If separation occurs 2.4 m of water absolute, Determine: 1. The speed at which separation may takes place at the beginning of suction stroke. 2. The speed of the pump if an air vessel is fitted on the suction side 2.4 m above the sump water level. Take atmosphere pressure head = 10.3 m of water and friction co-efficient, f = 0.01.

Given: $D = 0.3\hspace{0.05cm}m,\hspace{0.05cm}L = 0.5\hspace{0.05cm}m,\hspace{0.05cm}h_s = 3.2\hspace{0.05cm}m\\ d_s = 0.2\hspace{0.05cm}m,\hspace{0.05cm}l_s = 9\hspace{0.05cm}m,\hspace{0.05cm}h_{sep} = 2.4\hspace{0.05cm}m\hspace{0.05cm}(absolute)\\ H_{atm} = 10.3\hspace{0.05cm}m,\hspace{0.05cm}f = 0.01$

To Find:

1. N at which seperation takes place
2. N if the air vessel is filled on the suction side 2.4 m above the sump water level.

Solution: $1. h_{sep} = (h_s + h_{as})\hspace{0.05cm}m below atmospheric head\\ \hspace{1cm}= \textit{Atmospheric pressure head} - (h_s + h_{as})\hspace{0.05cm}m\hspace{0.05cm}(absolute)\\ \hspace{1cm} = H_{atm} - (h_s + h_{as})\hspace{0.05cm}m\hspace{0.05cm}(abs)\\ 2.4 = 10.3 - (3.2 + h_{as})\\ h_{as} = 4.7\hspace{0.05cm}m\\ h_{as} = \frac{l_s}{g}\hspace{0.05cm}\times\hspace{0.05cm}\frac{A}{a_s}\hspace{0.05cm}\times\hspace{0.05cm}w^2\hspace{0.05cm}\times\hspace{0.05cm}r\\ 4.7 = \frac{9}{9.81}\hspace{0.05cm}\times\hspace{0.05cm}\frac{0.3^2}{0.2^2}\hspace{0.05cm}\times\hspace{0.05cm}w^2\hspace{0.05cm}\times\hspace{0.05cm}0.1\\ w = 14.31rad/s\\ w = \frac{2 \pi N}{60} = \frac{2\hspace{0.05cm}\times\hspace{0.05cm}3.14\hspace{0.05cm}\times\hspace{0.05cm}N}{60} = 136.72\hspace{0.05cm}rpm = 137rpm\\ 2.h_s = 3.2\hspace{0.05cm}m$

$l^{'} =h_s - 2.4 = 3.2 - 2.4 = 0.8\hspace{0.05cm}m\\ l_s = 9 - l_s^{'} = 8.2\hspace{0.05cm}m\\ h_{sep} = (h_s + h_{as}^{'} + h_{fs}^{'} + h_{fs}) +\frac{v_5^2}{2g} below\hspace{0.05cm}H_{atm}\\ \hspace{0.5cm}= H_{atm} -(h_s + h_{as}^{'} + h_{fs}^{'} + h_{fs}) -\frac{v_5^2}{2g}\\ 2.4 = 10.3 - (3.2 + \frac{l_s^{'}}{g}\hspace{0.05cm}\times\hspace{0.05cm}\frac{A}{a_s^2}.w^2r\hspace{0.05cm} +\hspace{0.05cm}0\hspace{0.05cm}+\hspace{0.05cm}\frac{4fl_s}{d_s\times2g}(\frac{A}{a_s}\hspace{0.05cm}\times\hspace{0.05cm}\frac{wr}{\pi})^2 - \frac{1}{2g}(\frac{A}{a_s}\hspace{0.05cm}\times\hspace{0.05cm}\frac{wr}{\pi}^2))...(h_{fs} = 0 \hspace{0.1cm}as\hspace{0.1cm}\theta = 0)\\ 2.4 = 10.3 - (3.2 +0.0183w^2 + 4.2929\hspace{0.05cm}\times\hspace{0.05cm}10^{-4}w^2) - 2.617\hspace{0.05cm}\times\hspace{0.05cm}10^{-4}w^2\\ w^2 =247.5\\ w = 15.73\hspace{0.05cm}rad/s $