Collection

2005 ASEE Midwest Section Conference

Authors

Edgar C. Clausen; W. Roy Penney; Alison N. Dunn; Jennifer M. Gray; Jerod C. Hollingsworth; Pei-Ting Hsu; Brian K. McLelland; Patrick M. Sweeney; Thuy D. Tran; Christopher A. von der Mehden; Jin-Yuan Wang

Section Conference of the American Society for Engineering Education 5 Figure 3. Photograph of Insulated Wood Box used to Heat the Aluminum Plates Figure 4. Photograph of Experimental Horizontal Plate Heat Transfer ExperimentProceedings of the 2005 Midwest Section Conference of the American Society for Engineering Education 6Data Reduction 1. A heat balance on the cooling plate, with no heat generation, yields: − qOut = q Acc (1) 2

Collection

2005 ASEE Midwest Section Conference

Authors

Wangping Sun; J. M. Zhang; Z. J. Pei

minimize the student participation, leaving littleopportunity for effective learning 2. To prevent the “lecture-centered” situation from happeningin IMSE 250, Q & A interactions between the instructors and students took place quite often.These questions called forth the students’ attention, helped them better understand the lecture,and provided a good feedback through which the instructors could know how well the studentshad learnt in class.However, these interactions need to be concise. As observed by the instructors, it seemed betterif each Q & A could be finished in one minute. If the interaction was longer than that, then theclass was likely to lose the interests in the topic. If the question needed longer time to answer

Collection

2005 ASEE Midwest Section Conference

Authors

Ing-Chang Jong

the force or moment, respectively, during the action. It is theforce or moment, rather than the body, which does work. In teaching and learning the virtualwork method, it is well to refresh the following relevant basic concepts: Work of a forceIf a force F acting on a body is constant and the displacement vector of the body from positionA1 to position A2 during the action is q, then the work U1→ 2 of the force F on the body is1-4 U1→ 2 = F ⋅ q = Fq (1)where F is the magnitude of F, and q is the scalar component of q parallel to F. If the force isnot constant, then integration may be used to compute the work of the force. Work of a momentIf a moment M (or

Collection

2005 ASEE Midwest Section Conference

Authors

Ameya A. Chandelkar; Deepak G. Bhat

materials athigh temperatures. The present data, plotted as shown in Fig. 8, illustrates this trend for thecemented carbide samples as well.Deformation of materials at elevated temperatures is often described by a phenomenologicalequation of the form: ε& = Aσ n ⋅ e (− Q RT ) (4)where ε& is the strain rate, σ is the flow stress, Q is the activation energy for deformation, T is theabsolute temperature, R is the universal gas constant, and A and n are material constants9. Sincethe rate of loading ε& of the indenter in a hardness test is preset to a constant value, andconsidering the proportionality between hardness and applied stress, Eq. 4 may be modified as

Collection

2005 ASEE Midwest Section Conference

Authors

Andy Moore; C.J. Fisher; Pat Crosby; Wayne Helmer; Chih-Hao Wu

throughsenior level courses.Nomenclature C = electrical capacitance h = u + pv = enthalpy H = head in feet HP = horsepower ke = kinetic energy per unit mass "Proceedings of the 2005 Midwest Section Conference of the American Society for Engineering Education" 2 m = mass flow rate pe = potential energy per unit mass p = pressure ΔP = pressure differential in psi Q = heat transfer rate Qf = flow rate in gpm R = electrical resistance u = internal energy per unit mass v = specific volume V = volumetric flowrate W

Collection

2005 ASEE Midwest Section Conference

Authors

Ken French

100- 400 [ SCFH ] low flow rotameter 25-150 [ SCFH ] P air from compressor 85-110 [psi] Figure 6 Schematic of the testing Arena Team Name: Test Date & Time: P start P end Q T amb T in T cold T hot ( psi) (SCFH) (°F) Figure 7 Test Report SheetProceedings of the 2005 Midwest Section

Collection

2005 ASEE Midwest Section Conference

Authors

Edgar C. Clausen; W. Roy Penney; Cole E. Colville; Alison N. Dunn; Noor M. El Qatto; Crystal D. Hall; W. Brent Schulte; Christopher A. von der Mehden

Horizontal PlateProceedings of the 2005 Midwest Section Conference of the American Society for Engineering Education 6Data Reduction 1. A heat balance on the center plate, with no heat generation, yields: − qOUT = q ACC (1) 2. The plate is cooled by free convection and radiation as follows: qOUT = qCONV + qRAD = hAS (TSURFACE − T∞ ) + εσAS (TSURFACE − T∞4 ) 4