An Experimental Investigation of Impingement Heat Transfer in a Rib-Roughened Trailing-Edge Channel with Crossover Holes. A thesis presented

Transcription

1 An Experimental Investigation of Impingement Heat Transfer in a Rib-Roughened Trailing-Edge Channel with Crossover Holes A thesis presented by Xiao Huang to The Department of Mechanical and Industrial Engineering in partial fulfillment of the requirements for the degree of Master of Science in Mechanical Engineering In the field of Thermofluids Engineering Northeastern University Boston Massachusetts August 2012 i

2 Abstract Turbine cooling technology plays a critical role in increasing the thermal efficiency and power output of advanced gas turbines. The common cooling methods include the impingement cooling method, the film cooling method, and the rib roughened cooling method. This study focuses on the impingement cooling effect on a rib-roughened trailing edge channel with cross over holes and aims to find the influence of blocked exit holes to the heat transfer efficiency. A steady state liquid crystal technique was used to get the heat transfer coefficient. In this experiment, two different kinds of cross-over holes were used: five tilt angle and zero tilt angle. Both inline and staggered slot-arrangement were tested and investigated. There were total 16 geometry cases. The numbers of blocked exit holes are 0, 3, 4 and 5 for inline arrangement and 0, 2, 3 and 4 for staggered arrangement. For each geometry setting, a range of Reynolds numbers was tested. The graphs of the Nusselt number versus the Reynolds number were plotted. For each individual area, the Nusselt numbers with different number of blocked exit holes were compared. The results showed that the Nusselt numbers increase monotonically with increasing Reynolds numbers. It was also found that for staggered arrangement, the highest Nusselt numbers were found in the area two holes away from the nearest blocked exit hole. And for inline arrangement, conclusions are different between zero tilt angle and five tilt angle. With zero tilt angle, the Nusselt numbers of area1 and area2 both increased first and then decreased as the number of blocked holes increased. While with five tilt angle, Nusselt numbers of all the four areas decreased monotonically. i

3 Table of Contents Introduction..... Experimental set up and procedure..... Experimental set up. Experimental procedure.. Liquid crystal calibration Cold and heat transfer tests. Date reduction. Results and discussion Conclusion.. References Appendix A: Log sheets for each experiment... A-1 Appendix B: Source code for Ceck. f... B-1 Appendix C: Source code for Reduce. f C-1 Appendix D: Source code for intergarea. f... D ii

4 List of Figures Figure 1. Figure 2. Figure 3. Figure 4. Figure 5. Figure 6. Figure 7. Figure 8. Figure 9. Figure 10. Figure 11. Figure 12. Figure 13. Figure 14. Figure 15. Figure 16. Figure 17. Figure 18. Figure 19. Figure 20. Figure 21. Figure 22. Figure 23. Gas turbine engine. 1 Idealized Brayton Cycle The common cooling techniques for a modern gas turbine blade Schematics of the test section... 4 cross section of the supply channel and trailing edge channel togther The locations of the pressure taps Details of the cross-over holes (removable partition holes) Details of the trailing edge channel.. 9 Details of the trailing edge holes.. 9 The dimensions of the turbulators used in experiment Inline arrangement flow. Staggered arrangement flow Camera focuses and the four areas for staggered arrangement.. 13 Camera focuses and the four areas for inline arrangement Liquid crystal display of the reference color and the reference temperature. 15 Thermometer used in experiment.. True-rms Multimeter used in experiment.. Micro-manometer used in experiment 23 Orange-manometer used in experiment. Red-manometer used in experiment.. Electrical power panel used in experiment. 24 Nusselt Number versus Reynolds Number for geometry 1 (11crossover holes-12 exit holes, inline arrangement, 0 degree tilt angle case) at different pressure increments.. Nusselt Number versus Reynolds Number for geometry 2 (11crossover holes-9 exit holes, inline arrangement, 0 degree tilt angle case) at different pressure increments iii

5 Figure 24. Figure 25. Figure 26. Figure 27. Figure 28. Figure 29. Figure 30. Figure 31. Figure 32. Figure 33. Figure 34. Figure 35. Figure 36. Nusselt Number versus Reynolds Number for geometry 3 (11crossover holes-8 exit holes, inline arrangement, 0 degree tilt angle case) at different pressure increments.41 Nusselt Number versus Reynolds Number for geometry 4 (11crossover holes-7 exit holes, inline arrangement, 0 degree tilt angle case) at different pressure increments.42 Nusselt Number versus Reynolds Number for geometry 5 (11crossover holes-12 exit holes, inline arrangement, 5 degree tilt angle case) at different pressure increments.42 Nusselt Number versus Reynolds Number for geometry 6 (11crossover holes-9 exit holes, inline arrangement, 5 degree tilt angle case) at different pressure increments.43 Nusselt Number versus Reynolds Number for geometry 7 (11crossover holes-8 exit holes, inline arrangement, 5 degree tilt angle case) at different pressure increments.43 Nusselt Number versus Reynolds Number for geometry 8 (11crossover holes-7 exit holes, inline arrangement, 5 degree tilt angle case) at different pressure increments.44 Nusselt Number versus Reynolds Number for geometry 9 (11crossover holes-12 exit holes, staggered arrangement, 0 degree tilt angle case) at different pressure increments.44 Nusselt Number versus Reynolds Number for geometry 10 (11crossover holes-10 exit holes, staggered arrangement, 0 degree tilt angle case) at different pressure increments.45 Nusselt Number versus Reynolds Number for geometry 11(11crossover holes-9 exit holes, staggered arrangement, 0 degree tilt angle case) at different pressure increments.45 Nusselt Number versus Reynolds Number for geometry 12(11crossover holes-8 exit holes, staggered arrangement, 0 degree tilt angle case) at different pressure increments.46 Nusselt Number versus Reynolds Number for geometry 13 (11crossover holes-12 exit holes, staggered arrangement, 5 degree tilt angle case) at different pressure increments.46 Nusselt Number versus Reynolds Number for geometry 14 (11crossover holes- 10 exit holes, staggered arrangement, 5 degree tilt angle case) at different pressure increments...47 Nusselt Number versus Reynolds Number for geometry 15 (11crossover holes-9 exit holes, staggered arrangement, 5 degree tilt angle case) at different pressure increments.47 iv

6 Figure 37. Nusselt Number versus Reynolds Number for geometry 16(11crossover holes-8 exit holes, staggered arrangement, 5 degree tilt angle case) at different pressure increments.48 Figure 38. Nusselt Number versus Reynolds Number of four areas for geometry 1 (11crossover holes-12 exit holes, inline arrangement, 0 degree tilt angle) case 48 Figure 39. Nusselt Number versus Reynolds Number of four areas for geometry 2 (11crossover holes-9 exit holes, inline arrangement, 0 degree tilt angle) case 49 Figure 40. Nusselt Number versus Reynolds Number of four areas for geometry 3 (11crossover holes-8 exit holes, inline arrangement, 0 degree tilt angle) case 49 Figure 41. Nusselt Number versus Reynolds Number of four areas for geometry 4 (11crossover holes-7 exit holes, inine arrangement, 0 degree tilt angle) case 50 Figure 42. Nusselt Number versus Reynolds Number of four areas for geometry 5 (11crossover holes-12 exit holes, inline arrangement, 5 degree tilt angle) case 50 Figure 43. Nusselt Number versus Reynolds Number of four areas for geometry 6 (11crossover holes-9 exit holes, inline arrangement, 5 degree tilt angle) case 51 Figure 44. Nusselt Number versus Reynolds Number of four areas for geometry 7 (11crossover holes-8 exit holes, inline arrangement, 5 degree tilt angle) case 51 Figure 45. Nusselt Number versus Reynolds Number of four areas for geometry 8 (11crossover holes-7 exit holes, inline arrangement, 5 degree tilt angle) case 52 Figure 46. Nusselt Number versus Reynolds Number of four areas for geometry 9 (11crossover holes-12 exit holes, staggered arrangement, 0 degree tilt angle) case 52 Figure 47. Nusselt Number versus Reynolds Number of four areas for geometry 10 (11crossover holes-10 exit holes, staggered arrangement, 0 degree tilt angle) case 53 Figure 48. Nusselt Number versus Reynolds Number of four areas for geometry 11 (11crossover holes-9 exit holes, staggered arrangement, 0 degree tilt angle) case 53 Figure 49. Nusselt Number versus Reynolds Number of four areas for geometry 12 (11crossover holes-8 exit holes, staggered arrangement, 0 degree tilt angle) case 54 v

7 Figure 50. Nusselt Number versus Reynolds Number of four areas for geometry 13 (11crossover holes-12 exit holes, staggered arrangement, 5 degree tilt angle) case 54 Figure 51. Nusselt Number versus Reynolds Number of four areas for geometry 14 (11crossover holes-10 exit holes, staggered arrangement, 5 degree tilt angle) case 55 Figure 52. Nusselt Number versus Reynolds Number of four areas for geometry 15 (11crossover holes-9 exit holes, staggered arrangement, 5 degree tilt angle) case 55 Figure 53. Nusselt Number versus Reynolds Number of four areas for geometry 16 (11crossover holes-8 exit holes, staggered arrangement, 5 degree tilt angle) case 56 Figure 54. Figure 55. Figure 56. Figure 57. Figure 58. Figure 59. Figure 60. Figure 61. Figure 62. Figure 63. Figure 64. Figure 65. Nusselt Number versus Reynolds Number for geometry 1 (11crossover holes-12 exit holes, inline arrangement, 0 degree tilt angle) case...56 Nusselt Number versus Reynolds Number for geometry 2 (11crossover holes-9 exit holes, inline arrangement, 0 degree tilt angle) case...57 Nusselt Number versus Reynolds Number for geometry 3 (11crossover holes-8 exit holes, inline arrangement, 0 degree tilt angle) case...57 Nusselt Number versus Reynolds Number for geometry 4 (11crossover holes-7 exit holes, inline arrangement, 0 degree tilt angle) case...58 Nusselt Number versus Reynolds Number for geometry 5 (11crossover holes-12 exit holes, inline arrangement, 5 degree tilt angle) case...58 Nusselt Number versus Reynolds Number for geometry 6 (11crossover holes-9 exit holes, inline arrangement, 5 degree tilt angle) case...59 Nusselt Number versus Reynolds Number for geometry 7 (11crossover holes-8 exit holes, inline arrangement, 5 degree tilt angle) case...59 Nusselt Number versus Reynolds Number for geometry 8 (11crossover holes-7 exit holes, inline arrangement, 5 degree tilt angle) case...60 Nusselt Number versus Reynolds Number for geometry 9 (11crossover holes-12 exit holes, staggered arrangement, 0 degree tilt angle) case.60 Nusselt Number versus Reynolds Number for geometry 10 (11crossover holes- 10 exit holes, staggered arrangement, 0 degree tilt angle) case 61 Nusselt Number versus Reynolds Number for geometry 11 (11crossover holes-9 exit holes, staggered arrangement, 0 degree tilt angle) case.61 Nusselt Number versus Reynolds Number for geometry 12 (11crossover holes-8 exit holes, staggered arrangement, 0 degree tilt angle) case.62 vi

8 Figure 66. Figure 67. Figure 68. Figure 69. Figure 70. Figure 71. Figure 72. Figure 73. Figure 74. Figure 75. Figure 76. Figure 77. Figure 78. Figure 79. Figure 80. Figure 81. Figure 82. Figure 83. Nusselt Number versus Reynolds Number for geometry 13 (11crossover holes-12 exit holes, staggered arrangement, 5 degree tilt angle) case.62 Nusselt Number versus Reynolds Number for geometry 14 (11crossover holes-10 exit holes, staggered arrangement, 5 degree tilt angle) case.63 Nusselt Number versus Reynolds Number for geometry 15 (11crossover holes-9 exit holes, staggered arrangement, 5 degree tilt angle) case.63 Nusselt Number versus Reynolds Number for geometry 16 (11crossover holes-8 exit holes, staggered arrangement, 5 degree tilt angle) case.64 Nusselt Number versus Reynolds Number for Area 1 at different number of exit holes cases (inline arrangement, zero degree tilt angle)...64 Nusselt Number versus Reynolds Number for Area 2 at different number of exit holes cases (inline arrangement, zero degree tilt angle)...65 Nusselt Number versus Reynolds Number for Area 3 at different number of exit holes cases (inline arrangement, zero degree tilt angle)...65 Nusselt Number versus Reynolds Number for Area 3 at different number of exit holes cases (inline arrangement, zero degree tilt angle)...66 Nusselt Number versus Reynolds Number for Area 1 at different number of exit holes cases (inline arrangement, five degree tilt angle) 66 Nusselt Number versus Reynolds Number for Area 2 at different number of exit holes cases (inline arrangement, five degree tilt angle) 67 Nusselt Number versus Reynolds Number for Area 3 at different number of exit holes cases (inline arrangement, five degree tilt angle) 67 Nusselt Number versus Reynolds Number for Area 4 at different number of exit holes cases (inline arrangement, five degree tilt angle) 68 Nusselt Number versus Reynolds Number for Area 1 at different number of exit holes cases (staggered arrangement, zero degree tilt angle).68 Nusselt Number versus Reynolds Number for Area 2 at different number of exit holes cases (staggered arrangement, zero degree tilt angle).69 Nusselt Number versus Reynolds Number for Area 3 at different number of exit holes cases (staggered arrangement, zero degree tilt angle).69 Nusselt Number versus Reynolds Number for Area 4 at different number of exit holes cases (staggered arrangement, zero degree tilt angle).70 Nusselt Number versus Reynolds Number for Area 1 at different number of exit holes cases (staggered arrangement, five degree tilt angle)..70 Nusselt Number versus Reynolds Number for Area 2 at different number of exit holes cases (staggered arrangement, five degree tilt angle)..71 vii

9 Figure 84. Figure 85. Figure 86. Figure 87. Figure 88. Figure 89. Figure 90. Figure 91. Figure 92. Figure 93. Figure 94. Figure 95. Figure 96. Figure 97. Figure 98. Figure 99. Figure 100. Figure 101. Nusselt Number versus Reynolds Number for Area 3 at different number of exit holes cases (staggered arrangement, five degree tilt angle)..71 Nusselt Number versus Reynolds Number for Area 4 at different number of exit holes cases (staggered arrangement, five degree tilt angle)..72 Nu Vs Re (Area 1 for different arrangement of 11crossover holes-12exit holes).72 Nu Vs Re (Area 2 for different arrangement of 11crossover holes-12exit holes).73 Nu Vs Re (Area 3 for different arrangement of 11crossover holes-12exit holes).73 Nu Vs Re (Area 4 for different arrangement of 11crossover holes-12exit holes).74 Nu Vs Re (Area 1 for staggered arrangement 11crossover holes-10exit holes).74 Nu Vs Re (Area 2 for staggered arrangement 11crossover holes-10exit holes).75 Nu Vs Re (Area 3 for staggered arrangement 11crossover holes-10exit holes) Nu Vs Re (Area 4 for staggered arrangement 11crossover holes-10exit holes) 76 Nu Vs Re (Area 1 for inline arrangement 11crossover holes-9exit holes) Nu Vs Re (Area 2 for inline arrangement 11crossover holes-9exit holes) Nu Vs Re (Area 3 for inline arrangement 11crossover holes-9exit holes)...77 Nu Vs Re (Area 4 for inline arrangement 11crossover holes-9exit holes) Nu Vs Re (Area 1 for staggered arrangement 11crossover holes-9exit holes). 78 Nu Vs Re (Area 2 for staggered arrangement 11crossover holes-9exit holes). 79 Nu Vs Re (Area 3 for staggered arrangement 11crossover holes-9exit holes) Nu Vs Re (Area 4 for staggered arrangement 11crossover holes-9exit holes). 80 viii

10 Figure 102. Figure 103. Figure 104. Figure 105. Figure 106. Figure 107. Figure 108. Figure 109. Figure 110. Figure 111. Figure 112. Figure 113. Nu Vs Re (Area 1 for inline arrangement 11crossover holes-8exit holes).80 Nu Vs Re (Area 2 for inline arrangement 11crossover holes-8exit holes). 81 Nu Vs Re (Area 3 for inline arrangement 11crossover holes-8exit holes).81 Nu Vs Re (Area 4 for inline arrangement 11crossover-8exit holes). 82 Nu Vs Re (Area 1 for staggered arrangement 11crossover holes-8exit holes).82 Nu Vs Re (Area 2 for staggered arrangement 11crossover holes-8exit holes). 83 Nu Vs Re (Area 3 for staggered arrangement 11crossover holes-8exit holes).83 Nu Vs Re (Area 4 for staggered arrangement 11crossover holes-8exit holes). 84 Nu Vs Re (Area 1 for inline arrangement 11crossover holes-7exit holes) Nu Vs Re (Area 2 for inline arrangement 11crossover holes-7exit holes). 85 Nu Vs Re (Area 3 for inline arrangement 11crossover holes-7exit holes). 85 Nu Vs Re (Area 4 for inline arrangement 11crossover holes-7exit holes). 86 ix

11 Nomenclature m Air mass flow rate entering the test section A ven Throat area of the critical venturi P amb Ambient pressure P ven Air pressure at the inlet of the critical venture (gage) T ven Air temperature at the inlet of the critical venturi Re jet Reynolds Number based on the cross-over hole hydraulic diameter μ jet Air dynamic viscosity at the jet temperature m bleed Air mass flow rate from the bleed hole A bleed Total bleed area A cross Supply channel cross sectional area Q Total heat added to the air by the surface heaters V Voltage across the heater A Amperage through the heater Q" Heat flux at the Liquid crystal wall T F Film temperature x

12 T back Temperature of the back wall T m Air mixed mean temperature Nu jet Nusselt number based on the cross-over hole hydraulic diameter turb Heat transfer coefficient on the turbulated target wall D Cooling channel hydraulic diameter K Air thermal conductivity 0 Outer natural convection heat transfer coefficient K amb Air thermal conductivity at ambient temperature D e Hydraulic diameter of the outside of test T eater Channel heater temperature T liquid Reference temperature of the liquid crystal T ambient Ambient temperature R back Thermal resistance of the back wall R front Thermal resistance from the heating element center to the air inside the test section R 1 Thermal resistance for the front wall R top Thermal resistance for the top wall xi

13 R bot Thermal resistance for the bottom wall Fluxtop Heat flux from top wall Fluxf Heat flux from front wall Fluxbot Heat flux from bottom wall Fluxb Heat flux from back wall Fl back Heat loss from back wall F front Heat loss for front side inside the section Heat perloss The percentage of heat loss Fl top Heat loss from top wall Fl bot Heat loss from bottom wall Q wasted The total heat loss to the ambient air via radiation and conduction Q add The net heat added to the from the inlet to the point in question xii

14 Acknowledgements I would send out my greatest thanks to Professor Mohammed Taslim whose encouragement, guidance and support enabled me to finish this thesis. I want to thank Professor Mohammed Taslim for introducing me to his research. My deep sense of gratitude is to Sultan Al shehery, Nathaniel Risler Rosso, Mian Cao and Mehdi Abedi for their help and guidance throughout this work. xiii

15 Introduction Gas turbines are now widely used for aircraft propulsion, land-based power generation, and industrial applications. Gas turbine converts the fuel energy into practical power. Jet engine is one kind of the gas turbines. Figure1 shows the structure of one kind of jet engine and its working principle. Figure 1. Gas turbine engine In the jet engine, air is first compressed to high pressure and temperature, then mixed and burned with fuel in the combustion chamber. The hot exhaust gases then turn the turbine which drive the compressor, and provide forward thrust. 1

16 The basic thermodynamic cycle for gas turbine is known as Brayton cycle. It is shown in Figure 2. Figure 2. Idealized Brayton Cycle To improve thermal efficiency and power output of the gas turbine, high turbine inlet temperature is needed. Advanced gas turbine engines operate at high temperatures around However, as the turbine inlet temperature increases, the heat transfer to the turbine blade also increases which raises another issue, that is, the turbine blades life gets shorten. The level and variation in the temperature within the blade material, which cause thermal stresses, must be limited to achieve reasonable durability goals. The operating temperatures exceed the melting point temperatures of the turbine blade materials. Therefore, turbine cooling technology plays a critical role in increasing the thermal efficiency and power output of advanced gas turbines. Figure 3 shows the common cooling technology with three major internal cooling zones in a turbine blade. 2

17 Figure 3. The common cooling techniques for a modern gas turbine blade The leading edge is cooled by jet impingement with film cooling, the middle portion is cooled by serpentine rib-roughened passages with local film cooling, and the trailing edge is cooled by pin fins with trailing edge injection. This experiment focuses on the impingement cooling effect on a rib-roughened trailing edge channel with cross over holes and aims to find the influence of blocked exit holes to the heat transfer efficiency. 3

18 Experimental set up and Procedure Experimental set up To model the coolant flow in a typical gas turbine airfoil, we use a laboratory setting which is shown in Figure 4. Figure 4. Schematics of the test section This set up is made out of clear acrylic plastic, and has to endure the process of being taken apart, adjusted, put back together, and sealed again. In this experiment, the moving air is used to represent the coolant flowing in a real turbine airfoil. A compressor stores air in a reservoir, which is then dispensed into the pipe system through which air is sent to the laboratory room. Then the air travels through a heat exchanger system, the critical venture, and finally into the experiment setup. 4

19 The temperature of the inlet air plays a crucial role in this experiment. Lower inlet air temperature gives lower uncertainty of the test results which means that the results are more accurate. So before the air travels into the test section, a heat exchanger system is used to cool it down, especially when the atmospheric air temperature is high outside the laboratory. The reason is because of the size of the reservoir, it is emptied within an hour, whereas a typical experiment takes up to 5 hours. When the reservoir is emptied, the air comes directly from the compressor at a somehow higher temperature due to the hot weather outside. By using the heat exchanger, air is cooled at an appropriate temperature to make sure the results are acceptable. The heat exchanger system has an inner and outer diameter of 2.5 and 4, respectively. The critical venturi has a throat of diameter of 0.32 inches, and is used for measuring the total mass flow rate. A thermal couple and a pressure tap are attached to the critical venturi to get the temperature T ven and pressure P ven of the air inside. Finally air at specific pressure and low temperature travels to the test section which is shown in Figure4. The test section can be taken apart into three parts, which are the plenum, the supply channel and the trailing edge channel. First, air goes in the plenum and gets filtered. Then clean and dried air enters the supply channel and exits from the trailing edge channel. The plenum has a cubical shape of 20" 20" 20" with a wall thickness of 1. A honeycomb flow straightener is placed in a groove in the middle of the plenum. Two thermocouples and a pressure tap are placed on the plenum to measure the inlet air temperature T in 1, T in 2 and the plenum pressurep plen, which can be found in figure 4. 5

20 Figure 5 below shows the cross-section of the supply channel and the trailing edge channel. Figure 5. cross section of the supply channel and trailing edge channel togther The supply channel has a trapezoidal cross-section with a length of 49 inches, and is made of 0.5 inches thick acrylic plastic. The front, back and bottom walls of the supply channel are all rectangular. On the top, there is a removable partition wall which has 11 cross-over holes on it. Cooling air comes in from left side, travels through the cross-over holes on the top then enters in the trailing edge channel. In the middle of the bottom wall of the supply channel, a thermocouple is mounted to measure the jet temperature T jet. Two pressure taps are placed on the front wall to measure the pressures in the supply channel P sup 1 and P sup 2. P sup 1 locates 17 inches away from the plenum, and P sup 2 tap is inches away from P sup 1 tap. Another pressure tap is put on the right side wall to measure the pressurep sup 3. Their locations are shown in Figure 6 below. 6

21 Figure6. The locations of the pressure taps The removable partition wall is 36 inches long and made of clear acrylic plastic. The cross section is in a trapezoidal shape. It connects the supply channel and the trailing edge channel with 11 crossover holes. The first hole locates 8.5 inches from the left hand side end and other adjacent holes are 2 inches apart from each other. Figure 7 shows the details of the crossover holes. 7

22 Cross section view Upside view Figure 7. Details of the cross-over holes (removable partition holes) The trailing edge channel is the most important part in this experiment. It is 38.5 inches long and consists five parts: the front wall, the side wall, the removable wall which is bottom wall, the trailing edge wall and the heater. The front wall of the trailing edge channel has a rectangular shape while the two side walls are in trapezoidal shape. These three walls are made of 0.5 inches thick clear acrylic plastic. Two pressure taps are placed in the middle of the front wall from which the pressure difference of the air before and after flowing through the turbulator can be solved. Two thermal couples and two pressure taps are placed on both sides to measure the temperatures T end 1 & T end 2 and pressures P end 1 & P end 2. The removable wall was introduced before. More details of trailing edge channel are shown below in figure 8. 8

23 Figure 8. Details of the trailing edge channel The top wall of the trailing edge channel is called trailing edge wall. It is made of the clear plastic with a trapezoid shape cross section which can be seen from figure 5. This wall has 12 exit holes where the first hole locates 7.5 inches away from the left hand side end and other holes are 2 inches apart from each other. Details of the trailing edge exit holes are shown in figure 9. Figure 9. Details of the trailing edge holes 9

24 The back wall part of the trailing edge channel is the heater part which plays a crucial role in this experiment, because it provides the heat flux to measure the heat transfer coefficient. The heater part has couples of layers. The first layer is liquid crystal foil upon which 11 turbulators are pasted. These turbulators are placed at the center line of the liquid crystal foil right above each crossover holes (see Figure 11). The shape and dimensions of these turbulators are shown below in Figure 10. Figure 10. The dimensions of the turbulators used in experiment Liquid crystal is originally black and will turn to different colors as temperature changes. At any particular temperature, liquid crystals reflect a single wavelength of light. The colors can be calibrated to particular temperatures since the transition of color is sharp and precise. The calibration of this experimental investigation considers an appearance of specific green shade which corresponds to the temperature of The total thickness of this liquid crystal foil is inch. Under liquid crystal there are three heaters placed in a row called Heater1, Heater2 and Heater3 from the left to right respectively. Each 10

25 heater is 11 inches long and 2.5 inches wide. They are connected to a power supply with 8 different channels that controls the voltages across each heater (see Figure 21). Under three heaters, there is a 3 inches thick polyurethane layer with specific thermal conductivity which supports the heater and gets rid of the leaking of the air. A Panasonic Lumix camera was used to take the pictures of the liquid crystal foil in different temperatures. These photos will be digitalized in sigma scan pro5 software. After digitalizing, the areas of green color in each picture can be quantified, and then the average Nusselt number is solved. There are two kinds of holes arrangements in this experiment, that is, inline arrangement and staggered arrangement. Figure 11 shows inline arrangement while Figure 12 shows staggered arrangement. Figure 11. Inline arrangement flow 11

26 Figure 12. Staggered arrangement flow As the cross-over holes have two different angles, there are 4 kinds of geometries in this experiment: inline arrangement, zero tilt angle inline arrangement, five tilt angle staggered arrangement, zero tile angle staggered arrangement, five tilt angle And for this test, the camera only focused at the last four turbulators on the right hand side. This is because the aim of this experiment is to find out the influence of blocked exit holes on the right side to the heat transfer coefficient. 12

27 Camera focusing Area1 Area2 Area3 Area4 Figure 13. Camera focuses and the four areas for staggered arrangement Camera focusing Area1 Area2 Area3 Area4 Figure 14. Camera focuses and the four areas for inline arrangement 13

28 Experimental Procedure Liquid crystal calibration In this experiment, the liquid crystal is used to measure the heat transfer coefficient. Thus the property of the liquid crystal should be known, that is, which color represents what temperature. Fortunately, only the green color is calibrated in this experiment. This is because the green color is easy to identify and therefore is chosen to be the reference color for the liquid crystal. And the temperature it represents is called the reference temperance. According to the previous researches, the reference temperature of the liquid crystal is between 90 and 100. While doing the calibration, one piece of liquid crystal display sheet was merged in water starting at approximately 100. A thermocouple was used to test the temperature of the surface of the liquid crystal sheet. Both the temperature displayed on the screen and the color of the liquid crystal sheet were recorded by a video camera. During the whole process, the water was stirred carefully to make it cool fast and even. When the temperature dropped to 90, the calibration was done. The reference temperature can be decided from the video taken. For this experiment, the reference temperature is 94.6 which is shown below in Figure

29 Figure 15. Liquid crystal display of the reference color and the reference temperature. Cold and Heat Transfer Tests As to get accurate results, the test section must be hermetically sealed. Before running the test, the silicone was applied to any cracks and seals that were visible. Normally the silicone needs 24 hours to dry. After that, the compressor was turned on and the air was allowed to go through the system at a pressure equal or higher than 80 psi. The test was tested for leaks with a soap-water based method. If any more leaks were found, silicone was reapplied and let dry for another 24 hours. Once the leakage test had done, the experiment can be continued. There are total 16 different geometries in this experiment. 15

30 Geometry1: 11crossover holes-12 exit holes, inline arrangement, 0 degree tilt angle, no bleed Geometry2: 11crossover holes-9 exit holes, inline arrangement, 0 degree tilt angle, no bleed Geometry3: 11crossover holes-8exit holes, inline arrangement, 0 degree tilt angle, no bleed Geometry4: 11crossover holes-7 exit holes, inline arrangement, 0 degree tilt angle, no bleed Geometry5: 11crossover holes-12 exit holes, inline arrangement, 5 degree tilt angle, no bleed Geometry6: 11crossover holes-9exit holes, inline arrangement, 5 degree tilt angle, no bleed Geometry7: 11crossover holes-8 exit holes, inline arrangement, 5 degree tilt angle, no bleed Geometry8: 11crossover holes-7 exit holes, inline arrangement, 5 degree tilt angle, no bleed Geometry9: 11crossover holes-12 exit holes, staggered arrangement, 0 degree tilt angle, no bleed Geometry10: 11crossover holes-10 exit holes, staggered arrangement, 0 degree tilt angle, no bleed Geometry11: 11crossover holes-9 exit holes, staggered arrangement, 0 degree tilt angle, no bleed 16

31 Geometry12: 11crossover holes-8 exit holes, staggered arrangement, 0 degree tilt angle, no bleed Geometry13: 11crossover holes-12 exit holes, staggered arrangement, 5 degree tilt angle, no bleed Geometry14: 11crossover holes-10 exit holes, staggered arrangement, 5 degree tilt angle, no bleed Geometry15: 11crossover holes-9 exit holes, staggered arrangement, 5 degree tilt angle, no bleed Geometry16: 11crossover holes-8 exit holes, staggered arrangement, 5 degree tilt angle, no bleed For each case, two kinds of test are conducted: Cold Test Heat Transfer Test Cold Test: In the cold test, the experiment set up was running without using the heaters. There are 11 different pressures for cold test: 5, 10, 15, 20, 25, 30, 40, 50, 60, 70, 80 psi. This test was done at the beginning of each experiment. The original log sheets for the cold tests are located in Appendix A. The variables taken in both cold and heat transfer tests are explained below. Tven: The air temperature at the critical venturi inlet 17

32 Tin1: The temperature of the air flowing through the bell mouth opening on one side Tin2: The temperature of the air flowing through the bell mouth opening on the opposite side of Tin1 Tjet: The temperature of the air flowing through the channel Tend1: The temperature of the air at the end of the channel closest to the plenum Tend2: The temperature of the air at the opposite end of channel Tamb: Lab temperature V1 and A1: The voltage and amperage readings from heater 1 V2 and A2: The voltage and amperage readings from heater 2 V3 and A3: The voltage and amperage readings from heater 3 Pplen: The pressure reading from the plenum Psup1: The pressure reading from the cavity or supply channel closest to the plenum, also found in Figure 4 above Psup2: The pressure reading from the cavity or supply channel in the middle of the section, also found in Figure 4 above Psup3: The pressure reading from the cavity or supply channel from the opposite side of Psup1, also found in Figure 4 above Pend1: The pressure reading from the test section closest to the plenum, also found in Figure 4 above Pend2: The pressure reading from the test section opposite to Pend1, also found in Figure 4 above DP, orifice: Pressure difference across the orifice plate 18

33 Pamb: Lab pressure Cross-Over Hole Angle: Indicates the angle of the cross-over jet toward the target wall TE slot arrangement: Indicates whether or not the racetrack exit holes are in line or staggered with respect to the cross-over holes Bleed Valve: indicates whether or not the valve is open or closed Remarks: The comments of that specific experiment Heat transfer Test: For this experiment, two heaters were used in the heat transfer test. They are heater 2 and heater 3. The reason is because that heater 3 is the target zone while heater 2 is used to get rid of the heat transfer from heater 3 to heater 2 in horizontal direction. Unlike the cold test, there are 6 different venturi pressures in heat transfer test (13, 26, 40, 54, 69 and 80 psi) corresponding to six jet Reynolds Numbers in each experiment. When an experiment was chosen to be performed, the compressor was turned on. The venturi gauge pressure was set to a pre-determined pressure and the system was left on for approximately 30 minutes for it to reach equilibrium. In the meantime, the heat exchange system which using cold water was turned on to cool the incoming air. After the air system got equilibrium, make sure the venturi pressure is correct. Usually the venturi gauge pressure dropped a little. Then the electrical power panel, volt-meter, ampmeter, acquisition switch unit and micro-manometer were turned on. The last step before the data acquisition process was to determine the minimum and maximum voltages for the heaters. The minimum voltage is the voltage of the heater when the reference green 19

34 color first shows on the corresponding liquid crystal display sheet. The maximum voltage is the voltage of the heater when the reference green color is last seen on the liquid crystal. Then the voltage increment was got by the following formula V = (V max V min )/ is the desired number of pictures taken in each experiment. However a typical experiment would have pictures. According to this voltage increment, the heater progressed from the minimum voltage to the maximum voltage. A Panasonic Lumix digital camera was placed in front of the testing section focusing on the target zone which is the last four turbulators. Figure 13 and 14 show the target zone and the areas. The temperature was read in degrees Fahrenheit from 6 different channels. See Figure 16. The relationship between channel number and temperatures are shown below. Channel 101 reading for Tven Channel 102 reading for Tjet Channel 103 reading for Tin1 Channel 104 reading for Tin2 Channel 105 reading for Tend1 Channel 106 reading for Tend2 20

35 Figure 16. Thermometer used in experiment Figure 17. True-rms Multimeter used in experiment 21

36 The lab temperature Tamb was taken by a True-rms Multimeter which is shown in Figure 17. The pressures were taken by three types of manometers which are listed below. Micro-manometer Oil s specific gravity to water is Orange-manometer Oil s specific gravity to water is Red-manometer Oil s specific gravity to water is Micro-manometer is for low pressure measure but more accurate. Its accuracy is inches of distilled water. Its range is from 0 to inches of distilled water. The reading of Micro-manometer should be multiplied by 2 before used. Orange-manometer is for medium pressure measure. Its accuracy is 0.1 inches of oil which has a specific gravity to distilled water. It s range is from 0 to inches of orange oil. Redmanometer is for high pressure measure. Its accuracy is 0.1 inches of red oil which has a specific gravity to distilled water. It s range is from 0 to inches of red oil. Figure 18, 19 and 20 show these three types of manometers. 22

37 Figure 18. Micro-manometer used in experiment Figure 19. Orange-manometer used in experiment 23

38 Figure 20. Red-manometer used in experiment Figure 21. Electrical power panel used in experiment 24

39 Figure 21 shows the electrical power panel which controls the voltage of the heaters. This electrical power panel contains eight small rheostats. As two heaters were used in the experiment, only two small rheostats which are channel 3 and channel 4 were used. Channel 3 controlled heater 2 while channel 4 controlled heater 3. During the test, the main knob of the panel was kept at 80 Volt. Channel 3 and 4 were turned to the presolved minimum voltage first, and then for each step, the voltage of heater 3 which is channel 4 was increased by one voltage increment. Channel 3 which controlled heater 2 was turned to make sure the voltage of heater 2 is equal to that of heater 3. Each picture of the target zone was taken for each step. After all pictures were taken, a software sigma scan pro5 was used to digitalize them, the area of the green color for each picture in each area was calculated and put in an excel log sheet. These area values were used to calculate the average Nusselt Number for each area. Below is a summary of what was performed for each experiment. 1. Turn on the compressor to start air flow and keep track of time 2. Turn on the heat exchanger 3. Select the proper pressure from the pressure regulator 4. Wait for approximately 30 minutes for the system to reach operating equilibrium 5. Find the minimum and maximum voltages for the liquid crystal display, and then determine the voltage increments 6. Start with the minimum voltage, record all of the variables and including the picture 25

40 7. Finish with last picture, keep the compressor on to cool the whole system for 30 minutes 8. Start from step 1 again to start the new experiment 9. Using Sigma Scan Pro5 to digitalize the pictures and export the results to an excel file 26

41 Data reduction When all the experiments were completed and all the variables were collected, three different FORTRAN programs were run. They are check.f, reduce.f and integare.f. 1. Check.f Check.f program is used to check typographic errors. It read all the data taken in the experiment and compared them with the reference values. If any data was not in the specific range of values, it would be considered as a typographic error, and then the program alerted the user that there was a typing error in the input file. For example, for pressure P amb, the reference range of value is 28psi to 31psi. So if P amb < 28 or P amb > 31, it was seen as an error. For voltage and amperage, they were checked using the equation1 shown below. Error i = V i A i ( V i A I ) old ( V i A I ) old (1) Where i = 1,2,3 for heater 1, 2, 3 respectively. If Error i > , the program will alert the user to check it. The FORTRAN program (check.f) can be found in Appendix B. 2. Reduce.f After checking for errors, the data was taken to the next step, which is reduce.f program. The code for reduce.f is shown in Appendix C. This program takes all the values and calculates the energy balance. The total heat flux corresponding to each heater as well as 27

42 the local heat transfer coefficients, the total heat losses to the ambient, radiative fluxes and the Nusselt number were calculated. 2.1 Geometry calculation The reduce.f first performed all geometric calculations. Geometries for venturi, heaters, trailing-edge channel, cross-over slots and exit slots as well as the heat transfer area were calculated. The hydraulic diameter (D ) is determined by 4 times of the cross-section area divided by the cross-section s perimeter as shown in equation 2. D = 4A cross P (2) 2.2 Coefficient calcuation Then the jet Reynolds number is Re jet = ρ v Slot D μ jet = 4 m Peri meter μ jet (3) Where m = ρ v A cross. The Nusselt number is got by equation 4. Nu jet = turb Slot D K (4) Where the SlotD is the hydraulic diameter of the cross-over slot. K is the thermal conductivity of the air. And turb is the heat transfer coefficient which will be solved later by an energy balance equation. 28

43 The inlet temperature T in is the average of the T in1 and T in2. The air flow rate through the critical venturi is calculated by equation 5 which is the correlation for small critical venturi. m = A ven (P ven +P amb ) T ven +460 (5) The total heat added to the air by the heaters from the inlet to the point in question is solved by equation 6. Q = ( V 1 A 1 + V 2 A 2 + (V 3 A 3 )) (6) Where the heat flux fluxb is Q" = Fluxb = ( V 2 A 2 ) Heater Area (7) This variable will be used in the calculation later. 2.3 Energy balance-heating resistance calculation When all the geometries had been calculated, the program came to the energy balance part. Some variables used in energy balance are listed below. 0 = 0.36 K amb D e (8) Where D e = 7/12, R cov = 1/ 0 (9) 29

44 Where R cov is the heat resistance of the ambient air, T m = T jet (10) Here, T m = T jet is used for calculation, not for T m calculation which is a few degrees higher. In the energy balance, as the horizontal direction heat transfer got eliminated by using the same voltages for heater2 and heater3, only four directions of heat transfer were considered: back, front, top and bottom. While calculating the resistances, some abbreviations were used for convenience. For example, in statement like tadh1/kadh, tadh means the thickness of adhesive layer while kadh stands for the thermal conductivity of the adhesive layer. The materials thermal conductivities were got from relative references and are listed here: k kapton = BTU r. ft., k polyuret ane = BTU r. ft., k plexiglas = 0.11 BTU r. ft., k mylar = BTU. ft., k r adesive = BTU. ft., r k inconel = BTU r. ft., k blackground = BTU r. ft., k liquid crystal = BTU r. ft.. Back direction-back wall For the back direction, there was a back wall which is liquid crystal wall. The heating resistance was calculated from the center of the heating element to the ambient air. There were couples of layers between the heating element to the ambient air, which are 0.25 mil inconel heating element, 0.5 mil adhesive, 0.5 mil kapton, 2 mil adhesive, inches polyurethane, 2.0 inches Styrofoam. Their heating resistances are tinc1/kinc, tadh1/kadh, 30

45 tkap1/kkap, tadh4/kadh, tpoly/kpoly, tsty/ksty and 1/ho respectively. Thus the total heating resistance of the back direction was calculated by equation 11 below: R back = 0.5 R inc 1 + R ad1 + R kap 1 + R ad4 + R poly + R conv (11) Where R inc 1 = tinc1/kinc, R ad1 = tad1/kad, R kap 1 = tkap1/kkap, R ad4 = tad4/kad, and R poly = tpoly/kpoly. Front direction-front direction inside section and front wall For the front direction, there were two parts of processes. One is from the center of the heating element to the air inside the test section, and the other is from air inside the test section to the ambient air. For the first process, the layers are 0.25 mil inconel heating element, 1.0 mil adhesive, 2.0 mil kapton, 1.0 mil adhesive, 0.5 mil inconel spreader, 1.0 mil adhesive, 2.0 mil kapton, 1.5 mil adhesive, 3.0 mil absorptive black background, 2.0 mil liquid crystal, 5.0 mil mylar and air inside the test section. Their heating resistances are tinc1/kinc, tadh2/kadh, tkap2/kkap, tadh2/kadh, tinc2/kinc, tadh2/kadh, tkap2/kkap, tadh3/kadh, tblack/kblack, tliq/kliq, tmyl/kmyl and 1/hi. The second process was air inside the test section, 0.45 inches Plexiglas and ambient. Where the resistances are 1/hi, tplex/kplex and 1/ho. Thus the total heating resistance from the heating element center to the air inside the test section was calculated by equation 12 below: R front = R inc 1 + R ad2 + R kap 2 + R ad2 + R inc 2 + R ad2 + R kap 2 + R ad3 + R black (12) + R liq 31

46 Where R ad2 = tad2/kad, R kap 2 = tkap2/kkap, R inc 2 = tinc2/kinc, R ad2 = tad3/kad, R black = tblack/kblack and R liq = tliq/kliq. And the heating resistance for the front wall which is from the air inside the test section to the ambient air was calculated by: R 1 = R plex + R cov (13) Where R plex = tplex/kplex. Top direction-top wall For the top wall, R top = R plex + R conv (14) Bottom direction-bottom wall For the bottom wall, R bot = R plex + R sty + R cov (15) 2.4 energy balance-heat flux calculation The heat flux for all four directions is shown below: Fluxtop = 0 (16) Fluxf = 0 (17) Fluxbot = 0 (18) 32

47 And V 2 A 2 Fluxb = ( Heater Area ) see equation 7. Then the temperature at the center of the heating element was found by equation 17: T eater = Fluxb + T liquid + T amb R front R back R back R front (19) Where T liquid is the reference temperature of the liquid crystal. 2.5 energy balance-heat losses calculation When the heat resistances and heat flux for all four directions were solved, the losses were considered. Back direction-back wall The loss from back wall Fl back was Fl back = (T eater T ambient )/R back (20) Front direction-front direction inside the test section While the loss for front side inside the section F front was F front = (T eater T liquid )/R front (21) 33

48 Then the back wall temperature T back, the film temperature T f and the percentage of heat loss Heat perloss can be solved by equations below T back = T liquid F front R my 1 (22) T f = 0.5 (T m + T back ) (23) Heat perloss = 100 (Fl back /Q") = 100 (Fl back /Fluxb) (24) Top direction-top wall For the top wall, set T top = T m, and the loss of energy was found by Fl top = (T top T amb )/R top (25) Bottom direction-bottom wall For the bottom, set T bot = T in, then the loss from bottom was solved Fl bot = (T bot T amb )/R bot (26) Front direction-front wall For the front wall, the temperature T front and the loss Fl front were both unknown. However, both of them are important for solving the heat transfer coefficient turb. To solve this problem, Newton s method was used. 34

49 2.6 Newton s method The program first gave an initial guess for turb and T front, then using the relation of radiation heat loss to solve for new value of T front that is T frontn. And if T frontn satisfied the convergence condition, then set T front = T frontn. After this, the heat transfer coefficient turb can be solved. Initial guesses back = (Fluxb Fl back )/(T back T m ) (27) front = back (28) T front = T m (29) Here turb = back which was not a guess. And the convective resistance R 3 is equal to 1/ front which was also a truth. Solve for new value of T front From the radiation loss equation, T frontn = ( 1 R3 T m + 1 R1 T amb Fr front ) ( 1 R1 + 1 R3 ) (30) Convergence condition If T front T frontn 0.001, then T front = T frontn. 35

50 2.7 solve for heat transfer coefficient For the front wall, Fl front = (T front T amb )/R 1 (30) Then the total heat loss to the ambient air via radiation and conduction can be quantified, Q wasted = Fl top A top + Fl bot A bot + Fl front A front + Fl back A back (31) Then, the net heat added to the from the inlet to the point in question is solved by, Q add = Q Q waste (32) From the Newton law of cooling, the heat transfer coefficient can be found back = Fluxb Fl back Fr back T back T m (33) Then turb = back, using equation 4, the Nusselt number for the all target wall can be quantified. 2.8 uncertainty calculation The uncertainty for the entire experiment was found by using Kline and McClintock s method, and with help from further reading from Robert J Moffat [4]. 36

51 3. Integarea.f To find the area weighted average of the Nusselt number, integarea.f program was used. The average Nusselt number was calculated with equation 34, where A is the pixel area of the reference color from Sigma Scan Pro for each picture and n is the number of (34) pictures per experiment. Nu = i=n i=1 i=n i=1 A i Nu i A i The source code for integare.f can be found in Appendix D. After these three programs, graphs were made and analyzed. 37