Question

A horizontal pipe contains water at a pressure of 100 kPa flowing with a speed of 2.0 m/s . The pipe narrows to one-half its original diameter.

The speed of the water is 8 m/s

what is the pressure of the water?

Answer 1

We have the following

Pressure P₁=100kPa

Velocity V₁=2.0m/s

Assuming diameter of the first pipe to be D₁

Diameter of second pipe D₂ = D₁/2

Velocity in second pipe V₂ = 8.0m/s

We have formula P₁= ((Rho X Q₁)/(2A₁²))

where Q is flow rate Rho is density and remains constant.

Q₁ = V₁ X A₁

(P₁/P₂) = (V₁XA₁)/(V₂XA₂)

P₂ = (P₁V₂A₂)/(V₁A₁)

Now Since D₂ = D₁/2

A₂=A₁/4

Hence P₂ = (100 X 8 XA₁/4)/(2 X A₁)

P₂ = 100X8/(4X2)

P₂ = 100kPa

hence the pressure of water in the reduced pipe of half the diameter with velocity of water as 8m/s remains the same as 100kPa.

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Thanks

Bilbo