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.
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.
2 answers:
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 = = 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
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a) Mechanical efficiency ()=63.15% b) Temperature rise= 0.028ºC
Explanation:
For the item a) you have to define the mechanical power introduced (Wmec) to the system and the power transferred to the water (Pw).
The power input (electric motor) is equal to the motor power multiplied by the efficiency. Thus, .
Then, the power transferred (Pw) to the fluid is equal to the flow rate (Q) multiplied by the pressure jump . So
.
The efficiency is defined as the ratio between the output energy and the input energy. Then, the mechanical efficiency is
For the b) item you have to consider that the inefficiency goes to the fluid as heat. So it is necessary to use the equation of the heat capacity but in a "flux" way. Calling <em>H</em> to the heat transfered to the fluid, the specif heat of the water and the density of the water:
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Finally, the temperature rise is: