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1 year ago

Define waves as it applies to electromagnetic fields

1 answer:

5 0

Waves in the electric and magnetic fields are known as electromagnetic waves. You must first understand what a field is, which is just a technique of giving each square inch of space a numerical value. You may see that as a temperature field, for instance, when you look at the weather predictions and they mention the temperature in several locations. Every location on Earth has a unique temperature that can be quantified. Everywhere on Earth has its own wind velocity, which is another form of field. This field differs somewhat from the temperature field in that the wind velocity has both a direction and a magnitude, whereas the temperature just has a magnitude (how hot it is). A vector is a quantity that has both magnitude and direction, hence a field that contains vectors at every location is referred to as a vector field. Vector fields include the magnetic and electric fields. We may examine what would happen if we placed a charged particle at any given position in space. If the charged particle were to accelerate, we would state that the electric field there is the direction in which the particle is moving. In general, positively charged particles will move in the electric field's direction, whereas negatively charged particles will move in the opposite way. Because it is a vector field, the magnetic field exhibits comparable behavior. We discovered in the 19th century that the same interaction, electromagnetism, really produces both electric and magnetic fields. Like an electromagnet, a changing electric field will produce a magnetic field, and a changing magnetic field will induce an electric field (like in a generator). If your system is configured properly, you may have an electric field that fluctuates, which in turn produces a magnetic field, which in turn induces another electric field, which in turn generates another magnetic field, and so on indefinitely. At the speed of light, this oscillation between a strong magnetic field and strong electric field spreads out indefinitely. In reality, light is an electromagnetic wave—an oscillation in the electromagnetic fields. An electric or magnetic field may exist without a medium since they exist in a vacuum, which implies that waves in these fields don't require a medium like sound to flow through.

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Explanation:

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2 years ago

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2 years ago

Define factors that can change the performance of a polymer, such are additives

inna [77]

Answer:

The performance of the polymer is basically change by the various type of factors like shape, tensile strength and color.

The polymer based products are basically influenced by the environmental factors like light, acids or alkalis chemicals, salts and also heat.

The additives is one of the type of chemical polymer which basically include polymer matrix for improving the ability of processing of the polymer. It also helps to enhance the production and requirement of the polymer products in the environment.

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2 years ago

2. A counter flow tube-shell heat exchanger is used to heat a cold water stream from 18 to 78oC at a flow rate of 1 kg/s. Heatin

Anastaziya [24]

Answer:

a) $L = 220\,m$ , b) $U_{o} \approx 0.63\,\frac{kW}{m^{2}\cdot ^{\textdegree}C}$

Explanation:

a) The counterflow heat exchanger is presented in the attachment. Given that cold water is an uncompressible fluid, specific heat does not vary significantly with changes on temperature. Let assume that cold water has the following specific heat:

$c_{p,c} = 4.186\,\frac{kJ}{kg\cdot ^{\textdegree}C}$

The effectiveness of the counterflow heat exchanger as a function of the capacity ratio and NTU is:

$\epsilon = \frac{1-e^{-NTU\cdot(1-c)}}{1-c\cdot e^{-NTU\cdot (1-c)}}$

The capacity ratio is:

$c = \frac{C_{min}}{C_{max}}$

$c = \frac{(1\,\frac{kg}{s} )\cdot(4.186\,\frac{kW}{kg^{\textdegree}C} )}{(1.8\,\frac{kg}{s} )\cdot(4.30\,\frac{kW}{kg^{\textdegree}C} )}$

$c = 0.541$

Heat exchangers with NTU greater than 3 have enormous heat transfer surfaces and are not justified economically. Let consider that $NTU = 2.5$ . The efectiveness of the heat exchanger is:

$\epsilon = \frac{1-e^{-(2.5)\cdot(1-0.541)}}{1-(2.5)\cdot e^{-(2.5)\cdot (1-0.541)}}$

$\epsilon \approx 0.824$

The real heat transfer rate is:

$\dot Q = \epsilon \cdot \dot Q_{max}$

$\dot Q = \epsilon \cdot C_{min}\cdot (T_{h,in}-T_{c,in})$

$\dot Q = (0.824)\cdot (4.186\,\frac{kW}{^{\textdegree}C} )\cdot (160^{\textdegree}C-18^{\textdegree}C)$

$\dot Q = 489.795\,kW$

The exit temperature of the hot fluid is:

$\dot Q = \dot m_{h}\cdot c_{p,h}\cdot (T_{h,in}-T_{h,out})$

$T_{h,out} = T_{h,in} - \frac{\dot Q}{\dot m_{h}\cdot c_{p,h}}$

$T_{h,out} = 160^{\textdegree}C + \frac{489.795\,kW}{(7.74\,\frac{kW}{^{\textdegree}C} )}$

$T_{h,out} = 96.719^{\textdegree}C$

The log mean temperature difference is determined herein:

$\Delta T_{lm} = \frac{(T_{h,in}-T_{c, out})-(T_{h,out}-T_{c,in})}{\ln\frac{T_{h,in}-T_{c, out}}{T_{h,out}-T_{c,in}} }$

$\Delta T_{lm} = \frac{(160^{\textdegree}C-78^{\textdegree}C)-(96.719^{\textdegree}C-18^{\textdegree}C)}{\ln\frac{160^{\textdegree}C-78^{\textdegree}C}{96.719^{\textdegree}C-18^{\textdegree}C} }$

$\Delta T_{lm} \approx 80.348^{\textdegree}C$

The heat transfer surface area is:

$A_{i} = \frac{\dot Q}{U_{i}\cdot \Delta T_{lm}}$

$A_{i} = \frac{489.795\,kW}{(0.63\,\frac{kW}{m^{2}\cdot ^{\textdegree}C} )\cdot(80.348^{\textdegree}C) }$

$A_{i} = 9.676\,m^{2}$

Length of a single pass counter flow heat exchanger is:

$L =\frac{A_{i}}{\pi\cdot D_{i}}$

$L = \frac{9.676\,m^{2}}{\pi\cdot (0.014\,m)}$

$L = 220\,m$

b) Given that tube wall is very thin, inner and outer heat transfer areas are similar and, consequently, the cold side heat transfer coefficient is approximately equal to the hot side heat transfer coefficient.

$U_{o} \approx 0.63\,\frac{kW}{m^{2}\cdot ^{\textdegree}C}$

5 0

2 years ago

Risks can be changed.<br> True<br> False

Mrrafil [7]

Answer:

true

Explanation:

8 0

2 years ago