# transport phenomenon homework help

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• Q1: Problem 2 [10 points] Consider a specially designed capillary tube, sealed at both ends, with variable diameter, as shown in the illustration below. This capillary is used for various experiments on the ISS (International Space Station). As this capillary floats motionless in the ISS cabin, will the air bubble inside the capillary move? See Answer
• Q2: 4. Ethylene pipeline-E. Twenty lbm/s of ethylene gas (assumed to behave ideally) flow steadily at 60 °F in a pipeline whose internal diameter is 8 in. The pressure at upstream location 1 is p1 downstream location 2. Why does the pressure fall, and what are the velocities of the ethylene at the two locations?60 psia, and has fallen to p2=25 psia atSee Answer
• Q3: 5. Transient behavior of a stirred tank-E. The well-stirred tank of volume V = 2 m3 shown in Fig. P2.5 is initially filled with brine, in which the initial concentration of sodium chloride at t = 0 is co = 1 kg/m³. Subsequently, a flow rate of Q =0.01 m /s of pure water is fed steadily to the tank, and the same flow rate of brine leaves the tank through a drain. Derive an expression for the subsequent concentration of sodium chloride c in terms of co, t, Q, and V. Make a sketch of c versus t and label the main features. How long (minutes and seconds)will it take for the concentration of sodium chloride to fall to a final value of Cf = 0.0001 kg/m³? See Answer
• Q4:1. Rouse' Law and the SPM profile equation (40pts): The Rouse' Law approach to suspended particulate matter (SPM) conservation assumes that SPM concentration is conserved locally, in each vertical column of water; i.e., SPM concentration may move up and down in the water col- umn (depending on flow conditions), but horizontal transport of SPM has no effect on the con- centration profile, even though individual particles will travel more or less with the flow and the concentration profile (once known) is used to calculate horizontal transport. This Rouse' Law ap- proach is the basis of most practical calculations of suspended load, except in some 3-D numeri- cal models, where the total derivative (local time variations plus horizontal advection) is consid- ered, also. The specific form of SPM conservation used in Rouse' law is (after Reynold's averag- ing and simplification using physical reasoning): a ac =W, DC + 2 (KBC) Ĉz cz 0=W₁ Here, C is SPM concentration, and Kc is a vertical diffusivity for SPM. (a) Derive the law for SPM conservation (15 pts): ac ac ac ôt ax (2) where no Reynolds' averaging has been done so far. There is a summary of the derivation in the first file on the equations of motion, after the discus- sion of mass conservation. You can follow the approach mentioned there, by considering the flux of suspended sediment. That is, define SPM as a volume concentration C (e.g., μ-liter SPM/liter water), and then define fluxes in and out of a control volume dVol = dx dy dz. Note that the flux you are defining is Cxpxvelocity and you are looking at the time-change of Cxp in volume dVol, so you are going to use a control volume approach to define: dxdydz = 2(Cp) dvol +U. +V ++ (w-ws) a(pCV) a(Cp) ôt ôt dz (3) To do this, you will have to assume that the sediment velocity vector Us = {u, v, w - Ws), where (u, v, w) is the water velocity. The SPM settling velocity Ws is NOT a function of {x,y,z,t) and does not have turbulent fluctuations - it is a constant. (b) Reynold's average (2) (which is your result from (a)) to get (10 pts): ac ac ac +u. +V +(w-ws) ac dz (1) = ac côz (4) (c) Now reduce the global expression of SPM conservation in (4) to the form needed for Rouse' Law (1), specifying each necessary assumption (I've already told you most of what you need to know.); 10 pts d) Explain why horizontal mixing is neglected in (4); 5 pts. 1See Answer
• Q5:3) A problem on SPM profiles (40 pts): Consider a flow 12m deep with U- of 0.05 m/s and zo = 10 mm and Zref = 0.1 m (Zref is the location where the reference value of C is measured). A deep, rapidly flowing river like this would like have a continuum of sediment sizes that would have to be very simply "binned" for any sediment transport calculation. We will consider only three "size classes" (as they are called): silt d₁ = 25 µm, fine sand d₂ = 250 µm, and coarse sand d3 = 1000 μm (assume Corey shape factor =0.7), and information of the sort that one might be able obtain in a real system. (a) Find Ws, Ws/U., and the Rouse number P for each size class; 10 pts. (b) Assume that a log-layer velocity profile is adequate to describe the flow. What is that veloc- ity profile? 5 pts (c) What is the corresponding parabolic eddy diffusivity profile? 5 pts (d) Using the Rouse solution for the SPM profile in class notes (the solution applicable in a log- boundary layer), calculate for each size class the ratios: (Clz=10m/C|z=zref), (C|z= 1m/C|z=zref), and (C|z=0.5m/C|z=zref). If Clz- zref is the same for each of the three size classes, where do you expect the bulk of the sediment transport to occur for each size class? You don't need to know an actual value of Cz=zref to answer the question. But when working with observations, you want to have data (z= zref) as close to the bed as possible, to get good results. Otherwise, you'll miss the larger sizes; 20 pts 2See Answer
• Q6:CHE 3333 Introduction to Transport Phenomena Fall 2023 Problem 1. (3 pts) A viscous film drains uniformly down the side of a vertical rod of radius R. You may assume that the flow is fully developed and that the liquid forms a film with a constant thickness of h. You may also neglect the shear resistance due to the atmosphere. Use the shell balance approach to determine the velocity distribution in the film.See Answer
• Q7:Problem 2. (3 pts) Benzene, which is an incompressible Newtonian fluid, flows steadily and continuously at 100 °F through a 3,000 ft horizontal, 4" schedule 40 pipe. The pressure drop across the pipe under these conditions is 2 psi. You may assume fully developed, laminar flow. Use the shell balance approach to find the volumetric flow rate through the pipe in gallons per minute (gpm).See Answer
• Q8:1. An infinitely long cylinder with a diameter of 10 cm is filled with a fluid. At x=0 and at t=0, a delta function mass input of 2.0 gm CO₂. Assume 25°C. Write governing equation, initial and boundary conditions and solution. Find the time to reach a concentration of 2 ppm at x=100 cm for (i) Molecular diffusion in air [2 pts]. Use D=0.164 cm²/s. [Be careful about converting mass/volume to mass/mass for air since 1 mg/l is not equal to 1 ppm in air! You will need the density of air: 1180 g/m³] (ii) Molecular diffusion in water [2 pts]. Use D=2E-5 cm²/s. (iii) Turbulent diffusion in water [2 pts]. As we will derive later, substitute a value of 1 cm²/s for D as a reflection of uniformly generated (isotropic and homogeneous) turbulence in water.See Answer
• Q9:2. An infinitely long cylinder is filled with pure water on the right side of the partition, and water with an initial concentration of 100 ppm of dissolved CO₂ on the left. At time t=0 s, the partition is removed. Write governing equation, initial and boundary conditions and solution. (i) How long will it take the concentration of CO₂ at a point 50 cm to the right of the partition to reach 10 ppm? [2 pts] (ii) If the fluid was turbulent, how would that affect your answer above? As we will derive later, substitute a value of 1 cm²/s for D as a reflection of uniformly generated (isotropic and homogeneous) turbulence in water. How long will it now take to reach a CO₂ concentration of 10 ppm 50 cm from the partition? [2 pts]See Answer
• Q10:3. Compare the result of problem 1 with the following: The cylinder has the mass input of 2.0 gm CO₂ added over a width of 5 cm (not at a point). Compare the results of this solution to Problem 1 for part (iii) for a D=1 cm²/s. First you must derive the solution using superposition (see similar derivation in the lecture notes). [5 pts]See Answer
• Q11:4. Compare the result of problem 1(iii) with a numerical approach. Compute 2 cases for the numerical approach: Divide the grid into (a) 5 cm grid spaces and (b) 10 cm grid spaces and solve the governing equation using the numerical approach. Note that you will have to choose a numerical time step to ensure numerical "stability". Make sure you choose a value that Ax² satisfies this relationship: At <. Compare the numerical schemes with two different grids with the analytical solution in 1(iii) by plotting C vs x for each solution at t=100 s and comment on how well the numerical solution matches the analytical solution. [5 pts]See Answer
• Q12:1. Molecular diffusion in a bounded medium [12 pts] An instantaneous point of strength M is located at the position (0, 0, h) above a horizontal plane (z = 0) that forms a reflective boundary. With a uniform advective velocity field (U, 0, 0) and molecular diffusivity D, (1) determine the solution for the resulting concentration distribution, and (2) by differentiation normal to the boundary show that the normal diffusive flux is indeed zero. [2 pts each] Assuming an additional horizontal boundary, that behaves as a perfect absorber for the substance under consideration, is located at z = H (where H>h), determine the solution. [3 pts] Assume a 1-D governing equation for vertical diffusion in a vertical tube: a²c Әс at = D azz assuming homogeneity in the x-y plane. Assume the tube has a length of 100 cm and there is an injection of mass in the middle of the tube (z=50 cm) of 10 g. Assume D is 0.001 m²/s and diameter of the tube is 10 cm. Perform 2 simulations with 2 different boundary conditions at z=0 cm and z=100 cm: (1) a concentration of zero, and (2) a no flux boundary condition. Use the numerical approach with a grid spacing of 5 cm to solve for the vertical distribution of mass after 60 s (i.e., plot C vs z at t-60 s) for (1) and (2). Choose your time step based on this criterion: At < 0.44² [5 pts] a. b. C.See Answer
• Q13:1. Molecular diffusion in a bounded medium [12 pts] An instantaneous point of strength M is located at the position (0, 0, h) above a horizontal plane (z = 0) that forms a reflective boundary. With a uniform advective velocity field (U, 0, 0) and molecular diffusivity D, (1) determine the solution for the resulting concentration distribution, and (2) by differentiation normal to the boundary show that the normal diffusive flux is indeed zero. [2 pts each] Assuming an additional horizontal boundary, that behaves as a perfect absorber for the substance under consideration, is located at z = H (where H>h), determine the solution. [3 pts] Assume a 1-D governing equation for vertical diffusion in a vertical tube: дс a²c at = D assuming homogeneity in the x-y plane. Assume the tube has a əz² length of 100 cm and there is an injection of mass in the middle of the tube (z=50 cm) of 10 g. Assume D is 0.001 m²/s and diameter of the tube is 10 cm. Perform 2 simulations with 2 different boundary conditions at z=0 cm and z=100 cm: (1) a concentration of zero, and (2) a no flux boundary condition. Use the numerical approach with a grid spacing of 5 cm to solve for the vertical distribution of mass after 60 s (i.e., plot C vs z at t=60 s) for (1) and (2). Choose your time step based on this criterion: At < 0.44² [5 pts] a. b.See Answer
• Q14:2. Some diffusion problems in the atmosphere: [8 pts] a. A point source of 1000 g of SO₂ is released. If molecular diffusion (D = 0.12 cm²/s) takes place, (1) how long will it take to reach a standard deviation of 1 cm, 1 m and 100 m respectively, and (2) what are the maximum concentrations at these times? [2 pts each] b. C. By contrast, the molecular diffusion coefficient for a suspension of dust (1μm diameter) in air (Brownian motion) is approximately D = 2.2 x 10° cm²/s. Estimate the times now. Compare these to problem 2(a) and comment on the behavior of smog (made up of SO₂ and dust particles and other stuff!) in inversion layers over cites in which conditions are often without turbulence. [3 pts] An actual measurement during a SO₂ experimental point source release in an inversion layer shows that the half width as measured by the 10% level (i.e., where the concentration is 10% of the center point concentration) is 10 m at 1 hour after the release (see sketch). Estimate the diffusivity. Are these molecular conditions (compare to 2(a))? [3 pts] 021 44 Sxfel 10 mSee Answer
• Q17:Problem 1 A part of lubrication system consists of two circular disks between which a lubricant lows radially.See Answer
• Q18:Problem 2 Now, the AlaskaBreez group is interested in a detailed analysis of how well their spherical iglooSee Answer
• Q19:Problem 3 (15 Points)A catalyst pellet has a radius R and a thermal conductivity k. Because of the chemical reaction occurring within theSee Answer
• Q20:Species Accumulation and Transport within an Interface Accumulation and transport within an interface can be important for species that reside at phase boundaries, such as gases adsorbed on solids or surfactants at fluid-fluid interfaces. The objective is to derive more general interfacial conservation statements than those in Section 2.2. Consider a species that may be present at an interface or in either of the adjacent bulk phases. Let Cs and Ng be its surface concentration (moles m2) and surface flux (moles m¹s¹), respectively; note the differences in units from the corresponding three-dimensional quantities. The vector Ng is every- where tangent to the surface. (a) For part of an interface corresponding to surface S in Fig. A-2, state the macroscopic (integral) solute conservation equation. Assume that phase A is below S and phase B above it, such that n points toward the latter. Include the possibility of chemical reactions at the interface, trans- port to or from the bulk phases, and interfacial motion. The interface is not necessarily planar. (b) By reducing the integral equation of part (a) to a partial differential equation, show that at where N and C are the species flux and concentration, respectively, in a bulk phase. (Sub- scripts identifying the chemical species have been dropped for simplicity.) This result, which is valid instantaneously and locally, is equivalent to Eq. (5.2-2) of Edwards et al. (1991). How does it compare with what is obtained by applying Eq. (2.2-15) to a chemical species?See Answer

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