Given Data: Diameter of pipe at point 1 = 5.7 cm Radius of pipe at point 1, r1 = 5.7/2 = 2.85 cm Diameter of pipe at point 2 = 4.3 cm Radius of pipe at point 2, r2 = 4.3/2 = 2.15 cm Pressure at point 1, P1 = 34.0 kPa Pressure at point 2, P2 = 21.4 kPa We know that the volume rate of flow can be calculated by using the Bernoulli's equation. The Bernoulli's equation is: P1 + (1/2)ρv1² + ρgh1 = P2 + (1/2)ρv2² + ρgh2Here, ρ = density of water = 1000 kg/m³g = acceleration due to gravity = 9.81 m/s²h1 = h2 (since both points are in the same horizontal line)∴ ρgh1 = ρgh2 ⇒ h1 = h2 Cancelling out the terms which are same on both sides, we get :P1 + (1/2)ρv1² = P2 + (1/2)ρv2² -----
(1)Now, we can use the continuity equation, which states that the volume rate of flow of an incompressible fluid through any pipe or channel remains constant, if there is no source or sink of fluid along the length of the pipe. Mathematically, it can be written as: A1v1 = A2v2 where A is the cross-sectional area of the pipe and v is the velocity of the fluid at that point. Substituting the values, we get: (πr₁²)v₁ = (πr₂²)v₂ -----
(2)From equation (2), we get:v₁/v₂ = (r₂/r₁)²Also, we know that:A1 = πr1² and A2 = πr2²Substituting the values in equation (2), we get:v1 = (r2/r1)² v2 Substituting this in equation (1),
we get:P1 + (1/2)ρ[(r2/r1)² v2]² = P2 + (1/2)ρv2²
On solving the above equation, we get:v2 = √[(2(P1 - P2))/(ρ(1 - (r2/r1)⁴))] Substituting the given values, we get:v2 = 2.77 m/s Now, the volume rate of flow can be calculated as:Q = A2v2Q = (πr2²)(2.77)Q = (π(0.0215)²)(2.77)Q = 0.00106 m³/s
Therefore, the volume rate of flow is 0.00106 m³/s.
kPa : kPa (Kilopascal) is a unit of measure in Europe and other areas that use the metric system. BAR is a unit of measure in Europe and other areas that use the metric system and it is defined as 100 Kilopascals. It is about equal to the atmospheric pressure on Earth at sea level.