# gauge pressure at point B ?

The 3.0cm diameter water line in the figure splits into two 1.0cm diameter pipes. All pipes are circular and at the same elevation. At point A, the water speed is 2.0 m/s and the gauge pressure is 50 kpa

First of all, for branching tubes of flow …
… A₀ u₀ = A₁ u₁ + A₂ u₂ … where u denotes the speed of flow …
Since … A₁ = A₂ … we can assume that the water flows with the
same speed in the two smaller tubes with the same diameter …
… that is … A₀ u₀ = 2 A₁ u₁ —> [ π R₀² ] u₀ = 2 [ π R₁² ] u₁ …
… [ π ( ½ D₀ ) ² ] u₀ = 2 [ π ( ½ D₁ ) ² ] u₁ …
… ¼ π D₀ ² u₀ = 2 [ ¼ π D₁ ² ] u₁ —> D₀ ² u₀ = 2 D₁ ² u₁ …
Then, using Bernoulli’s equation, we get the following …
… ½ ρ u₀² + ρ g y₀ + p₀ = ½ ρ u₁² + ρ g y₁ + p₁ …
… same elevation —> y₀ = y₁ … so the equation reduces to …
… ½ ρ u₀² + p₀ = ½ ρ u₁² + p₁ … where … u₀ = 2.0 m / s …
… p₀ = 50 kPa … and solving for p₁ , we get the following …
… p₁ = ½ ρ [ u₀² – u₁² ] + p₀ … where … u₁ = ( D₀ ² / 2 D₁ ² ) u₀ …
… u₁² = ( D₀ ² / 2 D₁ ² ) ² u₀ ² = ( D₀ ⁴ / 4 D₁ ⁴ ) u₀ ²
……………..……..…………. = ¼ ( D₀ / D₁ ) ⁴ u₀ ² … and therefore
… p₁ = ½ ρ [ u₀² – u₁² ] + p₀ = ½ ρ [ u₀² – ¼ ( D₀ / D₁ ) ⁴ u₀ ² ] + p₀
……………..……..………… = ½ ρ [ 1 – ¼ ( D₀ / D₁ ) ⁴ ] u₀ ² + p₀ …
… 1 – ¼ ( D₀ / D₁ ) ⁴ = 1 – ¼ ( 3.0 cm / 1.0 cm ) ⁴ = – 19.25 …
… p₁ = ½ ( 1000 kg / m ³ ) ( – 19.25 ) ( 2.0 m / s ) ² + 50 kPa
……. = -38500 Pa + 50 kPa = -38.5 kPa + 50 kPa
……. = 11.5 kPa …

Scroll to Top