Physics of Fluids Questions Help

I just cannot get to the answers to these questions, anyone able to help? I need the working out shown.

QUESTION ONE. - A horizontal pipe contains water at a pressure of 250kPa, flowing with a speed of 2m/s. When the pipe narrows to 50% of its original radius, what is;
a) The Speed of the water?
b) The pressure of the water?

The answers should be:
a) 8 m/s
b) 220 kPA


QUESTION 2.
What is the lift force due to Bernoulli's Principle on a planes wing of area of 80m^2 if the air passes over the top of the wing at a speed of 240m/s and under the bottom of the wing at 175m/s? Take the density of air to be 1.3kgm^-3.

Answer should be approximately 1402700 N

Any help would be appreciated so much, thanks in advance.

(1).

v' = v(A/A')

A = ¼ π D²

v' = v(D/D')²

v' = 2(1/0.5)²

v' = 8 m/s




(2).

P₁ + ρ g h₁ + ½ ρ v₁² = P₂ + ρ g h₂ + ½ ρ v₂²

250 * 1000 + 0 + ½ * 1000 * 2² = P₂ + 0 + ½ * 1000 * 8²

P₂ = 220000 Pa





(2).

P₁ + ρ g h₁ + ½ ρ v₁² = P₂ + ρ g h₂ + ½ ρ v₂²

P₂ - P₁ = ½ ρ v₁² - ½ ρ v₂²

F/A = ½ ρ v₁² - ½ ρ v₂²

F/A = ½ ρ (v₁² - v₂²)

F/80 = ½ (1.3)(240² - 175²)

F = 1402700 Newton

Sorry, I only know the answer to Q #1-a.
So 'a' has to do with fluid flow continuity. A = area, v = velocity.
A1v1 = A2v2
So the area of a circle is A = pi*r^2
Pi is a constant, so it does not change anything. We can ignore it.
We shall say that the area of the original pipe is 1 for convenience. (Again, ignoring pi)
That means when the pipe narrows, the Area becomes A = (1/2)^2 = (1/4)

When we plug these numbers into the fluid flow continuity equation, we get:
2m/s * 1 m^2 = v2 * (1/4)
2 m^3/s = v2 * (1/4)m^2
-multiply by 4/m^2
8 m/s = v2
v = 8m/s
And there we go.

Sorry I can't help with more of it. I can't remember any pressure equations at the moment.

1a) By conservation of volume flow-rate .. A1.v1 = A2.v2

v2 = A1.v1 / A2 = r1².v1 / r2² .. .. π.r1² x 2.0m/s / π(0.50r1)²
v2 = 2 / 0.5² .. .. ►v2 = 8.0 m/s

1b) Using .. ∆P = ½ρ (v1² - v2²)

∆P = ½.1000 kg/m³(2² - 8²) = 500 x -60
∆P = -30000 Pa .. (30 KPa)

P2 = 250 KPa - 30 KPa .. .. .. ►P2 = 220 KPa

2) Using .. ∆P = ½ρ (v1² - v2²)

∆P = ½.1.3 (240² - 175²) = 17534 Pa .. (N/m²)

Lift force .. F = ∆P x A .. .. 17534 N/m² x 80 m² .. .. F = 1.403^6 N