Finding Shared Unknowns In Two Different Equations

[Hckyplayer8]

Hello! I'm seeking help on how to find two different unknowns in two different equations for an atmospheric dynamics homework problem.

It involves K Theory and Ekman layer where [math]ln z/z_o =ku/u_*[/math]
and

z = a fixed height
z sub o = roughness of the object
k = Von Karman's Constant = .4
u = wind speed
u sub * = friction velocity

The knowns are two heights, 10m and 2m where the wind is moving at 10 m/s at 10m and 4.5 m/s at 2m, where the objective is to find the unknowns z sub o and u sub *

Since the surface type and its roughness doesn't change, z sub o will be the same for both equations and seems like a good place to start. But I'm unaware what mathematical step to take to solve.

Equation 1
[math]ln 10/z_o =(.4)(5)/u_*[/math]
Equation 2
[math]ln 2/z_o =(.4)(4.5)/u_*[/math]

 

[Dr.Peterson]

[math]ln z/z_o =ku/u_*[/math][math]ln 10/z_o =(.4)(5)/u_*[/math][math]ln 2/z_o =(.4)(4.5)/u_*[/math]

First, we need to clarify. Are these the equations?

[math]\ln\frac{z}{z_o} =\frac{ku}{u_*}[/math][math]\ln\frac{10}{z_o} =\frac{(.4)(5)}{u_*}[/math][math]\ln\frac{2}{z_o} =\frac{(.4)(4.5)}{u_*}[/math]
Second, I would start by solving each of the two equations for [imath]u_*[/imath] and setting those equal, to eliminate that variable. Equivalently, you could divide one equation by the other.

You will probably also find it helpful to expand the logs into differences of logs.

 

[Hckyplayer8]

First, we need to clarify. Are these the equations?

[math]\ln\frac{z}{z_o} =\frac{ku}{u_*}[/math][math]\ln\frac{10}{z_o} =\frac{(.4)(5)}{u_*}[/math][math]\ln\frac{2}{z_o} =\frac{(.4)(4.5)}{u_*}[/math]
Second, I would start by solving each of the two equations for [imath]u_*[/imath] and setting those equal, to eliminate that variable. Equivalently, you could divide one equation by the other.

You will probably also find it helpful to expand the logs into differences of logs.

Those are the equations. Expanding into difference of logs means I can cancel the -ln sub 0 and the k term on the right side? So it reduces down to

[math]\ln{10} =\frac{(5)}{u_*}[/math][math]\ln{2} =\frac{(4.5)}{u_*}[/math]

 

[khansaheb]

Expanding into difference of logs means....

That means

ln(z/z₀) = ln(z) - ln(z₀) ..... then we get....

ln(10) - ln(z₀) = 2/u_*...... and,,,,,

ln(2) - ln(z₀) = 1.8/u_*

Do the algebra carefully ............................