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3 years ago

Explain the Corona discharge?

Engineering

1 answer:

rosijanka [135]3 years ago

8 0

Answer:

• Corona discharge means neutralization of charge action at a sharp point

example; a pear shaped conductor, lightening conductor, van de graaf generator.

Explanation:

• In corona discharge, When a sharp conductor is placed somewhere in open space. The charges of opposite chrage attract the opposite ions in the air around that conductor.

• This increases the charge density and electric field intensity at the sharp point, this neutralises the charge at sharp point hence a discharge known as corona discharge occurs.

$.$

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Using the characteristics equation, determine the dynamic behavior of a PI controller with τI = 4 applied to a second order proc

Sladkaya [172]

Answer:

The values of Kc that render this closed-loop process unstable are in the interval

(Kc < 0)

Explanation:

The transfer function of a PI controller is given as

Gc = Kc {1 + (1/sτI)}

τI = 4

Gc = Kc {1 + (1/4s)}

Gc = Kc {(4s+1)/(4s)}

Divide numerator and denominator by 4

Gc = Kc {(s+0.25)/(s)}

For a second order process, the general transfer function is given by

Gp = Kp {1/(τn²s² + 2ζτns + 1)}

Kp = 2, τn = 5 and ζ = 1.5

Gp = 2/(25s² + 15s + 1)

Divide numerator and denominator by 25

Gp = 0.08/(s² + 0.6s + 0.04)

Ga = 1

Gs = 1

We need to find the value(s) of Kc that makes the closed loop transfer function unstable. Gp*Ga*Gc*Gs + 1 = 0

The closed loop transfer function is unstable when the solution(s) of the characteristic equation obtained is positive.

Gp*Ga*Gc*Gs + 1 = 0

Becomes

[0.08/(s² + 0.6s + 0.04)] × [Kc (s+0.25)/(s)] + 1 = 0

[0.08Kc (s + 0.25)/(s³ + 0.6s² + 0.04s)] = - 1

0.08Kc (s + 0.25) = -s³ - 0.6s² - 0.04s

0.08Kc s + 0.02Kc = -s³ - 0.6s² - 0.04s

s³ + 0.6s² + 0.04s + 0.08Kc s + 0.02Kc = 0

s³ + 0.6s² + (0.08Kc + 0.04)s + 0.02Kc = 0

We will use the direct substitution method to evaluate the values of Kc that matter. The values of Kc at the turning points of the closed loop transfer function.

For the substitution,

We put s = jw into the equation. (frequency analysis)

Note that j = √(-1)

s³ + 0.6s² + (0.08Kc + 0.04)s + 0.02Kc = 0

(jw)³ + 0.6(jw)² + (0.08Kc + 0.04)(jw) + 0.02Kc = 0

-jw³ - 0.6w² + (0.08Kc + 0.04)(jw) + 0.02Kc = 0

we then collect terms with j and terms without.

(0.08Kcw + 0.04w - w³)j + (0.02Kc - 0.6w²) = 0

Meaning,

0.08Kcw + 0.04w - w³ = 0 (eqn 1)

0.02Kc - 0.6w² = 0 (eqn 2)

0.02 Kc = 0.6 w²

Kc = 15w²

Substituting this into eqn 1

0.08Kcw + 0.04w - w³ = 0

Kc = 15w²

0.08(15w²)w + 0.04w - w³ = 0

1.2w³ + 0.04w - w³ = 0

0.2w³ + 0.04w = 0

w = 0 or 0.2w² + 0.04 = 0

0.2w² = -0.04

w² = -0.2

w = ± √(-0.2)

w = ± 0.4472j or w = 0

Recall, Kc = 15w² = 15(-0.2) = -3 or Kc = 0

The turning points for the curve of the closed loop transfer function occur when

Kc = 0 or Kc = -3

To investigate, we pick values around these turning points to investigate the behaviour of the closed loop transfer function at those points.

Kc < -3, Kc = -3, (-3 < Kc < 0), Kc = 0 and Kc > 0

Note that, one positive characteristic root or pole is enough to make the system unstable.

We pick a value for Kc in that interval and evaluate the closed loop transfer function.

s³ + 0.6s² + (0.08Kc + 0.04)s + 0.02Kc = 0

- First of, let Kc = - 4 (Kc < -3)

s³ + 0.6s² - 0.28s - 0.08 = 0

Solving the polynomial

s = (-0.22002), 0.44223, (-0.82221)

One positive pole means the closed loop transfer function is unstable in this region

Let Kc = -3

s³ + 0.6s² - 0.20s - 0.06 = 0

s = 0.37183, (-0.21251) or (-0.75933)

One positive pole still means that the closed loop transfer function is still unstable.

Then the next interval

Let Kc = -1

s³ + 0.6s² - 0.04s - 0.02 = 0

Solving this polynomial,

s = 0.18686, (-0.1749) or (-0.61196)

The function is unstable in the region being investigated.

Let Kc = 0

s³ + 0.6s² + 0.04s = 0

s = 0, -0.0769, -0.5236

One zero, all negative roots, indicate that the closed loop transfer function is marginally stable at this point.

Let Kc = 1, Kc > 0

s³ + 0.6s² + 0.12s + 0.02 = 0

s = (-0.42894), (-0.08553 + 0.1983j) or (-0.08553 - 0.1983j)

All the real negative parts of the poles are all negative, this indicates stability.

Hence, after examining the turning points of the closed loop transfer function, it is evident that, the region's of Kc where the closed loop transfer function is unstable is (Kc < 0)

Hope this Helps!!!

8 0

3 years ago

Given the vector current density J = 10rho2zarho − 4rho cos2 φ aφ mA/m2:

Xelga [282]

Answer:

(a) Current density at P is $J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\$ .

(b) Total current I is 3.257 A

Explanation:

Because question includes symbols and formulas it can be misunderstood. In the question current density is given as below;

$J=10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}\\$

where $\textbf{a}_{\rho}$ and $\textbf{a}_{\phi}$ unit vectors.

(a) In order to find the current density at a specific point (P), we can simply replace the coordinates in the current density equation. Therefore

$J(P(\rho=3, \phi=30^o,z=2))=10.3^2.2.\textbf{a}_{\rho}-4.3.(\cos(30^o)^2).\textbf{a}_{\phi}\\\\J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\$

(b) Total current flowing outward can be calculated by using the relation,

$I=\int {\textbf{J} \, \textbf{ds}$

where integral is calculated through the circular band given in the question. We can write the integral as below,

$I=\int\{(10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}).(\rho.d\phi.dz.\textbf{a}_{\rho}})\}\\\\I=\int\{(10\rho^2z).(\rho.d\phi.dz)\}\\\\\\$

due to unit vector multiplication. Then,

$I=10\int\(\rho^3z.dz.d\phi$

where $\rho=3,\ 0$ . Therefore

$I=10.3^3\int_2^{2.8}\(zdz.\int_0^{2\pi}d\phi\\I=270(\frac{2.8^2}{2}-\frac{2^2}{2} )(2\pi-0)=3257.2\ mA\\I=3.257\ A$

4 0

3 years ago

Question #6 Fill in the Blank Complete the following sentence. The term describes initiatives and products that preserve the env

viktelen [127]

Answer:

The term "green" describes initiatives and products that preserve the environment

Explanation:

In order to preserve the environment the Environmental Protection Agency has provides tips for "Being Green on the Road", which highlights the topics including, Green Vehicle Guides, guides for Fuel Economy, guides for Clean Diesel and many more ways our environmental footprint can be reduced

A green economy is an economy that focuses on the reduction of risk of economic activity on the environment, such as the reduction of carbon emission and reduction in environmental pollution, biodiversity loss prevention, and promotes a sustainable development

A green tech aims to protect the environment and to reverse previous damage done to the environment, including technology for clean energy, water purification, and waste recycling.

6 0

3 years ago

Air flows through a device in which heat and work is exchanged. There is a single inlet and outlet, and the flow at each boundar

Pepsi [2]

Answer:

11548KJ/kg

10641KJ/kg

Explanation:

Stagnation enthalpy:

$h_{T} = c_{p}*T + \frac{V^2}{2}$