AME 517, Fall 2009

Problem set #4

Assigned: 10/21/2009

Due: 10/30/2009, 4:30 pm


Homework may be submitted by email (, by fax (213-740-8071) or put in my mailbox (down the hall from my office in the OHE 430 suites, in the small room with the copy machine).  DEN students should submit through the usual channels.


Chapter 8: 




8.3 (assume that the needle is a gray body with e = a = 0.8, donÕt worry about the non-gray effects.)




Other problem: 


Repeat problem 6.8 assuming that plate 2, instead of having a fixed temperature of 600K, has an unknown temperature but receives conductive heat transfer from the bottom side with a heat transfer coefficient of U2 = 10 W/m2K and T° = 600 K.


Notes on text problems:


Problems 8.1 and 8.3 are different from most of the problems weÕve done in previous chapters because they are not modeling radiative transfer via a finite number of discrete surfaces (² 4 for the cases we've been doing), but rather treating the antenna or needle as a continuous surface with conduction along the antenna or needle.  In this case you have to use Eq. 8.1 to solve the problem.  The difficulty is that qR(x) is not constant (though you might try qR(x) = constant for a reality check of your result.)  You need to compute qR(x) as a function of the distance (x) from the base of the antenna or needle, then integrate 8.1 with the appropriate boundary conditions.  Note that problems 8.1 and 8.3 are MUCH easier than the examples given in the text (sections 8.2 and 8.3) because in the text examples you don't know qR(x) beforehand - you have to determine it as part of the solution, which is why you wind up with the integro-differential equations in the text: 8.8 or 8.18+8.21.


Problem 8.4 can be treated via the spreadsheet as we've been doing in class since the bead is assumed to be isothermal; in part a there are only 2 surfaces, the tube and the thermocouple bead.  In part b there is a 3rd surface, namely the shield, which you may choose to model as 2 separate surfaces, one being the inside of the shield and the other being the outside.