Fluid Dynamics of Fires - Term Paper Example

Fluid Dynamics of Fires Student Name Student ID E-mail address Supervisor Name Program Academic Year Semester April 3, 2012 Outline Classical Mechanics of Fluids Heat Transfer and Thermochemistry Fluid Dynamics of Combustion Shock and Detonation Waves FLUIDS DYNAMICS 1. Classical mechanism of fluids 1.1f helium balloon volume is 1600m3 and has a total mass of 40kg then its density is calculated as Ρ= Density ==0.025 To find the altitude we use the formula ρA= ρ0-αh Where p=1.2kg/m3 and α=5x10-5kg/m4 0.025kg/m3= 1.2kg/m3 –(5x10-5kg/m4) h (5x10-5kg/m4) h = 1.2kg/m3 –0.025kg/m3 h = (1.175kg/m3)/ (5x10-5kg/m4) h= 23500m Pressure is a scalar quantity; it does not have a direction. The force acting on an object submerged in a fluid – or on some portion of the fluid itself- is a vector quantity; its direction is perpendicular to the contact surface. Pressure is defined as a scalar because, at a given location in a fluid, the magnitude of the force per unit area is the same for any orientation of the surface. The molecules in a static fluid are moving in a random direction, there can be no preferred direction since that would constitute fluid flow. There is no reason that a surface would have a greater number of collisions, or collisions with more energetic molecules, for one particular surface orientation compared with any other orientation. 1.2 Using the venture meter and the law of continuity the product of flow speed and area must be the same. Therefore = and the pressure will be calculated as follows: P1=ρgh1 P2=ρgh2 We have the areas of 2.7 cm2 and o.7cm2 Therefore the volume will be derived as follows V=Ax h = = 0.7h2 = 3.857 x 0.000567 h2= 31.2 P1=1000x9.8x2.1=20.58kPa P2=1000x9.8x0.312=3.057kPa The same volume of fluid that enters the pipe in a given time interval exits the pipe in the same time interval. Where the radius of the tube is large, the speed of the fluids is small, where the radius is small, the fluid sped is large. A familiar example is what happens when you use your thumb to partially block the end of a garden hose to make a jet of water. The water moves past your thumb, where the cross-section area is smaller, at a greater speed than it moves in the hose. Similarly, water travelling along a river speeds up, forming rapids, when the river bed narrows or is partially blocked by rocks and boulders. Streamlines are closed together where the fluid flows faster and further apart, where it flows more slowly. Thus, streamlines help us visualize fluid flow. The fluid velocity at any point is tangent to a streamline through that point. 1.3 γ= ρvsat=ψvpsat where ψv is the compressibility of the liquid. ρ=(1-γ)ρ01+(γψv+(1-γ)ψ1)psat +ψ(γ) (ρ-psat) The compressibility ψ, and choice of how it is modeled, were mentioned in section 2.2. For the single phase code, the considerations when choosing a model are the same, and for the simulations in paper 2 the linear model in equation 2.16 was chosen. For the mixtures viscosity, a simpler version of equation 2.2 can be used since only one compressible phase is then present. μf =γμv+(1-γ)μ1 When a linear model is used for the compressibility the equation of state can be simplified to ρ= (1-γ) ρ01+ψp The first term governs the liquid density when γ is low. If the fluid is cavitating , the second term becomes more dominant. The first term contains the property ρ01 which is: ρ01=ρ1sat -psatψ1 Where ρ1sat is the liquid density at standard conditions. The saturation density of the vapour , used earlier to calculate γ, is important for the liquid’s tendency to cavitate . If ρvsat is increased , ρ will not need to obtain a higher γ. The relationship between density and pressure are used in the continuity equation to transform it from a density equation to a pressure equation. For reference we have the compressible equation: +∆ .(Ρuu) =∆p+μf∆u +∆ . (ρu) =0 Since this is compressible flow, it is more straight forward than incompressible flow. –(ρ01 +(ψ1 -ψv)psat ) -psat +∆ .(ρU )= 0 1.4 The power of an aero plane with drag coefficient of 0.35 and a surface area of 9m2 going at a speed of 820km/h. To calculate the force we will use the formula Force=6AxVxη Where A is the area, η is the viscosity of the fluid, V is the speed If mair is the mass of displaced air, then Force=6AxVxη + mairg- maircraftg=0 Alternatively the formula will be written as, Drag force =drag coefficient x density x velocityx2 x area Drag force = (0.35x1.29kg/m3x (820km/h) 2 x 9m2)2 =1366.15N 2. Heat transfer and dam chemistry . Heat transfer and dam chemistry 2.1 Most solids expand as their temperature increases. The fractional range change caused by tensile is proportional to the stress that caused it. Similarly, the fractional length change caused by a temperature change is proportional to the temperature change, as long as the temperature change is not too great. If the length of a wire, rod, or pipe is Lo at temperature To then, =α∆T When ∆L=L- Lo and ∆T=T- To. The length at temperature T is L= Lo+∆L=(1+α∆T) Lo The constant of proportionality α is called the coefficient of thermal expansion of the substance. It plays a role in thermal expansion similar to that of the elastic modulus in tensile stress. If T is measured in Kelvin or in degrees Celsius, then α has a unit of K-1 or 0c-1. Since only the change in temperature is involved in equation. Either Celsius or Kelvin temperatures can be used to find ∆T; a temperature change of 1 k is the same as a temperature change of 10C. As is true for the elastic modulus, the coefficient of thermal expansion has different values for different solids and also depends to some extent on the starting temperature of the object. 2.2 The following formulas are wrong c) , d) , e) 2.3The equivalent ratio is defined as a ratio of fuel to air ratio. It is calculated as: Equivalent ratio= Equivalent ratio= =1.15 The mass fraction of gasoline in the mixture is calculated using the following formula: Balanced equations and the mass-mass relationships they convey help us answer several types of questions. How much of one reactant is needed to combine with a given amount of another reactant? For instance, how much silver chloride can be produced from 17.0 g of silver nitrate? There are other similar questions. Fortunately all these questions can be answered using the same procedure. The coefficients of a balanced equation give the relative amount (in moles) of reactants and products. ===0.938 3. Fluids dynamics of combustion 3Fluids dynamics of combustion 3.1 Sμ=suo()α()Б m/s Where ρ is the pressure, T is the temperature, the subscripts o and u denote reference and un burnt gas properties, respectively, and S is the laminar flame speed. The reference laminar flame speed Suo and the exponents α and Б depend on the equivalence ratio Ф of the fuel. 3.2 Sμ=suo()α()Б m/s Where ρb and ρu are the densities for burned gases and unburned gases. Flame velocity = Sμ =0.75()1.75()0.2 m/s Flame velocity =1.289 m/s 3.3 Sμ=suo()α()Б m/s Sμ=0.00075()1.75()Б = 0.1289 m/s 4 Shock and detonation waves Shock and detonation waves 4.1The ratio √s /√ is referred to as the Mach number, and the conical wave front produced when √s/ >√ (supersonic speed) is known as a shock wave. An interesting analogy to shock waves is the v-shaped wave fronts produced by a boat (the above wave) when the boats speed exceeds the speed of the surface –water waves. Jet airplanes travelling at supersonic speeds produce shock waves, which are responsible for the loud “sonic boom” one hears. The shock wave carries a great deal of energy concentrated on the surface of the cone, with corresponding great pressure variations. Such shock waves are unpleasant to hear and can cause damage to buildings when aircraft fly supersonically at low altitudes. In fact an airplane flying at supersonic speeds produces a double boom because two shock waves are formed, one from the nose of the plane and one from the tail. People near the path of a space shuttle as it glides towards its landing point often report hearing what sounds like two very closely spaced cracks of thunder. If the pressure in a pump jumps up 11 times, it means the pressure outside is lower 11 times. The Mach number is calculated as m=v When taking the jumping up of pressure, the equation can be simplified as =1+ () Whereas v is the velocity in shock fixed coordinate. m = Mach number. y = ratio of specific heat a = speed of sound Assuming that the ratio of specific heat is the same, then, the Mach number will be calculated as follows Therefore 11= 1+ () 11=1+ () 11= Mach number is equals to 3.3 4.2) ρg- +μ( + +)u(x,y,z)= ρ ρg - +μ ( + +) v(x, y,z)= ρ ρg - +μ ( + +) w(x, y, z) = ρ Whereas P= mass density of fluid g= gravitation acceleration p=driving pressure μ=dynamic viscosity of fluid t=time =2μ = 2μ 2μ = =μ ( + ) = =μ ( + ) = =μ ( + ) All stress components in a moving fluid can be expressed as follows: = 4.3. The wavelengths and frequencies of the standing waves can be found using either the pressure or the displacements descriptions of the wave. Using displacements, the fundamental has a node at the closed end, an antipode at the open end, and no other nodes or antinodes (see fig12.5). The distance from a node to the nearest anti node is always λ, so for the fundamental. L= λ OR λ =4L which is twice as large as the wavelength (2L) of the fundamental in a pipe of the same open at both ends. Two thin organ pipes of the same length, one open at both ends and one closed at one end, do not have the same wavelength twice as large and therefore a frequency half as large, assuming the pipes are thin. (For musicians: the pitch of the pipe closed at one end sounds an octave lower than the other, since the interval of an octave corresponds to a factor of two in frequency.) What are the other standing wave frequencies? The next standing wave mode is found by adding one node and one antinode. Then the length of the pipe is 3 quarter circles L= ¾λorλ = L. This is the wavelength of the fundamental and the frequency is 3 times that of the fundamental. Adding one more node and one more antinode, the wavelength is L. continuing the pattern; we find that the wavelengths and frequencies for standing waves are Standing sound waves (thin pipe closed at one end): λn= Fn = =n=nf1 Where n= 1, 3, 5, 7……….. Note that the standing wave frequencies for a pipe closed at one end are only odd multiples of the fundamental. 4.4 Rupture P1 = 14.5 psia P4 = 285 psia T1 = 534 R T4 = 534 R U1 = 0 U4 = 0 a1 = 1133 ft/sec a4 = 1133 ft/sec Assume  = 1.4 P2 = 53.6 psia P3 = 53.6 psia P4 = 285 psia P5 = 158 psia T2 = 826 R T3 = 331 R T4 = 534 R T5 = 1163 R U2 = 1200 ft/sec U3 = 1200 ft/sec U4 = 0 U5 = 0 a2 = 1409 ft/sec a3 = 892 ft/sec a4 = 1133 ft/sec a5 = 1672 ft/sec References list Eisner, H. 2008. Essentials of Project and Systems Engineering Management. Hoboken, N.J.: John Wiley & Sons. Euler, SD (ed.) 2002, Mine ventilation, Swets & Zeitlinger B.V, Lisse the Netherlands. Green, D. W., and Perry, R. H. (2008): Chemical Engineers' Handbook (8th Edition).. McGraw-Hill. Kohl, A & Nielsen, R 1997, Gas purification, Gulf publishing company, Houston Texas USA. Kreith, F, Manglik, RM & bohn, SM 2011, Principles of heat transfer, Cengage Learning, Stanford UK. Read More