Chapter 5. Principles of Unsteady - State Heat Transfer

Transcription

1 Suppleental Material for ransport Process and Separation Process Principles hapter 5 Principles of Unsteady - State Heat ransfer In this chapter, we will study cheical processes where heat transfer is taking place due to a teperature difference within the syste which is changing with tie. he following proble odules illustrate different exaples where unsteady state heat transfer processes are occurring in fuel cell vehicles and in the processes for producing fuel for fuel cells. 5.- ooling of a ylindrical Solid Oxide Fuel ell 5.- otal Aount of Heat in ooling of a Solid Oxide Fuel ell 5.3- Heat onduction in a Fuel ell Stack ransient Heat onduction in a ylindrical Solid Oxide Fuel ell wo Diensional onduction in a ylindrical Solid Oxide Fuel ell 5.4- Unsteady State onduction and the Schidt Nuerical Method Unsteady State onduction with onvective Boundary ondition Daniel López Gaxiola Student View

2 Principles of Unsteady-State Heat ransfer Exaple 5.-: ooling of a ylindrical Solid Oxide Fuel ell A cylindrical solid oxide fuel cell has an inner radius of 3., outer radius of 3.9 and a length of 0.. he fuel cell initially at a unifor teperature of K enters a ediu whose teperature is 60. Deterine how uch tie in inutes is required for the fuel cell to be cooled down to a teperature of K if the convective coefficient of the ediu is. K he properties of the ebrane electrode assebly for this type of fuel cell are estiated by Xue et al. [] to be: kg ρ p J kg K k.53 K Strategy his proble can be solved using the siplified equations for systes with negligible internal resistance. Solution he equation for the diensionless teperature as a function of tie is given by: 0 h t p x ρ In this equation: e eperature of the ediu 0 Initial teperature of the fuel cell x haracteristic diension of the body However, this equation is only applicable when the Biot nuber N Bi is less than 0.. Hence, we need to obtain this diensionless nuber first, defined as: N Bi For a cylindrical object, the characteristic length is calculated as follows: x r. Xue, X., ang, J., Saes, N., Du, Y., Journal of Power Sources, 4, (005) Daniel López Gaxiola Student View

3 Suppleental Material for ransport Process and Separation Process Principles In this case, since the fuel cell is a hollow cylinder, the characteristic diension will be obtained as shown below: x ( ) r r 000 o i Substituting this paraeter into the definition of Biot nuber, we get: N Bi ( ) K.53 K Since this value is less than 0., we can proceed to enter the corresponding values into the equation for the diensionless teperature, to yield: K K K K e K J kg kg K 3 t Solving for the tie, we have: t ln K K K K ( s ) t s t in Daniel López Gaxiola 3 Student View

4 Principles of Unsteady-State Heat ransfer Exaple 5.-: otal Aount of Heat in ooling of a Solid Oxide Fuel ell Deterine the total aount of heat reoved fro the fuel cell in Proble 5.- after 5 inutes. he characteristic length of the fuel cell is Strategy he equation for the total aount of heat in J can be used to solve this proble. Solution he equation for the total aount of heat is given by: Q pρv(0 ) e h t pρx he volue of the cylindrical fuel cell is calculated as follows: V π( ) L V π[ ] 0. V Substituting the calculated volue and the rest of the values into the equation for the total aount of heat transferred, yields: K J kg 4 3 s J kg 3 kg K Q ( )(873.5 K K) e kg K Q J Daniel López Gaxiola 4 Student View

5 Suppleental Material for ransport Process and Separation Process Principles Exaple 5.3-: Heat onduction in a Fuel ell Stack A stack of 0 fuel cells is initially at a teperature of K. his stack enters a roo at a teperature of K. Assuing all the sides but the front face of the stack are insulated, deterine the teperature of the 60 th fuel cell after 90 inutes. he thickness of the bipolar plates in this fuel cell stack is and it is uch thicker than the ebrane electrode assebly. he heat transfer coefficient of the air in this roo is 9 K he properties of the bipolar plates in this fuel cell stack are given below []: kg ρ 63 3 p J 44 kg K k 4.4 K A scheatic of this fuel cell stack is shown in the following figure: n cell n cell 60 n cell K x he sides and back face of the fuel cell stack are insulated Strategy x he teperature at a point inside the fuel cell stack can be obtained fro the charts for unsteady state heat conduction of a slab. Solution o use Figure of Geankoplis to deterine the unsteady state heat conduction in a flat plate, we need to find the paraeters X, and n, defined as: X k n hx. King, J.A., Lopez Gaxiola, D., Johnson, B.A., Keith, J.M., Journal of oposite Materials, 44 (7), (00). Daniel López Gaxiola 5 Student View

6 Principles of Unsteady-State Heat ransfer he characteristic length x in these equations corresponds to the distance fro the front face to the center of the fuel cell stack, obtained by ultiplying the nuber of fuel cells by the thickness of a bipolar plate, and dividing it by. hus, x 0 x he paraeter α is the diffusivity of the fuel cell stack defined as: k α Substituting the properties of the bipolar plates into this equation, we get: 4.4 α K kg J s 3 kg K o deterine the value of n, we need to know the distance at which the 60 th fuel cell is located fro the center of the stack. his distance can be calculated by ultiplying the thickness of a single bipolar plate by (0-60) which is the nuber of fuel cells fro the center. x ( 0 60 )( ) Now we can calculate the three values required to use Figure 5.3-5, as shown in the following steps: 60 s ( 90 in) s in X ( ) 4.4 K. 9 ( ) K n 0.45 Fro Figure we can read a value for a diensionless teperature Y. his value can be used to solve for the teperature in the 60 th fuel cell as follows: Daniel López Gaxiola 6 Student View

7 Suppleental Material for ransport Process and Separation Process Principles where: Y 0 eperature of the cooling ediu 0 Initial teperature of the fuel cell stack. e can enter the corresponding teperatures and solve for the teperature to get: K ( K K) K Daniel López Gaxiola 7 Student View

8 Principles of Unsteady-State Heat ransfer Exaple 5.3-3: ransient Heat onduction in a ylindrical Solid Oxide Fuel ell A cylindrical solid oxide fuel cell with a diaeter of 3.9 and a length of 0. is initially at a teperature of 50 K. he fuel cell is shut down in a roo where the air is at a teperature of 303 K. alculate the teperature at the center of the fuel cell after 5 inutes, assuing it is insulated on the flat ends. he heat transfer coefficient of the air is 0. he theral conductivity and K diffusivity of the fuel cell are Strategy.53 and K , respectively. s he solution to this proble can be found by using the charts for unsteady state heat conduction in a cylinder. Solution Since heat is only being transferred through the walls, the fuel cell can be considered as a long cylinder. he paraeters n, and X required to deterine the diensionless teperature Y fro Figure are calculated as shown in the following steps: At the center of the cylinder, x 0. hus, the value of n will also be equal to zero: x n x 0 n 0 o calculate the value of X, we need to substitute the diffusivity, radius of the fuel cell and the tie elapsed, as shown below: αt X x 60 s ( in) s in X ( ) Finally, we can deterine the paraeter : k hx Daniel López Gaxiola 8 Student View

9 Suppleental Material for ransport Process and Separation Process Principles K 3 ( ) K For these values, the corresponding Y fro Figure will be equal to. Now we can solve for the teperature at the center of the fuel cell fro the definition of the diensionless teperature Y: Y where: eperature of cooling ediu 0 Initial teperature of the fuel cell Substituting the teperature values in this equation, we get: 303 K ( 303 K K) K Daniel López Gaxiola 9 Student View

10 Principles of Unsteady-State Heat ransfer Exaple 5.3-4: wo Diensional onduction in a ylindrical Solid Oxide Fuel ell Deterine the teperature at the center of the solid oxide fuel cell fro Exaple now considering heat conduction also occurring through the ends of the cylinder. hat does this result indicate? Strategy he procedure to solve this proble consists of using the charts for unsteady state heat transfer in a cylinder for both radial and axial directions. Solution e need to calculate the required diensionless quantities X, and n for both radial and axial directions. For the radial direction, these will be given by: n radial n radial x x X radial radial k hx K 3 ( ) K αt radial x X 60 s ( 5 in) s in ( ) radial hese values will yield a diensionless teperature Y radial For heat conduction in the axial direction, we need to calculate the paraeters n, and X and locate the in Figure 5.3-6, applicable for two parallel planes. hus, n axial naxial 0 y 0 y 0. Daniel López Gaxiola 0 Student View

11 Suppleental Material for ransport Process and Separation Process Principles X axial axial k hy K ( ) K αt axial x X ( in )( ) s ( ) axial Locating these three paraeters in Figure gives a value of Y axial. Now that we have both Y values for both directions, we can obtain a Y for the overall heat transfer process as follows: Y Y Y axial radial Now we can solve for the teperature at the center of the cylinder to get: Y onclusion: 0 K ( ) K Daniel López Gaxiola Student View

12 Principles of Unsteady-State Heat ransfer Exaple 5.4-: Unsteady State onduction and the Schidt Nuerical Method A proton exchange ebrane fuel cell stack has a thickness of 0.3 and is initially at a unifor teperature of 60. he front face of the fuel cell stack suddenly exposed to an environental 6 teperature of he bulk theral diffusivity of the fuel cell stack is s Assuing the convective resistance is negligible and that the back face of the stack is insulated, deterine the teperature profile after 4 inutes using the Schidt nuerical ethod with M and dividing the fuel cell stack into slices with a thickness of Follow the special procedure for the first tie increent. he next figure illustrates the conditions in this cooling process: a Insulated Face 0 60 Strategy he equations we need to use to deterine the teperature profile will depend on the boundary conditions. Solution he nuber of tie steps to use in this proble will be deterined fro the definition of the paraeter M: ( x) M α t Solving for the tie increent yields: t and substituting the corresponding values into this equation, ( x) ( ) t M s α Daniel López Gaxiola Student View

13 Suppleental Material for ransport Process and Separation Process Principles t s he nuber of tie steps needed for this t is given by: t ( in) ntie steps 0 t s he front surface of the fuel cell stack corresponds to n. In this point, the teperature a used for the first tie increent is given by: a + a 0 where: 0 Initial teperature at point. a eperature of the environent Since there is no convective heat resistance at the interface, for the reaining tie increents: a he general equation for deterining the teperature for the slabs n to 6 is given below: tt n + t n t n+ e need an additional equation for the insulated face. his is the point where n 7 and its corresponding equation is given by: tt 7 ( M ) 7 + t 6 t M Now we can proceed to calculate the teperatures for the first tie increent. hus, for n, a 0 t+ t Since the proble indicates we should use the special procedure for the first tie increent, this teperature value we just obtained is equal to the abient teperature at the first tie increent: For n : t+ t a Daniel López Gaxiola 3 Student View

14 Principles of Unsteady-State Heat ransfer t+ t + t t 3 he special procedure used at n for the first tie increent also affects this equation: instead of using the teperature of we will use a. Hence, + + a t 3 t+ t he following four slabs corresponding to n 3 to 6 can be calculated with the general equation for n to 6, given in previous steps. Substituting the corresponding teperatures into this equation yields: t t 4 tt t 3 t 5 tt t 4 t 6 tt 5 60 tt t 5 t 7 60 For n 7, we use the equation for the insulated face: ( ) 7 + t 6 60 t tt 7 t 6 Now we can proceed to calculate the teperatures fro the second to the tenth tie increents, as shown in the following steps: For t: t+ t a tt t+ t t+ t t+ t t+ t 4 t+ t 3 t+ t t+ t 3 t+ t 5 4 Daniel López Gaxiola 4 Student View

15 Suppleental Material for ransport Process and Separation Process Principles t+ t 5 t+ t t+ t 4 t+ t t+ t 5 t+ t 7 For 3 t: t+ t 7 tt 6 t+3 t a 6.67 t+3 t tt tt tt tt 4 t+ 3 t 3 t+ 3 t tt 3 tt tt 4 tt 6 t+ 3 t 5 t+ 3 t tt 5 tt 7 t+ 3 t 7 t+ t 6 Daniel López Gaxiola 5 Student View

16 Principles of Unsteady-State Heat ransfer e can continue to repeat this procedure up to 0. he solutions for the all the teperatures (in ) are given in the following table: n t tt tt t+3 t t+4 t t+5 t t+6 t t+7 t t+8 t t+9 t t+0 t e can plot these teperature results to observe the change in teperature within the fuel cell stack: ( ) t t+δt t+δt t+3δt t+4δt t+5δt t+6δt t+7δt t+8δt n t+9δt t+0δt Daniel López Gaxiola 6 Student View

17 Suppleental Material for ransport Process and Separation Process Principles Exaple 5.4-3: Unsteady State onduction with onvective Boundary ondition Deterine the teperature profile for the sae fuel cell stack as in Exaple 5.4-, now with a convective coefficient of 3. he bulk theral conductivity of the fuel cell stack is 0. K K Use a value of M 4. a h 3 K Insulated Face 0 60 Strategy In this proble we need to use Schidt nuerical ethod depending on the boundary conditions. Solution Before we start applying Schidt ethod, we need to deterine the nuber of tie increents we need to use. Fro the definition of the paraeter M, we can solve for the tie increent t, as shown in the following steps: ( x) M α t ( x) (0.05 ) t s αm ( 4) s he nuber of tie steps needed for this t is given by: n tie steps 60 s ( in) t in t 7.9s As stated in Section 5.4B of Geankoplis, when the value of M is greater than 3, the value of the environental teperature a will be the sae for all tie increents. herefore, Daniel López Gaxiola 7 Student View

18 Principles of Unsteady-State Heat ransfer a 6.67 For the node n, corresponding to the front face of the fuel cell stack, the teperature can be calculated using Equation of Geankoplis: N + M (N + ) + M { [ ] } t t t a t t he value of N is a function of the convective heat transfer coefficient and the theral conductivity of the fuel cell, described by the following equation: h x N k Entering the corresponding values into this equation, yields: ( ) N K 0 K In order to use Equation 5.4-7, the value of M ust satisfy the following constraint: M N + Substituting nueric quantities into this constraint, we get: 4 (0.035) + 4 Hence, we can use Equation Fro this equation, we can find the teperature t t to be: ( ) 60 + (60 ) 4 t t { [ ] } t t For the points at n, 3, 4, 5, 6, we use Equation 5.4-: + ( M ) + M t t n t n+ t n t n By entering the value of M into this equation, we can get a general equation for t t n : t t n Daniel López Gaxiola 8 Student View

19 Suppleental Material for ransport Process and Separation Process Principles Now we need an equation for the insulated boundary at n 7, which is Equation of Geankoplis: tt 7 ( M ) 7 + t 6 t M tt 7 + t 7 t 6 e can now proceed to deterine the teperatures for n to 7 for the first tie increent. hus, t t t 3 t t t t t t 3 t 4 t t 3 t t t t 4 t 5 t 3 t 4 t t t t 5 t 6 t 4 t 5 t t t t 6 t 7 t 5 t 6 t t tt 7 + t 7 t 6 tt 7 ( ) + ( ) For t: [ ] t+ t t t a t t tt Daniel López Gaxiola 9 Student View

20 Principles of Unsteady-State Heat ransfer ( ) + ( ) + ( 60 ) 4 t+ t t t t t tt t+ t t t 3 t t tt t t t+ t 3 t t 4 t t tt 3 t t t+ t 4 t t 5 t t 3 tt 4 t t t+ t 5 t t 6 t t 4 tt 5 t t t+ t 6 t t 7 t t 5 tt 6 t t t+ t 7 + t t 7 tt 6 t+ t 7 ( ) + ( ) For 3 t: [ ] t+ 3 t t+ t a t+ t t+ t ( ) ( ) + ( ) 4 t+ 3 t t+ t t+ t t+ t t+ 3 t t+ t 3 t+ t t+ t t 3 t t+ 3 t 3 t+ t 4 t+ t t+ t 3 Daniel López Gaxiola 0 Student View

21 Suppleental Material for ransport Process and Separation Process Principles t 3 t t+ 3 t 4 t+ t 5 t+ t 3 t+ t 4 t 3 t t+ 3 t 5 t+ t 6 t+ t 4 t+ t 5 t 3 t t+ 3 t 6 t+ t 7 t+ t 5 t+ t 6 t 3 t t+ 3 t 7 + t+ t 7 t+ t 6 t+ 3 t 7 ( ) + ( ) Daniel López Gaxiola Student View

22 Principles of Unsteady-State Heat ransfer In a siilar way, we can continue to perfor this calculation for all the 0 tie increents. he results for all teperatures (in ) are shown in the following table: n t tt tt t+3 t t+4 t t+5 t t+6 t t+7 t t+8 t t+9 t t+0 t tt tt t+3 t t+4 t t+5 t t+6 t t+7 t t+8 t t+9 t t+0 t Daniel López Gaxiola Student View

23 Suppleental Material for ransport Process and Separation Process Principles he graphical representation of these teperatures is shown in the following figure: 60 t t+δt 59 t+δt t+3δt t+4δt t+5δt 58 t+6δt t+7δt ( ) 57 t+8δt t+0δt t+9δt t+δt 56 t+δt t+3δt 55 t+4δt t+6δt t+5δt t+7δt n t+8δt t+0δt t+9δt Daniel López Gaxiola 3 Student View