FUNDAMENTALS OF FLUID MECHANICS Chapter 2 Fluids at

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FUNDAMENTALS OF FLUID MECHANICS Chapter 2 Fluids at Rest - Pressure and its Effect

FUNDAMENTALS OF FLUID MECHANICS Chapter 2 Fluids at Rest - Pressure and its Effect 1

FLUID PROPERTIES PRESSURE Pressure is defined as a normal force exerted by a fluid

FLUID PROPERTIES PRESSURE Pressure is defined as a normal force exerted by a fluid per unit area. l Units of pressure are N/m 2, which is called a pascal (Pa). l Sphygmometer

Compressor Spray Gun Grease pump

Compressor Spray Gun Grease pump

PRESSURE l l Since the unit Pa is too small for pressures encountered in

PRESSURE l l Since the unit Pa is too small for pressures encountered in practice, kilopascal (1 k. Pa = 103 Pa) and megapascal (1 MPa = 106 Pa) are commonly used. 1 atm = 101, 325 Pa = 101. 325 k. Pa = 1. 01325 bar Other units include kgf/cm 2, lbf/in 2=psi. 1 kgf/cm 2 = 9. 807 N/cm 2 = 14. 223 psi

REFERENCE PRESSURE ATMOSPHERIC, Patm l GAGE (Vacuum when, Pgage is –ve) l ABSOLUTE l

REFERENCE PRESSURE ATMOSPHERIC, Patm l GAGE (Vacuum when, Pgage is –ve) l ABSOLUTE l

ABSOLUTE, GAGE, AND VACUUM PRESSURES Pabs = Patm - Pvac Pabs= Pgage + Patm

ABSOLUTE, GAGE, AND VACUUM PRESSURES Pabs = Patm - Pvac Pabs= Pgage + Patm Pgage (+ve) Patm Pabs Pvacuum = Pgage (-ve) Pabs Perfect Vacuum, Pabs = 0 atm

ATMOSPHERIC PRESSURE l Atmospheric pressure is the pressure at any point in the Earth's

ATMOSPHERIC PRESSURE l 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.

ATMOSPHERIC PRESSURE How does it change? Changes not only with elevation but also with

ATMOSPHERIC PRESSURE How does it change? Changes not only with elevation but also with weather condition Also known as “ Barometric Pressure” Pabs = Patm - Pvac Pabs= Pgage + Patm Pgage (+ve) Pat Pvacuum = Pgage (-ve) m Pabs Perfect Vacuum, Pabs = 0 atm

ABSOLUTE PRESSURE l Pressure measured relative to perfect vacuum is called ABSOLUTE PRESSURE, Pa(abs)

ABSOLUTE PRESSURE l 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. ) Pabs = Patm - Pvac Pabs= Pgage + Patm Pgage (+ve) Pat Pvacuum = Pgage (-ve) m Pabs Perfect Vacuum, Pabs = 0 atm

GAGE PRESSURE Pressure measured relative to atmospheric pressure, Pa (gage) or psi. Most pressure-measuring

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. Pabs = Patm - Pvac Pabs= Pgage + Patm Pgage (+ve) Pat Pvacuum = Pgage (-ve) m Pabs Perfect Vacuum, Pabs = 0 atm

GAGE PRESSURE l l A gage pressure above atmospheric pressure (relative to atmospheric) is

GAGE PRESSURE l l 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 Pabs = Patm - Pvac Pabs= Pgage + Patm Pgage (+ve) Pat Pvacuum = Pgage (-ve) m Pabs Perfect Vacuum, Pabs = 0 atm

EXAMPLE 1 Express a pressure of 155 k. Pa (gage) as an absolute pressure.

EXAMPLE 1 Express a pressure of 155 k. Pa (gage) as an absolute pressure. The local atmospheric pressure is 98 k. Pa (abs). ANSWER Absolute Pressure= Gage Pressure + Atmospheric Pressure = 155 k. Pa + 98 k. Pa = 253 k. Pa

EXAMPLE 2 Express a pressure of 225 k. Pa(abs) as a gage pressure. The

EXAMPLE 2 Express a pressure of 225 k. Pa(abs) as a gage pressure. The local atmospheric pressure is 101 k. Pa(abs). ANSWER 2 Solving algebraically for Pgage gives

EXAMPLE 3 Express a pressure of 75. 2 k. Pa (abs) as a gage

EXAMPLE 3 Express a pressure of 75. 2 k. Pa (abs) as a gage pressure. The local atmospheric pressure is 103. 4 k. Pa. ANSWER 3 Notice that the result is negative. This can also be read “ 28. 2 k. Pa below atmospheric pressure” or “ 28. 2 k. Pa vacuum. ”

EXAMPLE 4 Express a pressure of – 42. 7 k. Pa as an absolute

EXAMPLE 4 Express a pressure of – 42. 7 k. Pa as an absolute pressure. Assume, Patm=101 k. Pa: ANSWER 4

PRESSURE AT A POINT Pressure: Compressive force per unit area, and it gives the

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 P P

Variation of Pressure with Depth l The pressure of a fluid at rest increases

Variation of Pressure with Depth l The pressure of a fluid at rest increases with depth (as a result of added weight). l 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. l Force balance in z-direction gives

l. Where W = mg = is the weight of the fluid element. Dividing

l. Where W = mg = is the weight of the fluid element. 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

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

Pressure in a liquid at rest increases linearly with distance from the free surface.

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

l l Pressure in a fluid at rest is independent of the shape of

l l 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.

Pressure Head in Static Fluid 1/2 v The pressure difference between two points in

Pressure Head in Static Fluid 1/2 v The pressure difference between two points in a fluid at rest: p 1 - p 2 = γ(z 2-z 1)=γh h is called the pressure head and is interpreted as the height of a column of fluid of specific weight γrequired to give a pressure difference p 1 -p 2. 22

Scuba Diving and Hydrostatic Pressure

Scuba Diving and Hydrostatic Pressure

Scuba Diving and Hydrostatic Pressure 1 l Pressure on diver at 100 ft? l

Scuba Diving and Hydrostatic Pressure 1 l Pressure on diver at 100 ft? l Danger of emergency ascent? 100 ft 2 Boyle’s law If you hold your breath on ascent, your lung volume would increase by a factor of 4, which would result in embolism and/or death.

EXAMPLE 5 Calculate the change in water pressure from the surface (exposed to atmosphere)

EXAMPLE 5 Calculate the change in water pressure from the surface (exposed to atmosphere) to a depth of 5 m. ANSWER 5 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 k. Pa(gage). P = ρgh = (1000 kg/m 3)(9. 81 m/s 2)(5 m) Pa.

*GAS PRESSURE VARIATION WITH ELEVATION Air density at sea level, 15 degree Celcius is

*GAS PRESSURE VARIATION WITH ELEVATION Air density at sea level, 15 degree Celcius is 1. 225 kg/m 3. Pressure difference at 5 m height difference; ρgh = (1. 225 kg/m 3) * 9. 807 ms-2 * 5 m * (1 N/1 kgms-2) * ( 1 k. Pa/ 1000 N/m 2) = 0. 06 k. Pa 1 atm = 101. 325 k. Pa 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/s 2 at sea level, at elevation 14, 000 m above sea level, g = 9. 764 m/s 2, which is 0. 4% change. Therefore, g variation is so small and g can be considered constant.

*EXAMPLE 6* Figure below shows a tank of oil with one side open to

*EXAMPLE 6* 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.

EXAMPLE 6 Point A At this point, the oil is exposed to the atmosphere,

EXAMPLE 6 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

EXAMPLE 6 Now, the pressure at B is Point C The change in elevation

EXAMPLE 6 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

EXAMPLE 6 Point E Because point E is at the same level as point

EXAMPLE 6 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

PASCAL’S LAW l l Lifting of a large weight by a small force by

PASCAL’S LAW l l Lifting of a large weight by a small force by the application of 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:

EXAMPLE A force, F of 800 N is applied to the smaller cylinder of

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 cm 2 while the area, A of a larger piston is 200 cm 2. What mass can be lifted on the larger piston? F = 800 N W p 1 p 2 Area, A = 200 cm 2

Putting F = 800 N, a = 20/1000 m 2 , A = 200

Putting F = 800 N, a = 20/1000 m 2 , A = 200 / 1000 m 2 So that Mass lifted

EXAMPLE A hydraulic lift is to be used to lift a 2500 kg weight

EXAMPLE A hydraulic lift is to be used to lift a 2500 kg weight by putting a weight of 25 kg on a piston with a diameter of 10 cm. Determine the diameter of the piston (in meters) on which the weight is to be placed

SOLUTION

SOLUTION

PRESSURE MEASUREMENT Barometers l l l Atmospheric pressure is measured by a device called

PRESSURE MEASUREMENT Barometers l l l 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.

Example 2. 3 Barometric Pressure l A mountain lake has an average temperature of

Example 2. 3 Barometric Pressure l A mountain lake has an average temperature of 10 ℃ and a maximum depth of 40 m. For a barometric pressure of 598 mm Hg, determine the absolute pressure (in pascals) at the deepest part of the lake. Note: Barometric pressure is also known as atmospheric pressure From Table B. 2, at 10 0 C 38

Example 2. 3 Solution 1/2 The pressure in the lake at any depth, h

Example 2. 3 Solution 1/2 The pressure in the lake at any depth, h Note: Barometric pressure is also known as atmospheric pressure p 0 is the local barometric expressed in a consistent of units. 39

Example 2. 3 Solution 2/2 From Table B. 2, at 10 0 C 40

Example 2. 3 Solution 2/2 From Table B. 2, at 10 0 C 40

Manometry A standard technique for measuring pressure involves the use of liquid column in

Manometry A standard technique for measuring pressure involves the use of liquid column in vertical or inclined tubes. v 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. _ Inclined-Tube manometer. v 41

Piezometer Tube The fundamental equation is P = P 0 + γh >> PA

Piezometer Tube The fundamental equation is P = P 0 + γh >> PA = γ 1 h 1 PA : gage pressure ( P 0=0) γ 1 : the specific weight of the liquid in the container h 1: 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. 42

Simple U-Tube Manometer v The fluid in the manometer is called the gage fluid.

Simple U-Tube Manometer v The fluid in the manometer is called the gage fluid. A (1) (2) (3) Open P A + γ 1 h 1 – γ 2 h 2 = 0 >> PA =γ 2 h 2 –γ 1 h 1 If pipe A contains a gas then γ 1 h 1≒ 0 >> PA =γ 2 h 2 43

Example 2. 4 Simple U-Tube Manometer l A closed tank contains compressed air and

Example 2. 4 Simple U-Tube Manometer l A closed tank contains compressed air and oil (SGoil = 0. 90) as is shown in Figure E 2. 4. A U-tube manometer using mercury (SGoil = 13. 6) is connected to the tank as shown. For column heights h 1 = 36 in. , h 2 = 6 in. , and h 3 = 9 in. , determine the pressure reading (in psi) of the gage. 44

Example 2. 4 Solution 1/2 The pressure at level (1) is As we move

Example 2. 4 Solution 1/2 The pressure at level (1) is As we move from level (2) to the open end, the pressure must decrease by γHgh 3, and at the open end the pressure is zero. Thus, the manometer equation can be expressed as or 45

Exmaple 2. 4 2/2 The value for pair So that The pressure reading (in

Exmaple 2. 4 2/2 The value for pair So that The pressure reading (in psi) of the gage 46

Differential U-Tube Manometer A (1) (2) (3) (4) (5) B PA+γ 1 h 1-γ

Differential U-Tube Manometer A (1) (2) (3) (4) (5) B PA+γ 1 h 1-γ 2 h 2 -γ 3 h 3= PB The pressure difference is PA- PB=γ 2 h 2+γ 3 h 3-γ 1 h 1 47

Inclined-Tube Manometer v To measure small pressure change, an inclined-tube manometer is frequently used:

Inclined-Tube Manometer v To measure small pressure change, an inclined-tube manometer is frequently used: PA +γ 1 h 1 –γ 2 l 2 sinθ –γ 3 h 3 = PB PA – PB =γ 2 l 2 sinθ +γ 3 h 3 –γ 1 h 1 If pipe A and B contain a gas thenγ 3 h 3≒γ 1 h 1≒ 0 >> l 2 = ( PA – PB ) /γ 2 sinθ 48

Simple U-Tube Manometer A simple manometer is a tube bent in U-shape. One end

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 (next slide) 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. Now consider a simple manometer connected to a pipe containing a light liquid under high pressure. The high pressure in the pipe will force the heavy liquid, in the left-hand arm of the U-tube, to move downward. This downward movement of the heavy liquid in the left-hand arm will cause a corresponding rise of the heavy liquid in the right-hand arm. The horizontal surface, at which the heavy and light liquid meet in the left-hand arm is known as a common surface or datum line. Let B-C be the datum line, as shown (next slide). Gas: Low density, weight is so small to give significance pressure increase.

*MULTIFLUID MANOMETER-EX. 2 -6 * For same fluid (constant density) pressure does not vary

*MULTIFLUID MANOMETER-EX. 2 -6 * For same fluid (constant density) pressure does not vary in the horizontal direction. Pressure vary in vertical direction. l PA = P B l PA = Patm + ρmgh 3 l PB=PC + ρoilgh 2 l PC=Pair + ρwatergh 1 C B A

MULTIFLUID MANOMETER-EX. 2 -6 P 1+ ρgh 1+ ρ1 gh 2 - ρ2 gh

MULTIFLUID MANOMETER-EX. 2 -6 P 1+ ρgh 1+ ρ1 gh 2 - ρ2 gh 3 = P 2

Prob 2. 26

Prob 2. 26

Prob 2. 27

Prob 2. 27

Prob 2. 40

Prob 2. 40

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

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

Let h 1 = Height of the light liquid in the left-hand arm above

Let h 1 = Height of the light liquid in the left-hand arm above the common surface in cm. h 2 = Height of the heavy liquid in the right-hand arm above the common surface in cm. = Pressure in the pipe, expressed in terms of head of water in cm. = Specific weight of the light liquid = Specific gravity of the heavy liquid. The pressure in the left-hand arm and the right-hand arm above the datum line is equal. Pressure at B = Pressure at C

Pressure in the left-hand arm above the datum line P = Pressure, at A

Pressure in the left-hand arm above the datum line P = Pressure, at A + Pressure due to depth, of fluid P

Thus pressure in the right-hand arm above the datum line; = Pressure at D

Thus pressure in the right-hand arm above the datum line; = Pressure at D + Pressure due to depth , h 2 of liquid Q But, PD = Atmospheric Pressure = Zero gauge pressure And so , Since So,

EXAMPLE A U-tube manometer similar to that shown in Figure 3. 6 is used

EXAMPLE A U-tube manometer similar to that shown in Figure 3. 6 is used to measure the gauge pressure of water (mass density ρ = 1000 kg /m 3). If the density of mercury is 13. 6 × 103 kg /m 3, what will be the gauge pressure at A if h 1 = 0. 45 m and D is 0. 7 m above BC.

ANSWER Considering = Pressure, = = at A + Pressure due to depth, of

ANSWER Considering = Pressure, = = at A + Pressure due to depth, of fluid P

= Pressure at D + Pressure due to depth = = Since of liquid

= Pressure at D + Pressure due to depth = = Since of liquid Q

If negative pressure is to be measured by a simple manometer, this can be

If negative pressure is to be measured by a simple manometer, this can be measured easily as discussed below: In this case, the negative pressure in the pipe will suck the light liquid which will pull up the heavy liquid in the left-hand arm of the U-tube. This upward movement of the heavy liquid, in the left-hand arm will cause a corresponding fall of the liquid in the right-hand arm as shown in Fig. below.

In this case, the datum line B-C may be considered to correspond with the

In this case, the datum line B-C may be considered to correspond with the top level of the heavy liquid in the right column as shown in the Figure last slide. Now to calculate the pressure in the left- hand arm above the datum line. Let h 1 = Height of the light liquid in the left-hand arm above the common surface in cm. h 2 = Height of the heavy liquid in the right-hand arm above the common surface in cm. = Pressure in the pipe, expressed in terms of head of water in cm. = Specific weight of the light liquid = Specific gravity of the heavy liquid. Pressure PB at B = Pressure PC at C

Pressure in the left-hand arm above the datum line; PB = Pressure PA at

Pressure in the left-hand arm above the datum line; PB = Pressure PA at A + Pressure due to depth h 1 of fluid P + Pressure due to depth h 2 of liquid Q

Pressure in the right-hand arm above the datum line; PC = Pressure PD at

Pressure in the right-hand arm above the datum line; PC = Pressure PD at D But PD = Atmospheric pressure And so, PC = Patm Since PB = PC

Differential U-Tube Manometer It is a device used for measuring the difference of pressures,

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 The horizontal surface C-D, at which the heavy liquid meet in the left-hand arm, is the datum line.

We know that the pressures in the left-hand arm and right-hand arm , above

We know that the pressures in the left-hand arm and right-hand arm , above the datum line are equal. Pressure PC at C = Pressure PD at D Pressure in the left-hand arm above the datum line PC =Pressure PA at A + Pressure due to depth h of fluid P Pressure in the right-hand arm above the datum line PD = Pressure PB at A + Pressure due to depth h 1 of the fluid P + Pressure due to depth h 2 of liquid Q.

Since,

Since,

Example A U tube manometer measures the pressure difference between two points A and

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, h 2 = 0. 75 m and h 1 = 0. 5 m. The liquid at A and B is water ( ω = 9. 81 × 103 N/m 2) and the specific gravity of mercury is 13. 6.

Since C and D are at the same level in the same liquid at

Since C and D are at the same level in the same liquid at rest Pressure at C = For the left hand arm For the right hand arm since Pressure at D

Pressure difference

Pressure difference

Inverted Differential U-Tube Manometer An inverted differential manometer is used for measuring the 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.

Let h = Height of the heavy liquid in the left-hand arm below the

Let h = Height of the heavy liquid in the left-hand arm below the datum line, h 1= Height of the light liquid in the left-hand arm below the datum line , h 2= Height of the light liquid in the right-hand arm below the datum line, ωP= Specific weight of the light liquid ωQ= Specific weight of the heavy liquid We know that pressures in the left-hand arm and left-hand arm below the datum line are equal. Pressure PC at C = Pressure PD at D

Example 2. 5 U-Tube Manometer l As will be discussed in Chapter 3, the

Example 2. 5 U-Tube Manometer l As will be discussed in Chapter 3, the volume rate of flow, Q, through a pipe can be determined by means of a flow nozzle located in the pipes as illustrated in Figure. the nozzle creates a pressure drop, p. A - p. B, along the pipe which is related to the flow through the equation , where K is a constant depending on the pipe and nozzle size. The pressure drop is frequently measured with a differential U-tube manometer of the type illustrated. (a) Determine an equation for p. A - p. B in terms of the specific weight of the flowing fluid, γ 1, the specific weight of the gage fluid, γ 2, and the various heights indicated. (b) For γ 1= 9. 80 k. N/m 3 , γ 2 = 15. 6 k. N/m 3 , h 1 = 1. 0 m, and h 2 = 0. 5 m, what is the value of the pressure drop, p. A - p. B? 77

Example 2. 5 Solution we start at point A and move vertically upward to

Example 2. 5 Solution we start at point A and move vertically upward to level (1), the pressure will decrease by γ 1 h 1 and will be equal to pressure at (2) and (3). We can now move from (3) to (4) where the pressure has been further reduced by γ 2 h 2. The pressure at levels (4) and (5) are equal, and as we move from (5) to B the pressure will increase byγ 1(h 1 + h 2) (Ans) 78