Example 14.4-2. Radiant Heat Transfer Between Disks

Part a - for two disks with no other surfaces present:

Figure 3: Diagram of problem in part a

> restart;

> Q[12]:=A[1]*F[12]*sigma*(T[1]^4-T[2]^4); this is eqn. 14.4-9 which was also shown in the discussion section

Knowing the disk diameter is 2 ft. and the distance between them is 4 ft., it is possible to use curve 1 of the chart below and the ratio 2/4 to find the view factor. The view factor F12 is approximately 0.06.

Figure 4: View Factors for radiation between parallel, identical shapes, as calculated by Hottel

> r:=d/2; d:=2*ft; A[1]:=Pi*r^2; F[12]:=.06; sigma:=.1712e-8*Btu/hr/ft^2/R^4; T[1]:=2000*R; T[2]:=1000*R;

> simplify(Q[12]);

Part b - two disks connected by a cylindrical surface:

The only difference in this part is that rather than using F12, use F12bar, again using the chart above to find it with the ratio still at 0.5.

Figure 5: Diagram of problem in part b

> Fbar[12]:=.34; found from the chart using curve 5

> Q[12]:=A[1]*Fbar[12]*sigma*(T[1]^4-T[2]^4);

> simplify(Q[12]);

Which also agrees relatively closely to the book's value.