In a simple model of the wind speed associated with hurricane Emily, we assume there is calm eye 10.0 km in radius. The winds, which extend to a height of 5550 m, begin with a speed of 208.0 km/hr at the eye wall and decrease linearly with radial distance down to 0 km/hr at a distance of 150.0 km from the center. Assume the average density of the air from sea level to an altitude of 5550 m is 0.891 kg/m3. Calculate the total kinetic energy of the winds.
find KE as a function of radius slice dr
from r = 10^10^3 to r = 150*10^3
208 km/hr /3.6 = 57.8 m/s
Velocity = m r + b linear
57.8 = m (10*10^3) + b
0 = m( 150*10^3) + b
so solve for slope and intercept
m = -.413 * 10^-3
b = 61.9
so
v = 61.9 - .413*10^-3 r
the volume at radius r = 2 pi r dr(5550)
so the mass at radius r = .891*5550* 2 pi r dr
so the KE at radius r = (1/2) (.891*5550* 2 pi r dr) (61.9-.413*10^-3 r)^2
when you square that and multiply by r dr you will get terms like
a r + b r^2 + c r^3
which will integrate to
a r^2/2 + b r^2/3 + c r^4/4
evaluate that integral at r = 10*10^3 and at 150*10^3 and you have it
Where did 3.6 come from?
To calculate the total kinetic energy of the winds, we need to determine the kinetic energy at each radial distance from the center and then integrate those values over the entire area.
Let's break down the solution into steps:
Step 1: Convert the given units to SI units
- Wind speed at the eye wall: 208.0 km/hr = 208000 m/hr
- Radius of the eye: 10.0 km = 10000 m
- Distance where wind speed becomes 0: 150.0 km = 150000 m
- Air density: 0.891 kg/m3
Step 2: Calculate the area of the wind field
The area of the wind field is given by the formula for the area of a circular region:
A = π * r^2
where r is the radius of the wind field (distance to the point where the wind speed is 0).
In this case, r = 150000 m.
So, the area of the wind field is:
A = π * (150000)^2
Step 3: Determine the wind speed at each radial distance
Since the wind speed decreases linearly from the eye wall to 0 km/hr, we can use the equation of a straight line:
v = v0 - Δv/dr * r
where v is the wind speed at a specific radial distance r, v0 is the wind speed at the eye wall (208000 m/hr), Δv/dr is the rate of decrease in wind speed per unit radial distance (v0 / r0, where r0 is the radius of the eye wall), and r is the current radial distance.
In this case, v0 = 208000 m/hr and r0 = 10000 m.
Step 4: Calculate the kinetic energy at each radial distance
The kinetic energy of the winds at a specific radial distance is given by the formula:
KE = 0.5 * m * v^2
where KE is the kinetic energy, m is the mass of the air, and v is the wind speed at that radial distance.
To calculate the mass of the air, we need to multiply the air density (0.891 kg/m3) by the volume of the air column.
The volume of the air column is obtained by multiplying the area of the wind field (π * r^2) by the height of the wind column (5550 m).
So, the mass of the air is:
m = air density * (π * r^2 * height)
Step 5: Integrate the kinetic energy values to obtain the total kinetic energy
To find the total kinetic energy of the winds, we need to integrate the kinetic energy values over the entire wind field.
The integral of KE with respect to r will be:
Total KE = ∫ KE dr
Substituting the value of KE from Step 4, we have:
Total KE = ∫ 0.5 * m * v^2 dr
Now we can perform the integration using the given limits of integration (from 0 m to 150000 m).
I hope the explanation was clear. Let me know if you have any more questions or need further assistance with the calculations.
To calculate the total kinetic energy of the winds in the given scenario, we need to find the kinetic energy per unit volume and then integrate it over the volume of the winds.
First, let's find the equation for the wind speed as a function of radial distance. We are told that the wind speed decreases linearly from 208.0 km/hr to 0 km/hr at a distance of 150.0 km from the center of the eye. Since we know the speed at two points, we can use the equation of a straight line to find the equation of the wind speed:
v(r) = v0 - (v0/150)r
Here, v(r) is the wind speed at radial distance r, and v0 is the initial wind speed at the eye-wall, which is given as 208.0 km/hr.
Next, let's convert the wind speed into SI units (m/s). We can use the conversion factor 1 km/hr = 0.2778 m/s:
v(r) = (208.0 km/hr - (208.0 km/hr/150)r) * 0.2778 m/s
Now, we can calculate the kinetic energy per unit volume using the equation:
K.E. per unit volume = (1/2) * ρ * v^2
where K.E. is the kinetic energy, ρ is the density of air, and v is the wind speed.
Substituting the given density of air (0.891 kg/m^3) and the wind speed equation into the kinetic energy equation:
K.E. per unit volume = (1/2) * 0.891 kg/m^3 * [(208.0 km/hr - (208.0 km/hr/150)r) * 0.2778 m/s]^2
Now, we can integrate this equation over the volume of the winds. Since the winds extend to a height of 5550 m and have a radius of 10.0 km, we can integrate with respect to height (h) from 0 to 5550 m, and with respect to radial distance (r) from 0 to 10.0 km. Therefore, the total kinetic energy (K.E.) of the winds can be calculated as:
K.E. = ∫∫K.E. per unit volume * r dr dh
Integrating this equation with the given limits will give us the total value of the kinetic energy of the winds. Unfortunately, the integral is a complicated operation and requires numerical integration techniques to solve, which may be challenging to explain in detail using only text. We recommend using numerical integration software or a programming language with numerical integration capabilities, such as Python with libraries like SciPy, to evaluate this integral and find the total kinetic energy of the winds.