Pressure Pressure is force per unit area: F P = A Pressure Te direction of te force exerted on an object by a fluid is toward te object and perpendicular to its surface. At a microscopic level, te force is associated wit te atoms and molecules in te fluid bouncing elastically from te surfaces of te object. Te SI unit for pressure is te pascal. Pa = N/m Te arrows below illustrate te directions of te force of pressure acting on a sark at various point of its body. Tey sow tat te force is always pointing towards and perpendicular to te surface of te sark. Atmosperic pressure atm =.0 x 0 5 Pa = 4.7 psi = 760 torr (psi = pounds / square inc) At atmosperic pressure, every square meter as a force of 00,000 N exerting on it, coming from air molecules bouncing off it! Wy don t we, and oter tings, collapse because of tis pressure? We, uman, ave an internal pressure of atmospere. General solid objects do not collapse because tey possess an elastic modulus (i.e., rigidity) enabling tem to deform only 4 little wen compressed by te atm. pressure. Conceptual example : Blood pressure Conceptual example. Rank by pressure Question A typical reading for blood pressure is 0 over 80. Wat do te two numbers represent? Wat units are tey in? Ans. 0 mm Hg (millimeters of mercury) is a typical systolic pressure, te pressure wen te eart contracts. 80 mm Hg is a typical diastolic pressure, te blood pressure wen te eart relaxes after a contraction. 760 mm Hg is te typical atmosperic pressure. Te blood pressure readings are not absolute tey tell us ow muc above atmosperic pressure te blood pressure is. 5 A container, closed on te rigt side but open to te atmospere on te left, is almost completely filled wit water, as sown. Tree points are marked in te container. Rank tese according to te pressure at te points, from igest pressure to lowest.. A = B > C. B > A > C. B > A = C 4. C > B > A 5. C > A = B 6. some oter order See explanation on te next page. P B P C P B 6
Pressure in a static fluid Pressure in a Static Fluid A static fluid is a fluid at rest. In a static fluid: Pressure increases wit dept. Two points at te same vertical position experience te same pressure, no matter wat te sape of te container. If point is a vertical distance below point, and te pressure at point is P, te pressure at point is: P = P+ ρg Point does not ave to be directly below point - wat matters is te vertical distance. To understand te equation, P = P + ρg, consider a column of fluid wit area A and eigt as sown. Te weigt of te fluid column is ρga. Tis weigt will cause an extra pressure of ρga/a = ρg at te bottom of te fluid column and explains te equation. Area, A P P 7 8 Measuring pressure Conceptual example b. Rank by pressure Te relationsip between pressure and dept is exploited in manometers (or barometers) tat measure pressure. A standard barometer is a tube wit one end sealed. Te sealed end is close to zero pressure, wile te oter end is open to te atmospere. Te pressure difference between te two ends of te tube can maintain a column of fluid in te tube, wit te eigt of te column being proportional to te pressure difference. P = P+ρ g P (usually ~ 0) P 9 Wat is te pressure at tis level?. Cannot be determined. P A ρ fluid g(0.m). P A + ρ fluid g(0.m) 4. Atmosperic pressure 5. and 4 6. and 4 0 Example. Water pressure At te surface of a body of water, te pressure you experience is atmosperic pressure. Estimate ow deep you ave to dive to experience a pressure of atmosperes. Given tat density of water is 000 kg/m and g = 0 m/s (or N/kg). Solution: P = P+ ρg 00000 Pa = 00000 Pa + (000 kg/m ) (0 N/kg) works out to 0 m. Every 0 m down in water increases te pressure by atmospere. Te origin of te buoyant force P = P+ ρg Te net upward buoyant force is te vector sum of te various forces from te fluid pressure. Because te fluid pressure increases wit dept, te upward force on te bottom surface is larger tan te downward force on te upper surface of te immersed object. F = Δ P A= ρ g A= ρ gv net fluid fluid Tis is for a fully immersed object. For a floating object, is te eigt below te water level, so we get: F = ρ gv net fluid disp
Conceptual example 4. Wen te object goes deeper Question: If we displace an object deeper into a fluid, wat appens to te buoyant force acting on te object? You may assume tat te fluid density is te same at all depts. Te buoyant force Conceptual example 4. Wen te object goes deeper (cont d) Explanation: P = P+ ρg. increases. decreases. stays te same If te fluid density does not cange wit dept, all te forces increase by te same amount, leaving te buoyant force uncanged! 4 Fluid Dynamics Flowing fluids fluid dynamics We ll start wit an idealized fluid tat:. Has streamline flow (no turbulence). Is incompressible (constant density). Has no viscosity (flows witout resistance) 4. Is irrotational (no swirling eddies) 5 6 Te continuity equation We generally apply two equations to flowing fluids. One comes from te idea tat te rate at wic mass flowing past a point is constant, oterwise fluid builds up in regions of low flow rate. Δm ρδv ρaδx mass flow rate = = = = ρav Δt Δt Δt Te mass flow rate of a fluid must be continuous. Oterwise fluid accumulates at some points. So, we ave: ρ A v = ρ A v Density of fluid Av= Av (continuity equation) doesn t cange 8 An application of continuity Te concept of continuity (a constant mass flow rate) is applied in our own circulatory systems. Wen blood flows from a large artery to small capillaries, te rate at wic blood leaves te artery equals te sum of te rates at wic blood flows troug te various capillaries attaced to te artery. Te fluid flows faster in narrow sections of te tube. 7
Applying energy conservation to fluids Our second equation for flowing fluids comes from energy conservation. Bernoulli s equation U + K + W nc = U + K P A P A mgy + mv + W = mgy + mv nc ρgy + ρv + P = ρgy + ρv + P W = FΔx F Δ x = PAΔx P A Δx nc mgy + mv + P A Δ x = mgy + mv + P A Δx Dividing by volume: ρgy + ρv + P = ρgy + ρv + P (Bernoulli s equation) 9 0 Simple Interpretation of Bernoulli s equation W nc by te forces of pressure in moving a unit te fluid from point to point. ρ gy + ρv + P + ( P ) = ρgy + ρv question lower tan tat of points A and B. If te fluid is at rest, rank te points based on teir pressure. PE per unit point KE per unit point PE per unit point KE per unit point. Equal for all four. A>B=C>D v A > v B P A < P B question lower tan tat of points A and B. If te fluid is flowing from left to rigt, rank te points based on te fluid speed. question By similar argument, P C < P D lower tan tat of points A and B. If te fluid is flowing from left to rigt, rank te points based on te pressure.. Equal for all four. A>B=C>D. Equal for all four. A>B=C>D Use te concept of continuity to figure out tis problem. 4 4
In going from B to A, we still ave v > v P B > P A question 4 lower tan tat of points A and B. If te fluid is flowing from rigt to left, rank te points based on te pressure.. Equal for all four. A>B=C>D P B -P A By similar argument, PD > P C P C -P D Te answer ere being te same as tat to question is because Bernoulli s 5 equation, depending on v and v, does not depend on te sign of te velocities. Tree oles in a cylinder A cylinder, open to te atmospere at te top, is filled wit water. It stands uprigt on a table. Tere are tree oles on te side of te cylinder, but tey are covered to start wit. One ole is /4 of te way down from te top, wile te oter two are / and /4 of te way down. Wen te oles are uncovered, water soots out. Wic ole soots te water fartest orizontally on te table?. Te ole closest to te top.. Te ole alfway down.. Te ole closest to te bottom. 4. It's a tree-way tie. 6 Tree oles in a cylinder Te pressure at te top of te cylinder is te atmosperic pressure (P atm ) since te cylinder is open to te atmospere and normally te fluid speed tere is zero. Similarly, te pressure at point is also P atm. P atm At point : P = P atm + ρg v = 0 At point : P = P atm ρgy+ ρv + P = ρgy + ρv + P x x (y = y ) P atm Hence, v = (g) / It sows tat te deeper te ole is, te iger te speed of te fluid will be wen it emerges from te ole. 7 A flexible tube can be used as a simple sipon to transfer fluid from one container to a lower container. Te fluid as a density of 000. kg/m. See te dimensions given in te figure, and take atmosperic pressure to be 00 kpa and g = 0 m/s. If te tube as a cross-sectional area tat is muc smaller tan te cross sectional area of te iger container, wat is te absolute pressure at: (a) Point P? (b) Point Z 00 kpa 00 kpa Wat is te speed of : P (c) Point Z? (d) Point Y? (e) Point? All tree are te same equal.46 m/s Wat is te absolute pressure at: (e) Point Y? (f) Point? 9 kpa 94 kpa 8 (c) Solution for te speed of Z (v): First, note tat te pressures at P and Z are te same and equal atm. Apply te Bernoulli s equation to points P and Z and take =0 at point P: atm = atm + ρg(-0.6m) + (/)ρv (/)ρv = ρg(0.6m) v = (g(0.6)) / =.46 m/s P (e) Solution for pressure at Y (P Y ): Apply te Bernoulli s equation to points P and Y, and take =0 at point P: atm = P Y + ρg(0.m) + (/)ρv 00 kpa = P Y + (000)(0)(0.) + (000)(0)(0.6) P Y = (00 6) kpa= 9 kpa P 9 0 5