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1S, , mercury, , open to the, , The right limb of a simple U-tube manometer containing, 0.9 S flowing. The, amosphere while the left limb is connected to a vive in which a fluid of sp. gr., Find the pressure of fluid in, Cenire of the pipe is 12 cm below the level of mercury in the right limb., the pipe if the diference of mercury level in the two linmbs is 20 cm., Problem 2.9, , Solution. Given:, Sp. gr. of fluid,, , S1 = 0.9, , Density of fluid,, , P, , Sp. gr. of mercury,, , 20 cm, , P2=13.6 x 1000 kg/m, , Difference of mercury level,, Height of fluid from A-A,, Let p, , 12, , S2 = 13.6, , Density of mercury,, , Pressure of fluid in, , =, , = Sx 1000 = 0.9 x 1000 = 900 kg/m, , h, h, , =, , 20, , =, , 20, , cm, -, , =, , 12, , =, , 0.2, , m, , 8, , cm, , =, , 0.08, , m, , A, , A, , pipe, , Equating the pressure above A-A, we get, p+ P18h = Pz8h2, or, , p + 900 x 9.81 x 0.08, , 13.6 x 1000 >x 9.81 x .2, , p, , 13.6 x 1000 x 9.81 x .2, , 900 x 9.81 x 0.08, , 26683- 706, 25977 N/m, 2.597 N/cm. Ans., A simple U-tube manometer containing mercury is connected to a pipe in which a, =, , Problem 2.10, , Fig. 2.10, , =, , fuid of sp. gr. 0.8 and having vacuum pressure is flowing. The other end of the manometer is open to, atmosphere. Find the vacuum pressure in pipe, if the difference of mercury level in the two limbs is, , 40 cm and the height f fluid in the left from the centre ofpipe is 15 cm below., , Solution. Given:, Sp. gr. of fluid,, , S, , Sp. gr. of mercury,, , S, , Density of fluid,, , P 800, , Density of mercury,, , P2, , Difference of mercury level,, =, , = 0.8, , 15, , cm, , line A-A,, , =, , 0.15, , we, , m., , = 13.6, , H15 cm, , 13.6 x 1000, , 40 cm, , h, , 40 cm 0.4 m. Height of liquid in left limb,, Let the pressure in pipe =p. Equating pressure above datum, , h,, , =, , =, , get, P28h2 +Pi8h; +p, , =, , 0, , A, , A, , Fig. 2.11

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Pressu, 1000, , [13.6, , =, , [53366.4, , -, , x, , Ient, , 45, , N/cm2., , Ans., , P,ghl, , P=-I Pz8h2 +, =-, , and its Measure, , 9.81, , x, , 1177.2], , +, , A U-Tube manometer is used to, measure, fe in excess of atmospheric pressure. The right limb of the, , Drablem 2.11, , 0.4, , x, =, , -, , +, , 800, , 54543.6, , x, , 9.81, , N/m, , =, , x, -, , 0.151, 5.454, , the, , pressure of water in a pipe line, which, manometer contains mercury and is open of, to, atmosphere. The contact between water and mercury 1s n the left limb. Determine the pressure, water in the main line, if the difference in level of mercury in the limbs of U-tube is 10 cm and the, free surface of mercury is in level with the centre of the pipe. If the pressure of water in pipe line is, Teduced to 9810 N/nm', calculate the new difference in the level, , mercury., , in both cases., , Sketch the, , arrangements, , Solution. Given, Difference of mercury = 10 cm = 0.1 m, , The, , arrangement is shown in Fig. 2.11 (a), , Ist Part, , Let PA = (pressure of water in pipe line (i.e., at point A), The points B and C lie on the same horizontal line. Hence, pressure, at C. But pressure at B, , Pressure, , =, , at, , A, , +, , Pressure due, , to, , 10, , cm, , of wateer, , at, , B should be, , (or 0.1 m), , p = 1000 kg/m, , WATER, , C, , =Pressure, , at, , D, , +, , D, , A, , =PA +981 N/m, at, , pressure, , and h = 0.1 m, , PA+1000 x 9.81 x 0.1, Pressure, , to, , RIGHT LIMB, , PA +P X g xh, where, , equal, , ..i), , Pressure due, , to, , 10, , cm, , 0+ Po Xg xho, =13.6 x 1000 kg/m, ho10cm =0.1 m, , of mercury, , LEFT LIMB, , 10 cm, , =, , B, , where Po for mercury, and, , MERCURY, , Pressure at C = 0+ (13.6 x 1000) x 9.81 x 0.1, =, , 13341.6 N, , But pressure at B is equal, tions (i) and (ii), we get, , to, , C, , ..(ii), , pressure, , PA +981 =, , PA, , at C. Hence, , equating, , the equa-, , 13341.6, , Fig. 2.11 (a), , 13341.6, , 981, , = 12360.6 N2, , Ans, , Ans., , m, , Ilnd Part, Given,, , PA = 9810 N/m, , Find new difference of mercury level. The arrangement is shown in Fig. 2.11 (b). In this case the, , pressure at A is 9810 N/m which is less than the 12360.6 N/m. Hence mercury in left limb will rise., ne rise of mercury in left limb will be equal to the fall of mercury in right limb as the total volume or, , mercury remains same., Let x = Rise of mercury in left limb in cm, , Then fall of mercury in right limb = x cm, , final, , he points B,, conditions., , C and D show the initial conditions whereas, , points B*,, , C* and D* show the

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46 Fluid Mechanics, The pressure at B*, or, , Pressure, , =, , C*, , at, , Pressure at A + Pressure due to (10 - x) cm of water, = Pressure at D*+ Pressure due to, , (10, or, , 2) cm of mercury, , D, , A, , X, , PA+P1 Xg xhj = PD* + P2 Xg xh2, D, , 10-x, or, , 1910, , =, , +, , 0, , 1000, +, , x, , (13.6, , 9.81, , x, , x, , 10 cm, , |100, 100, , 1000), , x, , B, , 9.81, , x, , 10-2x, | 100, 100, , B, , Dividing by 9.81, we get, or, or, or, , 1000+100 10x =1360 -272x, , 272x, , 10x, , =, , 1360, , -, , 1100, , 262r = 260, , 260, , = 0.992 cm, , 262, New difference of mercury =, =, , 10 -, , 2x cm =10 -, , 8.016, , cm., , AnS., , 2 x, , 0.992, , Fig. 2.11 (b), , (10-2x