a). From the definition of pressure P, it can be expressed as, P= F/A, where F is the applied force and A is the cross sectional area of the surface of the material. Since the hydraulic cylinder has two radial cylinders the pressure at each cylinder can be expressed as taking the subscripts 1 & 2 to represent the cylinders, then at the smaller cylinder 1, P=F1/A1 and at larger cylinders 2, P= F2/A2. But A1= ? R12 and A2 =?? R22. From the law governing pressure in fluids, pressure at both cylinders is equal hence F1/A1= F2/A2. Thus force one can be represented in terms of force two as F1= F2 *A1/A2.

Substituting the area, then F1= F2? R12/? R22. This implies that, F1= F2R12/R22. And thus the force is, F1. = 6. 25×103 x9. 81/1004 = 61. 3125 N. b). The variation of pressure P, in static liquids is known to depend upon the acceleration due to gravity g, the density of the liquid r, and the elevation z. Use dimensional analysis to determine the hydrostatic law of pressure . Solution Directional breakdown can be affirmed as, [P] = ML-1T-2 … (I) where M stands for the units of mass, L stands for the units of length and T stands for the units of time respectively.

b) The power law of whirls dissipation in uniform turbulence. Solution. Turbulence or turbulent flow in a fluid system is characterized by disorderly changes resulting in fast deviations in pressure and velocity. This causes the swirling and turn around effect when a fluid flows past a barrier. This fluid in motion makes a space that is without a fluid flowing on the side of the downstream of the barrier. The fluid which is left behind the barrier flows into the null region making the eddy creation on the edges of the barrier possible.

These eddies formed vary in their power, span and energies. As eddies are created, they contain a lot of kinetic energies, because they are large initially, which continues to spill as they decrease in size. This whole process continues to make smaller and smaller formations thus producing a structured mechanism of eddies from big to smaller and smaller until eddies are created that are small enough to the extent that molecular diffusion can be singled out and dissipation of energy takes place. This stage is a scalerly and is often referred to as Kolmorogov length scale.

2. Heat Transfer a) Evaluate the formula for the Nusselt number through the estimation of the heat fluxes involved by their orders of magnitude. This number is considered using two sizes of magnitudes of heat fluxes, that is the convective and the conductive heat fluxes. It is a ratio of the former to the later. By Newtonian definition, the heat flux within a given area is approximately proportional equal the temperature difference (? T= TW -T*). Thus from the Newton’s law of cooling, q N W= h(TW -T*) where h is the coefficient of heat transfer, S. I units of w/m 2.

Now if you consider a fluid in motion with a layer of width L, that causes a convection, then the heat flux will be, q N W= h(TW -T*). On the other hand for a stationary layer of a similar fluid would provide a total conduction layer of the fluid. The heat flux of conduction to attribute to this stationary fluid, can be represented as, q N W= k (TW -T*)/L where k is the coefficient of heat of conduction and L is the width of the layer. Then from the definition of the number(Nu), the ratio of these two fluxes of heat generate the number, that is, flux heat of convection divided by the heat flux heat of conduction.

Arithmetically, Nu L = h (TW -T*)/ k (TW -T*)/L. = h L/k. 2. 2 The radiation heat flux formula. a) The formula of radiated heat flux is defined by the Stefan- Boltzman constant. This heat is equivalent to the radiated heat emitted by a surface of a body at some reference temperature per unit time. The Stefan – Boltzman cosnt. is given as, emissivity= ???? T 4 where J r stands for radiated heat and ?? is the emissivity of the black body. The emitted heat of the material always depends on the material itself.

For most materials the emissivity is always less than one and for a perfect black cavity has a unit emissivity. b) A flat steel plate of total area 1. 5 m2 at temperature 600 C radiates 30 KW/s. Calculate its emissivity coefficient? Relating the equation to the constant above, then we find the equation J r= ?? ????? T 4 as the equation that represent a general equation of a black body. Deriving the emissivity of a material from this equation, there an equation stated below can be reached at; ??? J r where J r is the radiated heat of the steel plate, which is a variable here is the emissivity of the steel plate, ? is the Stefan Boltzman constant and T is the given temperature of the steel plate and which in this case is 600 Co .

But since we are given an area, we have to adopt the above equation to include the area so that it can be solved using the given parameters. Hence, emissivity ?? J r A???? .? This implies that if the figures are inserted in the equation, then the answer can be arrived at. Therefore the solution can be computed as follows; Emissivity??? J r A???? = (1. 5x 3. 0x 10000)/ (5. 6696) e-8 x (600+273) =9. 09e-9 3. Fluid Dynamics of Combustion.

a) Explain what the normal flame speed is. The normal flame is the flame that when burnt in a premixed gas confined area creates a front region of unburnt gases then pushes these gases forward at a certain velocity while pushing behind the burn gases at a velocity opposite to that of the unburned gases. This push is so powerful that the flame actually travels at a very high speed through the enclosure. The difference if the flame velocity and that of the unburnt gases is the representation of the normal flame velocity with reference to the Cartesian co-ordinate system (Giovangigli, 1992, p. 118)

b) The normal flame in a pure stoichiometric mixture of very flammable gases like propane and oxygen at temperature 107 Co and normal atmospheric pressure using formulas formulae and data from the handouts. For a stoichiometric mixture ? =1 as depicted from the Metghalchi and Keck correlation. This implies that the other values can be deduced from the above clue. ?M= 1. 08, BM= 34. 22 and B2= -138. 65, Y dil = 0 for pure mixture. Putting these values in the Metghalchi and Keck equation I gather the following calculations and a solution below. = 34. 22 + 138. 65(1-1. 08) x (380/298)2. 18/100 =0. 77 m/s 2. a) Explain the turbulent burning velocity.

The burning turbulent velocity is gotten from a turbulent flame of mass m that has resulted from a combusted premixed combustible mixture enclosed in a pipe of a cross-sectional area A while the flame density is p. it can be related in the equation below. ST = m/pA. Where ST is the turbulence burning velocity. Under normal circumstances it is hard to give a direct definition of this terminology because the turbulent flames to be defined are very uneven.

Hence the need to use representations in terms of parameters. b) For the average amplitude of flow velocity fluctuations u? = 40 cm/s, compare the turbulent burning velocities S T of wrinkled flames calculated using the original Damkohler formula and correlations by Schelkin, Klimov and Clavin & Williams. Use L S calculated in the problem 3. 1. b Damkohler formula states that the turbulent burning velocity can be expressed as a function n of SL and u such that they sum up to give an equal of the turbulence burning velocity.

If this is represented parametrically, then S T = SL + u. From above, the calculated solutions of these parameters can then be inserted into the equation then it implies, 0. 77 = 0. 40=1. 17m/s. From Schelk, the turbulence burning velocity is represented as a combination of the normal flame velocity and the square root of the square of the ratio of the 2u to the normal flame velocity that has been added a unit. And thus from the above statement, S T was calculated and found to be; S T= 1. 11m/s. and finally according to Klimov’s equation S T= 3. 5sL [u’/ S L] 0. 7 = 3. 5×0. 77x (0. 40/0. 77)0. 7=1. 704m/s.

Clavin & William summarize their equation as shown below, S T=SL {? [1+v (1+8(u’/SL) 2)]} ?. Replacing the square part with the known values from the earlier equations, I get S T= 0. 907m/s 4. Diffusion flames. 4. 1. a) Explain the Richardson and Froude numbers. The Froude number is drawn from the magnitudes of forces two types, that is the magnitude of the force of gravity and the magnitude of the force of inertia. It is used to show the amount of resistance an object is undergoing which is in motion through the water and the n this resistance is compared to other objects of different sizes.

It is normally dimensionless that is it does not take proportions. Its foundation is actually based in the ratio of velocity per length, that is this equation can be represented parametrically as shown below; Fr= is a product of the velocity of the object Vo and the square root of both the acceleration due to gravity, and L (the specified length of the object) and Fr is the Froude number. The Richardson number is also a number without dimensions that seeks to compare the ratio of potential to kinetic energies. Because this number is much less than one, buoyancy is not considered as a very important variable here.

The initial expression of the number can be expressed as shown below; Ri= is the product of Lv (the vertical representation of the length (height)) and acceleration due to gravity then divided by u (the speed). However these two numbers are closely in a relationship with a kind of connection shown below; Fr-2= Ri. This implies that given either of these two numbers, one can possibly compute the other b) Evaluate the formula for the Richardson number through the estimation of the forces involved by their orders of magnitude.

The Richardson number as previously seen from above is the number that seeks to compare the ratio of potential to kinetic energies. This number can be used in computing forces in different magnitudes. The potential energy can also be referred to as a force of floating and the kinetic ability referred to as the inertial force. The up thrust force can be expressed as Fb?? ?? gV, where the symbols have their usual meanings, that is, g is the acceleration due to gravity, V is the volume of the fluid and ? is the density of the fluid itself.

Let us assume that the acceleration of a fluid is a, then given a time difference the velocity of this fluid can be calculated. Likewise if the velocity of this fluid is known, then the acceleration can be calculated. The order of the force of inertia defined as Fi = ?? V a. If this acceleration is achieved at speeds of order u and times of order of the same range, that is, then this acceleration can be calculated in terms of these two variables. In other words; a ? u/t and this is the acceleration order with a corresponding time order of; t ? L/u.

This implies that the acceleration order can be represented in terms of the components of the velocity and the order of time as illustrated below; a=u/ (L/u). This implies that the acceleration is specified in terms of only the velocity and the length, that is, a=u2/ L. Inserting this equation into the order of the force of inertia we get that, b Fi = pVa, changes to the expression Fi = p? V u2/ L. Having reached this result, it permits in the calculation of the Richardson’s number given as the ratio of the magnitudes of the upthrust to inertia and thus by definition;

Fb/Fi= ?? gV??? V u2/ L, =g L/ u2. This presents the whole equation as the previously defined Richardson number. 4. 2. a) The laminar diffusion flame height f L depends on the volumetric fuel flow rate Q & and diffusion coefficient D. Obtain a formula for f L using dimensional analysis. From the previous question above, dimensional analysis brings forward an equation in terms of only the units of given functions or variables. and the diffusion coefficient in terms of dimensional analysis yields the following, [D]=L2T-1 [Lf]=L, [Q]=L3T-1

where L represents units of the length, T stands for the units of time. From the equation of dimensional analysis and using the dimensional analysis symbols that is delta and Gama, the following equation can be obtained, L= (L 3T -1)? (L 2T-1) ? =L3?? +2 ? T -?? – ? , and equating the right hand side with the corresponding equations on the left hand side then separating them yields, for the values of L. – for the values of T. Solving the above simultaneous equations yield; ?????? ?implying that by substitution of ?? into the first equation yields that , putting this value into equation two solves for ?? as ? =-1. Having solved for the empirical values of alpha and gamma, these values are put in the definitive equation of proportionality. Hence the first expression transforms to Lf ?? Q /D. b) Using Roper & Roper model of the laminar diffusion flame at temperature = 2200 f T K, calculate its height for the diffusivity coefficient 3 10? D = m2 /s, volumetric flow rate 01. 0 = Q & m3 /s and stoichiometric molar oxidizer-fuel ratio S = 2. Fuel and oxidizer temperatures are equal to 40 CO

The Roper and Roper model defines a formula for the diffusive flame height with a consideration to the fuel used and the oxidizing agent this formula can be stated empirically as it appears in the model, that is, Rf= (Qx (KO/KF) x (KO/KF))/4? D ln (1+1/S) here using the notations Rf to denote Roper and Roper model and KO referring to the temperature of the oxidizing agent and KF referring to the fuel temperature. Given the values of S, D, and the KO and KF temperatures the, a solution can be obtained by substituting the values given above into the expression noting that KO/KF is equal to unity. Therefore it becomes; Rf = 0. 01x ((273+40)/2200)0. 67/4? x10-3x ln(1+1/2) ? 0. 53 m (Oppenheim, 2002, p. 89).

References. The Engineering Toolbox 200, pressure-gradient representations. Retrieved on August 23, 2008 from http://www. engineeringtoolbox. com Giovangigli, Smooke. (1992). Numerical modeling of symmetric laminar diffusion flames, Vladimir, G. (2001). Industrial Combustio: Fluid Dynamics,. Oppenheim, A. K. (2002). 1st ed. , Dynamics of Combustion Systems,

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