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// Chapter8
// Determine appropiate heat sink rating
// Page.No-296
// Example8_7
//Figure 8.34
// Given
clear;clc;
Tj=150; // in degree C
Ta=40; // in degree C
Qjc=3.0; // in C/W
Qcs=1.6; // in C/W
PD=6; //in W
Qsa=(Tj-Ta)/PD - Qjc-Qcs;
printf("\n Value of Qsa = %.2f C/W\n",Qsa); // Result
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