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Air Flow Rig – Pitot Static Tube & Venturi Meter
Introduction
In this report I will use a pitot-static tube and a Venturi meter set up within an air flow rig to demonstrate the application of Bernoulli’s Equation and the assumption of inviscid flow. The air flow rig was set up and an imaginary matrix was defined across the cross sectional area of the air flow at the pitot-static section. The matrix had 4 cells going across (A, B, C, and D) the width and 5 cells down the height of the rig (1, 2, 3, 4 and 5). The pitot-static tube will be moved around to the centre of each cell so we can then take a reading for the pressure at that particular point, which can then be used to calculate the velocity across the matrix. This will enable me*…show more content…*

To find the discharge coefficient (cd) we can divide the volumetric flow rate from Q=A1v1 by the volumetric flow rate from the venturi (Sherwin & Horsley, 1996). The value of the discharge coefficient on the venturi meter being used in the investigation is known to be 0.98. To take readings of the pressure differences, manometers were used. Manometers rely on the height of a column of liquid to measure the pressure difference. The columns were set to be inclined to decrease the errors in reading as it will increase the distance the fluid is displaced; we then use trigonometry to convert the readings into a vertical height. By knowing the density of the fluid, the distance displaced the angle of inclination and the acceleration due to gravity, I can use the following equation to find the pressure difference in the manometer: P=ρgh P=ρg(hsinθ) Results Air Pressure in the room = 76.1 cm of mercury The following tables show the results of the pressure difference inside the air flow rig in mmH2O displaced. Result 1 (mmH₂O) | Cell | A | B | C | D | 1 | 27.5 | 33.0 | 33.0 | 30.0 | 2 | 28.0 | 36.0 | 38.0 | 33.0 | 3 | 32.0 | 41.0 | 40.0 | 25.0 | 4 | 34.0 | 38.0 | 36.0 | 26.0 | 5 | 18.0 | 19.0 | 19.0 | 17.0 | | | | | | Result 2 (mmH₂O) | Cell | A | B | C | D | 1 | 25.0 | 31.0 | 32.0 | 29.0 | 2 | 31.0 | 38.0 | 38.0 | 33.0 | 3 | 31.0 | 39.0 | 39.0 | 32.0 | 4 | 33.0 | 38.0 | 38.0 | 29.0 | 5 |

To find the discharge coefficient (cd) we can divide the volumetric flow rate from Q=A1v1 by the volumetric flow rate from the venturi (Sherwin & Horsley, 1996). The value of the discharge coefficient on the venturi meter being used in the investigation is known to be 0.98. To take readings of the pressure differences, manometers were used. Manometers rely on the height of a column of liquid to measure the pressure difference. The columns were set to be inclined to decrease the errors in reading as it will increase the distance the fluid is displaced; we then use trigonometry to convert the readings into a vertical height. By knowing the density of the fluid, the distance displaced the angle of inclination and the acceleration due to gravity, I can use the following equation to find the pressure difference in the manometer: P=ρgh P=ρg(hsinθ) Results Air Pressure in the room = 76.1 cm of mercury The following tables show the results of the pressure difference inside the air flow rig in mmH2O displaced. Result 1 (mmH₂O) | Cell | A | B | C | D | 1 | 27.5 | 33.0 | 33.0 | 30.0 | 2 | 28.0 | 36.0 | 38.0 | 33.0 | 3 | 32.0 | 41.0 | 40.0 | 25.0 | 4 | 34.0 | 38.0 | 36.0 | 26.0 | 5 | 18.0 | 19.0 | 19.0 | 17.0 | | | | | | Result 2 (mmH₂O) | Cell | A | B | C | D | 1 | 25.0 | 31.0 | 32.0 | 29.0 | 2 | 31.0 | 38.0 | 38.0 | 33.0 | 3 | 31.0 | 39.0 | 39.0 | 32.0 | 4 | 33.0 | 38.0 | 38.0 | 29.0 | 5 |

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