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

Answer … 11.5 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 …

- Is the correct phrase “oh, bother” or “oh, brother”? Or do they mean different things? - January 27, 2023
- What is the average mass of a single sulfur atom in grams - January 27, 2023
- How many smaller rectangles are there in an area model that represents 27X83? - January 27, 2023