Water, which we can treat as ideal and incompressible, flows at 12 m/s in a horizontal pipe with a pressure of 3.0 x 10^4 Pa. If the pipe widens to twice its original radius, what is the pressure in the wider section?

p2 = 9.8*10^4 Pa

Explanation:

Total pressure is constant and PT = P = 1/2*ρ*v^2

So p1 + 1/2*ρ * (v1) ^2 = p2 + 1/2*ρ * (v2) ^2

from continuity we have ρ*A1*v1 = ρ*A2*v2

v2 = v1*A1/A2

and

r2 = 2*r1

then:

A2 = 4*A1

so,

v2 = (v1) / 4

then:

p2 = p1 + 1/2*ρ * (v1) ^2 - 1/2*ρ * (v2) ^2 = p1 + 1/2*ρ * (v1) ^2 - 1/2*ρ * (v1/4) ^2

p2 = 3.0*10^4 Pa + 1/2 * (1000 kg/m^3) * (12m/s) ^2 - 1/2 * (1000kg/m^3) * (12^2/16)

= 9.75*10^4 Pa

= 9.8*10^4 Pa

Therefore, the pressure in the wider section is 9.8*10^4 Pa