Technological Institute of the Philippines 938 Aurora Blvd., Cubao, Quezon City College of Engineering and Architecture Department of Civil Engineering Prepared by: Engr. Adams Royce A. Dionisio, RCE Structural Engineer CE 023 Fluid Mechanics Technological Institute of the Philippines 938 Aurora Blvd., Cubao, Quezon City College of Engineering and Architecture Department of Civil Engineering Pressure Prepared by: Engr. Adams Royce A. Dionisio, RCE Structural Engineer Pressure- or unit pressure, is defined as the normal force per unit area exerted by a solid, liquid or gas over a ππΉ surface. π = ππ΄ ππΉ π= ππ΄ Unit pressure ππΉ πΉ π= = ππ΄ π΄ Uniform or average pressure Technological Institute of the Philippines 938 Aurora Blvd., Cubao, Quezon City College of Engineering and Architecture Department of Civil Engineering Pascal’s Law Prepared by: Engr. Adams Royce A. Dionisio, RCE Structural Engineer Pascal’s Law- states that the intensity of the pressure acting at a point in a fluid at rest (or moving in constant velocity) is the same in all directions, stated by a French Mathematician Blaise Pascal in 17th century. πΉπ¦ − π − πΉπ cos π = 0 +↑ πΉπ = 0 π ππ₯(ππ§) − 0 − π ππ (ππ§) cos π = 0 π¦ π ππΉ π= ππ΄ πΉπ πΉπ₯ π ππ¦ ππ ππ₯ ππ₯ = ππ cos π ππ¦ = ππ sin π πΉπ¦ +→ ππ¦ (ππ cos π)(ππ§) − 0 − ππ ππ (ππ§) cos π = 0 ππ¦ − 0 − ππ = 0 ππ¦ = ππ πΉπ₯ − πΉπ sin π = 0 πΉπ» = 0 π ππ¦(ππ§) − π ππ (ππ§) sin π = 0 π₯ π ππ₯ (ππ sin π)(ππ§) − ππ ππ (ππ§) sin π = 0 ππ₯ − ππ = 0 π ππ₯ = ππ¦ = ππ ππ₯ = ππ Technological Institute of the Philippines Absolute and Gage Pressure 938 Aurora Blvd., Cubao, Quezon City College of Engineering and Architecture Department of Civil Engineering Prepared by: Engr. Adams Royce A. Dionisio, RCE Structural Engineer πππ πππ’π‘π ππππ π π’ππ (ππππ ) ππππ ππππ ππππ π π’ππ ππππ π π’ππ (πππππ ) (πππππ ) π£πππ’π’π ππ π π’ππ‘πππ Absolute pressure- is the intensity of pressure that is measured above the absolute zero pressure (lowest possible pressure) and can never be negative in value. Atmospheric pressure- is the prevailing pressure in the air surrounding which decreases with increasing altitude. Standard atmospheric pressure- is the absolute pressure that is measured at sea level and at a temperature of 15°C (59°F) equivalent to 101.325 kPa or 14.7 psi or 760 mm Hg or 29.9 in. Hg or 2116 lb/ft2 or 1 atmosphere (atm) Gage pressure- or relative pressure, is the pressure that is measured above or below the atmospheric pressure. πππππ = ππππ − πππ‘π ππ‘πππ πβππππ ππππ π π’ππ πππππ = ππππ − πππ‘π πππ πππ’π‘π π§πππ ππππ π π’ππ (πππππππ‘ π£πππ’π’π) Technological Institute of the Philippines 938 Aurora Blvd., Cubao, Quezon City College of Engineering and Architecture Department of Civil Engineering Prepared by: Engr. Adams Royce A. Dionisio, RCE Structural Engineer Static Pressure Variation in Incompressible Fluids πΉ1 1 π¦ π₯ π π β β cos π = πΏ +β πΉπ₯ = 0 2 ππΉ π= ππ΄ πΉ2 π πΎ= π πΉ1 + π cos π − πΉ2 = 0 π1 ππ΄ + πΎπΏ ππ΄ cos π − π2 ππ΄ = 0 π1 + πΎβ − π2 = 0 π2 − π1 = πΎβ ππ − ππ = πΈπ The difference in pressure between any two points in a homogeneous fluid at rest is equal to the product of the unit weight of the fluid and their difference in elevation(vertical distance between the points) ′ ′ π − π1 − ππ = πΎβ 2 π1 = π1 + ππ ′ π 2 = π1 + πΎβ + ππ [π2 − π1 = πΎβ] ′ π = π2 + ππ 2 π2 ′ − (π1 + ππ) = πΎβ ππ = ππ + πΈπ Any change in pressure applied at any point in a liquid at rest is transmitted equally and undiminished to every other point in the liquid. πΉ1 1 π2 = πππ‘π + πΎβ π2 = π1 + πΎβ π π2 = πΎβ [πππππ = ππππ − πππ‘π ] π π = πΈπ β The pressure at any point below a free liquid surface is equal to the product of the unit weight of the fluid β and its vertical distance from the free 2 cos π = πΉ2 πΏ liquid surface. Technological Institute of the Philippines 938 Aurora Blvd., Cubao, Quezon City College of Engineering and Architecture Department of Civil Engineering Prepared by: Engr. Adams Royce A. Dionisio, RCE Structural Engineer Static Pressure Variation in Incompressible Fluids πΉ1 1 π¦ π₯ π π¦ +β πΉπ₯ = 0 π₯ π π β β cos π = πΏ +→ 2 ππΉ π= ππ΄ πΉ1 2 1 πΏ πΉ2 π πΎ= π πΉ1 + π cos π − πΉ2 = 0 π1 ππ΄ + πΎπΏ ππ΄ cos π − π2 ππ΄ = 0 π1 + πΎβ − π2 = 0 π2 − π1 = πΎβ πΉ2 πΉ1 1 π π β β cos π = πΏ 2 πΉ2 πΉπ₯ = 0 π = ππΉ ππ΄ πΉ1 − πΉ2 = 0 π1 ππ΄ − π2 ππ΄ = 0 [π2 = π1 + πΎβ] π1 − π2 = 0 π2 = π1 + πΎ(0) π1 = π2 π1 = π2 The pressures along the same horizontal plane in a homogeneous fluid at rest are equal π2 = πππ‘π + πΎβ π2 = π1 + πΎβ π2 = πΎβ [πππππ = ππππ − πππ‘π ] π = πΈπ The pressure at any point below a free liquid surface is equal to the product of the unit weight of the fluid and its vertical distance from the free liquid surface. Technological Institute of the Philippines Pressure Head 938 Aurora Blvd., Cubao, Quezon City College of Engineering and Architecture Department of Civil Engineering Prepared by: Engr. Adams Royce A. Dionisio, RCE Structural Engineer Pressure Head- is the height of a column of homogeneous fluid that will produce a given intensity of pressure (gage). π π = πΎβ β= πΎ Given a pressure of magnitude of 50 kPa, find the pressure head of water with a unit weight of 9.81 kN/m3 and the pressure head of mercury with a unit weight of 133 kN/m3. π β= πΎ π 50 π₯ 103 π/π2 = 5.10 π βπ€ = = 3 3 πΎπ€ 9.81 π₯10 π/π βπ»π π 50 π₯103 π/π2 = 0.376 π = = 3 3 πΎπ»π 133 π₯10 π/π Technological Institute of the Philippines 938 Aurora Blvd., Cubao, Quezon City College of Engineering and Architecture Department of Civil Engineering Sample Problems: Prepared by: Engr. Adams Royce A. Dionisio, RCE Structural Engineer 1. The underground storage tank used in a service station contains gasoline filled to the level A. Determine the gage pressure at each of the five identified points. Note that point B is located in the stem, and point C is just ππ below it in the tank. Take ππ = 730 3. π Solution: ππ· = ππ΄ + πΎπ β π = 0 πππ π΄ [π2 = π1 + πΎβ] ππ ππ· = 0 + 730 3 π ππ΅ = ππ΄ + πΎπ β ππ π ππ΅ = 0 + 730 3 9.81 2 1 π π π π ππ΅ = 7161.3 2 = 7.1613 πππ π π 9.81 2 π 1π+2π π ππ· = 21,483.9 2 = 21.4839πππ π ππΈ = ππ· = 21.4839πππ ππΆ = ππ΅ = 7.1613 πππ Recall: Any change in pressure applied at any point in a liquid at rest is transmitted equally and undiminished to every other point in the liquid. The pressures along the same horizontal plane in a homogeneous fluid at rest are equal Technological Institute of the Philippines Measurement of Static Pressure 938 Aurora Blvd., Cubao, Quezon City College of Engineering and Architecture Department of Civil Engineering Prepared by: Engr. Adams Royce A. Dionisio, RCE Structural Engineer Barometer- is a device for measuring intensities of pressure exerted by the atmosphere. It was invented in the mid-17th century (1643) by Evangelista Torricelli, using mercury as a preferred fluid, since it has a high density and a very small vapor pressure. [π2 = π1 + πΎβ] ππ΅ = ππ΄ + πΎπ»π β ππ 101.325 πππ = 0 + 133.29 3 β π β = 0.760 π = 760 ππ = 29.9 ππ Technological Institute of the Philippines Measurement of Static Pressure 938 Aurora Blvd., Cubao, Quezon City College of Engineering and Architecture Department of Civil Engineering Prepared by: Engr. Adams Royce A. Dionisio, RCE Structural Engineer Manometer- is a device consists of a transparent tube that is used to determine the gage pressure in a liquid. Types of Manometer Open Type- a manometer with an atmospheric surface in one end and capable of measuring gage pressures. Piezometer- is the simplest type of manometer which consist of a tube which is open at one end to the atmosphere, while the other end is inserted into a vessel, where the pressure of a liquid is to be measured. U-tube Manometer- consist of a tube, usually bent in the form of a U, containing a liquid of known specific gravity, the surface of which moves proportionally to changes of pressure. Differential Type- a manometer without an atmospheric surface and capable of measuring only differences of pressure. Differential Manometer- is used to determine the difference in pressure between two points in a closed fluid system. Technological Institute of the Philippines Measurement of Static Pressure 938 Aurora Blvd., Cubao, Quezon City College of Engineering and Architecture Department of Civil Engineering Prepared by: Engr. Adams Royce A. Dionisio, RCE Structural Engineer Bourdon Gage- this gage consists of a coiled metal tube that is connected at one end to the vessel where the pressure is to be measured. The other end of the tube is closed so that when the pressure in the vessel is increased, the tube begins to uncoil and respond elastically. Using the mechanical linkage attached to the end of the tube, the dial on the face of the gage gives a direct reading of the pressure, which can be calibrated in various units, such as kPa or psi. Pressure Transducers- an electromechanical device that can be used to measure pressure as a digital readout. When end A is connected to a pressure vessel, the fluid pressure will deform the thin diaphragm. The resulting strain in the diaphragm is then measured using the attached electrical strain gage. Technological Institute of the Philippines 938 Aurora Blvd., Cubao, Quezon City College of Engineering and Architecture Department of Civil Engineering Sample Problems: Prepared by: Engr. Adams Royce A. Dionisio, RCE Structural Engineer 2. The funnel is filled with oil and water to the levels shown. Determine the depth of oil β′ that must be in the ππ funnel so that the water remains at a depth C, and the mercury level made h = 0.8 m. Take ππ = 900 3, ππ€ = ππ 1000 3, π Solution: ππ»π = π ππ 13,550 3 . π [π2 = π1 + πΎβ] ππ΅ = ππ΄ + πΎπ β′ ππΆ = ππ΅ + πΎπ€ (0.4) ππΆ = (ππ΄ + πΎπ β′) + πΎπ€ (0.4) ππ· = ππΆ − πΎπ»π [ 0.2 + β′ + 0.4 − 0.8] ππ· = 0= ππ΄ + πΎπ β′ + πΎπ€ 0.4 − πΎπ»π [ 0.2 + β′ + 0.4 − 0.8] ππ ′ ππ 0 + 900 3 β + 1000 3 π π β′ = 0.2458 π = 246 ππ 0.4 ππ − (13,550 3 )[ 0.2 + β′ + 0.4 − 0.8] π Technological Institute of the Philippines 938 Aurora Blvd., Cubao, Quezon City College of Engineering and Architecture Department of Civil Engineering Static Pressure Variation in Compressible Fluids [ππππ = ππ ππππ ] [πΎ = ππ] π π πβ π π ππ π =− πβ π π π ππ = − π1 1 π1 − π2 = ππ = −πΎ πβ πβ π2 2 ππππ π ππ πππ£ππ π1 π2 ππ = π β1 − β2 π πβ π π ππ π π’ππ π = ππππ π‘πππ‘(ππ ππ‘βπππππ) π ln π1 − ln π2 = − (β1 − β2 ) π π π1 π ln =− β1 − β2 π2 π π π1 − =π π2 − π1 = π2 π π π π β1 −β2 π π π β1 −β2 Prepared by: Engr. Adams Royce A. Dionisio, RCE Structural Engineer Technological Institute of the Philippines 938 Aurora Blvd., Cubao, Quezon City College of Engineering and Architecture Department of Civil Engineering Sample Problems: Prepared by: Engr. Adams Royce A. Dionisio, RCE Structural Engineer 3. The natural gas in the storage tank is contained within a flexible membrane and held under constant pressure using a weighted top that is allowed to move up or down as the gas enters or leaves the tank as shown in the figure. Determine the required weight of the top if the (gage) pressure at the outlet A is to be 600 kPa. ππ π½ The gas has a temperature of 20°C, ππ = 0.665 3 and π = 518.3 . π ππ−πΎ Solution: ππ π π’ππ πππππππ π ππππ πππ : +↑ πΉπ = 0 ππΉ π= ππ΄ πππ π = ππππ π‘πππ‘(ππ ππ‘βπππππ) ππ΅ = π − π π β1 −β2 ππ΄ π 9.81 − 30−0 3 101.325)10 )π (518.3)(20+273.15) π ππ΅ = ((600 + π = (202 )ππ΅ 4 ππ΅ = 598,642.881 ππ ππ π π’ππ πππππππππ π ππππ πππ : π ππ΄ = ππ΅ + πΎπ β π = 202 ππ΅ = 188,069.2077 ππ 4 ππ 600π₯103 ππ = ππ΅ + (0.665 3 )(9.81)(30 π) π ππ π π’ππ ππππ π‘πππ‘ πππ ππππ π π’ππ: ππ΅ = 599,804.2905 ππ ππ΄ = ππ΅ π π π = 202 ππ΅ = 188,434.0753 ππ π = 202 (600) = 188,495.5592 ππ 4 4 Technological Institute of the Philippines Sample Problems: 938 Aurora Blvd., Cubao, Quezon City College of Engineering and Architecture Department of Civil Engineering Prepared by: Engr. Adams Royce A. Dionisio, RCE Structural Engineer 4. Determine the difference in pressure ππ΅ − ππ΄ between the centers A and B of the pipes, which are filled with water. The mercury in the inclined-tube manometer has the level shown ππ»π = 13.55 Solution: [π2 = π1 + πΎβ] ππ΄ + 9810 0.1 + 13.55 9810 0.25 − 0.1 − 9810(0.25) = ππ΅ ππ΅ − ππ΄ = 18,467.325 ππ Technological Institute of the Philippines 938 Aurora Blvd., Cubao, Quezon City College of Engineering and Architecture Department of Civil Engineering Sample Problems: Prepared by: Engr. Adams Royce A. Dionisio, RCE Structural Engineer 5. A 0.5-in.-diameter bubble of methane gas is released from the bottom of a lake. Determine the bubble’s ππ diameter when it reaches the surface. The water temperature is 68°F and the atmospheric pressure is 14.7 2 ππ Solution: ππ‘ π‘βπ π π’πππππ: ππ‘ π‘βπ πππ‘π‘ππ: π [ππππ = ππ ππππ ] [π2 = π1 + πΎβ] π = π ππ ππ 14.7 2 + 62.4 3 20 ππ‘ ππ ππ‘ 1 ππ‘ 12 ππ. ππ ππ 14.7 2 + 62.4 3 20 ππ‘ ππ ππ‘ 1 ππ‘ 12 ππ. 2 π = π ππππ ππ 2 ππ = ππ ππππ ππ π 14.7 2 = π ππππ ππ ππ ππ 14.7 2 (ππ ) = ππ ππππ ππ ππ ππ 14.7 2 + 62.4 3 20 ππ‘ ππ ππ‘ 14.7 + 62.4 20 ππ = 0.5835 ππ 1 12 2 2 1 ππ‘ 12 ππ. 4 0.5 π 3 2 3 ππ ππ = 14.7 2 (ππ ) ππ 4 ππ = 14.7 π 3 2 3
