Question 15.3: A single acting reciprocating pump has a plunger diameter 0.......

Given: D = 0.15 m, L = 0.3 m, H_{s} =  4  m,  H_{at} = 10.3 m of water, N = 40 rpm, l_{s}  =  5  m,  d_{s} = 0.1 m
Now,

Angular speed of the crank, \omega={\frac{2\pi N}{60}}={\frac{2\times\pi\times40}{60}}=4.188\,{\mathrm{rad/s}}

Radius of the crank,  r={\frac{L}{2}}={\frac{0.3}{2}}=0.15\,\mathrm{m}

Area of the piston, A={\frac{\pi}{4}}\times\left(0.15\right)^{2}=0.01767\,{\mathrm{m}}^{2}

Area of the suction pipe,  a_{s}={\frac{\pi}{4}}\times(0.1)^{2}=0.007853\,{\mathrm{m}}^{2}

Thus, the equation of pressure head due to acceleration is

or        h_{s}=\frac{5}{9.81}\times\frac{0.01767}{0.007853}\times0.15\times(4.188)^{2}\cos\theta

or    h_{s}=3.01837\,\cos\theta

Again, at the beginning of stroke, θ = 0, cos θ = 1. Therefore,

h_{smax} = 3.01837 m

and at the middle of the stroke, θ = 90°, cos θ = 0. Therefore,
h_{s} = 0
Thus, the absolute pressure head in the cylinder at the beginning of the suction stroke is:
H_{at}  –  H_{s}  –  h_{smax}
= 10.3 – 4 – 3.01837 = 3.28163 m