Question Number 60421 by Tawa1 last updated on 20/May/19
Answered by MJS last updated on 20/May/19
$$\eta=\frac{\pi}{\mathrm{8}}×\frac{{pr}^{\mathrm{4}} }{{LQ}} \\ $$$$\eta_{{min}} =\frac{\pi}{\mathrm{8}}×\frac{.\mathrm{97}{p}×\left(.\mathrm{98}{r}\right)^{\mathrm{4}} }{\mathrm{1}.\mathrm{01}{L}×\mathrm{1}.\mathrm{005}{Q}}\approx.\mathrm{881}\frac{\pi{pr}^{\mathrm{4}} }{\mathrm{8}{LQ}} \\ $$$$\eta_{{max}} =\frac{\pi}{\mathrm{8}}×\frac{\mathrm{1}.\mathrm{03}{p}×\left(\mathrm{1}.\mathrm{02}{r}\right)^{\mathrm{4}} }{.\mathrm{99}{L}×.\mathrm{995}{Q}}\approx\mathrm{1}.\mathrm{132}\frac{\pi{pr}^{\mathrm{4}} }{\mathrm{8}{LQ}} \\ $$$$\Rightarrow\:\mathrm{max}.\:\%\:\mathrm{error}\:\mathrm{of}\:\eta\:\approx\mathrm{13}.\mathrm{2\%} \\ $$
Commented by Tawa1 last updated on 20/May/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$