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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?

Respuesta :

Karabo99

Answer:

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

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