Law of comparison is $pv^{1.25}$=constant. Mechanical efficiency is 80%. Find the power input to compressor, neglecting losses due to clearance, leakages due to cooling.

Given:

$\dot{m} = 20\hspace{0.05cm}kg/min\hspace{1cm}p_2 = 6 bar\\ p_1 = 1.1\hspace{0.05cm}bar\hspace{1cm}T_1 = 30^\circ = 303\hspace{0.05cm}K$

Solution:

i) Work done to compress and deliver 1kg of air,

$W = \frac{n}{n - 1}mRT_1[(\frac{p_2}{p_1})^\frac{n -1}{n} - 1]\\ W = \frac{1.25}{1.25 - 1}\hspace{0.05cm}\times\hspace{0.05cm}1\hspace{0.05cm}\times\hspace{0.05cm}0.287\hspace{0.05cm}\times\hspace{0.05cm}303[(\frac{6}{1.1})^\frac{1.25 - 1}{1.25} - 1]\\ W = 175.64\hspace{0.05cm}\textit{KJ/kg of air}$

ii) Power input to the compressor

$\textit{Indicated power}\hspace{0.05cm}(IP) = \dot{m}\hspace{0.05cm}\times\hspace{0.05cm}W\\ \hspace{03cm}IP = \frac{20}{60}\hspace{0.05cm}\times\hspace{0.05cm}175.64\\ \hspace{03cm}IP = 58.54\hspace{0.05cm}KW\\ \textit{Brake Power i.e Power Input}\\ \hspace{03cm}BP = \frac{IP}{\textit{mechanical efficiency}} = \frac{58.54}{0.8}\\ \hspace{03cm}\textbf{BP or Input Power} = 78.183\hspace{0.05cm}KW$