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Question 3.12: In grind polishing of a stainless-steel plate (316) of width...

In grind polishing of a stainless-steel plate (316) of width and length w = l = 20 cm, and thickness L = 0.05 cm, heat is generated by friction heating at an amount \dot {S}_{ m,F} = 4 × 10^{3} W. From this energy, the fraction flowing into the workpiece (i.e., plate) is a_{1} = 0.7.
(a) Draw the thermal circuit diagram.
(b) If the grinding surface of the stainless-steel (316) plate should have a temperature T_{1} = 350^{\circ }C, what should be the temperature at the opposite side of the plate T_{2}?
The thermal conductivity can be evaluated at T = 350^{\circ }C and the heat transfer may be assumed one dimensional.

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(a) The physical model of the problem is that shown in Figure (a) and the thermal circuit model is that shown in Figure (b).
(b) The temperature at surface 1 is found by (3.100), which is repeated here as

T_{1}=T_{2}+\frac{a_{1}\dot {S}_{m,F}L}{lwk}

and this gives

T_{2}=T_{1}+\frac{a_{1}\dot {S}_{m,F}L}{lwk}

The thermal conductivity of the stainless-steel (316) at T = 300 K is found from Table and is

k = 13 W/m-K        Table

Now substituting into the expression for T_{2}, we have

T_{2}=350(^{\circ }C)-\frac{0.7\times 4\times 10^{3}(W/m^{2})\times 0.05(m)}{0.2(m)\times 0.2(m)\times 13(W/m-k)} =80.77^{\circ }C
b
a
16.1PNG
16.2

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