Question 15.12: A double acting reciprocating pump has plunger diameter 17.5......

Given: D = 17.5 cm = 0.175 m, L = 35 cm = 0.35 m, $d_{s}$ = 15 cm = 0.15 m, N= 150 rpm, and r = 0.35/2 = 0.175 m
Area of the plunger  $A={\frac{\pi}{4}}(0.175)^{2}=0.02405\,\mathrm{m^{2}\quad}$

Angular velocity of crank for pump  ${{\omega}}={\frac{2\pi N}{60}}={\frac{2\pi\times150}{60}}=5\pi\,\operatorname{rad}/{\mathbf{s}}$

Discharge from the double acting pump  $Q=\,{\frac{2A L N}{60}}={\frac{2A\times2r\times60\,\omega}{60\times2\pi}}={\frac{2r A\omega}{\pi}}$

or            $Q={\frac{2\times0.175\times0.02405\times5\pi}{\pi}}=0.0420875\,{\mathrm{m}}^{3}/{\mathrm{s}}$             (i)

Discharge beyond the air vessel = Arω sin θ
= 0.02405 × 0.175 × 5π × sinθ
or                = 0.06611 sinθ                        (ii)
For no flow to or from the air vessel, discharges given by (i) and (ii) must be equal.
Hence,          0.06611 sin θ = 0.0420875
or              sin θ = 0.636628
Therefore,       θ = 39°32′