Pressure measurements

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Gabriella Miller
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1 Next Previous 8/6/01 Chapter six + seven Pressure and Flow easureents Laith Batarseh Hoe End Basic concepts Pressure is represented as a force per unit area Absolute pressure refers to the absolute value of the force per unit area exerted on the containing wall by a fluid Gage pressure represents the difference between the absolute pressure and the local atospheric pressure. Vacuu represents the aount by which the atospheric pressure exceeds the absolute pressure Fluid pressure results fro a oentu exchange between the olecules of the fluid and a containing wall. The interaction between fluid particles ay be staic (no otion) or dynaic (with otion) 1

2 8/6/01 Basic concepts Basic concepts Conversion factors 1 atosphere (at) = pounds per square inch absolute = newtons per square eter (Pa) = 116 pounds-force per square foot (lbf/ft) 1 N/ 1 pascal (Pa) 1 atosphere (at) = 760 illieters of ercury (Hg) 1 bar = 105 newtons per square eter (100 kpa)

3 8/6/01 Mechanical Pressure-Measureent Devices fluid anoeter P P a gh f Bourdon-Tube Pressure Gage Mechanical Pressure-Measureent Devices Diaphrag and Bellows Gages 3

4 8/6/01 Mechanical Pressure-Measureent Devices Diaphrag and Bellows Gages Mechanical Pressure-Measureent Devices Diaphrag and Bellows Gages

5 8/6/01 Exaple 6. A U-tube anoeter eploys a special oil having a specific gravity of 0.8 for the anoeter fluid. One side of the anoeter is open to local atospheric pressure of 9.3 inhg and the difference in colun heights is easured as 0 c±1.0 when exposed to an air source at 5 C. Standard acceleration of gravity is present. Calculate the pressure of the air source in Pascal and its uncertainty Exaple 6. Solution The anoeter fluid has a density of 8 percent of that of water at 5 C; so, ρ = 0.8ρ w = (0.8)(996 kg/ 3 ) = kg/ 3 The local atospheric pressure is p a = 9.3 inhg = Pa The fluid in this proble is the air which has a density at the above pressure and 5 C (98 K) of 5

6 8/6/01 Exaple 6. Solution For this proble the density is negligible copared to that of the anoeter fluid, but we shall include it anyway Positive-Displaceent Methods Flow Measureent The direct-weighing technique depends on collecting the fluid in a reservoir and easure the tie needed to finish this process and the flow is siply the volue of filled tank divided on the tie needed to fill that tank. Positive-displaceent flow-eters are used to weigh the ass or volue of liquid passing through it. Typical positive-displaceent device is the hoe water eter shown scheatically in Figs. 6

7 8/6/01 Flow Measureent Flow-Obstruction Methods Flow-Obstruction Methods Flow Measureent 7

8 8/6/01 Flow Measureent discharge coefficient No flow easureent channel is ideal,or frictionless, and soe losses are always present in the flow. The voluetric flow rate calculated fro Eq Is the ideal value, and it is usually related to the actual flow rate and an epirical discharge coefficient C by the following relation Exaple [] Assue that the venturi eter is used to find the flow rate in a pipe the carry water. If the pipe diaeter was 10c, the obstruction of vinturi is 5c. A ercury anoeter was used to find the pressure difference as 30cHg. If the water density was 1000 kg/ 3 and the discharge coefficient of this device is found as 0.85, find the value of the flow rate in 3 /s and the ass flow rate in kg/s. 8

9 8/6/01 Exaple [] Solution We can use the following equation to find the ideal flow A A x 1 d x d 10 kg N p1 p 30cHg kg 0.3 Pa Exaple [] Solution Substitute in Q 1.96x / 7.85 s 3 Q CvQideal s. kg Qactual s 3 actual 15 L s 9

10 8/6/01 Flow Measureent Drag Effects: Rotaeter The rotaeter is a very coonly used floweasureent device and is shown scheatically in Fig. The flow enters the botto of the tapered vertical tube and causes the bob or float to ove upward. The bob will rise to a point in the tube such that the drag forces are just balanced by the weight and buoyancy forces. A force balance on the bob gives But Drag Effects: Rotaeter Flow Measureent Cobine the previous equations and solve for u : Or Where: ρ f and ρ b are the densities of the fluid and bob V b is the total volue of the bob A b bob frontal area = (π/)d. Cd is drag coefficient A is the annular area and is found as: y is the vertical distance fro the entrance a is a constant indicating the tube taper 10

11 8/6/01 Exaple [3] A rotaeter is used to easure water (1000 kg/ 3 ) flow. If The bob frontal diaeter (d) is:.5c The bob volue (V b ) is 1 x The taper coefficient (a) is: 1 The tube diaeter (D) is: 3c The bob is ade of aluinu: ρ = 675 kg/ 3 drag coefficient (Cd) is: 0.01 The bob oves 3c (y) Find the flow rate in 3 /sof this device Exaple [3] Solution Ab d u A 5x x / s 5x x Q Au x10 / s L / s 11

Chapter 4: Temperature

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