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- 2. Ohm’s law In electrical circuits, Ohm's law states that the current through a conductor between two points is directly proportional to the potential difference or voltage across the two points, and inversely proportional to the resistance between them.
- 4. DEFINITION OF TRANSPORT PHENOMENA Transport phenomena are all irreversible processes of statistical nature stemming from the random continuous motion of molecules, mostly observed in fluids. They involve a net macroscopic transfer of matter, energy or momentum in thermodynamic systems that are not in statistical equilibrium.
- 6. Momentum, Heat & Mass Examples include heat conduction (energy transfer), viscosity (momentum transfer), molecular diffusion (mass transfer), radiation and electric charge transfer in semiconductors. Apart from conduction, another heat transfer mechanism is convection which is not a transport phenomenon per se, as it involves turbulent and bulk motion of fluids.
- 7. Analogies An important principle in the study of transport phenomena is analogy between phenomena. For example, mass, energy, and momentum can all be transported by diffusion: In any Dynamic System: Rate = Driving Force/ Resistance
- 8. EXAMPLES The spreading and dissipation of odors in air is an example of mass diffusion. The conduction of heat in a solid material is an example of heat diffusion. The drag experienced by a rain drop as it falls in the atmosphere is an example of momentum diffusion (the rain drop loses momentum to the surrounding air through viscous stresses and decelerates).
- 9. Driving Force & Resistance in Momentum Transfer Driving force : Velocity gradient Resistance : Viscosity ( μ) * Equation of Newton:
- 11. Driving Force & Resistance in Heat Transfer Driving Force: Temperature Gradient Resistance: 1/k (k=Material Conductivity) Fourier’s Law:
- 12. Example of Heat Conduction
- 13. Driving Force & Resistance in Mass Transfer Driving Force : Difference in Chemical Potential. When the pressure and temperature are the same in both sides, the driving force is the difference in concentration. In R.O, the pressure is different in the sides of the cell. Resistance: 1/ Diffusivity ( D)
- 14. Fick’s law:
- 15. Transport phenomena & unit operations The study of transport phenomena is the basis for most of unit operations. In many unit operations, such as distillation, all three transport phenomena ( i.e., fluid flow, heat transfer and mass transfer) occur often simultaneously. Empiricism is often required because the exact mathematical equations cannot be solved.
- 16. EQUILIBRIUM VS RATE PROCESSES A system in equilibrium means that all variables are constant with time ( Thermodynamics) Non- equilibrium means that at least one variable changes with time ( Transport Phenomena). In this case, we will have rate. They are four different rate processes: Rate of Momentum transfer Rate of Heat transfer Rate of Mass transfer Rate of Reaction ( Not covered in this course)
- 17. Example of momentum transfer Let’s consider two tanks with two different pressures separated by a valve. When we open the valve, we will observe a decrease of pressure in the tank having high pressure and increase in pressure in the tank with low pressure movement of molecules from high pressure to low pressure. Non – Equilibrium and ∆P= Driving Force. When the two tanks reach the same pressure. There will be no more transfer: (Equilibrium state: thermodynamics Study)
- 18. Example of molecular heat transfer If we put two solids having different temperatures side by side, we will observe a decrease of temperature of the hot solid and increase of temperature of the cold solid. Molecular heat transfer by conduction Driving force: Difference in temperature When the two solids reach the same temperature. No more movement of heat. Thermal Equilibrium.
- 19. Example of molecular mass transfer Let’s consider two tanks with two different gases ( nitrogen & oxygen separated by a valve. When we open the valve, we will observe a decrease of concentration of nitrogen in the tank having high concentration of nitrogen and increase in concentration of nitrogen in the tank with oxygen movement of molecules from high concentration to low concentration. Non –Equilibrium and ∆C= Driving Force. When the two tanks reach 50% of each gas, there will be no more mass transfer: (Equilibrium state: thermodynamics Study)
- 21. MAIN EQUATION
- 22. CASE OF HEAT TRANSFER Driving Force: ∆T Temperature gradient
- 23. Temperature Profile Consider a block of copper where all sides are insulated so that heat conduction can occur only in the x-direction ( Figure 2.2). At time t=t0 . The initial temperature of the block = 00C Not let’s put the block between ice (00C) at the bottom and steam ( 1000C) at the top. The top temperature of the top is instantaneously ( t=t0) 1000C. Some time later ( t=t1), we will observe a temperature profile between 00C and 1000C with a parabolic shape. At t=t2, the profile will still be parabolic but flatter At t=t∞ ( steady state),the profile will become a straight line.
- 24. Observation The linear temperature profile is an experimental observation and , provided enough time is allowed, this linear temperature profile will be observed as long as the temperature of the top and bottom remain respectively at 1000C and 00C. The observation is attributed to the French scientist Fourier and the following equation is named after him: (q/A)x= -k ( ∂T/∂x)
- 25. Fourier Equation In the Fourier Equation: 1)q is the amount of heat transferred per unit time 2)A is the area 3)x is considered the only direction of the heat flux 4)k is the thermal conductivity 5)Since ( ∂T/∂x) and (q/A) have opposite sign (Heat always flows from high temperatures to low temperatures), we need a negative sign.
- 26. Class work Calculate the steady state heat flux across a copper block 10 cm thick if one side is maintained at 00C and the other side at 1000C ( See figure 2.4). We also assume constant thermal conductivity k= 380 W.m-1.K-1
- 27. Solution (q/A)x= -k ( ∂T/∂x) Integration:
- 28. The integration will become: (q/A) x. ( x2 –x1)= -k ( T2-T1) After putting the numbers ( q/A)2 = - 3.8 x 105 J.m-2.s-1
- 29. Case of Mass Transfer Let’s study the concentration profile of a liquid A in a membrane. At time t=t0 , the top of the membrane has a high concentration of liquid A (C A in moles/ m3) and, in the rest of the membrane there is no liquid A ( figure 2.2). At time t=t1, material had diffused towards the bottom and the concentration is shown in Figure 2.2 . After some time ( t=t∞) we will observe a straight line ( Like for the case of heat transfer)
- 30. Observation The observation is attributed to the scientist Fick in 1855 and the following equation is named after him: Fick’s law: (JA/A)x= -D ( ∂CA/∂x)
- 31. Fick’s law In the Fick’s Equation: 1)JA is the amount of A diffused per unit time 2)A is the area of the transfer 3)x is considered the only direction of the mass transfer 4)D is the coefficient of Diffusion or Diffusivity 5)( ∂CA/∂x) is the concentration gradient 6) (JA/A)x is the molar flux. 7)Since they have opposite sign (Mass always flows from high Concentrations to low Concentrations), we need a negative sign.
- 32. Class Work CO2 gas and air diffuse from two sides of an iron tube. The concentrations of CO2 in the two sides of the tube are respectively 0.083 kmol/m3 and 0 kmol/m3. The coefficient of diffusion of CO2 is assumed 1.56x10-3 m2s-1 . A) Find the molar flux B) Find the number of lbs of CO2 that pass the tube of iron in one hour.
- 33. The case of momentum Transfer Let’s consider a fluid maintained between two plates having an area A. The lower plate is a stationary plate and a force is applied to the upper plate to move with a velocity V. Because of the internal friction in the fluid due to viscosity, the fluid will move between the two plates at different velocities varying from V to zero. In momentum transfer, we need two coordinates: 1)coordinate x is the direction of the velocity Vx 2) coordinate y is the direction of the change of velocity
- 34. Observation As shown in Figure 2.2 ,at time t=0, the upper plate has velocity V (due to force F) and the rest of the fluid has velocity equal to zero. With time, some fluid between the plates start to move . The layer of fluid which is closest to the upper plate starts to move first and then the others. At time t =t∞ , we will have a linear velocity gradient. The following observation is named : Newton’s law
- 35. Newton’s Law F is the force applied on the upper plate A is the area of the two plates F/A is called Shear Stress which is called the momentum flux ζ Newton’s law of viscosity is : ζ yx= F/A= -μ(∂Ux/∂y)
- 36. Class Work Two parallel plates are 10 cm apart. The bottom plate is stationary and the upper plate has a velocity of 30 cm.s-1 If the fluid has a viscosity of 1 centipoise ( 0.001 kg.m-1.s-1), calculate the momentum flux
- 37. MOMENTUM, HEAT AND MASS DIFFUSIVITIES
- 38. Introduction In the previous chapter, an analogy between the molecular transports has been developed. The Fourier, Fick and Newton equations are all empirical and have the same form. The mechanism is not the same: 1) Molecular diffusion occurs in multicomponent mixtures with a concentration gradient as the driving force. 2) Momentum transfer occurs perpendicular to the direction of the flow ( the direction of the pressure drop that causes the flow). 3) Heat transfer occurs by molecular transport ( conduction) in solids, does not involve flow or relative motion of the molecules
- 39. The Different Diffusivities The respective constants in these three equations are: 1) D: Diffusivity of Coefficient of diffusion which unit is m2.s-1 2) μ: Viscosity which unit is kg.m-1.s-1 3) k: Thermal conductivity which unit is W.m-1.K-1
- 40. D: Diffusivity I) The coefficient of diffusion is the simplest property because other properties are not involved. However, it is also the most difficult to measure because: A) there should be no velocity B) the diffusivity is very small in magnitude ( 10- 5). II) Diffusivity of binary mixtures increases with temperature but not linearly and decreases with pressure because pressure reduces movements of molecules. III) They are a large differences between diffusivities of gases, liquids and solids ( Table 2.2 page 49).
- 41. Effects of temperature and pressure of the Diffusivity They are a number of equations available in the literature for estimating diffusivities in gases. D0 known at P0 and T0 (K) and the exponent n varies from 1.75 and 2.0 over a range of normal temperatures and pressures. With pressure below 5 atm, there is no concentration dependence of the coefficient of diffusion.
- 42. Class Work Predict the diffusivity of water vapor in air at 2 atm and 75 0C, if the coefficient of diffusion is 0.219x10-4m2.s-1 at 1atm and 00C and assuming that the exponent n is equal to 1.75.
- 43. k= Thermal Conductivity The two important properties of materials in heat transfer are thermal conductivity and thermal diffusivity. We are interested in thermal conductivity. Table 2.2 gives some typical values. The thermal conductivity of gases can be predicted more accurately than the thermal conductivity of liquids and solids because in gases, the energy is directly carried by molecules while for liquids and solids, other mechanisms are involved. Since heat transfer is related to molecular collisions, the thermal conductivity of liquids and solids is much larger than gases. Water vapor at 1000C has k= 0.0248 W.m-1.K-1 and water liquid at 1000C has k= 0.68 W.m-1.K-1 ( Table 2.2)
- 44. μ= Viscosity In general, the viscosity of gases increases with temperature at low pressures while that of liquids usually decreases. For gases as rigid spheres and at low pressures ( < 10 atm), the variation of viscosity is related to the square root of temperature in absolute unit ( K or R). In reality, the power varies from 0.6 to 1. In general, we have Where A and B are constants.
- 45. For Liquids The viscosity of gases is independent of pressure. Of course the kinematic viscosity of gases depends both on pressure and temperature. For liquids, the theories are not very well developed as for gases. An approximate empirical observation for the temperature is Where A and B are empirical constants. Because liquids are incompressible, the viscosity does not depend on pressure. The fluids that follow Newton’s law are called Newtonian fluids. The others are called non- Newtonian fluids.
- 46. Class Work From Table 2.2 , estimate the constants A and B for air at low pressures using the viscosity at 280K and 400 K. From Table 2.2 , estimate A and B/R for water liquid using viscosity at 373.15K and 273.15K.
- 48. Momentum transfer In fluid mechanics, the most useful equations are based on the principle of mass balance or continuity equation. The equations are first written in differential form. The be useful Multi-dimensional flow:
- 49. Dρ/Dt= accumulation ( rate of density change) ∂u/ ∂x = variation of velocity in the x direction. For one –dimensional flow: dm= ρ.u.dS ρ is the density of fluid u is the velocity in the direction of motion dS is the cross section dm is the change of mass flow.
- 50. Total flow & Mass velocity To find the total flow, we need to integrate the equation: m= ρ∫u.dS Mass velocity G= ρ.V= m/S or m=ρ.V.S where V is the average velocity
- 51. Class work Crude oil with SG= 0.887 flows through two pipes. Pipe A has a diameter of 50 mm and pipe B has a diameter of 75 mm. Each pipe C has a diameter of 38 mm. The flow through pipe A is 6.65 m3/hr. Calculate: 1) the mass flow rate in each pipe 2) the average velocity in each pipe 3) the mass velocity in each pipe
- 52. Class Work Air at 200C and 2 atm enters a tube ( diameter=50 mm ) and an average velocity of 15ms-1.It leaves through a second pipe ( diameter 65 mm) at 900C and 1.6 atm. 1) Calculate the average velocity at the outlet
- 53. Heat transfer by conduction Fourier's law is the basic law. STEADY STATE CONDUCTION: 1) In figure 10.1a, a flat wall insulated tank contains a refrigerant at – 100C while the air is at 280C. 2) The temperature falls linearly with distance across the layer as heat flows from air to refrigerant. 3) In figure 10.1.b, a similar tank contains boiling water losing heat to the atmosphere at 200C. 4) As in Figure 10.1.a, there is a temperature profile but heat flows in the reverse direction.
- 54. Calculating heat flow Rearranging Fourier's law: Since T and x are the only variables, we will have: B is the thickness of the wall.
- 55. Calculating Heat Flow Taking ( B/k) = R, we will have: q is the rate R is the thermal resistance ∆T is the driving force.
- 56. k* =Average value of k
- 57. Class work A layer of a material of 6in ( 152 mm) thick is used for insulation. The temperature of the cold side is 400F and the warm side is 1800F. The thermal conductivity of the material is 0.021 Btu/ft.h.0F at 320F and 0.032 Btu/ft.h.0F at 2000F. The area of the wall is 25 ft2. 1) Calculate the heat flow through the pipe in Btu/hr.
- 58. Resistances in Series With an insulation having different layers, we will have different resistances to heat conduction. Assuming the thicknesses: BA,BB and BC The average conductivities will be : kA *,kB * and kC * The temperature drops will be : ∆TA; ∆TB and ∆TC Therefore: ∆T= ∆TA +∆TB+ ∆TC
- 59. Heat Flow For every layer, we can write: A is the cross sectional section.
- 60. Adding the resistances, we find the following equation: R = RA + RB+ RC Ri= Bi/ki *
- 61. Class work A flat furnace is composed of : 1) a Sil o Cel layer of 4.5 in with k= 0.08 Btu/ft.h.0F 2) Common brick layer of 9 in with k= 0.8 Btu/ft.h.0F The temperature of the inner face of the wall is 14000F and that of the outer wall is 1700F. Calculate: a) Heat loss through the wall b) Temperature of the interface c) Assuming that the contact between the two layers has a resistance of 0.50 0F .h.ft2/Btu, what is the new heat loss?
- 62. Mass Transfer The most common cause of diffusion ( driving force) is a concentration gradient. The driving force tends to move a component in a direction as to equalize the concentrations ( destroy the gradient). In R.O. the activity gradient is the driving force for diffusion of water from low concentration to higher concentration of water. Diffusion can be at molecular transfer through stagnant layers. Eddy diffusion is related to mixing ( Macroscopic level)
- 63. Theory of Diffusion As we have studied, there are similarities (analogy) between conduction of heat and transfer of mass by diffusion. The driving force in heat transfer is temperature gradient and in diffusion is concentration gradient. Differences between mass and heat transfer result from the fact that heat is not a substance but energy in transit whereas diffusion is the physical flow of material.
- 64. Four types of situations 1) Only one component A of the mixture is transferred to or from the interface ( Ex: Absorption of CO2 in amine solution) The diffusion of component A is balanced by an equal and opposite diffusion of a component B ( Ex: Distillation: There is no net volume flow in the gas A and B diffuse in the opposite direction with different flows ( from a surface of a catalyst) Different components diffuse in the same direction with different rates ( membrane
- 65. Diffusion quantities 1) velocity u defined usually as length.time-1 2) Flux across a plane N (mol.area-1 .time- 1 ) 3) Flux relative to a plane of zero velocity J ( mol.area-1 .time-1) 4) Concentration c and molar density ρM 5) concentration gradient dc/dt where b is the length of the path perpendicular to the area across which diffusion occurs.
- 66. Velocities , Molar flow, Flux Several velocities are needed to describe the movements of individual substances and the total phase. Since absolute motion has no meaning, any velocity should have a reference state of rest. The total molal flow N( moles by unit time and area) N= ρM .u0
- 67. ρM is the molar density of the mixture u0 is the volumetric average velocity For components A & B: NA= cA.uA NB= cB.uB Diffusivities are defined relatively to global movement: JA= cA.uA - cA.u0 = cA( uA-u0) JB= cB.uB - cB.u0 = cB( uB-u0)
- 68. Diffusion Flux: J Diffusion of A in B Diffusion of B in A: For ideal gases: DA,B= DB,A
- 69. For Ideal Gases Since molar density does not depend on composition: CA+CB= ρM= P/RT At constant pressure and temperature: dCA +dCB = dρM= 0 For equi-molar diffusion, we can write NA= JA= ( DA,B. ρM/∆x) ( yA,1- yA,2)
- 70. For Liquids CA.MA + CB.MB= ρ= Constant MA.dCA+ MB.dCB = 0 For Liquids: DA,B= DB,A
- 71. General Equation NA= CA.uA JA = CA.uA- CA.u0 = NA- CA.u0 For gases: CA= ρM. yA and u0= N/ ρM For gases and sometimes for liquids, we can apply: NA= yA.N- DA,B. (dCA/dx)
- 72. Diffusion of one component in a mixture of gases In this condition: N= NA Therefore NA = yA.NA - DA,B. (dCA/dx) NA( 1-yA) = - DA,B. ρM (dyA/dx) Rearranging: (NA .∆x/DA,B. ρM )= -∫[dyA/(1-yA)]= ln [( 1-yA2)/( 1-yA,1)] Finally: NA= ( DA,B. ρM/∆x). ln [( 1-yA2)/( 1-yA,1)]
- 73. Class Work For a diffusion of solute A through a layer of gas to an absorbing liquid, we have : yA,1= 0.20 and yA,2= 0.10. 1) Compare the transfer rate NA between one-way diffusion and the equi-molar diffusion ( hint: calculate the ratio of the two fluxes) 2) What is the composition of A half way through the layer for one-way diffusion and equi-molar diffusion?
- 75. Introduction: Chemical Engineers are always concerned with the flow of the fluids through the pipes, tubes and channels with non circular cross-section
- 76. Shear Stress & Skin Friction in Pipes Considering a steady state flow of fluid of constant density in fully developed flow trough a horizontal pipe. From Figure 5.1 page 95: A small element of the fluid is considered as disk-shaped element of fluid with the axis of the tube, of radius r and length dL. Let the fluid pressures be p and (p+dp) respectively. Since the element has a velocity, there is a shear force FS opposing the flow will
- 77. MOMEMTUM EQUATION Applying the momentum equation: ∑ F = pa.Sa – pb.Sb + FW –Fg Where: 1) pa and pb = inlet and outlet pressures Sa and Sb = inlet and outlet cross sections. Fw= net force of wall of channel on nfluid Fg = component of force of gravity
- 78. Applying the momentum equation between the two faces of the disk (Figure 5.1): ∑ F = pa.Sa – pb.Sb + FW –Fg Since the fluid is fully developed ∑ F = 0 . pa.Sa = ∏r2p pb.Sb = ∏r2 (p+dp) FW = 2∏.r.dL.ζ Fg = 0 ( horizontal tube)
- 79. Momentum Equation ∑ F= ∏r2p - ∏r2 (p+dp)- 2∏.r.dL.ζ = 0 Dividing by ∏r2dL, we will obtain: ( dp/dL) + (2ζ /r) = 0 In steady flow, either laminar or turbulent: the equation can be applied to the entire cross section: ( dp/dL) + (2ζw /rw) = 0 where ζw is the shear stree at the wall rw is the radius of the tube.
- 80. Subtracting the two above equations, we will obtain: ζw /rw = ζ /r At r=0 ζ =0 ( Figure 5.2 page 96)
- 81. Skin friction & wall shear Bernoulli equation: ( pA /ρ)+ g.ZA + (αA .VA 2 /2) = ( pB /ρ)+ g.ZB + (αB .VB 2 /2) + hf Applying Bernoulli equation over a definite length L of the complete stream in horizontal tube, we will have : ( pA/ρ) = ( pA- ∆ps)/ρ + hfs or (∆ps/ρ) = hfs We assume :same height ZA=ZB ; same velocity: VA=VB pA-pB= ∆p= pressure drop between two points
- 82. Relation between skin friction and shear stress Combining the momentum equation and Bernoulli equation, we will obtain: hfs = (2.ζw.L)/( ρ.rw)= (4/ρ). Ζw. (L/D)
- 83. Friction factor in turbulent flow For turbulent flow, the fanny factor (f) is usually used: f= 2.ζw/ρ.V2 V2/2 is the velocity head in Bernoulli equation.
- 84. Relationship between f and fhs hfs = (2.ζw.L)/( ρ.rw)= (4/ρ). ζw. (L/D) = (∆ps/ρ) hfs= 4.f . ( L.V2/ 2D) Where: f= ( ∆ps. D)/( 2L.ρ.V2) or : (∆ps/ L)= 2f..ρ.V2/D
- 85. Friction Factor Chart For design purposes, we need a friction factor chart. The chart is a log-log chart relating the friction factor f to the Reynolds number. For turbulent flow, it is more convenient to use the following equations than the chart. 50,000 < Re< 106 f= 0.046 Re-0.2 3000 < Re< 3.106 f= 0.0014 + ( 0.125/Re0.32) Re = D.V.ρ/ μ ( Dimensionless) 2
- 86. Class work ( SI UNITS) Water flows at 500F (μ= 1.307. 10-3 N.s / m2) through a long horizontal plastic pipe having an inside diameter of 3in. 1) If the velocity is 8ft/s, calculate the pressure drop in lbf/in2 per 100 ft of length. A) Reynolds number, B) f from equation for turbulent flow or from chart, C) Apply equation. 2) If the pressure drop must be limited to 2 lbf/in2 per 100 ft of pipe, what is the
- 87. Form frictions losses Bernoulli equation: ( pA /ρ)+ g.ZA + (αA .VA 2 /2) = ( pB /ρ)+ g.ZB + (αB .VB 2 /2) + hf Form friction losses or singularities should be incorporated in the formula for friction losses as: hf = ( 4.f.(L/D) + Kc + Ke + Kf) . ( V2/2) Kc .V2/2 is the contraction loss at the entrance of the pipe Ke.V2/2 is the expansion loss at the exit of the pipe.
- 88. Class Work ( SI UNITS) Crude oil having a specific gravity of 0.93 and a viscosity of 4 cP is draining by gravity from the bottom of a tank. The depth of liquid from the draw off connection is 6 ft. the line from the draw off is 3 in diameter schedule 40 pipe. Its length is 45 m and it contains one elbow and two globe valves. The oil discharges into atmosphere 9 m below the draw off connection of the tank. 1) What flow rate can be expected through the line.
- 90. Mach Number The Mach Number ( Ma) is defined as the ratio of u, speed of the fluid, to a, speed of the sound in the fluid under the conditions of flow. Ma = u/a Sonic Fluid: When the Mach number is equal to unity. Speed of the fluid is equal speed of the sound in the same at the pressure and temperature of the fluid.
- 91. Subsonic Flow: Mach number is less than unity Supersonic Flow: Mach number is higher than unity. The most important problem in compressible fluids flow lie in the high velocity range where Mach numbers are higher to unity and flow are supersonic.
- 92. Assumptions made The flow is steady. The flow is one-dimensional. Velocity gradients within cross section are neglected. Friction only in walls Shaft work is zero Gravitational effects are negligeable Fluid is an ideal gas.
- 93. Basic relations The continuity equation Steady flow total-energy balance The mechanical energy balance with wall friction The equation of the velocity of sound The equation of state of ideal gas
- 94. Continuity equation dm= ρ.u.dS can be rewritten in the logarithmic form: ln ρ + ln u + ln S = constant Differentiating the equation: (dρ/ρ) + ( du/u) + (dS/S) = 0
- 95. Total-Energy Balance A fluid in steady flow through a system, entering at station a with velocity ua and enthalpy Ha and leaving at velocity ub and enthalpy Hb. Assuming only heat added, we will have: (Q/m)= (Hb- Ha ) + [( u2 b – u2 a ) /2]
- 96. Mechanical Energy balance After using our assumptions, we will have: (dp/ρ) + d(u2/2) + (u2.f.dL)/2.rH) = 0
- 97. Velocity of Sound The velocity of sound through a continuous material is also called the acoustical velocity. The acoustical velocity is the velocity of a very small compression wave moving adiabatically and frictionlessly through the medium. Thermodynamically, the motion of a sound wave is isentropic.
- 98. Ideal Gas Equation p= (R/M).ρ.T For a pure substance or no change in composition, we will have: (dp/p)= (dρ/ρ) + (dT/T)
- 99. Enthalpy of a gas Assuming cp independent of temperature, the enthalpy of the gas at temperature T is: H= H0 + cp ( T-T0) Where T0 is the reference temperature. The differential form is: dH= cp.dT
- 100. Acoustic velocity in ideal gas γ = cp/cv
- 101. Mach Number in Ideal Gas Asterisk conditions: The conditions were u=a ( acoustic velocity) or Ma=1 are called asterisk conditions. They are denoted p*, T*, ρ* and H*.
- 102. Stagnation Temperature The stagnation temperature of a high speed fluid is defined as the temperature of the fluid brought adiabatically to a rest position ( v=0) without shaft work. Using the total energy equation without heat exchange Q, we will obtain: Hs- H = (u2/2)= where Hs is the stagnation enthalpy where vS=0
- 103. Using the enthalpy equation: H= H0 + cp. ( T-T0), we will get: Ts= T + (u2/2cp) and Hs= H + (u2/2)
- 105. Changes of gas properties Equations to be applied for subsonic and supersonic flows
- 106. Velocity in Nozzle In absence of friction, we can rewrite the mechanical energy balance: (dp/ρ)= -d( u2/2) replacing ρ by the previous equation, we will obtain:
- 107. Mach Number and pressure for isentropic flow
- 108. Critical Pressure Ratio rc
- 109. Mass Velocity
- 110. Class work Air enters a convergent-divergent nozzle at 555.6K and 20 atm. The throat area is one- half that of the discharge of divergent section. Assume isentropic flow. 1) Assume Mach number at the throat 0.8, what are the values of the following quantities at the throat: p, T, linear velocity, density and mass velocity. 2) What are the values of p*, T*, u*,G* 3) Assuming supersonic flow, what is the Mach number at the discharge . γ = 1.4; M=29 , R= 82.056 . 10-3 atm.m3/kgmol.K
- 112. INTRODUCTION Heat transfer from a warmer fluid to a cooler fluid, usually through a solid wall, is common in chemical engineering. Latent heat: heat transfer causing phase change such as condensers and evaporators. Sensible heat: rise in temperature in solid, liquid or gas phases without change of phase. Equipments: we have : a) Double pipe heat exchanger ( Figure 11.3 page 317) b) Shell and tube heat exchanger ( Figure 11.1 page 316)
- 113. Counter-Current & Parallel Flow Parallel Flow: When the two fluids enter at the same end of the heat exchanger ( Figure 11.4 b page 318) Counter-current Flow: When the two fluids enter at different ends of the heat exchanger ( Figure 11.4.a page 318)
- 114. Energy Balances ( steady state) Rates of heat transfer are based on energy balances. In heat exchangers, we can neglect any shaft work, mechanical, potential and kinetic energies compared to the heat exchange. Therefore, the total energy conservation for each stream can be written as: m ( Hb-Ha) = q m= mass flow rate of the considered stream Hb,Ha = enthalpies by unit mass at the exit and entrance of the considered stream. q= rate of heat transfer from or to the considered stream
- 115. Energy balances For the hot stream: mh ( Hhb –Hha ) = qh For the cold stream: mc ( Hcb –Hca ) = qc Since the hot stream gives heat: qh is negative and qc is positive qh= -qc
- 116. Energy balances mh ( Hha –Hhb )= mc ( Hcb – Hca ) = q
- 117. Energy balances for sensible heat mh.cph ( Tha –Thb )= mc.cpc ( Tcb –Tca ) = q
- 118. In a Total Condenser Condensation of a saturated vapor to saturated liquid and increase of temperature of the other fluid: mh. λ = mc.cpc ( Tcb –Tca ) = q if condensate leaves condenser at temperature lower than Thb, we will have: mh.[λ + cph (Th-Thb)] = mc.cpc(Tcb –Tca)= q
- 120. DEFINITIONS Heat flux: rate of heat transfer per unit area. Heat fluxes can be based on the internal or external area. The heat flux will not be the same. Average temperature of fluid stream: If the fluid is heated, the maximum temperature is located at the wall of the tube and decreases towards the center of the stream.
- 121. DEFINITIONS Overall Heat-Transfer coefficient: In a heat exchanger, the driving force is taken as ( Th-Tc), where Th is the average temperature of the hot fluid and Tc is the average temperature of the cold fluid. The difference (Th-Tc) is the overall local temperature difference ∆T. Using Figure 11.4, ∆T changes considerably from point to point along the tube. Since the flux depends on ∆T, the flux also varies along the tube. The local flux is therefore related to the local ∆T.
- 122. Local overall heat-transfer coefficient (dq/dA)= U.∆T= U.( Th-Tc) Or: U= (dq/dA)/∆T dq/dA= local flux ∆T= local difference of temperature
- 123. DEFINITION OF U For a tubular heat exchanger, it is necessary to specify the area ( inside or outside). For the outside area A0 U0 For the inside area Ai Ui Since ∆T and dq are independent: (U0/Ui) = ( Di/D0)
- 124. AFTER INTEGRATION ln ( ∆T1/ ∆T2)= [U(∆T2- ∆T1].AT/qT qT= [U(∆T2- ∆T1)].AT/ ln ( ∆T1/ ∆T2)= U.AT.∆TL ∆TL = ∆T2- ∆T1/ ln ( ∆T1/ ∆T2)
- 125. Individual heat transfer coefficient The film coefficient for the warm and cold fluids: hh = (dq/dA)/(Th-Tw); hc = (dq/dA)/( Tw- Tc). Conduction through the wall: dq/dA= -k( dT/dy)w Eliminating dq/dA h=-k (dT/dy)w/ (T-Tw)
- 126. dq/dA0 = (Th-Tc)/ [(D0/hiDi)+ xw.D0/kmDL+ 1/h0] DL= (D0-Di)/ln(D0/Di)]
- 127. CLASS WORK Methyl alcohol flowing in the inner pipe of a double pipe exchanger is cooled with water flowing in the jacket. The inner pipe is 1in ( 25mm) Schedule 40 steel pipe . The thermal conductivity could be taken from table 11.1 page 331. thermal conductivity of steel is 45 W/m.0C. 1) What is the overall coefficient based on the outside area of the inner pipe? 1in Schedule 40 : Do= 1.315 in ; Di= 1.049 in; thickness w= 0.133 in
- 128. Class work Aniline is to be cooled from 200 to 150F in a double-pipe heat exchanger having an area of 70ft2. Toluene is used for cooling at 8600lb/h at temperature of 100F. The heat exchanger consists of 1.25 in and 2 in schedule 40 pipe. The aniline flow is 10000lb/hr. 1) WHAT IS THE OUTLET TEMPERATURE OF TOLUENE, LMTD AND U FOR COUNTERCURRENT AND PARALLEL FLOWS.
- 130. CLASSIFICATIONS I) Convection can be classified as natural or forced depending on how the fluid motion is initiated. a) In forced convection; the fluid is forced to move over a surface or in a pipe by external means such as a pump or a fan. b) In natural convection, any fluid motion is caused by natural means such as buoyancy effect, which manifests itself as the rise of warmer fluid and the fall of cooler fluid.
- 131. II) Convection can also be classified as external or internal: a) External: Fluid forced to flow over a surface b) Internal: Fluid forced to flow inside a pipe
- 132. Heat transfer by convection Heat transfer by convection is similar to heat transfer by conduction because both mechanism require a presence of material medium. However, conduction requires a solid surface but convection requires a fluid in motion.
- 133. Newton’s law of cooling Despite the complexity of convection, the rate of heat transfer is observed to be proportional to the temperature difference and is expressed by the Newton’s law of cooling: Qconv= h.AS(TS-T∞) h= convection heat transfer coefficient ( W/m2.0C) AS= Surface area of heat transfer m2 TS= Temperature of the surface 0C T∞ = Temperature of the fluid far from the surface
- 134. No Slip Condition? Observations show that fluid in motion comes to a complete stop at the surface and zero velocity is assumed. Therefore, the fluid in direct contact with the surface sticks to the surface due to viscous effects and there is no slip. This is known as No-Slip Condition. An implication of no-slip condition is that heat transfer from the surface to the fluid is pure conduction and Fourier's law is applied. qconv=qcond= -k ( dT/dy)y=0
- 135. H=convection heat ransfer coefficient Equating the two equations, we will obtain: h= - k [( dT/dy)y=0/ ( TS-T∞)]
- 136. Nusselt Number? A dimensionless number that measures the effects of heat convection on heat conduction: Nu= h.LC/ k Lc= is a characteristic length. 1) High Nusselt number means convection more important than conduction. 2) Low Nusselt numbers means conduction more important that convection.
- 137. Nusselt Number qconv= h.∆T and qcond= k. ∆T/L (qconv/ qcond)= h.∆T.L/k.∆T= hL/k= Nu
- 138. Viscous & inviscid regions of flow Convection heat transfer is related to fluid mechanics and fluid flows need to be included. Two layers of a fluid in motion create friction or an internal resistance to flow called viscosity. There is no fluid with zero viscosity. Neglecting the viscous effects, we call the flow inviscid.
- 139. Internal & external Flow Internal flow: Fluid forced to flow in confined channel. Flow completely bounded by solid surfaces. Viscous effects important. External flow: fluid forced to flow over a surface. In external flow, viscous effects are limited
- 140. Compressible/incompressible fluids Compressible fluids (Gases): important density change with pressure. A pressure change of 0.01 atm can cause density of air changes by 1%. Of atmospheric air. Incompressible fluids ( Liquids): almost no density change with pressure. For example, a pressure of 210 atm causes the density change of water at 1 atm by just 1%. Incompressible gases approximation depends on the Mach Number Ma= u/a
- 141. Condition for Incompressible gases If density changes by a maximum of 5% which is usually the case at Ma < 0.3 Incompressibility effects of air can be neglected at speeds lower than 100 m/s.
- 142. Laminar & Turbulent Flows Laminar flow is smooth and parabolic because of low velocities ( high viscosity liquids). Turbulent is chaotic because of high velocities ( low viscosity gases).
- 143. Steady & unsteady state flows Steady means no change at a point with time dx/dt=0 Unsteady means change at a point with time dx/dt ≠ 0
- 144. Velocity boundary layer Velocity boundary layer develops when a fluid flows over a surface as a result of the fluid layer adjacent to the surface assuming zero velocity. Considering the movement of a flow over a flat plate. Fluid flows with velocity V Because of No-slip condition, velocity at the surface is zero. The velocity boundary layer is defined as a thickness δ from the surface, where the local velocity u = 0.99V.
- 145. Surface shear stress Newton’s law: ζS = -μ (du/dy)y=o
- 146. Thermal Boundary Layer In the same way as Velocity Boundary layer, Thermal boundary layer develops when a fluid at a specified temperature T∞ flows over a surface at different temperature TS Thermal boundary layer δT is defined as the position in the fluid from the surface where T-TS= 0.99 ( T-T∞ ) If TS=0 T=0.99. T∞
- 147. Prandtl Number The relative thickness of the velocity and the thermal boundary layers is best described by the dimensionless parameter Prandtl number defined as : Pr= ( molecular diffusivity of momentum)/ molecular diffusivity of heat) Pr= ν/α= μcp/k
- 148. Reynolds Number By definition: NRe= Inertia forces/ Viscous forces Inertia forces ρV2L2 Viscous forces: VL/ν with ν= μ/ρ Nre= ρVL/ μ Laminar flows at low Reynolds numbers Turbulent flows at high Reynolds numbers.
- 149. Continuity equation=Mass balance X-axis flow direction and y-axis the normal direction: ∂u/∂x + ∂v/∂y =0 if the flow is parallel then v=0 One-direction flow variation ∂u/∂x =0 u= f (y)
- 150. Momentum Equation We will study one –direction ( x-direction) ρ( u. ∂u/dx +v. ∂u/∂y) = μ∂2u/ ∂y2- ∂P/∂x
- 151. Energy Equation Steady state accumulation equal zero One direction 0= k∂2T/ ∂y2 +μ( ∂u/∂y)2 kd2T/dy2= -μ ( V/L)2
- 153. INTRODUCTION Heat transfer is governed by three distinct mechanisms: convection, conduction, and radiation. Unlike convection or conduction, heat transfer through radiation does not occur through a particular medium. To understand this phenomenon one must enter into the atomic or quantum realm. All atoms, at finite temperatures, are continuously in motion. Consequently, it may be understood that the mechanism of radiation is derived from the energetic vibrations and oscillations of these atomic particles, namely electrons.
- 154. METAL WORK BY HEAT
- 156. THEORY OF RADIATION James Clark Maxwell established in 1864 the theory of radiation. Thermal radiation is a direct result of the movements of atoms and molecules in a material. Since these atoms and molecules are composed of charged particles (protons and electrons), their movements result in the emission of electromagnetic radiation, which carries energy away from the surface. Since the amount of emitted radiation increases with increasing temperature, a net transfer of energy from higher temperatures to lower temperatures results.
- 157. Electromagnetic radiation Electromagnetic radiation or waves transport energy and all electromagnetic waves travel at the speed of light in a vacuum, which is c0= 2.9979. 108m/s. Electromagnetic waves are characterized by their frequency ν or wavelength λ.
- 158. C= SPEED OF PROPAGATION OF HEAT The speed of propagation of a wave in a medium is defined by : λ=c/ν the speed of propagation in a medium is related to the speed of light in a vacuum by: c=c0/n where n is the index of refraction of that medium. The refractive index is essentially: a)unity for air and most gases. b)1.5 for glass c)1.33 for water
- 159. ν=frequency of an electromagnetic wave The commonly used unit of wavelength is micrometer (μm). Unlike wavelength (λ) and speed of propagation (c), the frequency of an electromagnetic wave (ν) depends only on the source and independent on the medium through which the wave travels. the frequency ( oscillations per second) of an electromagnetic wave can range from less than a million (106) Hz to a septilion (1024) Hz or higher.
- 160. Max Plank: Energy of Photons In 1900, Max Planck proposed that Electromagnetic radiation is considered as the propagation of a collection of discrete packets of energy called photons or quanta. Each photon of frequency ν is considered to have an energy of : e= h.ν= h.c/λ h is the Plank’s constant = 6.626069 10-34J.s
- 161. Thermal Radiation Thermal radiation is continuously emitted by all matter whose temperature is above absolute zero. Thermal radiation is also defined as the portion of the electromagnetic spectrum that extends from about 0.1 to 100 μm. Thermal radiation includes all visible and Infrared (IR) radiation as well as a portion of ultraviolet (UV) radiation. Light is the visible portion of electromagnetic waves from 0.4 to 0.76 μm. Solar radiation falls in the range 0.3-3 μm
- 162. BLACKBODY RADIATION A blackbody is an idealized body to serve as standard against which the radiative properties of real surfaces may be compared. A blackbody is defined as perfect emitter and absorber of radiation. At a specified temperature and wavelength, no surface can emit more energy than a blackbody. A blackbody absorbs all incident radiation. a blackbody emits radiation uniformly in all directions.
- 163. RADIATION EMITTED BY A BLACKBODY In 1879, Joseph Stefan determined experimentally that the radiation emitted by a blackbody per unit time and unit surface can be described by the blackbody emisive power Eb(T) : Eb(T)= σ.T4 The Stefan-Boltzmann Constant σ = 5.67 108 W/m2.K4 Equation theoretically verified in 1884 by Ludwig Boltzmann.
- 164. Definition of Large capacity body Another type of body that closely resembles to a blackbody is a large capacity with a small opening ( Figure 12.8, page 667). Radiation coming through the small opening undergoes multiple reflections and thus it has several chances to be absorbed by the interior surfaces of the cavity before any part of it can escape.
- 165. Spectral blackbody emissive power It is defined as the amount of radiation energy emitted by a blackbody at a thermodynamic temperature T per unit time and unit area and per unit wavelength about the wavelength λ. Plank’s Law:
- 166. Plank’s Law Eb,λ (λ,T) is in ( W/m2.μm) C1= 2.π.h.c0 2= 3.74177 .108 W.μm4/m2 C2= hc0/k = 1.43878 .104 μm.K K ( Boltzmann’s Constant)= 1.38065. 10-23 J/K T is the absolute temperature of the surface λ =wavelength of the radiation emitted
- 168. From the figure The emitted radiation is a continuous function of wavelength At a specific temperature, it increases with wavelength, reaches a peak and then decreases with increasing wavelength. At any wavelength, the amount of emitted radiation increases with temperature. Radiation emitted by sun , blackbody, at 5780 K reaches its peak in the visible region.
- 169. Maximum Power As the temperature increases, the peak of the curves shifts to the shorter wavelength. The wavelength at which the peak occurs for a specified temperature is given by Wien’s displacement law: (λ.T)max.power= 2897.8 μm.K The peak for solar radiation occurs at 2897.8/ 5780 = 0.50 μm. The peak of radiation emitted by a surface at room temperature ( 298K) is 2897.8/298 = 9.7 μm
- 170. Total blackbody emissive power Integration is needed of the spectral blackbody emissive power Eb over the entire wavelength spectrum:
- 171. fλ = Blackbody radiation function fλ is useful because it is difficult to perform the integration knowing that each time we need the value Eb,0-λ fλ is dimensionless
- 172. Fλ1- λ2 = fλ2 (T)- fλ1 (T)
- 173. Class Work Example 12-1 and 12.2
- 175. Fick’s law Proposed in 1855, the law states that the rate of diffusion of a chemical species at a location in a gas mixture ( or liquid, or solid) is proportional to the concentration gradient of that species at that location. NA= - DAB.dCA/dx
- 176. Mass Basis Partial density of the species (i): ρi=mi/V Total density: ρ= m/V= Σ(mi)/V= Σρi Mass Fraction of (i)= ωi= mi/m= (mi/V)/(m/V)= ρi/ ρ ωi = ρi/ ρ Σωi=1
- 177. Molar Basis Partial molar concentration: Ci= Ni/V Total molar concentration: C=N/V Molar Fraction: yi= Ni/N= (Ni/V)/(N/V) yi= Ci/C Σyi=1
- 178. Relation between m and N m= N.M ρ = m/V= N.M/V C=N/V ρ= C.M & ρi=Ci.Mi
- 179. Molecular weight of the mixture M= m/N= ΣNi.Mi/N= Σ (Ni/N).Mi M= Σ yi.Mi
- 180. Relation between ωi and yi ωi=ρi/ρ = Ci.Mi/C.M = (Ci/C). ( Mi/M) ωi= yi.(Mi/M)
- 181. Ideal gas mixture yi= Ni/N= Pi/P
- 182. Boundary Conditions 1) At the surface of water, the partial pressure of water is the vapor pressure of water at the surface conditions. 2) using Henry’s law for ideal gas mixtures, the composition of the gas in the liquid side is defined by: xi= Pi/H Using Raoult’s law: Pi=yi.P= xi.Psat
- 183. Boundary Conditions The concentration of a gas in a solid is given by the solubility (S) : Ci= S.Pi
- 184. Stationary binary medium Mass Basis: ji= -ρ.Dij.(dωi/dx) kg/s.m2 ji= -Dij.(dρi/dx) at constant ρ Molar basis: Ji= -C.Dij.(dyi/dx) kmol/s.m2 Ji= -Dij.(dCi/dx) at constant C
- 185. Diffusivity D Because of the complex nature of mass diffusion, the diffusivity is determined experimentally: For water and air, Marrero and Mason (1972) proposed:280K< T< 450 K DH2O,Air= 1.87.10-10. T2.072/P (m/s) In general: DA,B,1/DAB,2= ( P2/P1). (T1/T2)1.5
- 186. Steady state diffusion through a wall, cylinder and sphere Knowing the boundary conditions: I) Wall m=ρ.Dij.A. (ωi1-ωi2)/L=Dij.A.(ρi1-ρi2) (kg/s) m= (ωi1-ωi2)/(L/ρ.Dij.A) or (ωi1-ωi2)/Rwall With Rwall= (L/ρ.Dij.A) II) Cylinder: m=2πL.ρ.Dij. (ωi1-ωi2)/ln(r2/r1) m =2πL.Dij.(ρi1-ρi2)/ln(r2/r1) (kg/s)
- 187. III) Sphere: m=2πr1r2 .ρ.Dij. (ωi1-ωi2)/(r2/r1) m =2πr1r2 .Dij.(ρi1-ρi2)/(r2/r1) (kg/s) SIMILAR EQUATIONS CAN BE FOUND FOR MOLAR FLOWS REPLACING ρ BY C AND ω BY y