Problem Statement:
A flat plate, 2 m x 0.4 m, is set parallel to a uniform flow. The air velocity is 30 km/h. Determine the approximate estimated thickness of the boundary layer at the downstream end of the plate. The air has a density of 1.2 kg/m³ and a viscosity of 16 centistoke.

Solution:

1. Understanding the Problem:
We are given a flat plate set parallel to a uniform flow of air. The plate has dimensions 2 m x 0.4 m. The air velocity is given as 30 km/h. We need to determine the approximate estimated thickness of the boundary layer at the downstream end of the plate. The air has a density of 1.2 kg/m³ and a viscosity of 16 centistoke.

2. Concept of Boundary Layer:
The boundary layer is a thin layer of fluid that develops along the surface of a solid body when it is in contact with a fluid flow. In this case, the boundary layer develops along the surface of the flat plate due to the flow of air.

3. Boundary Layer Thickness:
The boundary layer thickness is the distance from the surface of the flat plate to the point where the flow properties have reached 99% of their free-stream values. It represents the region where the flow is significantly influenced by the presence of the plate.

4. Calculation:
To calculate the approximate estimated thickness of the boundary layer at the downstream end of the plate, we can use the Blasius equation for laminar boundary layers. This equation relates the boundary layer thickness (δ) to the distance from the leading edge of the plate (x) and the properties of the fluid.

The Blasius equation is given by:
δ = 5.0 * sqrt((ν * x)/U)
where δ is the boundary layer thickness, ν is the kinematic viscosity (ν = μ/ρ), x is the distance from the leading edge of the plate, and U is the free-stream velocity.

5. Calculation Steps:
Let's calculate the boundary layer thickness at the downstream end of the plate using the given data:

Given:
- Plate dimensions: 2 m x 0.4 m
- Air velocity: 30 km/h = 8.33 m/s
- Air density: 1.2 kg/m³
- Air viscosity: 16 centistoke = 1.6 * 10^(-3) kg/(m*s) (since 1 centistoke = 1 * 10^(-3) kg/(m*s))

Using the Blasius equation:
δ = 5.0 * sqrt((ν * x)/U)

Let's assume the distance from the leading edge of the plate (x) to be 2 m (since the downstream end is not specified).

Substituting the given values:
δ = 5.0 * sqrt((1.6 * 10^(-3) * 2)/(8.33))

Simplifying the equation:
δ = 5.0 * sqrt(3.84 * 10^(-4)/8.33)
δ = 5.0 * sqrt(4.608 * 10^(-8))
δ ≈ 5.0 * 2.147 * 10