2 PRESSURE MEASUREMENT
2.1 Describe what is meant by (a) a piezometer; piezometer; (b) piezometric piezometric head; (c) piezometric piezometric level; (d) a manometer; (e) the equal level, equal pressure principle; (f) the gauge coeff icient; (g) surface tension, and (h) a meniscus. (a) Basically a piezometer is a vertical tube of small diameter connected to a pipe via a tapping about 3 or 4 mm in diameter. The pressure in the p ipe, P, forces liquid up the piezometer tube until the pressure generated by the height, h, of the column of liquid in the piezometer (plus atmospheric pressure) equals the pressure in the pipe (Fig 2.1a). (b) The piezometric head is the height, h, to which the liquid rises in a piezometer. (c) If there are several piezometers in a pipe, the line connecting their respective piezometric heads is the piezometric level. The position of this line may sometimes be calculated theoretically and drawn to ill ustrate the height to which a liquid will rise in a piezometer or standpipe if a tapping is made anywhere along the pipeline. It illustrates the variation of pressure along the pipeline. (d) A manometer is another device for measuring fluid pressure. It usually has two limbs, one of which is connected via a small diameter tapping to the pipeline. If a d ifferential pressure between two points in a pip eline is to be measured, then both bot h limbs of the manometer are tapped to the pipeline. The manometer may have either a U configuration and be located below the pipeline, or a ∩ configuration in which case it is located above. In either case, one other manometer liquid will be used in addition to the pipe liquid. The configuration and density of the second liquid is selected according to whether the pressure to be measured is large or small. Mercury is often used to measure large pressures, since it is the densest liquid, while a light oil may be used to measure small pressures. (e) The equal level, equal pressure principle is the basis of pressure measurement using manometers. It states that at the same elevation in a co ntinuous liquid of uniform density the pressure i s constant. This follows simply from the fact that P = ρ gh gh, so at a constant depth, h, below the surface of a continuous liquid of uniform density, , the pressure must always be P. This is explained in Box 1.9. ρ , (f) The gauge coefficient is a measure of a manometer’s sensitivity relative to a simple water gauge. This is useful when there is a choice of manometer liquids, such as water and mercury or water and oil. The coefficient can be written as 1/(1 – s) where s is the relative density of o f the manometer liquid (see Chapter 2.6). T he coefficient can be somewhat misleading when comparing manometers with a more complex configuration than a simple water gauge. (g) Surface tension is the force which makes a liquid contract inwards when it is in contact with a gas or another immiscible liquid. This cohesive force results in the formation of water droplets in ai r, and also the ‘skin’ which is sometimes apparent on the surface of a liq uid. (h) A meniscus is the curved surface of a liquid at the interface with the air. A typical example is the menicus in a glass piezometer or manometer. With water the curvature is concave upwards, since the adhesive force between the glass and water which pulls the water surface upward is stronger than the co hesive surface tension force which is pulling the surface inward and downward. 3
2.2 A piezometer measures measures the pressure in a pipeline carrying water (1000 kg/m ). The piezometer reading is 253 mm measured measured from the centreline of the pipe. At this point, what is the ga uge pressure and the absolute pressure in N/m2? (Take atmospheric pressure as the equivalent of 10. 3 m of water.) 3 2 (a) Gauge pressure, P = ρ = ρ gh gh = 1000 × 9.81 × 0.253 = 2.48 × 10 N/m . 2 If atmospheric pressure, PATM , is 10.3 m of water this = 1000 × 9.81 × 10.3 N/m . = 101.04 × 103 N/m2. Absolute pressure, PABS = P + P ATM 3 3 2 PABS = (2.48 + 101.04) × 10 = 103.52 × 10 N/m
2.3 A vertical tube contains 200 mm of water on top of 270 mm of mercury (relative density 13.6). What is the pressure at the bottom of the tube? The total pressure is the sum of the pressures generated by each of the two liquids individually. Thus:
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pressure at bottom of tube = 1000 × 9.81 × 0.200 + 13.6 × 1000 × 9.81 × 0.270 = (1.962 + 36.022) × 103 3 2 = 37.984 × 10 N/m 2.4 A U-tube manometer like that in Fig 2.2 is used to measure the pressure in a pipe which carries oil of relative density 0.88. The lower manometer liquid is mercury of relative density 13.6. The surface of separation between the oil and mercury is 0.93 m below the centreline of the pipe. (a) If the differential head, hM, is zero, what is the gauge and absolute pressure at the centreline of the pipe? (b) If z in Fig 2.2 remains at 0.93 m but hM is now 37 mm of mercury, what is the gauge pressure in the pipe? Take 2 atmospheric pressure as PATM = 101 040 N/m . (a) At the level XX the pressure in the left hand limb = P + ρ Ogz = P + 0.88 × 1000 × 9.81 × 0.93 = P + 8.03 × 103 N/m2 At the level XX the gauge pressure i n the right hand limb = PATM = 0 By the equal level, equal pressure p rinciple the pressures at level XX in the two limbs must be equal so: P + 8.03 × 103 = 0 P = – 8.03 ×103 N/m2 Note that if hM is zero then this means there is a negative pressure (suction) in the pipe, that is a pressure that is less than atmospheric. 3 3 2 Once again, PABS = P + P ATM so PABS = – 8.03 × 10 + 101.04 × 10 N/m = 93.01 × 103 N/m2 (b) The left limb is unchanged so the pressure at level XX = P + 8.03 × 103 N/m2 At level XX the gauge pressure in the right hand limb = 13 600 × 9.81 × 0.037 = 4.94 × 103 N/m2 Equating the two pressures: P + 8.03 × 103 = 4.94 × 103 P = – 3.09 × 103 N/m2 2.5 Calculate the value of ( P1 – P2) if the manometer in Fig 2.6 (Example 2.2) has exactly the same readings but now carries (a) oil of relative density 0.88; (b) fresh water. (a) Starting at the level XX in the left limb and working upward: pressure at XX = ρ Ogz1 + P1 = 0.88 × 1000 × 9.81 × 0.60 + P1 = 5180 + P1 Starting at the level XX in the right limb and working upward: pressure at XX = ρ MghM + ρ Ogz2 + P2 = 13 600 × 9.81 × 0.13 + 0.88 × 1000 × 9.81 × 0.73 + P2 = 23 646 + P2 Now the equal level, equal pressure principle tells us that the two pressures at the level XX are the same, so: 5180 + P1 = 23 646 + P2 (P1 – P2) = 18 466 N/m2 (b) Following the procedure above but with water replacing the oil: in the left limb pressure at XX = ρ gz1 + P1 = 1000 × 9.81 × 0.60 + P1 = 5886 + P1 in the right limb pressure at XX = ρ MghM + ρ gz2 + P2 = 13 600 × 9.81 × 0.13 + 1000 × 9.81 × 0.73 + P2 = 24 505 + P2 Equating the two pressures at level XX: 5886 + P1 = 24 505 + P2 2 (P1 – P2) = 18 619 N/m 2.6 A U-tube manometer has the readings shown in Fig Q2.6 (ie in the Revision Question). The pipe liquid is water, and the manometer liquid is mercury. Calculate ( P1 – P2). Draw a line XX through the lower surface of separatio n, that is the one in the right hand limb. The height of the column of mercury in the left limb measured upward from XX is: hM = (2.167 + 0.343) – 1.949 = 0.561 m Starting at the level XX in the left hand limb and working upward:
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pressure at XX = ρ MghM + ρ gz1 + P1 where z1 = 1.949 m = 13 600 × 9.81 × 0.561 + 1000 × 9.81 × 1.949 + P1 = 93 966 + P1 Starting at the level XX in the right hand limb and working upward: pressure at XX = ρ gz1 + P2 = 1000 × 9.81 × 2.167 = 21 258 + P2 Because of the equal level, equal pressure principle the two pressures at level XX are the same: 93 966 + P1 = 21 258 + P2 (P1 – P2) = 21 258 – 93 966 = –72 708 N/m2 2.7 (a) Fig Q2.7 shows an inverted U-tube manometer with oil of density 800 kg/m3 above the pipe liquid, which is water. The pipeline is horizontal. What is the value of ( P1 – P2) ? (b) If the manometer readings are the same but the oil is replaced by air, what is ( P1 – P2) now? (a) The differential head (between the water levels in t he two manometer limbs) is: h = 0.430 – 0.280 = 0.150 m Draw a line XX through the upper s urface of separation. For each limb the p ressure at the centreline of the pi pe should be calculated by starting in the pipe and working up to XX. The pressure of the oil at level XX is denoted by PX, which is the same in both limbs. P1 = 1000 × 9.81 × 0.280 + 800 × 9.81 × 0.150 + PX . . . . . . . (1) = 3924 + PX P2 = 1000 × 9.81 × 0.430 + PX = 4218 + PX. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2) Subtracting equation (2) from equation (1): (P1 – P2) = 3924 + PX – (4218 + PX) = 3924 + PX – 4218 – PX = –294 N/m2 (b) This follows the same procedure as above but the pressure resulting from the air in the left limb between the surface of separation and the level XX is assumed to be negligible and omitted. The pressure at level XX is PX. For this new configuration: P1 = 1000 × 9.81 × 0.280 + PX P1 = 2747 + PX . . . . . . . . . . . . . . . . . . . . (1) P2 = 1000 × 9.81 × 0.430 + PX P2 = 4218 + PX . . . . . . . . . . . . . . . . . . . . (2) Subtracting equation (2) from equation (1): (P1 – P2) = 2747 + PX – (4218 + PX) = 2747 + PX – 4218 – PX 2 = –1471 N/m There is a short cut for part (b). This configuration is the equivalent of two piezometers with a differential head of –150 mm. Thus the difference in pressure is: 2 (P1 – P2) = 1000 × 9.81 × –0.15 = –1472 N/m (negative since 430 mm > 280 mm) Note that the oil-water combination results in the greater sensitivity: the smaller pressure in part ( a) gives the same reading as the standard piezometers in (b). 2.8 An inverted U-tube manometer is connected to a pipeline which slopes upward, as shown in Fig Q2.8. The pipeline carries water. Calculate ( P1 – P2) when (a) the upper part of the manometer is filled with air, and (b) when oil of relative density 0.80 is introduced above the water. (a) Draw a horizontal line XX through the upper surface of separation. PX is the air pressure at level XX. Starting at the centreline of the p ipe and working up the left hand limb gives: P1 = 1000 × 9.81 × 0.57 + PX = 5592 + PX . . . . . . . . . . . . . . (1) Assuming the pressure caused by the air in the right hand limb below XX is negligible, then: P2 = 1000 × 9.81 × 0.18 + PX = 1766 + PX . . . . . . . . . . . . . . (2) Subtracting equation (2) from (1) gives: (P1 – P2) = 5592 + PX – (1766 + PX) = 3826 N/m2
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(b) With oil above the water the differential head, hD, between the water levels in the two limbs needs t o be calculated. hD = 0.57 – (0.21 + 0.18) = 0.18 m Following basically the same procedure as above but remembering to take into account the oil between the water and level XX in the right limb: P1 = 1000 × 9.81 × 0.57 + PX = 5592 + PX . . . . . . . . . . . . . . (1) P2 = 1000 × 9.81 × 0.18 + 0.80 × 1000 × 9.81 × 0.18 + PX = 3179 + PX . . . . . . . . . . . . . . (2) Subtracting equation (2) from (1) gives: (P1 – P2) = 5592 + PX – (3179 + PX ) = 2413 N/m2 2.9 An inclined piezometer measures the pressure in a pipeline carrying water. The piezometer is inclined at an angle of 30 degrees to the horizontal and has an inclined reading of 330 mm. What is the gauge pressure in the pipe? If the piezometer is inclined at 30 degrees to the horizontal then this is the same as 60 degrees to the vertical, so θ in Fig 2.11 is 60 °. Equation 2.13 gives the pressure in an inclined gauge as: P = ρ gy cos θ = 1000 × 9.81 × 0.330 × cos 60° 2 = 1619 N/m 2.10 A Bourdon gauge is being calibrated using a dead weight tester similar to that in Fig 2.14. The diameter of the piston is 20.5 mm and it has a mass of 1 kg (which is included in M below). Additional 3 2 masses are added to the piston and the indicated pressure (P) recorded. Note that P is in units of 10 N/m 2 (kN/m ). The masses are then removed and P recorded again. The readings are shown below. (a) Using equation 2.14 calculate the actual pressure ( PACT) corresponding to M . (b) For each value of M obtain the gauge error ( P – PACT). Draw a graph of M against ( P – PACT). (c) What is the largest percentage gauge error ie 100 × ( P – PACT) / PACT and the average percentage gauge error? (d) Is there any evidence of hysteresis? Loading: M (kg) 1 (piston) 2 3 4 5 6
3
2
P (10 N/m )
34 63 92 119 151 180
Unloading: M (kg) 5 4 3 2 1 (piston)
3
2
P (10 N/m )
154 123 93 63 34
(a) From equation 2.14, PACT = W / A where A = π × 0.02052 /4 = 3.301 × 10–4 m2. With g = 9.81 m/s2 then: –4 3 PACT = 9.81 M /3.301 × 10 or PACT = 29.722 M × 10 . These values are recorded in column 3 below. (b) Using the original data and the values in column 3 below, the values of (P – PACT) are as shown in column 4. A graph of M against (P – PACT) is shown in Fig Q2.10. (c) The percentage gauge error ie 100 × (P – PACT) / PACT is calculated in the last column of the table. The largest value is 14.4% and the average value 5.3%. (d) Yes, there is some evidence of hysteresis since readings are higher when unloading ie 154 > 151, 123 > 119, 3 2 93 > 92 (all × 10 N/m ).
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----------------------------------------------------------------------------------------------------- M kg
3
2
P 10 N/m
3
2
3
PACT 10 N/m
2
( P – PACT) 10 N/m % gauge error ------------------------------------------------------------------------------------------------------------1 34 29.722 4.278 14.4 2 63 59.443 3.557 6.0 3 92 89.165 2.835 3.2 4 119 118.886 0.114 0.1 5 151 148.607 2.393 1.6 6 180 178.329 1.671 0.9 5 154 148.607 5.393 3.6 4 123 118.886 4.114 3.5 3 93 89.165 3.835 4.3 2 63 59.443 3.557 6.0 1 34 29.722 4.278 14.4 ------Average 5.3%
Q2.10 Graph of M Fig against ( P – PACT)
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