to the first section due to different heat transfer mechanisms (i.e., boiling for section two but evaporation for section one) as discussed in relationship to Fig. 9. In the stable boiling range, at the same heat flux, the hybrid wick produces the lowest temperature differential. The biporous wick TGP then follows, and the monoporous wick leads to the highest temperature differential. This trend mirrors the thermal resistance order of the three TGPs. Once critical heat flux is reached, the dry out of the wick occurs, which is seen in the sudden increase of the temperature difference for each TGP in Fig. 11.

Based on the experimental results, there are four main findings:

• (1)

Within the monoporous wick, liquid supply and vapor escape shares the same flow path, which leads to early dry out.

• (2)

The biporous wick separates the liquid supply from the vapor flow which delays dry out.

• (3)

The hybrid wick is a two-level wick structure. The thin monoporous wick layer ensures a relatively low boiling thermal resistance, while the top biporous wick provides dual functions (liquid supply and vapor escape) leading to further delayed dry out.

• (4)

The vapor chamber with hybrid wick shows that delayed onset of boiling possibly due to the thin base layer of particles promotes a thin liquid film near the heated surface during evaporation.

Although the hybrid wick demonstrated improved thermal performance, lower thermal resistance, and delayed dry out, there is still room for improvement. One example is to use a thicker wick top layer to further increase the liquid supply capability. However, a thicker top layer will increase the wick thermal resistance and make vapor escape more difficult. One way around this issue is to decouple the two layers of the hybrid wick. The benefit is that the monoporous wick layer may be thinner to further decrease the thermal resistance, and the top layer may be thicker for better liquid supply. Then, to overcome the disadvantage of a thick liquid supply layer for vapor escape, artificial vapor vents may be constructed. Putting the above thinking into schematic form leads to a TGP evaporator wick design as shown in Fig. 12, which highlights a three-dimensional drawing of a proposed two-layer wick structure adapted from Ref. [24]. The condensed working fluid returns to this evaporator region from the periphery of the vapor chamber through a conventional bulk wick. Liquid is routed from the bulk wick surrounding the evaporator section into a relatively thick wick cap layer into an array of vertical liquid-feeding posts. These posts uniformly distribute the liquid to a thin base wick. Vapor generated by capillary-fed boiling in the base wick escapes through gaps between the posts and travels through vapor vents in the cap wick layer into the vapor core.

## Conclusions

In summary, TGPs with a biporous wick and hybrid wick were designed and fabricated based on wick level test data reported in the open literature. The two TGPs were compared with a commercial monoporous wick TGP from the perspectives of thermal resistance, boiling onset, and dry out heat flux. The two-layer wick structure of hybrid wick demonstrated the lowest thermal resistance, delayed boiling onset, and higher dry out heat flux. However, it is postulated that the intrinsic shortcomings of a monolithic hybrid two-layer wick limit its further performance improvement. Based on a similar idea, which in short leverages the separation of the local liquid supply path from the vapor flow path and utilizes a thinner base wick plus thicker top layer, a novel two-layer wick is targeted for future development. Following Ref. [24], the next step is to optimize the structure of the two-layer wick and apply it to a vapor chamber TGP. The performance of the TGP will be evaluated in a similar fashion to demonstrate its capability in spreading 1 kW/cm2 from a 10 mm × 10 mm heated area to a two-order of magnitude lower heat flux level for synergy with forced air cooling.

## Acknowledgment

The authors thank Purdue University, specifically Dr. Justin Weibel, Mr. Srivathsan Sudhakar, and Dr. Suresh V. Garimella, for their technical discussions and ongoing collaboration related to this work.

## Nomenclature

• d =

diameter of jet orifice, m

•
• H =

distance between jet orifice and heat sink, m

•
• Q =

power, W

•
• R =

thermal resistance, K/W

•
• T =

temperature, °C

### Subscripts

Subscripts

• b =

boundary

•
• c =

center

•
• h =

heater

•
• loss =

heat loss

•
• P =

power supply

•
• tot =

total

•
• w =

condenser side surface wall

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text:
Chemical Reactions ICSE Class-8th Concise Selina Chemistry Solutions Chapter-6. We Provide Step by Step Answer of Objective, True False , Fill in the blanks, Match the following , Short / Long Answer Type of Exercise-6 Chemical Reactions. Visit official Website CISCE for detail information about ICSE Board Class-8.

## Chemical Reactions ICSE Class-8th Concise Selina Chemistry Solutions

### Exercise – I

#### Question 1.

(a) Define a chemical reaction.
(b) What happens during a chemical reaction ?
(c) What do you understand by a chemical bond ?
(a) Any chemical change in matter which involves transformation into one or more substances with entirely different properties is called a chemical reaction.
(b) A chemical reaction involves breaking of chemical bonds between the atoms or groups of atoms of reacting substances and rearrangement of atoms making new bonds to form new substances.

(c) A chemical bond is the attractive force that holds the atoms of a molecule together, in a compound.

#### Question 2.

Give one example each of which illustrates the following characteristics of a chemical reaction:
(a) evolution of a gas
(b) change of colour
(c) change in state
(a) When Zinc reacts with dil. sulphuric acid. Hydrogen gas is evolved, with an effervescence

(b) When blue coloure d copper sulphate reacts with hydrogen sulphide gas, a black coloured substance copper sulphide is formed.

(c) The reaction between hydrogen sulphide and chlorine (both gases) produces sulphur (solid) and hydrogen chloride (gas).

#### Question 3.

How do the following help in bringing about a chemical change?
(a) pressure (b) light
(c) catalyst (d) heat.
(a) Some chemical reactions take place when reactants are subjected to high pressure.
e.g: Nitrogen and hydrogen when subjected to high pressure produce ammonia gas.

(b) Some chemical reactions can take place in the presence of light. Ex. Photosynthesis.

(c) A catalyst can either increases or decreases the rate of chemical reaction and some chemical reactions need a catalyst to change the rate of the reaction, in case it is too slow or too fast.

1. Positive catalyst: When a catalyst increase the rate of reaction finely divided iron is used as a positive catalyst in the manufacturing of ammonia from hydrogen and nitrogen.

1. Negative Catalyst: When a catalyst decreases the rate of reaction.
Ex. Phosphoric acid act as a negative catalyst to decrease the rate of the decomposition of hydrogen peroxide.

(d) Some chemical reactions take place only in the presence of heat.
e.g. When lead nitrate is heated, it breaks into lead monoxide, nitrogen dioxide and oxygen.

#### Question 4.

(a) Define catalyst.
(b) What are (i) positive catalysts and (ii) negative catalysts? Support your answer with one example for each of them.
(c) Name three biochemical catalysts found in the human body.
(a) Catalyst: A catalyst is a substance that either increases or decreases the rate of a chemical reaction without itself undergoing any chemical change during the reaction.

(b) (i) Positive catalyst: When a catalyst increases the rate of chemical reaction, it is called positive catalyst.
e.g. when potassium chlorate heated to 700°C decomposes to evolve oxygen gas, when MnO2 is added the decomposition takes place at 300°C

(ii) Negative catalyst: When a catalyst decreases the rate of chemical reaction it is called negative catalyst.
Example. Phosphoric acid acts as a negative catalyst to decrease the rate of the decomposition of hydrogen peroxide. Alcohol too acts as a negative catalyst in certain chemical reactions.

(c) Biochemical catalysts found in human body:

1. Pepsin
2. Tryspin
3. lipase.

#### Question 5.

What do you observe when
(a) dilute sulphuric acid is added to granulated zinc?
(b) a few pieces of iron are dropped in a blue solution of copper sulphate?
(c) silver nitrate is added to a solution of sodium chloride?

(d) ferrous sulphate solution is added to an aqueous solution of sodium hydroxide.
(e) solid lead nitrate is heated?
(f) when dilute sulphuric acid is added to barium chloride solution ?
(a) When Zinc reacts with dilute sulphuric acid, hydrogen gas is evolved with an effervesence.
Zn + dil. H2SO4 → Zn SO4 + H2.

(b) When a few pieces of iron are dropped into a blue coloured copper sulphate solution, the blue colour of the solution fades and eventually turns into green.

(c) When a solution of silver nitrate is added to a solution of sodium chloride, white insoluble ppt. of silver chloride is formed.
AgNO3 (aq) + NaCl (aq) → AgCl (ppt) + NaNO3 (aq)

(d) When ferrous sulphate solution is added to sodium hydroxide solution, a dirty green ppt. of ferrous hydroxide is formed.
FeSO4 (aq) + 2NaOH (aq) → Fe(OH)2 ↓ + Na2SO4(aq)

(e) When solid lead nitrate is heated, it decomposes to produce light yellow solid lead monoxide, reddish brown nitrogen dioxide gas and colourless oxygen gas.

#### Question 6.

Complete and balance the following chemical equations:

………….

### Exercise – II

#### Question 1.

1. Fill in the blanks.
(a) A reaction in which two or more substances combine to form a single substance is called a combination reaction.
(b) A catalyst is a substance which changes the rate of a chemical reaction without undergoing a chemical change.
(c) The formation of gas bubbles in a liquid during a reaction is called effervesence
(d) The reaction between an acid and a base is called neutralization reaction.
(e) Soluble bases are called alkalis.
(f) The chemical change involving iron and hydrochloric acid illustrates a displacement reaction.
(g) In the type of reaction called double decomposition reaction, ions two compounds exchange their positive and negative radicals ions respectively.
(h) A catalyst either increases or decreases the rate of a chemical change

but itself remains unchanged at the end of the reaction.
(i) The chemical reaction between hydrogen and chlorine is a combinaton reaction
(j) When a piece of copper is added to silver nitrate solution, it turns blue in colour.

#### Question 2.

Classify the following reactions as combination, decomposition, displacement, precipitation and neutralization. Also balance the equations.

#### Question 3.

Define:
(a) precipitation (b) neutralization (c) catalyst
(a) Precipitation: A chemical reaction in which two compounds in their aqueous state react to form an insoluble salt as one of the product.
Acid + Base → Salt + Water
Example.

here iron act as a catalyst and increases the rate of chemical reaction.

#### Question 4.

Explain the following types of chemical reactions giving two examples for each of them.
(a) combination reaction
(b) decomposition reaction
(c) displacement reaction
(d) double decomposition reaction
(a) Combination reaction: A reaction in which two or more substances combine to form a single substance is called combination reaction.
A + B → AB
e.g (i) When iron and sulphur are heated together, they combine to form iron sulphide.

c) Displacement reaction: A reaction in which a more active element displaces a less active element from a compound is called displacement reaction.
AB + C → CB + A
e.g. (i) Zinc, displaces copper from copper sulphate solution.
Zn + CuSO4 (aq) → ZnSO4 (aq) + Cu
(ii) Iron piece when added to copper sulphate solution, copper is displaced.
Fe + CuSO4 → FeSO4 + Cu.

(d) Double decomposition reaction: A chemical reaction in which two compounds in their aqueous state exchange their ions to form new compounds is called a double decomposition reaction.
AB + CD → CB + AD
e.g. (i) AgNO3 + HCl  AgCl + HNO3(aq)
(ii) NaOH (aq) + HCl (aq)  NaCl (aq) + H2O.

#### Question 5.

Write the missing reactants and products and balance the equations

#### Question 6.

How will you obtain?
(a) Magnesium oxide from magnesium.
(b) Silver chloride from silver nitrate.
(c) Nitrogen dioxide from lead nitrate.
(d) Zinc chloride from zinc.
(e) Ammonia from nitrogen.
Also give balanced equations for the reactions
(a) Magnesium when burnt in air (oxygen) Magnesium oxide is formed

#### Question 7.

What do you observe when
(a) Iron nail is kept in copper sulphate solution for sometime.
(b) Phenolphthalein is added to sodium hydroxide solution.
(c) Blue litmus paper is dipped in dilute hydrochloric acid.
(e) Magnesium ribbon is burnt in oxygen.
(f) Ammonia is brought in contact of hydrogen chloride.gas.
(a) A brown layer of copper gets deposited on iron nail. This is due to chemical reaction.
Fe (s) + CuSO4 (aq) → FeSO4 (aq) + Cu (s)
(b) Solution turns pink.
(c) Blue litmus turns red in an acid solution.
(d) The pale yellow solid is lead monoxide, the reddish brown gas is nitrogen dioxide and the colourless gas is oxygen.

(e) Magnesium ribbon bums with a dazzling white light and produces a white powder which is magnesium oxide.
The reaction can be represented as
2Mg + O2 → 2MgO (white powder)
(f) Ammonia and hydrogen chloride, both compounds, combine to form a compound, ammonium chloride.

#### Question 8.

Give reason:
(a) A person suffering from acidity is advised to take an antacid.
(b) Acidic soil is treated with quick lime.
(c) Wasp sting is treated with vinegar.
(a) An antacid neutralizes stomach acidity.
(b) If the soil is acidic it can be treated with base like quick lime, to make it neutral.
(c) Wasp stings are alkaline and they can be neutralized by vinegar which is a weak acid.

#### Question 9.

What is meant by the metal reactivity series ? State its importance, (any two points).
A list in which the metals are arranged in the decreasing order of their chemical reactivity is called the metal reactivity series.

Special features of the activity series:

1. The ease with which a metal in solution loses electron(s) and forms a positive ion decreases down the series, i.e. from potassium to gold.
2. Hydrogen is included in the activity series because, like metals do, it too loses an electron and becomes positively charged (H+) in most chemical reactions.
3. The series facilitates the comparative study of metals in terms of the degree of their reactivity.
4. The compounds of the metals (oxides, carbonates, nitrates and hydroxides) too can be easily compared.

#### Question 10.

What are oxides ? Give two examples of each of the following oxides.
(a) Basic oxide (b) Acidic oxide
(c) Amphoteric oxide (d) Neutral oxide
An oxide is a compound which essentially contains oxygen in its molecule, chemically combined with a metal or a non-metal.

#### Question 11.

Define exothermic and endothermic reactions. Give two examples of each.
Exothermic reactions: The chemical reaction in which heat is given out is called exothermic reactions. It causes rise in temperature. .
e.g. (i) When carbon bums in oxygen to form carbon dioxide, a lot of heat is produced.
C + O2 → CO2 + heat.
When water is added to quicklime a lot of heat is produced which boils the water.
CaO + H2O → Ca (OH)2 + Heat.

Endothermic reaction: A chemical reaction in which heat is absorbed is called endothermic reaction. It causes fall in temperature.
e.g. (i) When nitrogen and oxygen together are heated to a temperature of about 3000°C, nitric oxide gas is formed.
N2 + O2 + heat → 2NO (g)

(ii) Decomposition of calcium carbonate into carbon dioxide and calcium oxide when heated to a 1000°C.
CaCO3 + Heat → CaO (s) + CO2 (g)

#### Question 12.

State the effect of:
(a) an endothermic reaction
(b) an exothermic reaction on the surroundings.
(a) Carbon dioxide present in the atmosphere is trapped by infrared radiations, gives rise to temperature which is exothermic reaction.
(b) The melting of glaciers by global warming.

#### Question 13.

What do you observe when
(a) an acid is added to a basic solution.

(b) ammonium chloride is dissolved in water.
(a) A chemical reaction in which a base or an alkali reacts with an acid to produce a salt and water.
Acid + Base → Salt + Water
(b) Dissolution of ammonium chloride in water is an endothemic reaction in which heat energy is absorbed.

Thanks

$${}$$


text:
bug-cvs
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## Undelivered mail

 From: undelivered Subject: Undelivered mail Date: Sun, 23 May 2004 23:05:47 +0800

```I'm sorry to have to inform you that the message returned
below could not be delivered to one or more destinations.


text:
# Cylindrical waves

I am trying to solve the general equation for cylindrical symmetric waves: $$\frac1{c^2}\frac{\partial^2u}{\partial t^2}= \frac1r\frac{\partial}{\partial r}(r\frac{\partial}{\partial r}u)$$ with $u = u(r,t)$.

I was expecting that by plugging a function of the form $u(r,t) = \frac{1}{\sqrt{r}}f(r,t)$ I would arrive with something a nice plane wave form for this $f(r,t)$ but I don't. My idea was that for spherically symmetric waves we would put $u = \frac1R f$ as the energy goes as $\frac{1}{r^2}$, but with cylindrical propagation it goes as $\frac1r$. What am I thinking or doing wrong? Does someone know the common solution for this equation?

Edit:

After having thought a bit more about it, I am now trying to get the eigenmodes, so to resolve: $$\frac{\omega^2}{c^2}u + \frac{\partial}{\partial r^2}u + \frac1r \frac{\partial}{\partial r}u=0$$ which, after multiplying by $r^2$ looks like a Bessel diff. equation: $$x^2y''+xy'+(x^2-p^2)y=0$$ for $p=0$. But which operation can I do to get rid of the $\frac{\omega^2}{c^2}$ factor. Even after that, I am not sure how to ling the eigenvalues of the modes to the zeros of the Bessel functions, given that I have the boundary condition $u(R,t) = 0$ for $R$ the diameter of my cylinder.

-

## migrated from physics.stackexchange.comJan 30 '13 at 2:34

This question came from our site for active researchers, academics and students of physics.

I pretty much found my answer in there: vixra.org/pdf/0908.0045v1.pdf An amplitude decreasing as $\frac{1}{\sqrt{r}}$ is true asymptotically, once you can neglect the energy spread through the basis of your cylinder. –  Learning is a mess Jan 29 '13 at 14:54
Look up the fundamental solution to the wave equation in various dimensions. The "cylindrical symmetry" case is essentially the same as the situation in 2 spatial dimensions. When the spatial dimension is even strong Huygens' principle fails and there is no simple reduction to the plane-wave form. When the spatial dimension is odd, there is a reduction, but the formula is much more complicated in general compared to the 3 dimensional case. –  Willie Wong Jan 29 '13 at 16:14
Thanks for the input Willie. I just updated my question and sharpened my question. –  Learning is a mess Jan 29 '13 at 18:32
To see that this is a Bessel DE, try rescaling your $r$. (i.e. substitute $r' = kr$ where $k = \omega/c$) –  Wouter Jan 29 '13 at 18:53
That's right. On a sidenote: in this case there's no problem applying the boundary conditions because they are zero. However in general, if the BC's are e.g. $u(R,t) = f(t)$, you'd have to fourier transform those as well before applying them. (or inverse-fourier transform your solution and apply the original BC's) –  Wouter Jan 29 '13 at 19:04


text:
# Transforming range of variable to 0 to 1 and custom ranges

Created:18 Jul 2020 00:39:07 , in  Maths

One useful skill when it comes to dealing with variables is to be able to change their range. One of the most useful and desirable ranges is [0,1]. Occasionally you might need a different range like [-1,1], too. This article describes a couple of methods of scaling and descaling a variable's range.

## Quick and easy transformation to [0,1]

One way to scale range of variable x to range from 0 to 1 is based on the following function:

$y=x/\left(1+x\right)$

scaled variable x, called y, will be in the range between 0 and 1.

Scaling a to custom range

If a range from a to b, where a is the minimum and b is the maximum allowed value, is required, one of the three following formulas can be used:

1)

$y=\frac{\left(\mathit{ax}+b\right)}{\left(1+x\right)}$

2)

$y=a+\left(b-a\right)\left(1-{e}^{-x}\right)$

where e is the Euler's number

3)

$y=a+2\cdot \left(b-a\right)\left(\mathrm{arctan}\left(x\right)\right)/\mathrm{\pi }$

## Scaling variable from [0,1] to a custom range

If a variable y is in range from 0 to 1, it can be scaled to a variable z which is in range between a and b as follows:

$z=a+\left(b-a\right)\ast y$

## Min-max scaling

Min-max scaling, also known as min-max normalization, is another easy way, which is frequently used for transforming variable's range.

The formula for min-max scaling is as follows:

$y=\frac{\left(x-a\right)}{\left(b-a\right)}$

where a and b denote minimum and maximum values of x in a set of data respectively.

Descaling variable y back to x can be done using:

$x=y\cdot \left(b-a\right)+a$

## Min-max scaling in PHP

Two functions given below perform min-max scaling and descaling as described in the previous section.

function minMaxNormalize($x,$a, $b) {$y = false;
try {
$y = ($x - $a) / ($b - $a); }catch(DivisionByZeroError$e){
echo $e -> getMessage(),"\n"; } return$y;
}

function minMaxDenormalize($y,$a, $b) {$x = ($y * ($b - $a) +$a);
return \$x;
}

Here is how to use these functions

Suppose that the range of values of x is from -10 to 10, but we want it to be between 0 and 1. To transform the range, do the following:

minMaxNormalize(7, -10, 10);
// => 0.85

To transform variables to be back in the range [-10,10] :

minMaxDenormalize(0.85,-10,10);
// => 7

## Conclusion

As can be seen in this article, there are quiet a few ways of transforming a range of a variable both to [0,1] or to a custom range. I hope you have found at least one of them sufficient for your purposes. If you have something to add, you are welcome to post a comment below.

This post was updated on 18 Jul 2020 01:30:45

Tags:  php

### Author

Author of the above article, Sylwester Wojnowski, enjoys sWWW writing computer code in PHP, JavaScript and BASH, and some other things he wrote more on on the About page of this website.

©Copyright, 2020 Sylwester Wojnowski. This article may not be reproduced or published as a whole or in parts without permission from the author. If you share it, please give author credit and do not remove embedded links.

Computer code, if present in the article, is excluded from the above and licensed under GPLv3.

### Citation

Wojnowski, Sylwester. "Transforming range of variable to 0 to 1 and custom ranges." From sWWW - Code For The Web . https://wojnowski.net.pl//main/index/transforming-range-of-variable-to-0-to-1-and-custom-ranges

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## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

We can find the initial momentum of the system consisting of the five train cars. Let $M$ be the mass of each train car. $p_0 = 3M(2.0~m/s)+M(4.0~m/s)+ M(0)$ $p_0 = (10M)~N~s$ We can use conservation of momentum to find the final speed $v_f$ of the five cars when they are moving together. $p_f = p_0$ $(5M)~v_f = (10M)~N~s$ $v_f = 2.0~m/s$ The speed of the five-car train is 2.0 m/s.


text:
What is the smallest graph that contains all non-isomorphic 4-node and 5-node connected graphs as induced subgraphs?

I'm looking for an undirected graph $G$ which contains as induced subgraphs:

• the six $4$-node connected graphs and,
• the twenty one $5$-node connected graphs.

Ideally, I would also like it to be as small as possible (least number of nodes). If we take the union of these graphs, it has $6 \times 4+21 \times 5=129$ nodes. If we take the graphs and identify them at a point, it has $6 \times 4+21 \times 5-(6+21)+1=103$ nodes.

This would be helpful for testing our program for network motif detection program (NetMODE), which searches for connected induced subgraphs in networks.

If it helps, I've attached a picture of the graphs below:

-
Each of the graphs with 4 nodes can be produced from one of the graphs of 5 nodes by removing a node, therefore any graph containing all 5 node connected graphs will also contain all 4 node connected graphs. If a 5 node graph has x arcs and x<9 then a 5-node graph with x+1 arcs can be made by removing a node of degree n and replacing with a node of degree n+1. Sometimes more than one x+1 graph can be made from one x graph, or more than one x graph makes the same x+1 graph, but at least I can deduce that the smallest G has at most 5*(21-(10-4))+(10-4)+5=86 nodes. –  Angela Richardson Oct 24 '11 at 5:38
The smallest graph which contains all of the 4-vertex graphs has 7 vertices, and there are actually 77 such graphs (from a calculation with Maple). –  jp26 Oct 24 '11 at 15:59
Plus there is a graph with 8 vertices and 16 edges which contains 19 of the 21 5-vertex graphs, so I would conjecture that there will be a graph with 9 vertices which meets your needs... –  jp26 Oct 24 '11 at 17:20

Surprisingly, the answer is 7 vertices for all 4-vertex subgraphs and 9 vertices for all 5-vertex subgraphs. I checked all graphs with 8 vertices and this was the first one (and only one so far) with 9 vertices that I found:

This is the listing of each of the 21 different subgraphs with 5 vertices:

This is a graph which contains all the graphs with 4 vertices:

That is, $K_4 : \{a,b,c,d\}$, $K_4-e : \{ a,b,c,e \}$, $C_4 : \{a,b,e,f \}$, $P_4 : \{ c,d,e,f \}$, $K_{1,3} : \{b,e,f,g\}$ and the final one $\{b,c,d,f\}$.

And this is a graph with 10 vertices which contains all connected graphs with 5 vertices:

This is all of the different subgraphs of the latter graph:

I'll continue the search for a 9 vertex graph overnight, but I'm now doubting it can exist. I can also update my answer with the Maple code if anyone is interested.

Even with the exponential increase in number of small subgraphs, there are so many graphs that I'd like to believe that there would be a graph with all subgraphs on $k$ vertices which isn't exponential in $k$. (but I would be wrong!)

-
I wouldn't mind seeing the maple code. Feel free to put it somewhere and link from the answer if it would clutter the answer. –  Jack Schmidt Oct 24 '11 at 20:47
Not sure how to make code look nice, but here's the base code. Call it for a maple graph G with: DrawGraph(fiveinlist(G, Glist5)); with(GraphTheory): Glist5:=[NonIsomorphicGraphs(5,output=graphs,outputform=graph,restrictto=connected)]: fiveinlist := proc (G, Glist) local S, bl, ts, l, ns; ts := []; for S in Glist do ns := 0; bl := 0; for lsubs in combinat[choose](Vertices(G), NumberOfVertices(Glist[1])) do if IsIsomorphic(InducedSubgraph(G, lsubs), S) then ns := 1; bl := lsubs end if end do; if ns = 1 then ts := [op(ts), InducedSubgraph(G, bl)] end if end do; RETURN(ts) end proc; –  jp26 Oct 24 '11 at 21:41
thanks. Do you have the code to make your specific G as well? (Sorry part of this is just to learn how to do reasonable things with graphs in maple.) Did you just search through random graphs on 10 vertices and are now searching 9 vertices exhaustively? –  Jack Schmidt Oct 24 '11 at 21:48
(ok, got the "four" one done at least) –  Jack Schmidt Oct 24 '11 at 21:54
I made that graph with: G5 := Graph(8); G5 := AddEdge(G5, {{1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {1, 7}, {2, 3}, {2, 4}, {2, 5}, {2, 6}, {2, 8}, {3, 4}, {3, 5}, {4, 7}, {6, 7}, {7, 8}}); G5b := AddVertex(G5, 9, inplace = false); AddEdge(G5b, {{5, 9}, {8, 9}}); G5c := AddVertex(G5b, 10, inplace = false); AddEdge(G5c, {{1, 10}, {2, 10}, {3, 10}, {4, 10}}); –  jp26 Oct 24 '11 at 22:28

The size of a graph that contains all $k$-vertex subgraphs as induced subgraphs is exponential in $k$.

There are more than $2^{\binom{k}2}/k!$ graphs on $n$ vertices. So if $X$ is a graph on $n$ vertices that contains all $k$ vertex subgraphs as induced subgraphs then $$\binom{n}{k} \ge 2^{\binom{k}2}/k!$$ and as $$\binom{n}{k} \le \frac{n^k}{k!}$$ we have $$n^k \ge 2^{\binom{k}2}.$$ So $$k\, \log_2(n) \ge k(k-1)/2$$ and therefore $n \ge 2^{(k-1)/2}$.

Let $p$ be a prime congruent to 1 mod 4. The vertices of the Paley graph are the integers mod $p$; two integers are adjacent if their difference is a square mod $p$. The Paley graph on $p$ vertices contains all graphs on $k$ vertices as induced subgraphs, where $k\approx c\,\log_2(p)$.

-


text:
# Routing tests in ZF2 / ZF2 equivalent of Zend_Test_PHPUnit_Controller_TestCase?

Thus far I've been testing my ZF2 controllers as follows:

namespace Application\Controller;

use Application\Controller\IndexController;
use Zend\Http\Request;
use Zend\Http\Response;
use Zend\Mvc\MvcEvent;
use Zend\Mvc\Router\RouteMatch;
use PHPUnit_Framework_TestCase;

class IndexControllerTest extends PHPUnit_Framework_TestCase
{
public function testIndexActionCanBeAccessed()
{
$this->routeMatch->setParam('action', 'index');$result   = $this->controller->dispatch($this->request);
$response =$this->controller->getResponse();

$this->assertEquals(200,$response->getStatusCode());
$this->assertInstanceOf('Zend\View\Model\ViewModel',$result);
}

protected function setUp()
{
\Zend\Mvc\Application::init(include 'config/application.config.php');

$this->controller = new IndexController();$this->request    = new Request();
$this->routeMatch = new RouteMatch(array('controller' => 'index'));$this->event      = new MvcEvent();
$this->event->setRouteMatch($this->routeMatch);
$this->controller->setEvent($this->event);
}

protected $controller = null; protected$event = null;
protected $request = null; protected$response = null;
protected $routeMatch = null; }  This allows me to test that the ViewModel is having the correct data (if any) assigned to it before the view is rendered. This serves that purpose just fine, but what it doesn't do is test that my routing is working correctly like the ZF1 Zend_Test_PHPUnit_Controller_TestCase tests would. In those, I'd kick off the test by running $this->dispatch('/some/relative/url') and only get positive test results if the routes were set up correctly. With these ZF2 tests, I'm specifically telling it which route to use, which doesn't necessarily mean that a real request will be routed correctly.

How do I test that my routing is working correctly in ZF2?

-

I'm late to the party, but it still could be useful for newcomers. The solution nowadays would be to inherit from \Zend\Test\PHPUnit\Controller\AbstractControllerTestCase, so the usage would be very similar to ZF1:

class IndexControllerTest extends \Zend\Test\PHPUnit\Controller\AbstractHttpControllerTestCase
{
public function setUp()
{
$this->setApplicationConfig( include __DIR__ . '/../../../../../config/application.config.php' ); parent::setUp(); } public function testIndexActionCanBeAccessed() {$this->dispatch('/');

$this->assertResponseStatusCode(200);$this->assertModuleName('application');
$this->assertControllerName('application\controller\index');$this->assertControllerClass('IndexController');
$this->assertMatchedRouteName('home');$this->assertQuery('html > head');
}
}

Note: This uses \Zend\Test\PHPUnit\Controller\AbstractHttpControllerTestCase which includes assertQuery($path) as well as other web related methods. - That actually doesn't allow to test the return value of the controller dispatch since AbstractHttpControllerTestCase::dispatch() doesn't return anything. – Eric MORAND Aug 19 '13 at 21:46 @EricMORAND, doesn't$this->getResponse()->getContent() allow you to test whatever needs to be tested ? –  PowerKiKi Aug 31 '13 at 16:08

EDIT: ZF2 has been updated since I self-answered this. PowerKiki's answer is better.

-


text:
# Micromouse Lab 4

Micromouse Lab 3
Odometry

Contents: Checkoff 1

Previously, we learned how to read the encoders on our mouse. Encoders are a type of sensor that tells us how many times each wheel is rotated. Using this information, we can calculate how far our mouse travelled and how much it has turned. This is a form of odometry, the use of data from motion sensors to estimate change in position over time.

Why is odometry important? Suppose we want our mouse to move forward by one maze cell. We could turn on both motors for a fixed amount of time, and hope the mouse moves exactly the distance we want it to. In practice, this doesn’t work due to friction, changing battery voltage, and many other factors. The difference between the desired movement and the actual movement is called error, and knowing the error is the first step towards correcting it.

Differential Drive Robot Odometry

Now that we have odometry information (distance and velocity) for each wheel, let’s figure out how to compute odometry for the entire mouse. Our mouse is a differential drive robot, which means that it has two drive wheels located side by side. Both wheels turning in the same direction causes forward (or backward) motion, and both wheels turning in opposite directions causes rotation.

Sidenote
When we describe the position and velocity of a robot, we are actually describing the position and velocity of a single point somewhere on the robot. For a differential drive robot like our mouse, this point is usually taken to be halfway between the wheels. Since a point can’t have a rotational velocity, the “point” is actually a coordinate frame. You can think of a coordinate frame as a set of coordinate axes that are “attached” to the mouse and move/rotate with it.

Mouse coordinate frame

Consider the coordinate frame in the diagram above. When the mouse moves forward, it moves in the positive x direction. When the mouse rotates, it rotates around the z axis (which points upwards). The concept of coordinate frames is fundamental to robotics, but we won’t be able to cover it in much depth (try EECS C106A if you're interested in this type of stuff!). Fortunately, we only need to know some basics.

We are mostly interested in the forward velocity and rotational velocity of our mouse. Intuitively, the center of the mouse moves with a velocity that is the average of the wheel velocities. If both wheels move forward at the same speed, the center of the mouse moves forward with that speed. If the two wheels move in opposite directions at the same speed, the center of the mouse stays in place.

We can describe the forward velocity using the following formula:

$$\huge{v_{f} = \frac{1}{2}(v_{1} + v_{r})}$$

Where vl is the left wheel velocity and vr is the right wheel velocity

Forward velocity

We can also derive a formula for the rotational velocity of the mouse. Just as the forward velocity only depends on the sum (or average) of the two wheel velocities, the rotational velocity only depends on the difference of the two wheel velocities.

The speed of each wheel relative to the center of the mouse is half of the difference between their individual speeds. Rotational velocity is arc velocity divided by radius. We can describe the rotational velocity using the following formula:

$$\huge{\omega = \frac{\frac{1}{2}(v_{1} + v_{r})}{\frac{1}{2}d} = \frac{v_{r} - v_{l}}{d}}$$

Where $$\omega$$ is rotational velocity and $$d$$ is the distance between the wheels (wheel track) is stored as the WHEELBASE_DIAMETER in the code.

You will need to translate these formulas into Arduino code.

Arduino Code
#include <Wire.h>
#include <VL53L0X.h>

#define PIN_ENCODER_LEFT_A 7
#define PIN_ENCODER_LEFT_B 2
#define PIN_ENCODER_RIGHT_A 8
#define PIN_ENCODER_RIGHT_B 3

// Mouse physical parameters
const float ENCODER_TICKS_PER_REVOLUTION = /* TODO: fill this in! */;
const float WHEELBASE_DIAMETER = /* TODO: measure this and fill it in! */; // mm
const float WHEEL_DIAMETER = 34.0; // mm

// Constant to compute velocities with (mm/sec)
const float VELOCITY_COEFF = WHEEL_DIAMETER * PI / ENCODER_TICKS_PER_REVOLUTION * 1000000.0;

// Encoder helper variables
unsigned long prev_pulse_time_right;
unsigned long prev_pulse_time_left;

// Encoder state variables
long ticks_left = 0; // ticks
long ticks_right = 0; // ticks
double velocity_left = 0; // millimeters/sec
double velocity_right = 0; // millimeters/sec

double velocity_forward;
double velocity_turn;

int count = 0;

void setup() {
Serial.begin(9600);

// Encoder setup
pinMode(PIN_ENCODER_LEFT_A, INPUT);
pinMode(PIN_ENCODER_LEFT_B, INPUT);
pinMode(PIN_ENCODER_RIGHT_A, INPUT);
pinMode(PIN_ENCODER_RIGHT_B, INPUT);
attachInterrupt(digitalPinToInterrupt(PIN_ENCODER_LEFT_B),
leftEncoderRisingEdge, RISING);
attachInterrupt(digitalPinToInterrupt(PIN_ENCODER_RIGHT_B),
rightEncoderRisingEdge, RISING);

/*The checkEncodersZeroVelocity() function sets wheel velocities to zero
if it hasn’t seen an encoder pulse in 100 milliseconds. Why do we need this?
Will the interrupt callbacks (leftEncoderRisingEdge or rightEncoderRisingEdge)
ever set a wheel velocity to 0? */
}

void loop() {
checkEncodersZeroVelocity();

velocity_forward = /* YOUR CODE HERE */;
velocity_turn = /* YOUR CODE HERE */;

if (count % 1000 == 0) {
// Print debug info every 1000 loops
Serial.print(velocity_turn);
Serial.print(" ");
/* Scale the forwards velocity so it's easier to visualize next to
the turning velocity. */
Serial.println(velocity_forward / 100.0);
}
count++;
}

//////////////////////
// Helper functions //
//////////////////////

void checkEncodersZeroVelocity(void) {
// Sets the wheel velocity to 0 if we haven't seen an edge in a while
unsigned long curr_time = micros();
if (curr_time - prev_pulse_time_left > 100000) {
velocity_left = 0;
}
if (curr_time - prev_pulse_time_right > 100000) {
velocity_right = 0;
}
}

void leftEncoderRisingEdge(void) {
unsigned long curr_time = micros();
int direction;

direction = 1;
} else {
direction = -1;
}

if (direction * velocity_left < 0) {
velocity_left = 0;
} else {
// Otherwise, convert the period of our pulse in mm/second
velocity_left = direction * VELOCITY_COEFF / (curr_time - prev_pulse_time_left);
}
ticks_left += direction;
prev_pulse_time_left = curr_time;
}

void rightEncoderRisingEdge(void) {
unsigned long curr_time = micros();
int direction;

direction = -1;
} else {
direction = 1;
}

if (direction * velocity_right < 0) {
velocity_right = 0;
} else {
// Otherwise, convert the period of our pulse in mm/second
velocity_right = direction * VELOCITY_COEFF / (curr_time - prev_pulse_time_right);
}

ticks_right += direction;
prev_pulse_time_right = curr_time;
}

Checkoff #1

1. How many millimeters does a wheel travel per encoder tick?
2. The checkEncodersZeroVelocity() function sets wheel velocities to zero if it hasn’t seen an encoder pulse in 100 milliseconds. Why do we need this? Will the interrupt callbacks (leftEncoderRisingEdge or rightEncoderRisingEdge) ever set a wheel velocity to 0?
3. In the code above, use the individual wheel velocities (in mm/sec) to calculate the forward velocity (in mm/sec) and turning velocity (in rad/sec). Visualize these in your Serial Plotter.