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Answer:
Explanation:
Given that:
x(t) = 10 sin(10t) . sin (15t)
the objective is to find the power and the rms value of the following signal square.
Recall that:
sin (A + B) + sin(A - B) = 2 sin A.cos B
x(t) = 10 sin(15t) . cos (10t)
x(t) = 5(2 sin (15t). cos (10t))
x(t) = 5 × ( sin (15t + 10t) + sin (15t-10t)
x(t) = 5sin(25 t) + 5 sin (5t)
From the knowledge of sinusoidial signal Asin (ωt), Power can be expressed as:
For the number of sinosoidial signals;
Power can be expressed as:
As such,
For x(t), Power
For the number of sinosoidial signals;
For x(t), the RMS value is as follows:
Answer:
q = (ρg/μ)(sin θ)(h³/3)
Explanation:
I've attached an image of a figure showing the coordinate system.
In this system: the velocity components v and w are equal to zero.
From continuity equation, we know that δu/δx = 0
Now,from the x-component of the navier stokes equation, we have;
-δp/δx + ρg(sin θ) + μ(δ²u/δy²) = 0 - - - - - (eq1)
Due to the fact that we have a free surface, it means we will not have a pressure gradient in the x-component and so δp/δx = 0
Then our eq 1 is now;
ρg(sin θ) + μ(δ²u/δy²) = 0
μ(δ²u/δy²) = -ρg(sin θ)
Divide both sides by μ to get;
(δ²u/δy²) = -(ρg/μ)(sin θ)
Integrating both sides gives;
δu/δy = -(ρg/μ)(sin θ)y + b1 - - - - (eq2)
Now, the shear stress is given by the formula;
τ_yx = μ[δu/δy + δv/δx]
From the diagram, at the free surface,τ_yx = 0 and y = h
This means that δu/δy = 0
Thus, putting 0 for δu/δy in eq 2, we have;
0 = -(ρg/μ)(sin θ)h + b1
b1 = h(ρg/μ)(sin θ)
So, eq 2 is now;
δu/δy = -(ρg/μ)(sin θ)y + h(ρg/μ)(sin θ)
Integrating both sides gives;
u = -[(y²/2) × (ρg/μ)(sin θ)] + h(ρg/μ)(sin θ)y + b2 - - - eq3
Because u = 0 when y = 0, it means that b2 = 0 also because when we plug 0 for u and y into eq3, we will get b2 = 0.
Thus, we now have:
u = -[(y²/2) × (ρg/μ)(sin θ)] + h(ρg/μ)(sin θ)y
Factorizing like terms, we have;
u = (ρg/μ)(sin θ)[hy - y²/2] - - - (eq 4)
The flow rate per unit width is gotten by Integrating eq 4 between the boundaries of h and 0 to give;
∫u = (h,0)∫(ρg/μ)(sin θ)[hy - y²/2]
q = (ρg/μ)(sin θ)[hy²/2 - y³/6] between h and 0
q = (ρg/μ)(sin θ)[h³/2 - h³/6]
q = (ρg/μ)(sin θ)(h³/3)
