Given information:

- Two-dimensional incompressible steady flow around an airfoil
- Stream lines are 2 cm apart at a great distance from the airfoil, where the velocity is 30 m/sec
- Stream lines are 1.5 cm apart near the airfoil

To find:

- Velocity near the airfoil where the stream lines are 1.5 cm apart

Solution:

The continuity equation for incompressible flow is:

Q = AV

where Q is the flow rate, A is the cross-sectional area, and V is the velocity.

Since the flow is steady and incompressible, the flow rate is constant along a stream line. Therefore, we can write:

Q1 = Q2

where subscripts 1 and 2 refer to two different points along a stream line.

The cross-sectional area A is given by:

A = bh

where b is the width of the stream and h is the distance between two adjacent stream lines.

Since the flow is two-dimensional and incompressible, the velocity components along the stream lines and normal to the stream lines do not change. Therefore, the velocity can be written as:

V = Vt cos(θ)

where Vt is the tangential velocity along the stream line, and θ is the angle between the stream line and the x-axis.

Using trigonometry, we can write:

h = Δy = Δs sin(θ)

b = Δx = Δs cos(θ)

where Δs is the distance along the stream line.

Substituting these expressions for b and h in the continuity equation, we get:

Q = AV = bΔy Vt cos(θ) = hΔx Vn sin(θ)

where Vn is the normal velocity to the stream line.

Dividing both sides by Δs and taking the limit as Δs approaches zero, we get:

dQ/ds = Vt cos(θ) = Vn sin(θ)

where dQ/ds is the rate of change of flow rate along the stream line.

Since the flow is incompressible and the stream lines are equidistant, the flow rate is constant along each stream line. Therefore, we can write:

dQ/ds = 0

Substituting the given values, we get:

Vt cos(θ1) = Vn sin(θ1)

Vt cos(θ2) = Vn sin(θ2)

where subscripts 1 and 2 refer to two different points along the same stream line.

Dividing these equations, we get:

tan(θ1) = tan(θ2)

Since the stream lines are parallel, the angles are equal. Therefore, we can write:

θ1 = θ2

Using the given values, we can write:

tan(θ) = h/b = 1.5/2 = 0.75

Taking the inverse tangent, we get:

θ = 36.9°

Substituting this value in the continuity equation, we get:

Vt = V cos(θ) = 30 cos(36.9°) = 24.0 m/s

Therefore, the velocity near the airfoil where the stream lines are 1.5 cm apart is:

Vn = Vt/tan(θ) = 24.0/0.75 = 32.0 m/s

Converting to cent