written 6.0 years ago by
teamques10
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modified 6.0 years ago
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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:
- N at which seperation takes place
- 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 $