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Diffusion Flames and Fluid Dynamics of Combustion - Coursework Example

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The paper "Diffusion Flames and Fluid Dynamics of Combustion" concern such issues as a mass on the smaller piston needed to balance the system, the relationship between velocity, density, and pressure gradient in a pitot-static tube, the cruising speed of the aircraft at the sea level, etc…
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1. Classical Mechanics of Fluids. 1.1. A) Given data is Working cylinder radii ratio, R1/ R2 = 100 Mass on bigger piston, F2 = 62.5 tonnes Thus the ratio of area for system, A1/A2 = 100, 00 Now applying the formula, F1/A1 = F2/A2 F1 = F2 X (A1/A2) = 62.5 X 100, 00 = 62.5 X 104 tonnes Thus mass on the smaller piston needed to balance the system is 62.5 X 104 tonnes. 1.1. B) Given that velocity V measured by the Pitot-static tube is proportional to the density ρ and pressure gradient ∆р Velocity α (density) a X (Pressure Gradient) b V α ρa X ∆рb Comparing the dimensions on both sides M0 L1 T-1 α (M1L3T0) a X (M1L-1T-2) b Solving for a & b we get a = -1/2 and b = ½ Putting back the value we get Velocity α (density) -1/2 X (Pressure Gradient) ½ Thus relationship between velocity, density and pressure gradient in a pitot-static tube is V α √ ∆р/ρ 1.2. A) In laminar flow the motion of the fluid particles follows a proper orderly pattern such that all the particles move parallel to the pipe walls in straight linei. Turbulent flow is regarded by disordered and stochastic property changes. In turbulent flow the pressure and velocity varies very rapidly with respect to space and timeii. Reynolds number is defined as the ratio of inertial forces to viscous forces and it quantifies the importance of these forces in fluid flow. Reynolds number, Re, which is a dimensionless number is given byiii Where р = density of fluid, u = mean velocity of the fluid, d = diameter of the pipe and µ = viscosity of the fluid Laminar flow: Re < 2000 Transitional flow: 2000 < Re < 4000 Turbulent flow: Re > 4000 1.2. B) Given data is Aircraft’s cross-section area, A = 1.47 m2 Steady thrust = 48.29 KN Aerodynamic coefficient, C d = 0.73 Cursing speed at sea level is to be determined Using formula for aerodynamic drag Fd = -1/2 X density X A X C d X V2 Assuming density of air at the sea level to be 1 kg/m3 Putting the values we get 48.29 X 103 = -1/2 X 1 X 1.47 X 0.73 X V2 V = 300 m/sec Thus the cruising speed of the aircraft at the sea level is 300 m/sec. 2. Heat Transfer 2.1. A) Dimension of a physical quantity is its characteristic or quality which can be quantified with help of units. Dimensions beak up any physical quantity to its fundamental level and helps in establishing relationship among different physical quantities. The fundamental dimensions of physical quantity are mass, length, time, and temperature. Area A = a × b, [A] = L2, e.g. m2, Volume V =a×b×c, [V] = L3, e.g. m3, Velocity V = ∆l/∆t, [V] = L/T, e.g. m/s 2.1. B) Time of temperature equalisation in a solid body of length l and thermal diffusivity К Dimensions are [t] = T, [L] = L, and [К] = L2/T. Looking for the solution in the form t α La X Кb This yields T = L a × (L2/T) b, L0T1 = La+2b X T-b. Comparing both sides we get a=2 & b=-1 Putting back we get t α L2/К 2.2. A) Radiation flux emitted by a body at temperature T is given by the Stefan-Boltzmann law. The Stefan–Boltzmann law is also known as Stefan's law and it highlights the heat transfer related to black bodies. This law states that the total energy radiated per unit surface area of a black body in unit time which is know as the emissive power, is directly proportional to the fourth power of the black body's absolute temperature T. The formula for heat transfer per unit surface area of black body is: Q = ε σ T4 σ is the Stefan-Boltzmann Constant = 5.67 x 10-8 Watts-m-2-°K-4 Emissivity of any material is a measure of its ability to radiate the absorbed energy. It can be defined as the ratio of energy radiated by a particular material to energy radiated by a black body under the same conditions. It is expressed by ‘ε’ or ‘e’. Since it is a ratio of same physical quantity so it has got no unit and is thus dimensionless. For a true black body, ε =1 For real objects, ε 1, the flow is said to be supercritical. This means that the current is strong but the water is not flowing deep. On the contrary if Fr < 1, the flow is said to be subcritical. This means that the flow becomes thins when it approaches an obstacle. Richardson number is also a dimensionless number which is defined as the ratio of potential to kinetic energy. It is given by the following formula. Where g is the acceleration due to gravity, h a representative vertical length scale, and u a representative speed. 4.1. B) Unimportance of buoyancy in the flow is represented by lesser than unity. But on the contrary, if Richardson number is much greater than unity then it signifies the dominance of buoyancy. In third case when the Richardson number is unity, the flow is buoyancy-driven which means that the energy of the flow is derived from the potential energy present in the system originally. 4.2. A) The laminar diffusion flame height Lf depends on the volumetric fuel flow rate Q and diffusion coefficient D Lf α Q a X D b Putting the dimensions L1 α L3aT-a x L2bT-b Now comparing both sides, we get a= 1 & b = -1 Thus Lf α Q 1 X D -1 Lf α Q/D 4.2. B) Given data is Flame temperature Tf = 2186 K Diffusivity coefficient, D = 1.21 X 10−3 m2 /s Volumetric flow rate, Q = 0.012 m3/s Stoichiometric molar oxidizer-fuel ratio S = 2 Fuel temperature, TF = 44 oC = 317.15 K Oxidizer temperatures, TO = 44 oC = 317.15 K Flame height is to be determined Applying Roper & Roper model of the laminar diffusion flame Lf = 0.012 X (317.15 / 317.15 ) X ( 317.15/2186)0.67 4X 3.14X1.21 X 10−3X ln1.5 Thus Lf = 0.53 m Read More
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