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LECTURE NOTES - I « FLUID MECHANICS » Prof. Dr. Atıl BULU Istanbul Technical University College of Civil Engineering Civil Engineering Department Hydraulics Division CHAPTER 1 FUNDAMENTALS 1.1. INTRODUCTION Man’s desire for knowledge of fluid phenomena began with his problems of water supply, irrigation, navigation, and waterpower. Matter exists in two states; the solid and the fluid, the fluid state being commonly divided into the liquid and gaseous states. Solids differ from liquids and liquids from gases in the spacing and latitude of motion of their molecules, these variables being large in a gas, smaller in a liquid, and extremely small in a solid. Thus it follows that intermolecular cohesive forces are large in a solid, smaller in a liquid, and extremely small in a gas. 1.2. DIMENSIONS AND UNITS Dimension = A dimension is the measure by which a physical variable is expressed quantitatively. Unit = A unit is a particular way of attaching a number to the quantitative dimension. Thus length is a dimension associated with such variables as distance, displacement, width, deflection, and height, while centimeters or meters are both numerical units for expressing length. In fluid mechanics, there are only four primary dimensions from which all the dimensions can be derived: mass, length, time, and force. The brackets around a symbol like [M] mean “the dimension” of mass. All other variables in fluid mechanics can be expressed in -2 terms of [M], [L], [T], and [F]. For example, acceleration has the dimensions [LT ]. Force [F] is directly related to mass, length, and time by Newton’s second law, F = ma (1.1) Force= Mass×Acceleration -2 From this we see that, dimensionally, [F] = [MLT ]. 1 kg-force = 9.81 Newton of force = 9.81 N 1 Prof. Dr. Atıl BULU Primary Dimensions in SI and MKS Systems Primary Dimension MKS Units SI Units Force [F] Kilogram (kg) Newton (N=kg.m/s2) 2 Kilogram Mass [M] M=G/g = (kgsec/m) Length [L] Meter (m) Meter (m) Time [T] Second (sec) Second (sec) Secondary Dimensions in Fluid Mechanics Secondary Dimension MKS Units SI Units 2 2 2 Area [L ] m m 3 3 3 Volume [L ] m m -1 Velocity [LT ] m/sec m/sec -2 2 2 Acceleration [LT ] m/sec m/sec Pressure or stress 2 2 -2 -1 -2 kg/m Pa= N/m (Pascal) [FL ] = [ML T ] -1 -1 -1 Angular Velocity [T ] sec sec Energy, work kg.m J = Nm (Joule) 2 -2 [FL] = [ML T ] Power kg.m/sec W = J/sec (Watt) -1 2 -3 [FLT ] = [ML T ] Specific mass (ρ) 2 4 3 -3 2 -4 kg.sec /m kg/m [ML ] = [FT L ] Specific weight (γ) 3 3 -3 -2 -2 Kg/m N/m [FL ] = [ML T ] Specific mass = ρ = The mass, the amount of matter, contained in a volume. This will be expressed in mass-length-time dimensions, and will have the dimensions of mass [M] per 3 unit volume [L ]. Thus, SpecificMass = Mass Volume 2 Prof. Dr. Atıl BULU ⎡M⎤ ⎡FT2⎤ 2 4 [] () ρ =⎢ 3⎥=⎢ 4 ⎥, kgsec m ⎣L ⎦ ⎣ L ⎦ Specific weight= γ = will be expressed in force-length-time dimensions and will have 3 dimensions of force [F] per unit volume [L ]. Specificweight = Weight Volume ⎡ F ⎤ ⎡ M ⎤ 3 () []γ = ⎢ 3⎥ = ⎢ 2 2⎥, kg m ⎣L ⎦ ⎣LT ⎦ Because the weight (a force), W, related to its mass, M, by Newton’s second law of motion in the form W=Mg In which g is the acceleration due to the local force of gravity, specific weight and specific mass will be related by a similar equation, γ = ρg (1.2) o 3 EXAMPLE 1.1: Specific weight of the water at 4 C temperature is γ = 1000 kg/m . What is its the specific mass? SOLUTION: ( 3) γ = ρg =1000 kg m 1000 2 4 () ρ = 9.81 =101.94 kgsec m EXAMPLE 1.2: A body weighs 1000 kg when exposed to a standard earth gravity 2 g = 9.81 m/sec . a) What is its mass? b) What will be the weight of the body be in Newton if it is exposed to the Moon’s standard acceleration g = 1.62 m/sec2? c) How fast will the moon body accelerate if a net force of 100 kg is applied to it on the Moon or on the Earth? SOLUTION: a) Since, ( ) W=mg=1000kg W 1000 2 4 () M= g = 9.81 =101.94kgsec m 2 b) The mass of the body remains 101.94 kgsec /m regardless of its location. Then, 3 Prof. Dr. Atıl BULU
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