Chapter 2 Fluid Static - Pressure

Clinical mercury Manometer PRESSURE Pressure is defined as a normal force exerted by a fluid per unit area. Units of pressure are N/m2, which is called a pascal (Pa). Since the unit Pa is too small for pressures encountered in practice, kilopascal (1 kPa = 103 Pa) and megapascal (1 MPa = 106 Pa) are commonly used. 1 atm = 101,325 Pa = 101.325 kPa = 1.01325 bar Other units include kgf/cm2, lbf/in2=psi. 1 kgf/cm2 = 9.807 N/cm2 = 14.223 psi A few pictures of Pressure applications Sphygmometer Compressor Spray Gun Clinical mercury Manometer

REFERENCE PRESSURE ATMOSPHERIC, Patm GAGE (Vacuum when, Pgage is –ve) ABSOLUTE Perfect Vacuum, Pabs = 0 atm Patm Pabs Pvacuum = Pgage (-ve) Pgage (+ve) Pabs = Patm - Pvac Pabs= Pgage + Patm

ATMOSPHERIC PRESSURE Atmospheric pressure is the pressure at any point in the Earth's atmosphere. In most circumstances atmospheric pressure is closely approximated by the hydrostatic pressure caused by the weight of air above the measurement point. Source: en. Wikipedia.com.

ATMOSPHERIC PRESSURE How does it change? Changes not only with elevation but also with weather condition Also known as “ Barometric Pressure” Perfect Vacuum, Pabs = 0 atm Patm Pabs Pvacuum = Pgage (-ve) Pgage (+ve) Pabs= Pgage + Patm Pabs = Patm - Pvac Density and Temperature If the air mass is dense, the pressure is higher. Warm air is not as dense as cool air. That is the reason why warm air goes up and cool air does not. Altitude and temperature interact, and so they affect air pressure.

ABSOLUTE PRESSURE Pressure measured relative to perfect vacuum is called ABSOLUTE PRESSURE, Pa(abs) or psi. A perfect vacuum is the lowest possible pressure. Absolute pressure (always +ve.) Perfect Vacuum, Pabs = 0 atm Patm Pabs Pvacuum = Pgage (-ve) Pgage (+ve) Pabs= Pgage + Patm Pabs = Patm - Pvac A perfect vacuum is defined as a region in space without any particles. It cannot be achieved in a laboratory.

GAGE PRESSURE Pressure measured relative to atmospheric pressure, Pa (gage) or psi. Most pressure-measuring devices are calibrated to read zero in the atmosphere, and gage pressure, Pgage=Pabs - Patm. Perfect Vacuum, Pabs = 0 atm Patm Pabs Pvacuum = Pgage (-ve) Pgage (+ve) Pabs= Pgage + Patm Pabs = Patm - Pvac

GAGE PRESSURE A gage pressure above atmospheric pressure (relative to atmospheric) is +ve. A gage pressure below atmospheric pressure (relative to atmospheric) is –ve. It is also called vacuum pressure Perfect Vacuum, Pabs = 0 atm Patm Pabs Pvacuum = Pgage (-ve) Pgage (+ve) Pabs= Pgage + Patm Pabs = Patm - Pvac

EXAMPLE Express a pressure of 155 kPa (gage) as an absolute pressure. The local atmospheric pressure is 98 kPa (atm). ANSWER Absolute Pressure= Gage Pressure + Atmospheric Pressure = 155 kPa + 98 kPa = 253 kPa

PRESSURE AT A POINT Pressure: Compressive force per unit area, and it gives the impression of being a vector Pressure at any point in a fluid is the same in all directions. It has Magnitude, but not a specific direction, and thus it is a scalar quantity. Other words, Pressure at any point in fluid has the same magnitude in all directions. P

Variation of Pressure with Depth In the presence of a gravitational field, pressure increases with depth because more fluid rests on deeper layers. To obtain a relation for the variation of pressure with depth, consider rectangular element of height Dz, length Dx, and unit depth (Dy = 1 unit into the page) in equilibrium. Force balance in z-direction gives The pressure of a fluid at rest increases with depth (as a result of added weight).

Where W = mg = is the weight of the fluid element and Δz = z2 – z1 .Dividing by Δy Δx and rearranging gives In constant density and constant gravitational acceleration Free-body diagram of a rectangular fluid element in equilibrium.

The pressure in a tank containing a gas, can be considered to be uniform since the weight of the gas is too small to make a significant. The pressure in a room filled with air can be assumed to be constant In a room filled with a gas, the variation of pressure with height is negligible. Take point 1 to be at the free surface of a liquid open to the atmosphere, where the pressure is the atmospheric pressure Patm , then the pressure at a depth h from the free surface becomes

When the variation of density with elevation is known Pressure in a liquid at rest increases linearly with distance from the free surface. When the variation of density with elevation is known

Pressure in a fluid at rest is independent of the shape of the container. Pressure is the same at all points on a horizontal plane in a given fluid. The pressure at points A, B, C, D, E, F, and G are the same since - same depth - same static fluid Pressure at points H and I are not the same since these two points cannot be interconnected by the same fluid.

P = ρgh = (1000kg/m3)(9.81m/s2)(5m) Pa. EXAMPLE Calculate the change in water pressure from the surface (exposed to atmosphere) to a depth of 5 m. ANSWER If the surface of the water is exposed to the atmosphere, the pressure there is 0 Pa(gage). Descending in the water (decreasing elevation) produces an increase in pressure. Therefore, at 5 m the pressure is 49.05 kPa(abs). P = ρgh = (1000kg/m3)(9.81m/s2)(5m) Pa.

*GAS PRESSURE VARIATION WITH ELEVATION Air density at sea level, 15 degree Celcius is 1.225 kg/m3. Pressure difference at 5m height difference; ρgh = (1.225 kg/m3) * 9.807 ms-2 * 5m * (1 N/1 kgms-2) * ( 1 kPa/ 1000 N/m2) = 0.06 kPa 1 atm = 101.325 kPa For gas, the variation of pressure with height is negligible, because of their low density. Also, weight is too small. However is accuracy is desired, it becomes significant. Gravitational Effect g = 9.807 m/s2 at sea level, at elevation 14,000 m above sea level, g = 9.764 m/s2, which is 0.4% change. Therefore, g variation is so small and g can be considered constant.

EXAMPLE Figure below shows a tank of oil with one side open to the atmosphere and the other side sealed with air above the oil. The oil has a specific gravity of 0.90. Calculate the gage pressure at points A, B, C, D, E, and F and the air pressure in the right side of the tank.

Point A At this point, the oil is exposed to the atmosphere, and therefore Point B The change in elevation between point A and point B is 3.0 m, with B lower than A. The specific weight of the oil: Then we have

Now, the pressure at B is Point C The change in elevation from point A to point C is 6.0 m, with C lower than A. Then, the pressure at point C is Point D Because point D is at the same level as point B, the pressure is the same. That is, we have

Point E Because point E is at the same level as point A, the pressure is the same. That is, we have Point F The change in elevation between point A and point F is 1.5 m, with F higher than A. Then, the pressure at F is

Air Pressure in the right side of the tank is exposed to the surface of the oil, where pF = -13.2kPa the air pressure is also -13.2kPa or 13.2 kPa below atmospheric pressure.

PASCAL’S LAW Pascal’s Law : “A pressure applied to a confined fluid increases the pressure throughout by the same amount”. In picture, pistons are at same height:  Consider a “pierced” vessel full of water and provided of a piston: a force acts on the piston and if the force increases, the water comes out of the vessel with bigger intensity. So the rise in pressure is the same anywhere in water Lifting of a large weight by a small force by the application of Pascal’s law.

Hydraulic Brake System Jack System Hydraulic grease gun

EXAMPLE A force, F of 800 N is applied to the smaller cylinder of a hydraulic jack. The area, a of a small piston is 20 cm2 while the area, A of a larger piston is 200 cm2. What mass can be lifted on the larger piston? W F = 800 N p2 p1 Area, A = 200 cm2

Putting F = 800 N, a = 20/1000 m2 , A = 200 / 1000 m2 So that Mass lifted

PRESSURE MEASUREMENT 1)Barometers Atmospheric pressure is measured by a device called a barometer; thus, atmospheric pressure is often referred to as the barometric pressure. PC can be taken to be zero since there is only Hg vapor above point C, and it is very low relative to Patm. Change in atmospheric pressure due to elevation has many effects: Cooking, nose bleeds, engine performance, aircraft performance.

2)Manometer Piezometer Tube. A standard technique for measuring pressure involves the use of liquid column in vertical or inclined tubes. Pressure measuring devices based on this technique are called manometers. The mercury barometer is an example of one type of manometer, but there are many other configuration possible, depending on the particular application. Piezometer Tube. U-Tube manometer. -Simple U-Tube Manometer -Differential U-Tube Manometer -Inverted Differential U-Tube Manometer Inclined U-Tube manometer.

Piezometer Tube The fundamental equation is >> PA : gage pressure ( P0=0) :the specific weight of the liquid in the container h1: measured from the meniscus at the upper surface to point(1) Only suitable if the pressure in the container is greater than atmospheric pressure, and the pressure to be measured must be relatively small so the required height of the column is reasonable. The fluid in the container must be a liquid rather than a gas.

Simple U-Tube Manometer A simple manometer is a tube bent in U-shape. One end of which is attached to the gauge point and the other is open to the atmosphere as shown in figure below. The liquid used in the bent tube or simple manometer is generally mercury which is 13.6 times heavier than water. Hence, it is also suitable for measuring high pressure. A(1)(2)(3)Open PA + ρ1gh1 – ρ2g h2 = 0 >> PA = ρ2gh2 –ρ1gh1 If pipe A contains a gas then ρ1gh1≒0 >> PA =ρ2gh2 Gas: Low density, weight is so small to give significance pressure increase.

Digital Manometer A volunteer for the RAF blowing into a 'U-tube manometer' to check his lung power.

EXAMPLE A U-tube manometer that shown above is used to measure the gauge pressure of water (mass density ρ = 1000 kg /m3). If the density of mercury is 13.6 × 103 kg /m3, what will be the gauge pressure at A if h1 = 0.45 m and D is 0.7 m above BC.

ANSWER Since

Differential U-Tube Manometer It is a device used for measuring the difference of pressures, between two points in a pipe, or in two different pipes. A differential manometer consists of a U-tube, containing a heavy liquid with two ends connected to two different points A(1)(2)(3)(4)(5)B PA＋ρ1gh1－ρ2gh2 －ρ3gh3＝ PB The pressure difference is PA－ PB＝ρ2gh2＋ρ3gh3－ρ1gh1

EXAMPLE A U tube manometer measures the pressure difference between two points A and B in a liquid. The U tube contains mercury. Calculate the difference in pressure if h =1.5 m, h2 = 0.75 m and h1 = 0.5 m. The liquid at A and B is water ( γ = 9.81 × 103 N/m2) and the specific gravity of mercury is 13.6.

Pressure difference

Inverted Differential U-Tube Manometer An inverted differential manometer is used for measuring the difference of low pressure, where accuracy is the prime consideration. It consists of an inverted U-tube, containing a light liquid.

Inclined-Tube Manometer To measure small pressure change, an inclined-tube manometer is frequently used: PA +γ1h1 –γ2l2sinθ –γ3h3 = PB PA – PB =γ2l2sinθ +γ3h3 –γ1h1 If pipe A and B contain a gas then γ3h3≒γ1h1≒0

*MULTIFLUID MANOMETER EXAMPLE * For same fluid (constant density) pressure does not vary in the horizontal direction. Pressure vary in vertical direction. PA = PB PA = Patm + ρmgh3 PB=PC + ρoilgh2 PC=Pair + ρwatergh1 C B A

P1+ ρgh1+ ρ1 gh2 - ρ2 gh3 = P2