Question Number 8886 by vuckintv last updated on 04/Nov/16 | ||
$${solve}\:{for}\:\beta \\ $$$${T}\left({h}\right)={T}_{{s}} −\frac{\sqrt{\pi}}{\mathrm{2}}×\frac{\beta}{\xi}×\left[{erf}\left(\xi{H}\right)−{erf}\left(\xi{h}\right)\right] \\ $$$$ \\ $$ | ||
Answered by Rasheed Soomro last updated on 04/Nov/16 | ||
$${T}\left({h}\right)={T}_{{s}} −\frac{\sqrt{\pi}}{\mathrm{2}}×\frac{\beta}{\xi}×\left[{erf}\left(\xi{H}\right)−{erf}\left(\xi{h}\right)\right] \\ $$$$\frac{\sqrt{\pi}}{\mathrm{2}}×\frac{\beta}{\xi}×\left[{erf}\left(\xi{H}\right)−{erf}\left(\xi{h}\right)\right]={T}_{{s}} −{T}\left({h}\right) \\ $$$$\frac{\beta}{\xi}=\frac{\mathrm{2}\left({T}_{{s}} −{T}\left({h}\right)\right)}{\sqrt{\pi}\left(\left[{erf}\left(\xi{H}\right)−{erf}\left(\xi{h}\right)\right]\right)} \\ $$$$\beta=\frac{\mathrm{2}\xi\left({T}_{{s}} −{T}\left({h}\right)\right)}{\sqrt{\pi}\left(\left[{erf}\left(\xi{H}\right)−{erf}\left(\xi{h}\right)\right]\right)} \\ $$ | ||