Fluid Dynamics of Fire - Lab Report Example

1. ical Mechanics Q1 a) Mass (m2) on the smaller working cylinder (radius R2) will balance the mass (m1 = mass of Challenger 2 tank = 62.5 ton) kept on the large working cylinder (radius R2) by equating the pressure of the fluid, which is common and connected in the two working cylinders according to Pascal’s Law. Therefore, i.e. kg Q1.1. b) Let us assume that flow velocity (V) is directly proportional to fluid density () raised to a power ‘a’ and pressure drop (p) raised to power ‘b’. Then, V  ra*(p)b i.e., V = k* ra*(p)b; where ‘k’ is a dimensionless number Now, both sides, i.e. LHS and RHS should have identical dimensions. Putting the dimensions of the physical quantities in this equation, [LT-1] = [ML-3]a*[ML-1T-2]b = [Ma+bL-3a-bT-2b] Comparing, the exponents of individual basic quantities like M, L and T yields, Comparing exponents of T on either side gives, -1 = -2b i.e., b = ½ Comparing exponents of M on either side gives, 0 = a + b i.e., a = - ½ Putting the values of ‘a’ and ‘b’ for getting exponents of L on the either sides yields, Exponent of L on RHS = -3a – b = - 3*(-1/2) – ½ = 1 = Exponent of L on LHS This confirms that values of ‘a’ and ‘b’ are correct. Putting these values of ‘a’ and ‘b’ in the equation of ‘V’ yields, Q1.2. a) ‘Laminar Flow’ is the flow in which velocity vectors at a point do not change its direction with time and the velocity vector of individual fluid elements are parallel to each other. In case of a ‘Turbulent Flow’, the velocity vectors at a point keep changing its direction with time and also velocity vectors of individual fluid elements are not parallel to each other, rather intersect. Transitional Regime is one in which the flow condition has a mixed character i.e. it is laminar in some section and turbulent in another. Whether a flow is Laminar or Turbulent or in Transitional state is dictated by ‘Reynolds Number’ (Re). This number is defined as where,  = density of the fluid, V = flow velocity, d = diameter of the tube and  = viscosity of the fluid If Reynolds Number (Re) is less than 2300, then the flow remains laminar; the flow is in transition state for Re in the range of 2300 to 4000 and for Re greater than 4000, flow becomes turbulent. Q1.2. a) Magnitude of Newton’s aerodynamic drag (Fd) is given by the following equation, Fd = Where, Fd = Drag Force = 48.29 kN = 48290 N R = Density of air = 1.2 kg/m3 at 20 oC at sea level V = velocity of the aircraft = ? (to be calculated) A = Area of cross-section of the aircraft = 1.47 m2 Cd = Aerodynamic drag coefficient = 0.73 Putting these values in the above equation, m/s 2. Heat Transfer Q2.1. a) A physical quantity can be expressed as a product of fundamental quantities (Mass, Length, Time etc) raised by certain exponent and this expression is termed as ‘Dimension’ of that physical quantity. Let us take an example of Force. Force = mass * acceleration = mass*(change in velocity/time) = mass*(change in position/time)/time Therefore, Dimension of force will be [F] = [MLT-2] Similarly, energy is a very useful physical quantity. Its dimension can be calculated in the following manner. Change in energy = work done = Force * Displacement Therefore, dimension of energy will be, [E] = [F]*[d] = [MLT-2]*[L] = [ML2T-2] Q2.1. b) Let us assume that the time of temperature equalization (t) is directly proportional to length (l) of the solid body raised to a power ‘a’ and diffusivity (k) raised to power ‘b’. Then, t  la*(k)b i.e., V = c* la*(k)b; where ‘c’ is a dimensionless number Now, both sides, i.e. LHS and RHS should have identical dimensions. Putting the dimensions of the physical quantities in this equation, [T] = [L]a*[L2T-1]b = [La+2bT-b] Comparing, the exponents of individual basic quantities like L and T yields, Comparing exponents of T on either side gives, 1 = -b i.e., b = -1 Comparing exponents of L on either side gives, 0 = a + 2b i.e., a = 2 Putting these values of ‘a’ and ‘b’ in the equation of ‘t’ yields, Q2.2. a) Radiation heat flux from a hot body at temperature T is given by the famous Stefan-Boltzman equation, which is JR = T4 Where,  is emissivity of the body and its value lies between 0 and 1  = 5.6696*10-8 Wm-2K-4 is Stefan – Boltzmann constant And T = Absolute temperature of the body This law states that radiation heat flux is directly proportional to emissivity and directly proportional to the absolute temperature of the body raised to its fourth power. Value of emissivity is maximum ‘1’ for a black body, this means a black body radiates maximum as compared to any other body at the same temperature. Q2.2. b) Net Radiation heat flux from a hot body at temperature T kept in a surrounding at temperature Ts is given by JR = T4 – Ts4) Where, JR = Raditive Heat Flux = Q/A = 27000 / 1.47 W/m2 = 18367.35 W/m2  = Emissivity of the body = ? (To be calculated)  = 5.6696*10-8 Wm-2K-4 is Stefan – Boltzmann constant T = Absolute temperature of the body = 631 oC = 904 K Ts = Absolute temperature of the surrounding = 300 K Putting these values in the above equation of net radiative heat flux, 3. Fluid Dynamics of Combustion Q3.1. a) Normal laminar flame velocity is the planar flame front speed with respect to the un -burnt gas mixture (fuel – oxidizer mixture). It is expressed as SL = uv − uu Where, SL is normal laminar flame velocity; uv is flame speed and uu is speed of un-burnt gases. Q.3.1. b) Metghalchi and Keck correlation for normal flame speed (SL) is the following SL,ref = reference flame speed = BM + B2( – M) = 34.22 + (-138.65)(1 – 1.08) = 45.312 Here, values of BM, B2 and M (for propane) have been taken from the lecture handout and  = 1 as the fuel – oxidizer ratio is stoichiometric. Tu,ref = 298 K Tu = 48 oC = 321 K  = 2.18 – 0.8( – 1) = 2.18 – 0.8(1 – 1) = 2.18 Pref = 1 Atm P = 1.87 Atm B = -0.16 + 0.22(( – 1) = -0.16 – 0.22(1 – 1) = -0.16 Ydil = mass fraction of residue gas = 0 as this is stoichiometric mixture Putting these values in the expression of the normal flame velocity (SL) = 48.2 cm/s Q3.2. a) Damköhler’s formula for turbulent burning velocity is where, ST = Turbulent burning velocity SL = Normal flame velocity u’ = average amplitude of flow velocity fluctuation Q3.2. b) Damköhler’s formula for turbulent burning velocity is Here, ST = Turbulent burning velocity = ? (to be calculated) SL = Normal flame velocity = 48.2 cm/s (from Q3.1. b) u’ = average amplitude of flow velocity fluctuation = 33 cm/s Putting these values in the above equation, cm/s = 81.2 cm/s Incorporating Schelkin correction results in Therefore, ST = 1.2135*SL = 1.2135*48.2 cm/s = 58.5 cm/s Incorporating Klimov correction results in Therefore, ST = 2.6847*SL = 2.6847*48.2 cm/s = 129.40 cm/s Incorporating Clavin & Williams correction results in Therefore, ST = 1.261*SL = 1.261*48.2 cm/s = 60.77 cm/s 4. Diffusion Flames Q4.1. a) Richardson Number (Ri) is a dimensionless number giving ratio of potential energy to kinetic energy. It is represented as Ri = This number is very useful in solving problems in the field of thermal convection, where Richardson number is a ratio of natural convestion to the force convection. It is defined as Ri = where g is the gravitational acceleration, β is the coefficient of thermal expansion, DT is the temperature difference between hot wall and the surrounding, L is the characteristic length, and u is the characteristic velocity. The Richardson number can also be expressed by using a combination of the Grashof number and Reynolds number i.e. Ri = Froud Number (Fr) is a dimensionless number giving ratio of characteristic velocity to water wave velocity. It is expressed as Fr = u/c Where, u is characteristic velocity and c is water wave velocity This number is also defined as the ratio of centrifugal force to inertial force. It is expressed as Fr = Where, u is characteristic velocity, g is acceleration due to gravity and L is characteristic length. Q 4.1. b) Richardson Number (Ri) is a dimensionless number giving ratio of potential energy to kinetic energy. It is represented as Ri = For Ri < 0.1, natural convection is negligible, for Ri > 10 forced convection is negligible and for the intermediate values neither is negligible. Q 4.2. a) Let us assume that the laminar diffusion flame height (Lf) is directly proportional to volumetric fuel flow rate (Q.) raised to the power ‘a’ and diffusion coefficient (D) raised to power ‘b’. Then, Lf  Q.a*(D)b i.e., Lf = c* Q.a*(D)b; where ‘c’ is a dimensionless number Now, both sides, i.e. LHS and RHS should have identical dimensions. Putting the dimensions of the physical quantities in this equation, [L] = [L3T-1]a*[L2T-1]b = [L3a+2bT-a-b] Comparing, the exponents of individual basic quantities like L and T yields, Comparing exponents of L on either side gives, 3a + 2b = 1 Comparing exponents of T on either side gives, a + b = 0 Solving for a and b gives, a = 1 and b = -1 Putting these values of ‘a’ and ‘b’ in the equation of ‘Lf’ yields, Q4.2. b) Roper and Roper model of laminar diffusion flame gives the following equation for height of the laminar diffusion flame. where, Q. = 0.012 m3/s is volumetric flow rate D = 1.21*10-3 m2/s is diffusivity coefficient S = 2 is stoichiometric molar oxidizer-fuel ratio, Tf = 2186 K is temperature of the flame, TO = 44 oC = 317 K is temperature of the oxidizer and TF = 44 oC = 317 K is temperature of the fuel Putting these values in the expression of Lf m = 53.4 cm Read More