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Principles of Steady State Heat Transfer, Example 4 1 1 Heat Loss through a Stainless Steel Bipolar Plate. Calculate the heat flux through a stainless steel bipolar plate in a polymer electrolyte membrane fuel. cell with a thickness of 4 5 mm The fuel cell is operating at a temperature of 80 C during the. summer season in Houghton Michigan where the temperature is 70 F. The equation for the heat flux obtained from Fourier s Law can be used to obtain the solution to this. Equation 4 1 10 of Geankoplis is defining the heat transfer per unit area as follows. A x 2 x1 1 2, We can substitute the values given in the problem statement into this equation but first we need to. convert the temperature outside the fuel cell to C. T F 32 70 F 32, Entering the temperatures inside and outside the fuel cell stack into the heat transfer equation as. well as the thickness of the bipolar plate represented by x 2 x1 we get. q m K K 294 25 K, The thermal conductivity of steel was obtained from Table 4 1 1 of Geankoplis. Daniel L pez Gaxiola 2 Student View, Jason M Keith.

Supplemental Material for Transport Process and Separation Process Principles. Example 4 3 1 Cooling of a Fuel Cell, Air at a temperature of 25 C is being used for cooling a single cell fuel cell The convective heat. transfer coefficient of the air is 61 2 2 and is capable of removing heat at a rate of 183 6 W. What would be the dimensions of the square surface of the fuel cell if its temperature must not. exceed 50 C, Air 25 C Air 25 C, The heat transfer rate by convection can be obtained using Newton s Law of Cooling. The heat flux q when heat is being transferred by forced convection is defined as follows. q heat transfer rate, h convective heat transfer coefficient. TS temperature on the surface of the object C, T temperature of the air C. A surface area of the fuel cell m2, Daniel L pez Gaxiola 3 Student View.

Jason M Keith, Principles of Steady State Heat Transfer. To determine the dimensions of the surface of the fuel cell we can substitute the given temperatures. and heat transfer rate and solve for the area A to yield. W 50 C 25 C, A 0 12 m 2, Since the heat is being removed from the fuel cell through both the left and right faces of the fuel. cell this value of A must be divided by 2 Thus, 0 12 m 2 cm 2. A fuel cell, A fuel cell cm 2, The dimensions of a fuel cell with a square surface could be obtained as follows. L A fuel cell cm 2, Therefore for a heat transfer rate of 183 6 W air at 25 C can be used to keep the surface area of a.

cm x cm fuel cell at a temperature of C, Daniel L pez Gaxiola 4 Student View. Jason M Keith, Supplemental Material for Transport Process and Separation Process Principles. Example 4 3 2 Heat Loss in Fuel Reforming Applications. A pipe made of 308 stainless steel schedule number 80 with a nominal diameter of 1 5 is carrying. methane at a temperature of 400 C in a steam methane reforming process for producing hydrogen. The pipe is insulated with a layer of glass fiber with a thickness of 1 Determine the temperature at. the interface between the pipe and the glass fiber and the heat loss through the insulated pipe with a. length of 15 m The surface of the insulating material is at a temperature of 25 C. A schematic of the pipe is shown below, Natural Gas. The equation for the heat loss through a pipe can be applied to the different layers in the pipe. The heat loss through the walls of a cylinder is given by. Tin temperature at the inner wall of the pipe K, Tout temperature at the outer wall of the pipe K. R resistance of the pipe to the heat transfer through its walls. Daniel L pez Gaxiola 5 Student View, Jason M Keith.

Principles of Steady State Heat Transfer, In this problem we need to apply this equation for both the steel pipe and the insulated pipe The. overall heat loss will be obtained using this equation for the insulated steel pipe The resistance to. heat transfer in cylindrical coordinates is calculated with the following equation. in this equation, rout outer radius of the cylinder m. rin inner radius of the cylinder m, k thermal conductivity of the material. Alm log mean area m2, The log mean area of the pipe is defined as. A out A in, where Aout and Ain are the outer and inner surface areas of the cylinder respectively.

Applying the equations for resistance and the log mean areas to the steel and the overall pipe we. Steel Pipe Overall, T1 T2 T1 T3, q 1 2 q 1 3, R1 2 R 1 2 R 2 3. k steel A lm 1 2, A lm 1 2 ln 3, Daniel L pez Gaxiola 6 Student View. Jason M Keith, Supplemental Material for Transport Process and Separation Process Principles. To determine the log mean areas we need to look for the radius of the steel pipe in Appendix A 5 of. Geankoplis For the 1 5 pipe, The radius of the pipe including the insulation is obtained by adding the thickness of 1 to the outer. radius of the steel pipe Hence, r3 m 0 0254 m m, With these diameter values and the length of the pipe the areas A1 A2 and A3 can be calculated as.

shown in the following steps, A1 2 r1L 2 0 01905 m m 1 795 m2. A2 2 r2L 2 m m m2, A3 2 r3L 2 m m m2, The thermal conductivities for the glass fiber and the steel can be found in Appendix A 3 of. Geankoplis and are shown below The conductivity of the glass fiber was selected at the highest. temperature available in Table A 3 15 The thermal conductivity of steel was obtained from Table. k glass fiber, Substituting the values we obtained into the equations for the individual layers yields. Steel Pipe, T1 T2 C T2, R 1 2 1 17 10 4, A lm 1 2 2 025 m 2. Daniel L pez Gaxiola 7 Student View, Jason M Keith.

Principles of Steady State Heat Transfer, q 1 3 2695 6 W. m 0 02415 m C, R 2 3 0 139, A lm 2 3 m 2, Hence the heat loss through the insulated pipe is 2695 6 W. Since this answer represents the amount of heat lost per unit time if we assume that the system is at. steady state The heat loss per unit time will be the same in the individual layers Thus we can use. the equation for the heat loss through the steel pipe to determine the temperature at the steel glass. fiber interface, q 1 2 2695 6 W, Solving for the temperature T2 we get. T2 C 2695 6 W, As it can be seen the temperature at the pipe insulator interface is almost the same as the. temperature of the inner wall of the steel pipe This is because most of the heat is lost through the. metal pipe due to the high thermal conductivity of steel in comparison to the thermal conductivity of. the insulating material, Daniel L pez Gaxiola 8 Student View.

Jason M Keith, Supplemental Material for Transport Process and Separation Process Principles. Example 4 3 3 Heat Loss by Convection and Conduction in a Steam Methane. Reforming Process, Natural gas at 400 C is flowing inside a steel pipe with an inner diameter of 1 5 in and an outer. diameter of 1 9 in The pipe is insulated with a layer of glass fiber with a thickness of 1 in The. convective coefficient outside the insulated pipe is 1 23 2 The temperature on the external. surface of the pipe is 43 4 C, Calculate the convective coefficient of natural gas and the overall heat transfer coefficient U based. on the inside area Ai if heat is being lost at a rate of 7115 in a pipe with a length of 49 2 ft. To determine the heat transfer coefficients we will use the equation for heat loss for a multilayer. The heat loss through a cylinder with different layers is defined by the following equation. Ti To Ti To, R Ri RA RB Ro, Ti Temperature on the internal surface of the pipe. To Temperature on the external surface of the pipe. Ri Convective resistance inside the pipe, RA Conductive resistance through the steel pipe.

RB Conductive resistance through the insulation layer. Ro Convective resistance outside the pipe, The resistance to heat transfer due to convection is defined as follows. Daniel L pez Gaxiola 9 Student View, Jason M Keith. Principles of Steady State Heat Transfer, h Convective heat transfer coefficient. A Area of heat transfer, The resistance of a cylinder to heat conduction is calculated as follows. rout Outer radius of the cylinder, rin Inner radius of the cylinder.

k thermal conductivity of the material, Alm log mean area of the cylinder. We can enter the definitions of the resistances due to conduction and convection into the equation. for the heat loss to yield, 1 r1 ri ro r1 1, h i A i k steel A A lm k glass fiber A B lm h o A o. In this equation, ri inner radius of the steel pipe in. r1 outer radius of the steel pipe in, ro outer radius of the insulated pipe in. With these values we can calculate the log mean areas AA lm and AB lm and the inner and outer areas. of the insulated pipe Thus, A i 2 ri L 2 0 75 in ft ft.

A1 2 r1L 2 in ft 24 48 ft, Daniel L pez Gaxiola 10 Student View. Jason M Keith, Supplemental Material for Transport Process and Separation Process Principles. A o 2 ro L 2 in ft ft, A1 A i 24 48 ft 2 ft 2, A A lm 21 80 ft 2. A 24 48 ft 2, A o A1 ft 2 24 48 ft 2, A B lm ft 2, A1 24 48 ft 2. The temperature of the inner and outer surfaces of the pipe are given in C and therefore they must. be converted to F, Ti F 400 C 1 8 32 F, To F 43 4 C 1 8 32 F.

Now we can substitute all the values into the heat loss equation and solve for the convective. coefficient hi, 1 Ti To r r ro r1 1, Ai q k steel A A lm k glass fiber A B lm h o A o. The thermal conductivity values can be obtained from Appendices A 3 15 and A 3 16 of. Geankoplis However since the values are given in the SI system they must be converted to the. English system Hence, J 1 btu m 3600 s 1 C btu, k steel 45 26. m s C J 1 ft 1 hr F ft hr F, J 1 btu m 3600 s 1 C btu. k glass fiber 0 0317, m s C J 1 ft 1 hr F ft hr F, The thermal conductivities of steel and glass fiber were obtained at the highest temperature available. in Appendix A 3, Daniel L pez Gaxiola 11 Student View.

Jason M Keith, Principles of Steady State Heat Transfer. Now we can calculate the convective heat transfer coefficient hi as shown below. in 0 75 in, 1 F 110 1 F in, in 0 95 in, 1 hr F hr F hr F hr F. 2 95 10 5 0 0162, ft btu btu btu btu, To calculate the overall heat transfer coefficient U we need to use the equation for the heat loss in. terms of U Thus, q U i A i Ti To, We can solve this equation for the coefficient U and substitute the corresponding values to get. A i Ti To ft F 110 1 F, Daniel L pez Gaxiola 12 Student View.

Jason M Keith, Supplemental Material for Transport Process and Separation Process Principles. Example 4 3 4 Heat Generation in a Tubular Solid Oxide Fuel Cell. A tubular solid oxide fuel cell with an outer diameter of 2 2 cm and a length of 150 cm is operating. at a current density of 202 6 2 Determine the heat generation rate in 3 if the voltage of the. fuel cell is 1 V Assume that the thickness of the electrodes and electrolyte membrane are small. compared to the overall diameter of the fuel cell, The following figure shows a tubular solid oxide fuel cell. Cathode interconnection, Electrolyte, Air Flow Anode. The heat generation rate of the fuel cell can be obtained from the power of the fuel cell which. depends on the current and the voltage, The heat generated by the fuel cell in terms of the power is given by the equation shown below. The power of the fuel cell can be obtained by multiplying the current by the voltage of the fuel cell. Daniel L pez Gaxiola 13 Student View, Jason M Keith.

Principles of Steady State Heat Transfer, Substituting this equation into the equation for the heat generation rate yields. The problem statement is not giving the value of the current However if we calculate the cross. surface area of the fuel cell we can determine the value of the current in A. Asurface 2 RL, Solving for the current I and substituting the dimensions of the fuel cell into this equation we get. I DLi cm cm 202 6 2, cm 1000 mA, Entering this value into the heat generation equation we have. q 3 68 105, Daniel L pez Gaxiola 14 Student View, Jason M Keith. Supplemental Material for Transport Process and Separation Process Principles. Example 4 5 1 Heating of Natural Gas in Steam Methane Reforming Process. Natural gas at a temperature of 310 C is flowing inside a steel pipe Schedule 80 with an inner. diameter of 1 5 in at a rate of 7 79 10 3 The natural gas is being heated by the product of the. reforming process at 850 C The convective heat transfer coefficient of the reformate is 1025 2. Calculate the heat transfer rate in W through a pipe with a length of 7 m The properties of natural. gas are given in the following table, 1 942 10 5, 2 909 10 5.

The equation for heat transfer through a pipe will be used to determine the heat flux. When heat is being transferred through a fluid the heat flux is given by. q Heat transfer rate in W, Tr Temperature of the heating medium reformate C or K. Tn Temperature of the fluid inside the pipe natural gas C or K. R Sum of resistances to heat transfer through the pipe W. Daniel L pez Gaxiola 15 Student View, Jason M Keith. Supplemental Material for Transport Process and Separation Process Principles Daniel L pez Gaxiola 1 Student View Jason M Keith Chapter 4 Principles of Steady State Heat Transfer Heat transfer is occurring in many chemical and separation processes as a consequence of a temperature difference In Chapter 4 the following problem modules