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
You might be interested in
The local convection heat transfer coefficient at 1 m from the leading edge is , the average convection heat transfer coefficient over the entire plate is
and the total heat flux transfer to the plate is
.
Explanation:
It is case of heat and mass transfer in which due to temperature difference between gas and surface. Further temperature boundary layer will developed on flat plate in longitudinal direction.
Hot carbon dioxide exhaust gas
physical properties
υ =
Apart from these other data arr given below,
To find the local convection heat transfer coefficient at 1 m from the leading edge, we use correlation used for laminar flow over flat plate,
where h= Average heat transfer coefficient
L= Length of a plate
k= Thermal Conductivity of carbon dioxide
Re = Reynold's Number
Pr = Prandtle Number
(a) Convection heat transfer coefficient at 1 m from the leading edge
is referred as local convection heat transfer coefficient.
To find convection heat transfer coefficient at 1 m from leading edge,
Here, first we have to find Re and Pr,
Pr number is take from physical property data and Pr is 0.725.
Putting value of Re and Pr in main equation,
we get
(b) To find average convection heat transfer coefficient,
it can be find out as case (a), only difference is that instead of L=1 m, L=1.5 m would come,
Therefore,
Finally,
(C) Total heat flux transfer to the plate is found out by,