Questions: A square 400 km on each side is embedded in an easterly flow that increases in magnitude toward the north as a rate of 8 m s^-1 per 100 km and southerly flow that increases toward the east at 3 m s^-1 per 200 km. What is the mean relative vorticity in the square? Show this using two methods, and do NOT use natural coordinates.

A square 400 km on each side is embedded in an easterly flow that increases in magnitude toward the north as a rate of 8 m s^-1 per 100 km and southerly flow that increases toward the east at 3 m s^-1 per 200 km. What is the mean relative vorticity in the square? Show this using two methods, and do NOT use natural coordinates.

Transcript text: 1. ( 5 pts). A square 400 km on each side is embedded in an easterly flow that increases in magnitude toward the north as a rate of $8 \mathrm{~m} \mathrm{~s}^{-1}$ per 100 km and southerly flow that increases toward the east at $3 \mathrm{~m} \mathrm{~s}^{-1}$ per 200 km . What is the mean relative vorticity in the square? Show this using two methods, and do NOT use natural coordinates.

Solution

Solution Steps

Step 1: Understanding the Problem

We are given a square region with side length 400 km. The flow has two components:

An easterly flow ($u$) that increases northward at a rate of $8 \, \mathrm{m/s}$ per 100 km.
A southerly flow ($v$) that increases eastward at a rate of $3 \, \mathrm{m/s}$ per 200 km.

We need to find the mean relative vorticity in the square.

Step 2: Define the Velocity Components

The velocity components can be expressed as: \[ u = 8 \times \frac{y}{100} \, \mathrm{m/s} \] \[ v = 3 \times \frac{x}{200} \, \mathrm{m/s} \] where $x$ and $y$ are the coordinates in km.

Step 3: Calculate the Partial Derivatives

The relative vorticity ($\zeta$) is given by: \[ \zeta = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \]

Calculate the partial derivatives: \[ \frac{\partial v}{\partial x} = \frac{\partial}{\partial x} \left( 3 \times \frac{x}{200} \right) = \frac{3}{200} \, \mathrm{s}^{-1} \] \[ \frac{\partial u}{\partial y} = \frac{\partial}{\partial y} \left( 8 \times \frac{y}{100} \right) = \frac{8}{100} = 0.08 \, \mathrm{s}^{-1} \]

Step 4: Compute the Relative Vorticity

Substitute the partial derivatives into the vorticity formula: \[ \zeta = \frac{3}{200} - 0.08 \]

Convert $\frac{3}{200}$ to a decimal: \[ \frac{3}{200} = 0.015 \, \mathrm{s}^{-1} \]

Thus, \[ \zeta = 0.015 - 0.08 = -0.065 \, \mathrm{s}^{-1} \]

Step 5: Mean Relative Vorticity

Since the flow is linear and the vorticity is constant throughout the square, the mean relative vorticity is the same as the calculated vorticity.