Daniel López Gaxiola 1 Student View Jason M. Keith

Transcription

1 Suppleental Material for Transport Process and Separation Process Principles Chapter Principles of Moentu Transfer and Overall Balances In fuel cells, the fuel is usually in gas or liquid phase. Thus, the student ust be failiar with the principles of fluid echanics or oentu transfer, which will be covered in the following proble odules. In later sections of this chapter, situations cobining oentu transfer and heat transfer will be illustrated..- Conversion of Pressure to ead of a Fluid.-1 Diffusivity of ydrogen inside a Fuel Cell.5-1 Reynolds Nuber of ydrogen flowing into a Fuel Cell.6-1 Flow of ydrogen into Fuel Cells.6- Velocity of ydrogen in Bipolar Plate Channel.7-1 Energy Balance on Ethanol boiler.10-1 Methanol Flow in Fuel Cell.10- Use of Friction Factor in Lainar Flow.10- Use of Friction Factor in Turbulent Flow.10-4 Trial-and-Error Solution to Calculate Pipe Diaeter.10-5 Flow of Gas in Line and Pressure Drop.10-8 Entry Length for a Fluid in a Rectangular Channel.11-1 Copressible Flow of a Gas in a Pipe Line.11- Maxiu Flow for Copressible Flow of a Gas Daniel López Gaxiola 1 Student View

2 Principles of Moentu Transfer and Overall Balances Exaple.-: Conversion of Pressure to ead of a Fluid ydrogen is stored in a copressed gas tank with a volue of 50 L at a pressure of 140 at and a teperature of 5 C. What is the pressure in the tank in g and inches of water? Strategy To deterine the pressure of hydrogen as head of a fluid we need to use the equation for calculating pressure of a fluid as a function of height. Solution The equation for pressure as a function of height is shown below. where: P ρgh ρ density of fluid g acceleration due to gravitational force h head or height of fluid We can solve this equation for the head h to yield: P h ρg To deterine the pressure in of g we need to use the density of ercury, which can be obtained fro Table -1 Perry s Cheical Engineers andbook, 8 th Edition to be: ρ g Since the units in the expression for the head of a fluid need to atch, the value of g ust be in s and the pressure ust be converted to Pa N. fter entering the values into the equation for h we get: N 140 at s 1 at 1 N h 9.81 s Daniel López Gaxiola Student View

3 Suppleental Material for Transport Process and Separation Process Principles h g This value can be converted to of g by ultiplying by the conversion factor fro to : h g h g To calculate the head in inches of water, we need to follow a siilar procedure. The difference is that we need to use the density of water instead of the density of ercury, and use a conversion factor to convert eters to inches. Thus, N 140 at s 1 at 1 N h 9.81 s h 1446 O The density of water was obtained fro Table.- of Geankoplis at a teperature of 5 C. Converting this value to inches we get: 1 in O h 1446 O O h in O Daniel López Gaxiola Student View

4 Principles of Moentu Transfer and Overall Balances Exaple.-1: Diffusivity of ydrogen inside a Fuel Cell ydrogen in a bipolar plate is diffusing in the anode side of a fuel cell. The current density of a stack of 440 fuel cells is 600. c The hydrogen is entering the fuel cell at a pressure of at and a teperature of 5 C. Deterine the diffusivity of hydrogen through the gas-diffusion layer with a thickness of 100 µ in fuel cell perforance is liited by ass transfer. s The aount of hydrogen reacted as a function of current is described by the following equation: if the n, reacted IN F where: Strategy I current in aperes () N nuber of cells in the fuel cell stack F Faraday s constant C ol e The diffusivity of hydrogen can be deterined using Fick s Law of diffusion and the definition of the consuption rate of hydrogen in ters of the current. Solution Fick s first law relates the diffusive flux to the difference in concentration of a substance. For this proble, Fick s first law is given by: dc n D, reacted dx where: D diffusivity of hydrogen through the gas diffusion layer cross sectional area of the gas diffusion layer dc change in the concentration of hydrogen along the thickness of the gas diffusion layer dx Daniel López Gaxiola 4 Student View

5 Suppleental Material for Transport Process and Separation Process Principles The left hand side of Fick s first law can be re-written in ters of the current. ence, D dc dx The proble stateent is giving the value of the current density defined as the current of the fuel cell divided by the area. Thus, this equation can be rewritten in ters of the current density I as follows: D dc dx ssuing the diffusivity reains constant along the gas-diffusion layer, this equation can be separated and integrated as follows: x µ x 0 µ x 100µ C x dx D dc Since all the hydrogen entering the fuel cell is reacting when ass transfer is liiting the fuel cell perforance, the concentration of hydrogen in the anode will be given by: C x 100µ The concentration of a substance is defined as the nuber of oles of substance in a volue of solution. Thus, C n V The concentration of hydrogen entering the channels in the bipolar plate can be obtained using ideal gas law as shown in the following steps: PV nrt Solving for the concentration C n V n V fro the ideal gas law, we have: Daniel López Gaxiola 5 Student View

6 Principles of Moentu Transfer and Overall Balances Substituting the corresponding quantities in the right side of this equation yields: C C 1015 Pa at 1 at Pa 98.15K ol K ol ( ) Substituting the concentration values into the integral equation we get: dx D dc 0 ol We can integrate this equation and substitute the values for the current density, nuber of fuel cells and Faraday s constant to give: ( cells) c ( 4 ol 1 10 ) 0 D C ol e Finally we can solve for the diffusivity D to obtain: 1 C/s c ( cells) c D ( 1 10 ) C ol ol e D s This diffusivity value can be converted to. s by ultiplying it by a conversion factor fro to 5 D s D s Daniel López Gaxiola 6 Student View

7 Suppleental Material for Transport Process and Separation Process Principles Exaple.5-1: Reynolds Nuber of ydrogen flowing into a Fuel Cell copressed gas tank contains hydrogen at roo teperature and a pressure of 140 at. The valve on this tank is opened and the hydrogen enters a fuel cell at a rate of 0.75 through a steel pipe hr with an inner diaeter of Deterine the type of pipe connecting the fuel cell to the hydrogen tank and deterine if the flow of hydrogen is lainar or turbulent. Strategy To deterine the type of pipe used to connect the hydrogen tank to the fuel cell we need to find the pipe corresponding to the inner diaeter given in the proble stateent. The flow regie ay be deterined depending on the value of the Reynolds nuber. Solution ppendix.5 of Geankoplis is showing the properties of different types of standard steel pipes. For the inner diaeter of 5.46, the noinal size of the pipe is 1 in. with a Schedule Nuber of For flow inside a pipe, the Reynolds nuber is given by: Dυρ Re µ where: D inner diaeter of the pipe in eters () υ velocity of the fluid inside the pipe in s ρ density of fluid in µ viscosity of fluid in s The velocity of hydrogen circulating in the pipes can be obtained by dividing the voluetric flow rate of hydrogen by the inner cross-sectional area of the pipe. owever, since we are given the ass flow rate, we will have to calculate the voluetric flow rate using the ideal gas law. RT PV M Solving for the voluetric flow rate V yields: Daniel López Gaxiola 7 Student View

8 Principles of Moentu Transfer and Overall Balances RT V MP Substituting the corresponding quantities into this equation we get: V Pa 0.75 ( 98 K) hr ol K 1015 Pa 1 kol 140 at 1 at kol 1000 oles ( ) Note that conversion factors for the pressure and nuber of oles were used in order to get the correct units for the voluetric flow rate. V hr To calculate the velocity of hydrogen in the pipe, we also need to deterine the cross-sectional area of the pipe. This value can be obtained directly fro Table.5 of Geankoplis. Now we can obtain the velocity of hydrogen as shown in the following equations. Note that we are ultiplying the velocity equation by the conversion factor fro hours to seconds, hence to obtain the velocity in s. V υ 1 hr hr s υ 0.77 s The density of hydrogen at the pressure of 140 at and the teperature of 5 C (98 K) can be calculated using the ideal gas equation of state: RT PV M Since the density r is equal to dividing the ass flow rate by the voluetric flow rate, we can solve for the density as shown in the following step. ρ V Daniel López Gaxiola 8 Student View

9 Suppleental Material for Transport Process and Separation Process Principles We can enter the nueric values into the right side of this equation to get: 1 kol ( at ) 1 kol 1000 oles ρ Pa 8.14 ( 98 K) ol K The only paraeter left to be deterined before calculating Reynolds nuber is the viscosity, which can be obtained fro Figure.- for gases at a pressure of 1 at. owever, for hydrogen at the pressure and teperature conditions in this proble, the viscosity does not depend on pressure. Thus, it is valid to use Figure.-. To obtain the viscosity of hydrogen we need to locate the coordinates for hydrogen in Table.-8 and draw a line that passes through these coordinates and the teperature value of 5 C. ence, the viscosity will be estiated to be: µ s Now we can enter the quantities we found for velocity, density and viscosity into the equation for the Reynolds nuber to yield: Re Conclusion: s s ( ) Re Daniel López Gaxiola 9 Student View

10 Principles of Moentu Transfer and Overall Balances Exaple.6-1: Flow of ydrogen into Fuel Cells ydrogen is exiting the fuel tank in a vehicle through a stainless steel pipe (schedule nuber 80) L with a noinal diaeter of ¼. The hydrogen is leaving at a flow rate of at roo hr teperature and at a pressure of at. This flow rate is equally distributed between the 5 fuel cells required to power this vehicle. The hydrogen is being distributed to the fuel cells by a pipe of 1/16 inner diaeter. The hydrogen flow rate entering the channels of the bipolar plate of a fuel cell is equally distributed between the channels of the bipolar plate. (Note: The channels in the bipolar plate will be assued to be seicircular with an inner diaeter of 1/16 ) a) Deterine the ass flow rate of hydrogen entering each cell and each one of the channels in the bipolar plate. ydrogen Tank n 1 n 5 Strategy To deterine the ass flow rate fro the voluetric flow rate we need to use ideal gas law. Solution We can solve the ideal gas law for the ass flow rate as shown in the following steps: overall PV M overall RT Substituting the corresponding quantities into the right hand side of this equation yields: Daniel López Gaxiola 10 Student View

11 Suppleental Material for Transport Process and Separation Process Principles overall overall L g ( at) hr ol L at ol K g hr ( K) To deterine the flow rate of hydrogen entering each fuel cell, we divide the ass flow rate of hydrogen leaving the tank by the total nuber of fuel cells in the stack. Thus, fuel cell n fuel cells g hr cells fuel cell g 0.5 hr This value can by divided by (nuber of channels on each bipolar plate) to deterine the flow of hydrogen to each channel: fuel cell channel nchannels g 0.5 hr channels channel g hr b) What is the average velocity of the hydrogen leaving the tank? Strategy The velocity of hydrogen can be deterined by applying the definition of velocity in ters of the flow rate. Solution The velocity of a fluid can be deterined with the following equation: V υ First we need to deterine the voluetric flow rate of hydrogen to each fuel cell and to each channel in the bipolar plates. This can be done in a siilar way to part a) of this proble. Daniel López Gaxiola 11 Student View

12 Principles of Moentu Transfer and Overall Balances V V fuel cell fuel cell L V hr n 5 cells fuel cells L hr V V fuel cell channel nchannels L hr channels V channel L hr Now we need to calculate the cross-sectional areas of different pipes through which hydrogen is circulating. First, the area of the ¼ pipe connected to the fuel tank can be found in Table.5-1 of Geankoplis to be: 1/4" pipe The cross-sectional area of the pipe entering the fuel cells can be calculated using the equation for the area of a circle, where the diaeter will be 1/16. ence, 1/16" pipe 1/16" pipe 1 in πd π in Converting this value to, we have: 1/16" pipe 1/16" pipe 1 in in Since the channels on the bipolar plate are assued to be seicircular, the cross-sectional area of the channel can be calculated as follows: channel πd 4 Daniel López Gaxiola 1 Student View

13 Suppleental Material for Transport Process and Separation Process Principles Substituting the corresponding quantities into the right hand side of this equation yields: channel 1 π in 16 1 in 8 channel Now we can obtain the velocities in the three different sections, as shown in the following equations: L hr 1000 L υ in 1/4" pipe υ in 1/4" pipe.68 s L 1 1 hr hr 1000 L 600 s υ in fuel cell υ in fuel cell s L 1 1 hr hr 1000 L 600 s υ in channel υ in channel s c) Calculate the ass flux of hydrogen circulating through each channel in. hr Strategy The flux of hydrogen refers to the aount of hydrogen flowing through an area during a period of tie. Daniel López Gaxiola 1 Student View

14 Principles of Moentu Transfer and Overall Balances Solution The flux of hydrogen can be obtained by dividing the ass flow rate of hydrogen entering each channel, by the cross sectional area of the channel. Thus, G channel channel Entering nueric values into this equation we get: G g hr G hr Daniel López Gaxiola 14 Student View

15 Suppleental Material for Transport Process and Separation Process Principles Exaple.6-: Velocity of ydrogen in Bipolar Plate Channel The following figure is showing the flow of hydrogen along the horizontal channel of length L in a bipolar plate. x B x B y B y B y z L Reactant air channels ydrogen channels x The velocity of hydrogen along a square channel in a bipolar plate is given by the following equation: x y υ z υax 1 1 B B Deterine the average velocity of the hydrogen flowing through the channel. Strategy The average velocity can be calculated fro the expression of the velocity as a function of the position in the x and y diensions. Solution The average velocity can be calculated using Equation.6-17 of Geankoplis, shown below: υ 1 av υ z d In Cartesian coordinates d ay be written as dxdy. The cross-sectional area of the channel is obtained by ultiplying the diensions of the channel. ence, 1 B B x y B B ax 1 1 υ av υ ( B) B B dxdy Daniel López Gaxiola 15 Student View

16 Principles of Moentu Transfer and Overall Balances Integrating this equation in the x-direction, we get: B υ B ax y av x 1 4B B B B υ Evaluating the integrated expression for x yields: υ B ax y av B 1 B υ dy 4B B We can reduce this equation to get: υ B ax y υ av 1 B 4B dy B dy υ B ax y υ av 1 B 4B dy B υ B ax υ av 1 B y dy B Now we can proceed to integrate and evaluate the equation for the average velocity with respect to y as shown in the following steps: υ ax υ av υ ax υ av B B Reducing this equation we deterine the expression for the average velocity of hydrogen in the channels: υ ax υ av B υ ax υ av υ ax 4υ υ av 9 ax Daniel López Gaxiola 16 Student View

17 Suppleental Material for Transport Process and Separation Process Principles Exaple.7-1: Energy Balance on Ethanol Boiler liquid solution of ethanol and water is entering a boiler before undergoing a reforing reaction for producing hydrogen for proton-exchange ebrane fuel cells. diagra of the process is shown below. Liquid Ethanol/Water Solution υ l s kj 116 l Ethanol/Water gas ixture υ g 1.71 s kj g What is the power required to produce 1 hr of vapor? Strategy The overall energy balance for a steady-state flow syste can be used to deterine the aount of heat required. Solution The energy balance equation is given by Equation.7-10 of Geankoplis and is shown below: where: ( υ υ 1 ) + g( z z 1) Q Ws α Enthalpy of the substance exiting the syste 1 Enthalpy of the substance entering the syste α Kinetic-energy velocity correction factor υ Velocity of the substance exiting the syste υ Velocity of the substance entering the syste 1 g cceleration due to gravity z z Difference in height between the outlet and inlet points in the syste 1 Q eat added (+) or reoved ( ) to the syste W s External work applied by (+) or to ( ) the syste Daniel López Gaxiola 17 Student View

18 Principles of Moentu Transfer and Overall Balances For this proble we will assue turbulent flow, so the value of the kinetic-energy velocity correction factor α is close to 1. The external work ter W s will be neglected in the balance equations as there are no echanical parts perforing work to the syste. Since no inforation is given about the height of the inlet and outlet points in the syste, we will assue the values are the sae. Thus, the difference z z is equal to zero. 1 fter applying these assuptions, the energy balance equation will be given by: 1 Q + 1 ( ) Where the tilde represents the aount per unit ass. The values given for the ass enthalpies of the ethanol/water ixtures and the velocities of the liquid and gas can be now substituted into this equation to obtain the aount of heat required. Thus, 1 Q + 1 ( ) kj kj 1 Q s s kj Q 4 To deterine the power required, we need to ultiply the heat per kilogra of vapor by the flow rate: kj Q 4 hr Q kw Daniel López Gaxiola 18 Student View

19 Suppleental Material for Transport Process and Separation Process Principles Exaple.10-1: Methanol Flow in Fuel Cell solution of 40 wt. % ethanol and 60 wt. % water is entering the channels of a bipolar plate in a direct ethanol fuel cell. The viscosity and density of this solution are shown below: µ Pa s ρ 91.5 The pressure drop along a rectangular channel is shown by Bahrai et al. [1] to be: where: P c µ Lυ 1 64ε π tanh 5 π ε P Pressure drop, Pa µ Viscosity of the fluid, Pa s L Length of the channel, υ Velocity of the fluid in the channel, s The paraeters c and ε are related to the diensions of the channel as shown in the following figure: Gas-Diffusion Layer c 1 b 1 L 0 c ε b What is the pressure drop in at and g along a single channel if the ethanol flowing in the fuel cell has a Reynolds nuber of 00? Strategy The pressure drop in the channel can be calculated using the equation for the pressure drop given in the proble stateent. 1. Bahrai, M., Yovanovich, M. M., Culha, J. R., Pressure Drop of Fully-Developed, Lainar Flow in Microchannels of rbitrary Cross Section, Journal of Fluids Engineering, 18, (006) Daniel López Gaxiola 19 Student View

20 Principles of Moentu Transfer and Overall Balances Solution The equation given in the proble stateent for the pressure drop depends on the velocity of the fluid in the channel. Since we are not given this value directly, we will have to calculate it fro the definition of the Reynolds nuber. For an open channel, the Reynolds nuber is defined as follows: Re D µ υρ where D is the hydraulic diaeter, given by: D 4 P In this equation, is the cross-sectional area of the channel, and P is the perieter of the channel in contact with the fluid. Since the fluid in the channel is in contact with the gas diffusion layer, we need to consider all the diensions of the channel when calculating the wetted perieter. ence, the hydraulic diaeter can be calculated using the diensions of the channel as shown in the following step: D ( ) 4 b c Substituting the values of b and c into this equation yields: D D ( ) Now we can enter this value for the hydraulic diaeter into the equation for Reynolds nuber and solve for the velocity to get: Reµ s υ D ρ ( 1 10 ) 91.5 υ s The only reaining values that need to be calculated before being able to deterine the pressure drop are the paraeters b, c and ε : Daniel López Gaxiola 0 Student View

21 Suppleental Material for Transport Process and Separation Process Principles b c 1 b c 1000 c ε 4 b 5 10 ε Finally, the pressure drop along the length of the channel of 0 can be calculated as follows: ( 0.0 ) s s 1 at P 1 64( 1) π 1015 Pa ( ) tanh 5 π ( 1) P at Converting this value to g, we have: P at P g Daniel López Gaxiola 1 Student View

22 Principles of Moentu Transfer and Overall Balances Exaple.10-: Use of Friction Factor in Lainar Flow n aqueous solution of 40 ol % ethanol is flowing along the channels in a bipolar plate of a direct-ethanol fuel cell at a velocity of and a teperature of 7 C. s Bahrai et al.[1] are showing the following equation to deterine the friction factor in a rectangular channel: where: f Re 1 19ε π 1 tanh 5 ( 1+ ε) ε π ε Re υρ µ The paraeter ε is a function of the diensions of the channel, as shown in the following figure. Gas-Diffusion Layer c 1 b 1 L 0 c ε b Use the definition of friction factor to calculate the pressure drop in at along a single channel. Strategy The pressure drop in the channel can be calculated using the equation for the friction factor given in Geankoplis. Solution Equation.10-4 of Geankoplis gives the definition of Fanning friction factor: f P w ρυ 1. Bahrai, M., Yovanovich, M. M., Culha, J. R., Pressure Drop of Fully-Developed, Lainar Flow in Microchannels of rbitrary Cross Section, Journal of Fluids Engineering, 18, (006) Daniel López Gaxiola Student View

23 Suppleental Material for Transport Process and Separation Process Principles where: Cross sectional area of the channel. w Surface area of the channel being wetted by the fluid. P Pressure drop along the channel of length L ρ Density of the fluid circulating through the channel υ Velocity of the fluid in the channel We can start by solving for the pressure drop Equation.10-4 of Geankoplis. P fro the definition of friction factor, given in P For the rectangular channel in the bipolar plate, the cross sectional area and the wetted surface area are given by: ( b)( c) ( ) w L Note that for the wetted perieter we are considering the area of the channel in contact with the gas-diffusion layer. Since b c, we can rewrite the area and the wetted perieter as follows and substitute the diensions of the channel in these equations to get: 1 4 ( ) 1 w 8 ( )( ) the value of ε and enter the obtained value into the equation for f Re. Thus, ε c 0.5 b Daniel López Gaxiola Student View

24 Principles of Moentu Transfer and Overall Balances f Re 1 19 π 1 tanh 5 ( 1 + ) π f Re The Fanning friction factor f can be obtained fro this equation but first we need to calculate the Reynolds nuber Re using the equation given in the proble stateent as shown in the following steps: Re υρ µ ( ) The density of ethanol at the teperature of 7 C can be obtained fro Table -4 of Perry s Cheical Engineers andbook, 8 th Edition to be: ρ C O ol L Converting this value to, we have: ol g C O 1 L ρ C O L 1 ol C O 1000 g 1 ρ C O Since the ethanol entering the fuel cell is diluted in water, the density will depend on the concentration of the ethanol solution. Thus, the density can be calculated as shown below: ρ ρ + ρ C O O The density of water is given in Table.- of Geankoplis. Since the value of the density at the teperature of 7 C, we can use the value in the Table at the teperature closest to 7 C, which is 5 C: ρ O Daniel López Gaxiola 4 Student View

25 Suppleental Material for Transport Process and Separation Process Principles Substituting the individual densities of water and ethanol into the equation for the density of the solution ρ yields: ρ ρ The viscosity of ethanol can be deterined using Figure.-4 of Geankoplis and the coordinates in Table.-1. Thus, µ s Now we can obtain the value of the Reynolds nuber Re Re as follows: ( ) ( ) υρ s µ s Re Fro the value calculated for f Re we can solve for the Fanning friction factor to yield: f f Re Finally we can deterine the value of the pressure drop along the channel in the direct-ethanol fuel cell by entering all the corresponding values into the equation for P ( ) P s 1 at P Pa Pa P at Daniel López Gaxiola 5 Student View

26 Principles of Moentu Transfer and Overall Balances Exaple.10-: Use of Friction Factor in Turbulent Flow Pure hydrogen is exiting a pressure swing adsorption unit at a rate of 6.1 hr and a pressure of 061 kpa. The hydrogen is entering the storage tank through a pipe with an inner diaeter of 100 J and a length of 10. If the axiu friction loss peritted along the pipe is.60, what aterial would you propose for the pipe transporting the hydrogen fro the pressure swing adsorption unit to the storage tank? The gas leaving the pressure swing adsorption unit has the following properties: µ Pa s ρ 1.45 Strategy To deterine what type of pipe to use in this process, we need to deterine the paraeter ε for the flow conditions in this process. Solution The paraeter D ε ay be obtained fro Figure.10- of Geankoplis. To do this, we need to know the values of the Reynolds nuber and the Fanning friction factor. The friction factor f can be calculated fro the definition of the friction loss, as shown in the following equation: L υ F 4f f D Solving for the Fanning friction factor f, we get: f The value of the velocity of hydrogen inside the pipe is not given. owever, its value can be calculated by dividing the voluetric flow rate of hydrogen by the cross-sectional area of the pipe. ence, V υ Daniel López Gaxiola 6 Student View

27 Suppleental Material for Transport Process and Separation Process Principles The voluetric flow rate is calculated by dividing the ass flow rate by the density. The calculation of the voluetric flow rate and the cross-sectional area of the pipe are shown in the following steps: ( 0.1 ) πd π hr 1 hr υ ρ 600 s 1.45 ( ) υ s Substituting this velocity value into the equation for the friction factor we obtained, we have: f ε The other paraeter required to locate the paraeter in Figure.10- is the Reynolds nuber, D which is obtained as follows: Dυρ Re µ Substituting the corresponding values into this equation yields: Re s ( 0.1 ) 1.45 Re s Now we can locate the value of D ε in the graph for the Reynolds nuber and the friction factor. Fro Figure.10-, this paraeter is estiated to be: ε D Daniel López Gaxiola 7 Student View

28 Principles of Moentu Transfer and Overall Balances Solving for the pipe roughness ε, we get: Conclusion: ε D ( 0.1 ) ε Daniel López Gaxiola 8 Student View

29 Suppleental Material for Transport Process and Separation Process Principles Exaple.10-4: Trial-and-Error Solution to Calculate Pipe Diaeter ydrogen in a fuel cell vehicle is flowing fro the tank to the fuel cell stack through a coercial g steel pipe with a length of. The hydrogen is being consued at a rate of 1.17 and is entering s the fuel cell stack at a teperature of 5 C and a pressure of.5 at. Deterine the diaeter of the pipe connecting the fuel tank to the fuel cell stack if a head of 40 in g is available to copensate for the friction loss. Strategy Fro the value of F f given as head of fluid we can deterine the required diaeter for feeding hydrogen to the fuel cells. Solution The friction loss can be calculated by ultiplying the head of fluid in of O by the acceleration due to gravity. Thus, O F 40 in g s 1 in g f F f J To deterine the pipe diaeter we have to obtain the diaeter fro the definition of friction force, given by the following equation: L υ F 4f f D Solving for the diaeter D, we have: L υ D 4f F f s it can be seen in this equation, we need to obtain the Fanning friction factor f and the velocity υbefore being able to calculate the diaeter. The friction factor can be obtained fro Figure.10- of Geankoplis. To do this, we need to calculate Reynolds nuber first and obtain the relative roughness D ε of coercial steel. ence, ε D D Daniel López Gaxiola 9 Student View

30 Principles of Moentu Transfer and Overall Balances Before obtaining the velocity of the fluid first we need to convert the ass flow rate to voluetric flow rate. To do this, we need to calculate the density of the fluid using the ideal gas equation of state. The velocity can then be deterined by dividing the voluetric flow rate of hydrogen by the cross-sectional area of the pipe as shown in the following steps: g 1 (.5 at) ol PM 1000 g ρ RT L at 1 ( K) ol K 1000 L g s 1000 g V ρ s πd 4 4 V s υ s πd υ D The other paraeter needed to obtain the friction factor fro Figure.10- is the Reynolds nuber, calculated as follows: D( ) Dυρ D s Re µ s 169. Re D The viscosity was obtained fro ppendix. of Geankoplis. s we can see, the diaeter of the pipe D appears in all the expressions required for deterining the friction factor. Therefore, we will select a diaeter value and copare the result obtained to the calculated friction force. The initial guess for the diaeter will be: Daniel López Gaxiola 0 Student View

31 Suppleental Material for Transport Process and Separation Process Principles Trial 1 D Substituting this value in the equations for Reynolds nuber, velocity and relative roughness, we get: Re D υ D (0.010) s ε D D (0.010 ) Locating the values of the relative roughness and the Reynolds nuber in Figure.10- we find: f Substituting this result as well as the velocity and diensions of the pipe into the equation for the friction force we get: L υ F 4f f D ( ) s F f F f J It can be seen that this value does not atch the calculated friction force. For a second trial, we will select a diaeter of Selecting a higher value for the diaeter, will result in an decrease in the velocity of the fluid, thus reducing the value of the friction force. Trial D 0.00 Substituting this value in the equations for Reynolds nuber, velocity and relative roughness, we get: Re D 0.00 Daniel López Gaxiola 1 Student View

32 Principles of Moentu Transfer and Overall Balances υ D (0.00) s ε D D (0.00 ) f F f F f ( ) 4 J s 0.00 s we can see, the value obtained for a diaeter of 0.00 yields a friction force of J. The error between this value and the friction force calculated using the head of fluid available is given by: J J Error % (100%) J Conclusion: Error % Daniel López Gaxiola Student View

33 Suppleental Material for Transport Process and Separation Process Principles Exaple.10-5: Flow of Gas in Line and Pressure Drop ydrogen at roo teperature and standard pressure is entering a fuel cell stack through a sooth pipe with an inner diaeter of 1 c and a length of. Calculate the pressure of hydrogen in the g fuel tank. The hydrogen consuption rate is.05. s Strategy The initial pressure of the gas can be deterined using the equation for the pressure drop of a gas inside a tube. Solution We will start by assuing turbulent flow of hydrogen in the pipe. The initial pressure of hydrogen can be calculated using Equation of Geankoplis shown below: P P 4f 1 DM L G RT where: P 1 Initial pressure of the fluid in the pipe, Pa P Final pressure of the fluid in the pipe, Pa f Fanning friction factor L Length of the tube, G Mass flux of the fluid in the pipe, s R Ideal gas constant, Pa ol K T Teperature of the fluid, K D Inside diaeter of the pipe, M Molecular weight of the fluid, g ol The flux of hydrogen ay be calculated by dividing the ass flow rate of hydrogen by the crosssectional area of the pipe as shown in the following steps. G Daniel López Gaxiola Student View

34 Principles of Moentu Transfer and Overall Balances ( ) πd π 4 4 Substituting this value and the ass flow rate into the equation for the ass flux of hydrogen we get: G g 1.05 s 1000 g G s The friction factor can be obtained fro Figure.10- of Geankoplis. To do this, first we need to ε calculate Reynolds nuber and the relative roughness of the pipe. Thus, D DG Re µ Substituting the corresponding values for the paraeters on the right side of this equation yields: 0.01 s Re s The viscosity of hydrogen was obtained fro ppendix. of Geankoplis. We can see that this value for the Reynolds nuber indicates that the flow of hydrogen to the fuel cell is turbulent. Therefore, the equation we selected for calculating the pressure P 1 is valid for this proble. Now with the Reynolds nuber value we can obtain the friction factor for a sooth pipe to be given by: f Solving Equation of Geankoplis for the pressure P 1 and substituting all the known quantities into this equation, we can obtain the pressure of hydrogen leaving the fuel tank as shown in the following steps. L G RT + 1 P 4f P DM Daniel López Gaxiola 4 Student View

35 Suppleental Material for Transport Process and Separation Process Principles Pa ( K) s ol K 1015 Pa P 4 1 ( ) + 1 at g 1 1 at ( 0.01 ) ol 1000 g 1 at P Pa Pa P 1 at Daniel López Gaxiola 5 Student View

36 Principles of Moentu Transfer and Overall Balances Exaple.10-8: Entry Length for a Fluid in a Rectangular Channel Deterine if the velocity profile for the gas flowing through the channels of a fuel cell bipolar plate is fully developed for the following cases: a) ydrogen in a Proton-Exchange Mebrane Fuel Cell at a teperature of 5 C with lainar g flow. The fuel consuption rate is of s b) Carbon onoxide in a Solid - Oxide Fuel Cell at a teperature of 800 C with lainar flow. The CO fuel cell is converting carbon onoxide into products at a rate of hr The viscosity of carbon onoxide was found in ppendix.- to be: 5 µ.8 10 s c) Sae gases fro parts a) and b) with turbulent flow. The diensions of the channels in the bipolar plate are shown in the following figure: Gas-Diffusion Layer 0 c Strategy 5 Fuel Gas The entry length is the distance required for the establishent of fully-developed velocity profile. Solution a) To deterine if the velocity profile is fully established in the channel of a bipolar plate, we need to calculate the entry length L e, defined by the following equation: L e Re D For this proble, the diaeter D of the channel will be equal to the hydraulic diaeter, i.e. the perieter of the channel 'wetted' by the fluid flowing through it. Thus, Daniel López Gaxiola 6 Student View

37 Suppleental Material for Transport Process and Separation Process Principles L e Re D (1) To deterine the entry length, we need to calculate the Reynolds nuber, as shown in the following steps: D G Re µ The ass flux of hydrogen G is obtained by dividing the ass flow rate of hydrogen by the crosssectional area of the channel. ence, g s 1000 g G 1 ( )( ) s The hydraulic diaeter is calculated using the following equation D ( ) ( + ) 4 4 P where: Cross-sectional area of the channel P Perieter of the channel wetted by the fluid. Substituting this value and the ass flux in the equation for Reynolds nuber we get: ( ) s Re s The viscosity of hydrogen was obtained fro ppendix. of Geankoplis. Now we can solve equation (1) for the entry length L e and substitute the calculated values to yield: e L e Conclusion: ( )( ) L Re D Daniel López Gaxiola 7 Student View

38 Principles of Moentu Transfer and Overall Balances b) The entry length for the carbon onoxide flowing in the solid-oxide fuel cell will be deterined in a siilar way to part a) of this proble. L e D Re The calculations of the ass flux and Reynolds nuber of carbon onoxide are shown in the following steps: CO 1 hr hr 600 s G 1 ( )( ) s The hydraulic diaeter fro part a) is given by: D Substituting the hydraulic and the ass flux of CO in the equation for Reynolds nuber yields: ( ) s Re s Solving for the entry length, we get: e ( )( ) L Re D Le Conclusion: 0.01 c) For turbulent flow, the entry length is relatively independent fro Reynolds nuber and estiated to be: L e 50D Substituting the hydraulic diaeter of the channel into this equation yields: Le L e Conclusion: ( ) 50 Daniel López Gaxiola 8 Student View

39 Suppleental Material for Transport Process and Separation Process Principles Exaple.11-1: Copressible Flow of a Gas in a Pipe Line gas ixture of 1.5 ol % ethanol and 87.5 % water is leaving a boiler at a teperature of 400 C and a pressure of at. The ethanol/water ixture enters a reforer in a large-scale ethanol reforing plant at a rate of s through a coercial steel pipe with a length of 10 and an inner diaeter of 75 c. Deterine the pressure of the ethanol/water vapor ixture entering the reforer if the viscosity of 5 the gas is Pa s []. Strategy The pressure of the gas entering the reforer can be calculated using the equation for the pressure drop for isotheral copressible flow. Solution Equation.11-9 of Geankoplis can be solved for the pressure P at the end of the pipe to yield: where: P 1 P P ln 1 DM P f Fanning friction factor L Length of the pipe, G Mass flux of gas in the pipe, s R Gas constant, Pa ol K D Inner diaeter of the pipe M Molecular weight of the gas flowing through the pipe, ol P 1 Pressure at the beginning of the pipe segent, Pa P Pressure at the end of the pipe segent, Pa. DOE ydrogen Progra: DOE nalysis Production Case Studies, ha_prod_studies.htl. ccessed: July 010. Daniel López Gaxiola 9 Student View

40 Principles of Moentu Transfer and Overall Balances To use this equation, the friction factor and the ass flux of the ethanol ixture ust be calculated. Before deterining the friction factor we need to calculate the relative roughness of the pipe and the Reynolds nuber. Thus, Dυρ Re µ The velocity of the gas can be calculated with the following equation: V υ owever, to deterine the voluetric flow rate, we need to divide the ass flow rate by the density of the fluid. The density of this ixture is calculated using ideal gas law equation of state. ρ Substituting the corresponding quantities into this equation, after calculating the olecular weight of the gas ixture, we get: M x M + x M Ethanol Ethanol O O ol ethanol g ol O g M ol ol ethanol ol ol O g M ol g 1 ( at) 1.5 ol 1000 g ρ L at ( K) ol K 1000 L ρ Now the voluetric flow rate can be deterined as follows: V s ρ s Substituting this value into the equation for the velocity of the fluid yields: Daniel López Gaxiola 40 Student View

41 Suppleental Material for Transport Process and Separation Process Principles V υ s π( ) s 4 The value of the Reynolds nuber required to deterine the friction factor can now be calculated as shown below: Re s s ( 0.75 ) Re The relative roughness D ε of a coercial steel pipe is given by: ε 6 10 D 5 The friction factor can be obtained fro Figure 6-9 of Perry s Cheical Engineers andbook, 8 th Edition to be: f The only value left to be calculated before being able to solve for the pressure of the gas entering the reforer is the ass flux, defined as: G Substituting the ass flow rate and the cross-sectional area of the pipe into this equation, gives: G s s π( ) 4 s Daniel López Gaxiola 41 Student View

42 Principles of Moentu Transfer and Overall Balances Finally we can substitute all the corresponding values into the equation for P P to get: Pa Pa s ol K at 1 at g 1 ( 0.75 ) ol 1000 g ( )( ) ( ) Pa 1015 Pa s ol K ( 15 ) at 1 at ln g 1 P ol 1000 g 11 Pa P Pa Pa Pa ln P This equation can be solved using coputer software or trial and error to yield: P Pa Daniel López Gaxiola 4 Student View

43 Suppleental Material for Transport Process and Separation Process Principles Exaple.11-: Maxiu Flow for Copressible Flow of a Gas Deterine the axiu velocity that can be obtained for the ethanol/water ixture fro Exaple.11-1 and copare to the actual velocity of the fluid fed to the ethanol reforer. Strategy The axiu velocity of the fluid is obtained using the definition of the velocity of sound in an isotheral fluid. Solution The axiu velocity of the ethanol/water ixture can be deterined using Equation.11-1 of Geankoplis, as shown below: υ ax RT M Substituting the ideal gas constant, as well as the teperature of the gas in the pipe lines and the olecular weight into this equation yields: υ ax υ ax Pa ol K g 1 ol 1000 g s ( K) The velocity of this gas in the process at the entrance to the ethanol reforer is given by Equation.11-1 of Geankoplis: RTG υ P M Substituting the pressure found in Exaple.11-1 and the ass flux of ethanol vapor, we find that the velocity is given by: υ Pa ol K s g 1 Pa ol 1000 g ( K ) ( ) υ s Daniel López Gaxiola 4 Student View