question_answer

What horse power engine is required to lift 18.24 quintals of coal per minute from a mine 50 m deep?

Answer:

\[m=18.24\,qu\operatorname{int}al\]                 \[=1824\,\,kg\] \[t=0.1\,\,mm=60\,\,\sec \] \[h=50\,m\] \[g=9.8\,m/s\] \[P=?\] \[P=\frac{mgh}{t}=\frac{1824\times 9.8\times 50}{60}w\]                 \[=\frac{1824\times 9.8\times 50}{60\times 746}hp=20\,hp\]