quiz_solution - PROBLEM 3.71 KNOWN Temperature distribution...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PROBLEM 3.71 KNOWN: Temperature distribution in a composite wall. FIND: (a) Relative magnitudes of interfacial heat fluxes, (b) Relative magnitudes of thermal conductivities, and (0) Heat flux as a function of distance x. SCHEMATIC: X ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties. ANALYSIS: (a) For the prescribed conditions (one-dimensional, steady-state, constant k), the parabolic temperature distribution in C implies the existence of heat generation. Hence, since dT/dx increases with decreasing x, the heat flux in C increases with decreasing x. Hence, 61% > £13 However, the linear temperature distributions in A and B indicate no generation, in which case <13 = 015 (b) Since conservation of energy requires that qu = qic and dT/dx)B < dT/dx)c , it follows from Fourier’s law that kB > kc. Similarly, since £15, A = qi’B and dT/dx) A > dT/dx)B, it follows that k A < kB. (0) It follows that the flux distribution appears as shown below. If X X; X 3 X4 0 x 92 COMMENTS: Note that, with dT/dx)4,(3 = 0, the interface at 4 is adiabatic. PROBLEM 3.46 KNOWN: Conditions associated with a composite wall and a thin electric heater. FIND: (a) Equivalent thermal circuit, (b) Expression for heater temperature, (c) Ratio of outer and' inner heat flows and conditions for which ratio is minimized. SCHEMATIC: "2r ' Mp ' (1,-4— / \ +qo / In (r21r1) In (r 3If2) (\hozflrar'l h-Z r '1 ( ' 1' 1) 21th anA ASSUMPTIONS: (1) One-dimensional, steady-stare conduction, (2) Constant properties, (3) Isothermal heater, (4) Negligible contact resistance(s). ANALYSIS: (a) On the basis of a unit axial length, the circuit, thermal resistances, and heat rates are as shown in the schematic. (b) Performing an energy balance for the heater, Bin = Bout , it follows that Th“ -00T Th_Tcoo qh(2”12)=CIi+Cli-i= ——m——’—-—+ ’ < —11nr2/I1 —1 Inn/r2 (hum 11) 22,1; (hams) +——*,E,kA) (c) From the circuit, _ —1 1n(r2/r1) ’ ' , q_, a, w (W) m .423— $1" in __ ,, _ q; (Th— 4,0,) (huh 131+)—+111(13/12) ’ ”r g, 1 r3 ZflkA To reduce qb /q§ , one could increase k3, h,, and r3/r2, while reducing 19,, he and r2/r1. COMMENTS: Contact resistances between the heater and materials A and B could be important. ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern