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# Manometers.pptx

Pressure measurement.

Pressure measurement.

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### Manometers.pptx

1. 1. M.405.8 0 DEPARTMENT OF TECHNICAL EDUCATION ANDHRA PRADESH Name : Dr. P. Ugandhar Designation : Sr. Lecturer in Mechanical Engg. Branch : Mechanical Engineering Institute : Govt. Polytechnic, SKLM Year / Semester : IV Subject : Hydraulics & Hydraulic Machinery Subject Code : M-405 Topic :Fluid Properties Duration :50 Minutes Sub Topic : Problems on Manometers Teaching Aids : PPT Animations
2. 2. M.405.6 1 OBJECTIVES On completion of this period, you would be able to • Know the various pressure measuring devices • Know the Difference between simple and differential manometers
3. 3. M.405.6 2 Measurement of Pressure Differential (measure pressure difference) Fig. 1
4. 4. M.405.6 3 Hydrostatic Law: · The pressure at any depth ‘h’ from the free surface in a liquid of density  is given by P = gh · If ‘h’ is measured in upward direction P = -gh The pressure of liquid if expressed in terms of the height (h) of liquid column, it is known as pressure head of liquid. Pressure head h = P/ g
5. 5. M.405.6 4 Piezometer is used to measure low liquid pressures What is Piezometer Fig. 2 Piezometer
6. 6. M.405.6 5 Manometer • It is pressure measuring device • It is used to measure difference of pressures between two points. • Its one column of liquid is balanced by another column of liquid
7. 7. M.405.6 6 Types of Manometers • Simple U – tube Manometer • Differential U – tube Manometer • Simple U-tube manometer is used to measure pressure at a point (gauge or vacuum) • Differential U-tube manometer is used to measure the difference of pressure between two points
8. 8. M.405.6 7 How to find pressure at a point using U-tube Manometer Fig 3 U-tube Manometer • The Pressures at two points P and Q must be equal (being the same liquid)
9. 9. M.405.6 8 U-tube Manometer: • Equating pressures at P and Q, we have P1+Ag(y+x) = patm +Bgx P1-Patm = (B-A) gx - Agy • Mercury is generally used as a manometric fluid because of its high density and low vapour pressure
10. 10. M.405.6 9 To find Vaccum Pressure: • The Pressures at two points P and Q must be equal. • Equating pressures at P and Q, we have Fig 4 U-tube Manometer
11. 11. M.405.6 10 Conversion of one fluid column into other fluid column • The pressure is given by P =  gh where h is the height in terms of fluid of density  • The same pressure can be expressed in terms of different fluid columns. i.e. P = 1gh1 = 2gh2 Where h1 and h2 are heights in terms of fluids of density  1 and  2 respectively.  1h1 = 2h2 Or s1h1 = s2h2
12. 12. M.405.6 11 SUMMARY • In this class, we have discussed about • Hydrostatic Law • Piezometers • Simple U-tube manometer to measure positive gauge pressures. • Simple U-tube manometer to measure Vaccum Pressures
13. 13. M.405.6 12 QUIZ • A U-tube manometer with both limbs open to atmosphere contains two immiscible liquids of densities 1 and 2 as shown in figure. Under equilibrium ‘h’ is given by (a) h =0 (b) h = L  2/ 1 (c) h = L(1- 1/ 2) (d) h = L ( 2/ 1 –1) Fig-5
14. 14. M.405.6 13 • A U-tube manometer is connected to a pipeline conveying water as shown in figure. Calculate the pressure head of water in the pipeline is Assignment Fig-6
15. 15. M.405.7 14 OBJECTIVES On the completion of this period, you would be able to • Know the principle of differential manometer • Working principle of Bourdon’s pressure gauge
16. 16. M.405.7 15 Differential Manometer:   1 2 w w m P y x g P y g x g           1 2 m w P P gx      • The Pressures at two points P and Q must be equal. • Equating pressures at P and Q, we have Fig-1
17. 17. M.405.7 16 Inverted Differential Manometer • When a small difference of pressure between two points is to be measured, then inverted differential manometer is used • A liquid of specific gravity less than that of the flowing liquid is used
18. 18. M.405.7 17 Fig-2 Inverted differential manometer Pressure at D = Pressure at C PA -  1gb- mgc = PB - 2g(a+b+c) PB-PA =  2g(a+b+c) -  1gb -  mg c
19. 19. M.405.7 18 Bourdon’s Pressure Gauge: ∙ It is a mechanical gauge and is used for measuring pressure in steam boilers. This gauge has an elastic metallic tube of elliptical section bent into circular arc Fig-3 Bourdon´s Pressure Guage
20. 20. M.405.7 19 Bourdon’s Pressure Gauge: •When fluid enters the bent tube at one end, it tends to straighten the tube. • This causes the other end to move, causing pointer to move on a graduated scale giving pressure reading As the tube is surrounded by atmospheric pressure, the movement of the tube is against atmospheric pressure and hence this gauge measures pressure above or below atmospheric pressure (gauge pressure)
21. 21. M.405.7 20 SUMMARY In this class, we have discussed about • Differential Manometers • Differential inverted U-tube Manometers • Bourdon’s Pressure Gauge
22. 22. M.405.7 21 QUIZ 1. The manometer shown in figure connects two pipes carrying oil and water respectively. The difference between oil and water pressure is (a) ( Water+  oil)gH (b) ( Water / oil)gH (c) ( Water -  oil)gH (d)  Oil gH Fig-4
23. 23. M.405.7 22 2. The manometer shown in figure connects two pipes carrying oil and water respectively. From this figure it can be concluded that (a) pressure in pipes are equal (b) the pressure in oil pipe is higher (c) the pressure in water pipe is higher (d) The pressure in two pipes can not be compared (for want of Sufficient data) QUIZ Fig-5
24. 24. M.405.7 23 Frequently Asked Questions 1. What is differential manometer? 2. Why is mercury used as a manometric liquid? 3. How does Bourdon’s pressure gauge work?
25. 25. M.405.8 24 RECAP In the previous class, we have discussed about • Principle of manometers • Finding pressure at a point by using simple manometer • Finding pressure difference between two points by using differential manometer
26. 26. M.405.8 25 OBJECTIVES • On the completion of this period, you would be able to •solve simple problems on manometers
27. 27. M.405.8 26 Problem –1: • Find the pressure of a fluid of specific gravity 0.9 at the point A as shown in figure. Fig-1
28. 28. M.405.8 27 Solution • Let x-x be the datum line  Density of oil (0) = 0.9 x 1000 = 900 kg/m3 • Pressure in the left limb at x-x = Pressure in the right limb at XX 0 20 12 20 0 100 100 m g g                  • PA+
29. 29. M.405.8 28 Solution 3 8 20 900 9.81 13.6 10 9.81 100 100 A P                  • PA = 13.6  103 9.81  0.2 – 900 9.81  0.08 • PA = 25976.88 N/m2 • PA+
30. 30. M.405.8 29 • Problem – 2 A differential manometer is connected as shown in Fig. Find the pressure difference between A and B in N/m2 and in cm of water. • Let us choose xx as datum line. • Pressure in left limb at xx = Pressure in right limb at xx Fig-2
31. 31. M.405.8 30 Solution 2 2 0 38 23 (0.53) (0.23) 100 A H o B H o P g g P g                  2 0 0.61 0.53 (0.23) A B H o P P g g       2 0 (0.08) (0.23) A B H o P P g            9.81 1000(0.08) 810(0.23) A B P P      3 0.81 1000 810 / o kg m     2 1042.8 / A B P P N m    . . . . . . .
32. 32. M.405.8 31 • Pressure difference in terms of water P = gh   • h = -0.1063 m of water or h = -0.1063 100 cm  h = -10.63 cm of H20 2 A B P = P P 1042.8 / N m     . .
33. 33. M.405.8 32 ASSIGNMENT 1. The pressure at certain point is observed to be 40 m of oil of specific gravity 0.6. Find the corresponding height in terms of mercury of specific gravity 13.6 at that point. 2. Pressures have been observed at four different point in different units of measurements of follows: (a) 150 kpa (b) 1800milli bar (c) 20m of water (d) 1240mm of mercury Arrange the points in the descending order of magnitude of pressure