Question 15.10: A single acting reciprocating pump has a stroke length 15 cm......

Given: L = 15 cm, l_{s} = 7.5 m, D/d_s= 4/3, H_{s}  =  2.5  m,  d_{s} = 7.5 cm = 0.075 m, N = 75 rpm, f = 0.01, and r = L/2 = 15/2 = 75 cm = 0.075 m

Now,  {\frac{A}{a_{s}}}=\left({\frac{D}{d_{s}}}\right)^{2}=\left({\frac{4}{3}}\right)^{2}={\frac{16}{9}}

and           \omega={\frac{2\pi N}{60}}=2\pi\times{\frac{75}{60}}=2.5\,\pi\,\mathrm{rad/s}

Hence,          Acceleration head h_{a s}={\frac{l_{s}}{g}}\,{\frac{A}{a_{s}}}r\omega^{2}={\frac{7.5}{9.81}}\times{\frac{16}{9}}\times0.075\times(2.5\pi)^{2}

or                        h_{as} = 6.288 m
(a) Pressure head at the beginning = H_{a t}-(H_{s}+h_{a s})

= 10.3 – (2.5 + 6.288)
= 1.512 m of water absolute
or                                  = 8.78 m of water vacuum
(b) Pressure head in the middle of the stroke = H_{a t}-(H_{s}+h_{fs})

=\,10.3-2.5-\frac{f l_{s}}{2g d_{s}}\left\lgroup\frac{A}{a_{s}}r\omega\right\rgroup ^{2}

= 7.774 m of water
= 2.52 m of water vacuum

(c) At the end of the stroke, h_{fs} is zero, hence
Pressure head = H_{at}- (H_{s}- h_{as})
= 10.3 – (2.5 – 6.288)
= 14.088 m of water absolute
= 3.788 m of water gauge