Unit1 principle concepts of fluid mechanics
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This document discusses key concepts in fluid mechanics including temperature scales, pressure measurements, and fluid properties. It defines temperature scales like Celsius, Fahrenheit, Kelvin and Rankine and shows conversions between them using formulas. It describes different pressure terms like atmospheric pressure, gauge pressure, absolute pressure and vacuum. Atmospheric pressure is the pressure at sea level of about 101 kPa. Gauge pressure is measured relative to atmospheric pressure and can be positive or negative. Absolute pressure is the sum of gauge and atmospheric pressures. Vacuum refers to a perfect empty space with zero pressure. Formulas are provided to convert between these pressure terms and examples are given to demonstrate conversions and calculations.
Unit1 principle concepts of fluid mechanics
- 1. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/1 PRINCIPLE CONCEPTS OF FLUID MECHANICS OBJECTIVES General Objective : To know, understand and apply the measurement of temperature, pressure and physical properties of fluid. Specific Objectives : At the end of the unit you should be able to : list the temperature scales. convert the temperature to Celsius (ºC), Fahrenheit (ºF), Rankine (R), Kelvin (K) scales. define Pressure ( P ) , Atmospheric Pressure (Patm) , Gauge Pressure ( PG ) , Absolute Pressure ( PA ) and Vacuum ( Pv ) calculate pressure gauge using the formula given. differentiate mass density ( ρ ), specific weight ( ω ), specific gravity ( s ) , specific volume ( v ) and viscosity. calculate fluid properties using the formula given. UNIT 1
- 2. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/2 INTRODUCTION TO FLUID MECHANICS Fluid Mechanics - Introduction Fluid Mechanics is a section of applied mechanics, concerned with the static and dynamics of liquids and gases. Knowledge of fluid mechanics is essential for the chemical engineer, because the majority of chemical processing operations are conducted either partially or totally in the fluid phase. The handling of liquids is much simpler, cheaper, and less troublesome than handling solids. Even in many operations a solid is handled in a finely divided state so that it stays in suspension in a fluid. Fluids and their Properties Fluids In everyday life, we recognize three states of matter: solid, liquid and gas. Although different in many respects, liquids and gases have a common characteristic in which they differ from solids. Both are fluids, but lack the ability of solids to offer a permanent resistance to a deforming force. A fluid is a substance which deforms continuously under the action of shearing forces, however small they may be. Conversely, if a fluid is at rest, there can be no shearing forces acting and, therefore, all forces in the fluid must be perpendicular to the planes upon which they act.
- 3. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/3 1.1 TEMPERATURE UNITS Temperature scales are defined by the numerical value assigned to a standard fixed point. By international agreement the standard fixed point is the easily reproducible triple point of water. These are represented by the state of equilibrium between steam, ice and liquid water. In this unit we learn how to convert temperatures into Celsius, Fahrenheit Kelvin and Rankine scales. The Celsius temperature scale uses the unit degree Celsius (ºC), which has the same magnitude as the Kelvin. Thus the temperature differences are identical on both scales. However, the zero point on the Celsius scale is shifted to 273K, as shown by the following relationship between the Celsius temperature and the Kelvin temperature: ( ) ( ) 273−=° KTCT …(1) By definition, the Rankine scale, the unit of which is the degree Rankine (R) is proportional to the Kelvin temperature according to ( ) ( )KTRT 8.1= …(2) A degree of the same size as that on the Rankine scale is used in the Fahrenheit scale, but the zero point is shifted according to the relation ( ) ( ) 460−=° RTFT …(3) INPUTINPUT
- 4. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/4 substituting Eqs. (1) and (2) into Eq. (3), it follows that ( ) ( ) 328.1 +°=° CTFT Example 1.1 Convert 200ºC to K. Solution to Example 1.1 273+°= CK 273200 += K473= Example 1.2 Convert 250 ºC to ºF Solution to Example 1.2 CF °+=° 8.132 ( )2508.132 += F°= 482 Example 1.3 Convert 365 ºF to R Solution to Example 1.3 FR °+= 460 365460 += R825= Example 1.4
- 5. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/5 Convert 200 ºF to R Solution to Example 1.4 FR °+= 460 200460 += R660= Example 1.5 Convert 450 R to K Solution to Example 1.5 8.1 R K = 8.1 450 = K250= Example 1.6 Convert 410 K to ºF Solution to Example 1.6 ( ) 322738.1 +−=° KF ( ) 322734108.1 +−= F°= 6.278 ACTIVITY 1A
- 6. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/6 TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 1.1 Write the formula to convert the following temperature scales. 1.2 Solve the problems below: i) Air entering a wet scrubber is at 153 ºC. What is the temperature expressed in degree Rankine? ii) The gas stream temperature entering a fabric filter is 410 ºF. What is the temperature expressed in degree Kelvin? FEEDBACK ON ACTIVITY 1A ºC to K ºF to R R to K
- 7. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/7 1.1 1.2 i) Convert temperature from Celsius to Fahrenheit. ºF = 32 + 1.8ºC ºF = 32 + 1.8(153) = 307 ºF Convert temperature from Fahrenheit to Rankine. R = ºF + 460 R = 307 + 460 = 767 R ii) Convert temperature from Fahrenheit to Rankine. R = ºF + 460 R = 410 + 460 = 870 R Convert temperature from Rankine to Kelvin. ºC to K ºF to R R to K K = ºC + 273 R = 460 + ºF K =
- 8. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/8 8.1 R K = 8.1 870 = K483= 1.2 DEFINITION OF PRESSURE INPUTINPUT
- 9. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/9 Fluid will exert a normal force on any boundary it is in contact with. Since these boundaries may be large and the force may differ from place to place it is convenient to work in terms of pressure, p, which is the force per unit area. Units : Newton’s per square metre, Nm-2 ,kgm-1 s-2 . (The same unit is also known as Pascal, Pa i.e 1 Pa = 1Nm-2 ) Also frequently used is the alternative SI unit the bar, where 1 bar = 105 Nm-2 1.3 MEASUREMENT OF PRESSURE GAUGES Absolute Pressure, Gauge Pressure, and Vacuum In a region such as outer space, which is virtually void of gases, the pressure is essentially zero. Such a condition can be approached very nearly in a laboratory when a vacuum pump is used to evacuate a bottle. The pressure in a vacuum is called absolute zero, and all pressures referenced with respect to this zero pressure are termed absolute pressures. Many pressure-measuring devices measure not absolute pressure but only difference in pressure. For example, a Bourdon-tube gage indicates only the difference appliedisforcethewhichoverArea Force pressure =
- 10. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/10 between the pressure in the fluid to which it is tapped and the pressure in the atmosphere. In this case, then, the reference pressure is actually the atmospheric pressure. This type of pressure reading is called gauge pressure. For example, if a pressure of 50 kPa is measured with a gauge referenced to the atmosphere and the atmospheric pressure is 100 kPa, then the pressure can be expressed as either p = 50 kPa gauge or p = 150 kPa absolute. Whenever atmospheric pressure is used as a reference, the possibility exists that the pressure thus measured can be either positive or negative. Negative gauge pressure is also termed as vacuum pressure. Hence, if a gauge tapped into a tank indicates a vacuum pressure of 31 kPa, this can also be stated as 70 kPa absolute, or -31 kPa gauge, assuming that the atmospheric pressure is 101 kPa . 1.3.1 Atmospheric Pressure, patm - The earth is surrounded by an atmosphere many miles high. The pressure due to this atmosphere at the surface of the earth depends upon the head of the air above the surface. - The air is compressible, therefore the density is different at different height. - Due to the weight of atmosphere or air above the surface of earth, it is difficult to calculate the atmospheric pressure. So, atmospheric pressure is measured by the height of column of liquid that it can support. - Atmospheric pressure at sea level is about 101.325 kN/m2 , which is equivalent to a head of 10.35 m of water or 760 mm of mercury approximately, and it decreases with altitude. 1.3.2 Gauge Pressure, pG - It is the pressure, measured with the help of a pressure measuring instrument, in which the atmospheric pressure is taken as datum; in other words the atmospheric pressure at the gauge scale is marked zero. - The gauge pressure can be either positive or negative depending on whether the pressure is above atmospheric pressure (a positive value) or below atmospheric pressure (a negative value). 1.3.3 Absolute Pressure, pA - It is the pressure equal to the algebraic sum of the atmospheric and gauge pressures.
- 11. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/11 pressurecAtmospheripressureGaugepressureAbsolute += atmGA ppp += 1.3.4 Vacuum, pv - In a perfect vacuum which is a completely empty space, the pressure is zero. Example 1.7 Define the following terms : i. Pressure (p ) ii. Atmospheric Pressure ( patm) iii. Gauge Pressure ( pG ) iv. Absolute Pressure ( pA ) v. Vacuum ( pv ) Solution to Example 1.7 Local atmospheric pressure reference absolute pressure gauge pressure +ve pressure –ve absolute pressure Figure 2.1 atmospheric, gauge, absolute and vacuum pressures Pgauge Figure 1.2 Pressure Gauges
- 12. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/12 i. Pressure ( p ): Pressure is force ( F ) per unit area ( A ). ii. Atmospheric Pressure ( patm) : The pressure due to atmosphere at the surface of the earth depends upon the head of the air above the surface. iii. Gauge Pressure ( pG ): It is the pressure, measured with the help of a pressure measuring instrument, in which the atmospheric pressure is taken as datum. iv. Absolute Pressure ( pA ) It is the pressure that equals to the algebraic sum of the atmospheric and gauge pressures. v. Vacuum ( pv ) A completely empty space where the pressure is zero. Example 1.8 What is the pressure gauge of air in the cylinder if the atmospheric gauge is 101.3 kN/m2 and absolute pressure is 460 kN/m2 . Solution to Example 1.8 pA = 460 kN/m2 patm = 101.3 kN/m2 pG = ? With reference to the formula below :
- 13. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/13 Absolute pressure, pA = Gauge pressure, pG + atmospheric pressure, patm Therefore , pG = pA – patm = 460 – 101.3 = 2 /7.358 mkN ACTIVITY 1B TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 1.3 Label the diagram below :
- 14. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/14 1.4 A Bourdon pressure gauge attached to a boiler located at sea level shows a reading pressure of 7 bar. If atmospheric pressure is 1.013 bar, what is the absolute pressure in that boiler (in kN/m2 ) ? FEEDBACK ON ACTIVITY 1B 1.3 Local atmospheric pressure reference a b c d
- 15. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/15 1.4 pA = ? patm = 1.013 bar pG = 7 bar With reference to the formula below : Absolute pressure,pA = Gauge pressure,pG + atmospheric pressure,patm Therefore , pA = pG + patm Local atmospheric pressure reference a = absolute pressure b = gauge pressure +ve d = pressure –ve d = absolute pressure
- 16. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/16 = 7 x 105 + 1.013 x 105 = 801300 N/m2 = 2 /3.801 mkN 1.4 PHYSICAL PROPERTIES OF FLUID INPUTINPUT
- 17. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/17 Fluid properties are intimately related to fluid behaviour. It is obvious that different fluids can have grossly different characteristics. For example, gases are light and compressible, whereas liquids are heavy and relatively incompressible. To quantify the fluid behaviour differences certain fluid properties are used. The fluid properties are mass density, specific weight, specific gravity, specific volume and viscosity. 1.4.1 Mass density, ρ is defined as the mass per unit volume. ( SI units, kg/m3 ) Vvolume mmass , , =ρ 1.4.2 Specific weight, ω is defined as the weight per unit volume. ( SI units, N/m3 ) Vvolume Wweight , , =ω V mg =ω gρω = (where g = 9.81m/sec2 ) In SI units the specific weight of water is 9.81 x 1000 = 9810 N/m3 1.4.3 Specific gravity or relative density, s is the ratio of the weight of the substance to the weight of an equal volume of water at 4 ºC. water cesubs s ω ω tan = water cesubs s ρ ρ tan = 1.4.4 Specific volume, v is defined as the reciprocal of mass density. It is used to mean volume per unit mass. (SI units, m3 /kg ). ρ ν 1 =
- 18. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/18 mmass Vvolume , , =ν 1.4.5 Viscosity A fluid at rest cannot resist shearing forces but once it is in motion, shearing forces are set up between layers of fluid moving at different velocities. The viscosity of the fluid determines the ability of the fluid in resisting these shearing stresses. Example 1.9 What is the mass density, ρ of fluid (in kg/m3 ) if mass is 450 g and the volume is 9 cm3 . Solution to Example 1.9 V m =ρ 6 3 109 10450 − − × × = 33 /1050 mkg×= Example 1.10 What is the specific weight, ω of fluid (in kN/m3 ) if the weight of fluid is 10N and the volume is 500 cm2 . Solution to Example 1.10 V W =ω 6 3 10500 1010 − − × × = = 20 000 N/m3 3 /20 mkN= Example 1.11 What is the specific gravity of fluid in Example 1.10.
- 19. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/19 Solution to Example 1.11 water cesubs s ω ω tan = ( )81.91000 1020 3 × = 039.2= Example 1.12 What is the specific volume, v of fluid in Example 1.9. Solution to Example 1.12 m V v = 3 6 10450 109 − − × × = kgm /102 35− ×= ACTIVITY 1C TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 1.5 Match the following ρ ω s v Vvolume Wweight , , mmass Vvolume , , Vvolume mmass , , forwater cesubsfor ω ω tan m3 /kg kg/m3 N/m3 Specific gravity Specific volume Mass density Specific weight
- 20. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/20 FEEDBACK ON ACTIVITY 1C 1.5 ρ ω s v Vvolume Wweight , , mmass Vvolume , , Vvolume mmass , , forwater cesubsfor ω ω tan m3 /kg kg/m3 N/m3 Specific gravity Specific volume Mass density Specific weight
- 21. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/21 SELF-ASSESSMENT You are approaching success. Try all the questions in this self-assessment section and check your answers with those given in the Feedback on Self-Assessment. If you face any problems, discuss it with your lecturer. Good luck. 1.1 Convert the temperatures below according to the specified scales: a) 175 ºC to ºF, K and R. b) 518 ºR to ºC, ºF and K. c) 214 ºF to ºC, R and K.
- 22. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/22 d) 300 K to ºC, R and ºF. 1.2 Assume the density of water to be 1000 kg/m3 at atmospheric pressure 101 kN/m2 . What will be: a) the gauge pressure b) the absolute pressure of water at a depth of 2000 m below the free surface? 1.3 Determine in Newton per square metre, the increase in pressure intensity per metre depth in fresh water. The mass density of fresh water is 1000 kg/m3 . 1.4 Given specific weight of fluid is 6.54 kN/m3 and its mass is 8.3 kg, calculate the following: a) volume of fluid b) specific volume of fluid c) density of fluid 1.5 Given oil specific gravity is 0.89, find : a) density of oil b) specific weight of oil c) specific volume of oil FEEDBACK ON SELF-ASSESSMENT Answers : 1.1 a) 347 F,807 R,448.3 K , b) 58 °F,287.7 K,14.7 C , c) 674 °C,374.4 R,101.4 K d) 27 °C,80.6 °F,540.6 R 1.2 a) 117.72 kN/m2 ,
- 23. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/23 b) 218.72 kN/m2 1.3 9.81 x 103 N/m2 1.4 a) 0.072 m3 b) 0.0015 m3 /kg c) 691.67 kg/m3 1.5 a) 0.89 x 103 kg/m3 b) 8730.9 N/m3 c) 0.00112 m3 /kg