Question Number 8886 by vuckintv last updated on 04/Nov/16
![solve for β T(h)=T_s −((√π)/2)×(β/ξ)×[erf(ξH)−erf(ξh)]](https://www.tinkutara.com/question/Q8886.png)
$${solve}\:{for}\:\beta \\ $$$${T}\left({h}\right)={T}_{{s}} −\frac{\sqrt{\pi}}{\mathrm{2}}×\frac{\beta}{\xi}×\left[{erf}\left(\xi{H}\right)−{erf}\left(\xi{h}\right)\right] \\ $$$$ \\ $$
Answered by Rasheed Soomro last updated on 04/Nov/16
![T(h)=T_s −((√π)/2)×(β/ξ)×[erf(ξH)−erf(ξh)] ((√π)/2)×(β/ξ)×[erf(ξH)−erf(ξh)]=T_s −T(h) (β/ξ)=((2(T_s −T(h)))/( (√π)([erf(ξH)−erf(ξh)]))) β=((2ξ(T_s −T(h)))/( (√π)([erf(ξH)−erf(ξh)])))](https://www.tinkutara.com/question/Q8887.png)
$${T}\left({h}\right)={T}_{{s}} −\frac{\sqrt{\pi}}{\mathrm{2}}×\frac{\beta}{\xi}×\left[{erf}\left(\xi{H}\right)−{erf}\left(\xi{h}\right)\right] \\ $$$$\frac{\sqrt{\pi}}{\mathrm{2}}×\frac{\beta}{\xi}×\left[{erf}\left(\xi{H}\right)−{erf}\left(\xi{h}\right)\right]={T}_{{s}} −{T}\left({h}\right) \\ $$$$\frac{\beta}{\xi}=\frac{\mathrm{2}\left({T}_{{s}} −{T}\left({h}\right)\right)}{\:\sqrt{\pi}\left(\left[{erf}\left(\xi{H}\right)−{erf}\left(\xi{h}\right)\right]\right)} \\ $$$$\beta=\frac{\mathrm{2}\xi\left({T}_{{s}} −{T}\left({h}\right)\right)}{\:\sqrt{\pi}\left(\left[{erf}\left(\xi{H}\right)−{erf}\left(\xi{h}\right)\right]\right)} \\ $$