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DIA [1.3K]
3 years ago
9

Suppose you want to find the sum of two sinusoidal voltages, given as follows: v1(t)=V1 cos(ωt+ϕ1) and v2(t)=V2 cos(ωt+ϕ2)v1(t)=

V1 cos(ωt+ϕ1) and v2(t)=V2 cos(ωt+ϕ2).
If you stay in the time domain, you will have to use trigonometric identities to perform the addition. But if you transform to the frequency domain, you can simply add the phasors V1V1 and V2V2 as complex numbers using your calculator. Your answer will be a phasor, so you will need to inverse phasor-transform it to get the answer in the time domain. This is an example of a problem that is easier to solve in the frequency domain than in the time domain.

Use phasor techniques to find an expression for v(t) expressed as a single cosine function, where v(t)=[100cos(300t+45∘)+500cos(300t−60∘)] V. Enter your expression using the cosine function. Round real numbers using two digits after the decimal point. Any angles used should be in degrees.

Engineering
2 answers:
nydimaria [60]3 years ago
5 0

Answer:

Rectangular form V = 320.71 - j362.3

Polar form V = 483.85 < -48.48°

Phasor form V = 483.85cos(300t - 48.48°)

Explanation:

We are given a sinusoidal function

V(t) = 100cos(300t + 45°) + 500cos(300t - 60°)

We are required to find the v(t) expressed as a single cosine function using phasor technique.

In polar form,

100cos(300t + 45°) = 100 < 45°

500cos(300t - 60°)  = 500 < -60°

In rectangular form,

100 < 45° = 70.71 + j70.71

500 < -60° = 250 - j433.01

Adding the two signals

(70.71 + j70.71) +  (250 - j433.01)

In rectangular form,

V = 320.71 - j362.3

In polar form

V = 483.85 < -48.48°

Therefore, the answer is  

in rectangular form V = 320.71 - j362.3

in polar form V = 483.85 < -48.48°

in phasor form V = 483.85cos(300t - 48.48°)

Conversion from Rectangular to Polar form:

V = X + jY to Magnitude < Angle

V = 320.71 - j362.3

Magnitude = \sqrt{(320.71)^2 +(362.3)^2} = 483.85

Angle = tan⁻¹(Y/X) = tan⁻¹(-362.3/320.71) = -48.48°

V = 483.85 < -48.48°

Conversion from Polar to Rectangular form:

V = 483.85 < -48.48°

X =  Magnitude*cos(Angle) and jY = Magnitude*sin(Angle)

X = 483.85*cos(-48.48°) and jY = 483.85*sin(-48.48°)

X = 320.71 and jY = -362.3

V = 320.71 - j362.3

seropon [69]3 years ago
3 0

Answer:

Attached is the full solution.

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professor190 [17]

Answer:

Option (d) 2 min/veh

Explanation:

Data provided in the question:

Average time required = 60 seconds

Therefore,

The maximum capacity that can be accommodated on the system, μ = 60 veh/hr

Average Arrival rate, λ = 30 vehicles per hour

Now,

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⇒ \frac{1}{\mu(1-\frac{\lambda}{\mu})}

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Average time spent by the vehicle = \frac{1}{60(1-\frac{30}{60})}

or

Average time spent by the vehicle = \frac{1}{60(1-0.5)}

or

Average time spent by the vehicle = \frac{1}{60(0.5)}

or

Average time spent by the vehicle = \frac{1}{30} hr/veh

or

Average time spent by the vehicle = \frac{1}{30}\times60 min/veh

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7 0
4 years ago
Air enters a compressor steadily at the ambient conditions of 100 kPa and 22°C and leaves at 800 kPa. Heat is lost from the comp
telo118 [61]

Answer:

a) 358.8K

b) 181.1 kJ/kg.K

c) 0.0068 kJ/kg.K

Explanation:

Given:

P1 = 100kPa

P2= 800kPa

T1 = 22°C = 22+273 = 295K

q_out = 120 kJ/kg

∆S_air = 0.40 kJ/kg.k

T2 =??

a) Using the formula for change in entropy of air, we have:

∆S_air = c_p In \frac{T_2}{T_1} - Rln \frac{P_2}{P_1}

Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K

Solving, we have:

[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]

-0.40= 1.005(ln T_2 - 5.68697)- 0.5968

Solving for T2 we have:

T_2 = 5.8828

Taking the exponential on the equation (both sides), we have:

[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]

b) Work input to compressor:

w_in = c_p(T_2 - T_1)+q_out

w_in = 1.005(358.8 - 295)+120

= 184.1 kJ/kg

c) Entropy genered during this process, we use the expression;

Egen = ∆Eair + ∆Es

Where; Egen = generated entropy

∆Eair = Entropy change of air in compressor

∆Es = Entropy change in surrounding.

We need to first find ∆Es, since it is unknown.

Therefore ∆Es = \frac{q_out}{T_1}

\frac{120kJ/kg.k}{295K}

∆Es = 0.4068kJ/kg.k

Hence, entropy generated, Egen will be calculated as:

= -0.40 kJ/kg.K + 0.40608kJ/kg.K

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3 years ago
Which of the following identifies the limitations of green engineering?
olga_2 [115]

Answer:

Resources and Cost

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Hope it helps!<3

4 0
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A silicon diode has a saturation current of 6 nA at 25 degrees Celcius. What is the saturation current at 100 degrees Celsius?
Illusion [34]

Answer:

0.0659 A

Explanation:

Given that :

I_{0}  =  6nA ( saturation current )

at 25°c = 300 k ( room temperature )

n = 2  for silicon diode

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Diode equation at room temperature = I = Io \frac{V}{e^{0.025*n} }

next we have to determine the value of V at 373 k

q / kT = (1.6 * 10^-19) / (1.38 * 10^-23 * 373) = 31.08 V^-1

Given that I is constant

Io = \frac{e^{0.025*2} }{31.08} =  0.0659 A

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Answer:

Some general principles are given below in the explanation segment.

Explanation:

Sewage treatment seems to be a method to extract pollutants from untreated sewage, consisting primarily of domestic sewage including some solid wastes.

<u>The principles are given below:</u>

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  • Inertial influence.
  • Sieving seems to be an excellent method to distinguish particulates.

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